[{"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "Let's find the volume of a few more solid figures, and then if we have time, we might be able to do some surface area problems. So let me draw a cylinder over here. So that is the top of my cylinder, and then this is the height of my cylinder. This is the bottom right over here. If this was transparent, maybe you could see the backside of the cylinder. So you can imagine this kind of looks like a soda can. And let's say that the height of my cylinder H is equal to 8."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "This is the bottom right over here. If this was transparent, maybe you could see the backside of the cylinder. So you can imagine this kind of looks like a soda can. And let's say that the height of my cylinder H is equal to 8. I'll give it some units. 8 centimeters. That is my height."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "And let's say that the height of my cylinder H is equal to 8. I'll give it some units. 8 centimeters. That is my height. And then let's say that the radius of one of these, of the top of my cylinder, of my soda can, let's say that this radius over here, is equal to 4 centimeters. So what is the volume here? What is the volume going to be?"}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "That is my height. And then let's say that the radius of one of these, of the top of my cylinder, of my soda can, let's say that this radius over here, is equal to 4 centimeters. So what is the volume here? What is the volume going to be? And the idea here is really the exact same thing that we saw in some of the previous problems. You can find the surface area of one side and then figure out how deep it goes. You'll be able to figure out the volume."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "What is the volume going to be? And the idea here is really the exact same thing that we saw in some of the previous problems. You can find the surface area of one side and then figure out how deep it goes. You'll be able to figure out the volume. So what we're going to do here is figure out the surface area of the top of this cylinder, or the top of this soda can. And then we're going to multiply it by its height, and that'll give us a volume. This will tell us essentially how many square centimeters fit in this top."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "You'll be able to figure out the volume. So what we're going to do here is figure out the surface area of the top of this cylinder, or the top of this soda can. And then we're going to multiply it by its height, and that'll give us a volume. This will tell us essentially how many square centimeters fit in this top. And then if we multiply that by how many centimeters we go down, then that will give us the number of cubic centimeters in this cylinder or soda can. So how do we figure out this area up here? Well, the area of the top, this is just finding the area of a circle."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "This will tell us essentially how many square centimeters fit in this top. And then if we multiply that by how many centimeters we go down, then that will give us the number of cubic centimeters in this cylinder or soda can. So how do we figure out this area up here? Well, the area of the top, this is just finding the area of a circle. You could imagine drawing it like this. If we were to just look at it straight on, that's a circle with a radius of 4 centimeters. The area of a circle with a radius 4 centimeters, area is equal to pi r squared."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "Well, the area of the top, this is just finding the area of a circle. You could imagine drawing it like this. If we were to just look at it straight on, that's a circle with a radius of 4 centimeters. The area of a circle with a radius 4 centimeters, area is equal to pi r squared. So it's going to be pi times the radius squared, times 4 centimeters squared, which is equal to 4 squared is 16, times pi. And our units now are going to be centimeters squared. Or another way to think of these are square centimeters."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "The area of a circle with a radius 4 centimeters, area is equal to pi r squared. So it's going to be pi times the radius squared, times 4 centimeters squared, which is equal to 4 squared is 16, times pi. And our units now are going to be centimeters squared. Or another way to think of these are square centimeters. So that's the area. The volume is going to be this area times the height. So the volume is going to be equal to 16 pi centimeters squared, times the height, times 8 centimeters."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "Or another way to think of these are square centimeters. So that's the area. The volume is going to be this area times the height. So the volume is going to be equal to 16 pi centimeters squared, times the height, times 8 centimeters. And so when you do multiplication, the associative property, you can kind of rearrange these things and the commutative property. Doesn't matter what order you do if it's all multiplication. So this is the same thing as 16 times 8."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "So the volume is going to be equal to 16 pi centimeters squared, times the height, times 8 centimeters. And so when you do multiplication, the associative property, you can kind of rearrange these things and the commutative property. Doesn't matter what order you do if it's all multiplication. So this is the same thing as 16 times 8. Let's see, 8 times 8 is 64. 16 times 8 is twice that. So it's going to be 128 pi."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "So this is the same thing as 16 times 8. Let's see, 8 times 8 is 64. 16 times 8 is twice that. So it's going to be 128 pi. And you have centimeters squared times centimeters. So that gives us centimeters cubed. Or 128 pi cubic centimeters."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "So it's going to be 128 pi. And you have centimeters squared times centimeters. So that gives us centimeters cubed. Or 128 pi cubic centimeters. Remember, pi is just a number. We write it as pi because it's kind of a crazy, irrational number that if you were to write it, you could never completely write pi 3.14159. Keeps going on, never repeats."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "Or 128 pi cubic centimeters. Remember, pi is just a number. We write it as pi because it's kind of a crazy, irrational number that if you were to write it, you could never completely write pi 3.14159. Keeps going on, never repeats. So we just leave it as pi. But if you wanted to figure it out, you can get a calculator. And this would be 3.14 roughly times 128."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "Keeps going on, never repeats. So we just leave it as pi. But if you wanted to figure it out, you can get a calculator. And this would be 3.14 roughly times 128. So it would be close to 400 cubic centimeters. Now, how would we find the surface area? How would we find the surface area of this figure over here?"}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "And this would be 3.14 roughly times 128. So it would be close to 400 cubic centimeters. Now, how would we find the surface area? How would we find the surface area of this figure over here? Well, part of the surface area of the two surfaces, the top and the bottom. So that would be part of the surface area. And then the bottom over here would also be part of the surface area."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "How would we find the surface area of this figure over here? Well, part of the surface area of the two surfaces, the top and the bottom. So that would be part of the surface area. And then the bottom over here would also be part of the surface area. So if we're trying to find the surface area, let's do surface. Let's find the surface area of our cylinder. It's definitely going to have both of these areas here."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "And then the bottom over here would also be part of the surface area. So if we're trying to find the surface area, let's do surface. Let's find the surface area of our cylinder. It's definitely going to have both of these areas here. So it's going to have the 16 pi centimeters squared twice. This is 16 pi. This is 16 pi square centimeters."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "It's definitely going to have both of these areas here. So it's going to have the 16 pi centimeters squared twice. This is 16 pi. This is 16 pi square centimeters. So it's going to have 2 times 16 pi centimeters squared. I'll keep the units still. So that covers the top and the bottom of our soda can."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "This is 16 pi square centimeters. So it's going to have 2 times 16 pi centimeters squared. I'll keep the units still. So that covers the top and the bottom of our soda can. And now we have to figure out the surface area of this thing that goes around. And the way I imagine it is, imagine if you were trying to wrap this thing with wrapping paper. So let me just draw a little dotted line here."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "So that covers the top and the bottom of our soda can. And now we have to figure out the surface area of this thing that goes around. And the way I imagine it is, imagine if you were trying to wrap this thing with wrapping paper. So let me just draw a little dotted line here. So imagine if you were to cut it just like that, cut the side of the soda can. And if you were to kind of unwind this thing that goes around it, what would you have? Well, you would have something."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "So let me just draw a little dotted line here. So imagine if you were to cut it just like that, cut the side of the soda can. And if you were to kind of unwind this thing that goes around it, what would you have? Well, you would have something. You would end up with a sheet of paper where this length right over here is the same thing as this length over here. And then it would be completely unwound. And then these two ends, let me do this in magenta, these two ends used to touch each other."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "Well, you would have something. You would end up with a sheet of paper where this length right over here is the same thing as this length over here. And then it would be completely unwound. And then these two ends, let me do this in magenta, these two ends used to touch each other. And let me do it in a color that I haven't used yet. I'll do it in pink. These two ends used to touch each other when it was all rolled together."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "And then these two ends, let me do this in magenta, these two ends used to touch each other. And let me do it in a color that I haven't used yet. I'll do it in pink. These two ends used to touch each other when it was all rolled together. And they used to touch each other right over there. So the length of this side and that side is going to be the same thing as the height of my cylinder. So this is going to be 8 centimeters."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "These two ends used to touch each other when it was all rolled together. And they used to touch each other right over there. So the length of this side and that side is going to be the same thing as the height of my cylinder. So this is going to be 8 centimeters. And then this over here is also going to be 8 centimeters. And so the question we need to ask ourselves is, what is going to be this dimension right over here? And remember, that dimension is essentially how far did we go around the cylinder?"}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "So this is going to be 8 centimeters. And then this over here is also going to be 8 centimeters. And so the question we need to ask ourselves is, what is going to be this dimension right over here? And remember, that dimension is essentially how far did we go around the cylinder? Well, if you think about it, that's going to be the exact same thing as the circumference of either the top or the bottom of the cylinder. So what is the circumference? The circumference of this circle right over here, which is the same thing as the circumference of that circle over there, it is 2 times the radius times pi, or 2 pi times the radius, 2 pi times 4 centimeters, which is equal to 8 pi centimeters."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "And remember, that dimension is essentially how far did we go around the cylinder? Well, if you think about it, that's going to be the exact same thing as the circumference of either the top or the bottom of the cylinder. So what is the circumference? The circumference of this circle right over here, which is the same thing as the circumference of that circle over there, it is 2 times the radius times pi, or 2 pi times the radius, 2 pi times 4 centimeters, which is equal to 8 pi centimeters. So this distance right over here is the circumference of either the top or the bottom of the cylinder is going to be 8 pi centimeters. So if you want to find the surface area of just the wrapping, just the part that goes around the cylinder, not the top or the bottom, when you unwind it, it's going to look like this rectangle. And so its area, the area of just that part, is going to be equal to 8 centimeters times 8 pi centimeters."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "The circumference of this circle right over here, which is the same thing as the circumference of that circle over there, it is 2 times the radius times pi, or 2 pi times the radius, 2 pi times 4 centimeters, which is equal to 8 pi centimeters. So this distance right over here is the circumference of either the top or the bottom of the cylinder is going to be 8 pi centimeters. So if you want to find the surface area of just the wrapping, just the part that goes around the cylinder, not the top or the bottom, when you unwind it, it's going to look like this rectangle. And so its area, the area of just that part, is going to be equal to 8 centimeters times 8 pi centimeters. So let me do it this way. It's going to be 8 centimeters times 8 pi centimeters. And that's equal to 64 pi."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "And so its area, the area of just that part, is going to be equal to 8 centimeters times 8 pi centimeters. So let me do it this way. It's going to be 8 centimeters times 8 pi centimeters. And that's equal to 64 pi. 8 times 8 is 64. You have your pi centimeters squared. So when you want the surface area of the whole thing, you have the top, you have the bottom."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "And that's equal to 64 pi. 8 times 8 is 64. You have your pi centimeters squared. So when you want the surface area of the whole thing, you have the top, you have the bottom. We already threw those there. And then you want to find the area of the thing around. We just figured that out."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "So when you want the surface area of the whole thing, you have the top, you have the bottom. We already threw those there. And then you want to find the area of the thing around. We just figured that out. It's going to be plus 64 pi centimeters squared. And now we just have to calculate it. So this gives us 2 times 16 pi is going to be equal to 32."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "We just figured that out. It's going to be plus 64 pi centimeters squared. And now we just have to calculate it. So this gives us 2 times 16 pi is going to be equal to 32. That is 32 pi centimeters squared plus 64 pi. Let me scroll over to the right a little bit. Plus 64 pi centimeters squared."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "So this gives us 2 times 16 pi is going to be equal to 32. That is 32 pi centimeters squared plus 64 pi. Let me scroll over to the right a little bit. Plus 64 pi centimeters squared. And then 32 plus 64 is 96 pi centimeters squared. So it's equal to 96 pi square centimeters, which is going to be a little bit over 300 square centimeters. And notice, when we did surface area, we got our answer in terms of square centimeters."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "Plus 64 pi centimeters squared. And then 32 plus 64 is 96 pi centimeters squared. So it's equal to 96 pi square centimeters, which is going to be a little bit over 300 square centimeters. And notice, when we did surface area, we got our answer in terms of square centimeters. That makes sense, because surface area, it's a two-dimensional measurement. We think about how many square centimeters can we fit on the surface of the cylinder. When we did the volume, we got centimeters cubed, or cubic centimeters."}, {"video_title": "Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "And notice, when we did surface area, we got our answer in terms of square centimeters. That makes sense, because surface area, it's a two-dimensional measurement. We think about how many square centimeters can we fit on the surface of the cylinder. When we did the volume, we got centimeters cubed, or cubic centimeters. And that's because we're trying to calculate how many 1 by 1 centimeter cubes can we fit inside of this structure. And so that's why it's cubic centimeters. Anyway, hopefully that clarifies things up a little bit."}, {"video_title": "Proof perpendicular radius bisects chord.mp3", "Sentence": "We're also told that segment OD is perpendicular to this chord, to chord AC or to segment AC. And what we wanna prove is that segment OD bisects AC. So another way to think about it, it intersects AC at AC's midpoint. So pause this video and see if you can have a go at that. All right, now let's go through this together. And the way that I'm going to do this is by establishing two congruent triangles. And let me draw these triangles."}, {"video_title": "Proof perpendicular radius bisects chord.mp3", "Sentence": "So pause this video and see if you can have a go at that. All right, now let's go through this together. And the way that I'm going to do this is by establishing two congruent triangles. And let me draw these triangles. So I'm gonna draw one radius going from O to C and another from A to O. Now we know that the length AO is equal to OC because AO and OC both radii. In a circle, the length of the radius does not change."}, {"video_title": "Proof perpendicular radius bisects chord.mp3", "Sentence": "And let me draw these triangles. So I'm gonna draw one radius going from O to C and another from A to O. Now we know that the length AO is equal to OC because AO and OC both radii. In a circle, the length of the radius does not change. So I can put that right over there. And then we also know that OM is going to be congruent to itself. It's a side in both of these triangles."}, {"video_title": "Proof perpendicular radius bisects chord.mp3", "Sentence": "In a circle, the length of the radius does not change. So I can put that right over there. And then we also know that OM is going to be congruent to itself. It's a side in both of these triangles. So let me just write it this way. OM is going to be congruent to OM. And this is reflexivity, reflexivity, kind of obvious."}, {"video_title": "Proof perpendicular radius bisects chord.mp3", "Sentence": "It's a side in both of these triangles. So let me just write it this way. OM is going to be congruent to OM. And this is reflexivity, reflexivity, kind of obvious. It's going to be equal to itself. It's going to be congruent to itself. So you have it just like that."}, {"video_title": "Proof perpendicular radius bisects chord.mp3", "Sentence": "And this is reflexivity, reflexivity, kind of obvious. It's going to be equal to itself. It's going to be congruent to itself. So you have it just like that. And now we have two right triangles. How do I know they're right triangles? Well, they told us that segment OD is perpendicular to segment AC and our assumptions in our given."}, {"video_title": "Proof perpendicular radius bisects chord.mp3", "Sentence": "So you have it just like that. And now we have two right triangles. How do I know they're right triangles? Well, they told us that segment OD is perpendicular to segment AC and our assumptions in our given. Now, if you just had two triangles that had two pairs of congruent sides, that is not enough to establish congruency of the triangles. But if you're dealing with two right triangles, then it is enough. And there's two ways to think about it."}, {"video_title": "Proof perpendicular radius bisects chord.mp3", "Sentence": "Well, they told us that segment OD is perpendicular to segment AC and our assumptions in our given. Now, if you just had two triangles that had two pairs of congruent sides, that is not enough to establish congruency of the triangles. But if you're dealing with two right triangles, then it is enough. And there's two ways to think about it. We had thought about the RSH postulate, where if you have a right triangle or two right triangles, you have a pair of sides are congruent, a pair and the hypotenuses are congruent. That means that the two triangles are congruent. But another way to think about it, which is a little bit of common sense, is using the Pythagorean theorem."}, {"video_title": "Proof perpendicular radius bisects chord.mp3", "Sentence": "And there's two ways to think about it. We had thought about the RSH postulate, where if you have a right triangle or two right triangles, you have a pair of sides are congruent, a pair and the hypotenuses are congruent. That means that the two triangles are congruent. But another way to think about it, which is a little bit of common sense, is using the Pythagorean theorem. If you know two sides of a right triangle, the Pythagorean theorem would tell us that you could determine what the other side is. And so what we could say is, and let's just use RSH for now, but you could also say, we can use the Pythagorean theorem to establish that AM is going to be congruent to MC. But let me just write it this way."}, {"video_title": "Proof perpendicular radius bisects chord.mp3", "Sentence": "But another way to think about it, which is a little bit of common sense, is using the Pythagorean theorem. If you know two sides of a right triangle, the Pythagorean theorem would tell us that you could determine what the other side is. And so what we could say is, and let's just use RSH for now, but you could also say, we can use the Pythagorean theorem to establish that AM is going to be congruent to MC. But let me just write it this way. I will write that triangle AMO is congruent to triangle CMO by RSH. And if the triangles are congruent, then the corresponding sides must be congruent. So therefore, we know that AM, AM, segment AM is going to be, I'm having trouble writing congruent, is going to be congruent to segment CM, that these are going to have the same measure."}, {"video_title": "Proof perpendicular radius bisects chord.mp3", "Sentence": "But let me just write it this way. I will write that triangle AMO is congruent to triangle CMO by RSH. And if the triangles are congruent, then the corresponding sides must be congruent. So therefore, we know that AM, AM, segment AM is going to be, I'm having trouble writing congruent, is going to be congruent to segment CM, that these are going to have the same measure. And if they have the same measure, we have just shown that M is the midpoint of AC or that OD bisects AC. So let me just write it that way. Therefore, OD bisects AC, segment OD bisects segment AC."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "It looks something like this. ABC. I want to think about the minimum amount of information. I want to come up with a couple of postulates that we can use to determine whether another triangle is similar to triangle ABC. So we already know that if all three angles, all three of the corresponding angles are congruent to the corresponding angles on ABC, then we know that we're dealing with congruent triangles. For example, if this is 30 degrees, this angle is 90 degrees, and this angle right over here is 60 degrees, and we have another triangle that looks like this, that looks like this. It's clearly a smaller triangle, but its corresponding angles, so this is 30 degrees, this is 90 degrees, and this is 60 degrees, we know that XYZ, in this case, is going to be similar to ABC."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "I want to come up with a couple of postulates that we can use to determine whether another triangle is similar to triangle ABC. So we already know that if all three angles, all three of the corresponding angles are congruent to the corresponding angles on ABC, then we know that we're dealing with congruent triangles. For example, if this is 30 degrees, this angle is 90 degrees, and this angle right over here is 60 degrees, and we have another triangle that looks like this, that looks like this. It's clearly a smaller triangle, but its corresponding angles, so this is 30 degrees, this is 90 degrees, and this is 60 degrees, we know that XYZ, in this case, is going to be similar to ABC. We would know from this, because corresponding angles are congruent, we would know that triangle ABC is similar to triangle XYZ. You've got to get the order right to make sure that you have the right corresponding angles. Y corresponds to the 90 degree angle, X corresponds to the 30 degree angle, A corresponds to the 30 degree angle, so A and X are the first two things, B and Y, which are the 90 degrees, are the second two, and then Z is the last one."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "It's clearly a smaller triangle, but its corresponding angles, so this is 30 degrees, this is 90 degrees, and this is 60 degrees, we know that XYZ, in this case, is going to be similar to ABC. We would know from this, because corresponding angles are congruent, we would know that triangle ABC is similar to triangle XYZ. You've got to get the order right to make sure that you have the right corresponding angles. Y corresponds to the 90 degree angle, X corresponds to the 30 degree angle, A corresponds to the 30 degree angle, so A and X are the first two things, B and Y, which are the 90 degrees, are the second two, and then Z is the last one. That's what we know already, if you have three angles. But do you need three angles? If we only knew two of the angles, would that be enough?"}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "Y corresponds to the 90 degree angle, X corresponds to the 30 degree angle, A corresponds to the 30 degree angle, so A and X are the first two things, B and Y, which are the 90 degrees, are the second two, and then Z is the last one. That's what we know already, if you have three angles. But do you need three angles? If we only knew two of the angles, would that be enough? Sure, because if you know two angles for a triangle, you know the third. For example, if I have another triangle, if I have a triangle that looks like this, and if I told you that only two of the corresponding angles are congruent, maybe this angle right here is congruent to this angle, and that angle right there is congruent to that angle. Is that enough to say that these two triangles are similar?"}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "If we only knew two of the angles, would that be enough? Sure, because if you know two angles for a triangle, you know the third. For example, if I have another triangle, if I have a triangle that looks like this, and if I told you that only two of the corresponding angles are congruent, maybe this angle right here is congruent to this angle, and that angle right there is congruent to that angle. Is that enough to say that these two triangles are similar? Sure, because in a triangle, if you know two of the angles, then you know what the last angle has to be. If you know that this is 30 and you know that that is 90, then you know that this angle has to be 60 degrees. Whatever these two angles are, subtract them from 180, and that's going to be this angle."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "Is that enough to say that these two triangles are similar? Sure, because in a triangle, if you know two of the angles, then you know what the last angle has to be. If you know that this is 30 and you know that that is 90, then you know that this angle has to be 60 degrees. Whatever these two angles are, subtract them from 180, and that's going to be this angle. In general, in order to show similarity, you don't have to show three corresponding angles are congruent. You really just have to show two. This will be the first of our similarity postulates."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "Whatever these two angles are, subtract them from 180, and that's going to be this angle. In general, in order to show similarity, you don't have to show three corresponding angles are congruent. You really just have to show two. This will be the first of our similarity postulates. We've called it angle-angle. If you could show that two corresponding angles are congruent, then we're dealing with similar triangles. For example, just to put some numbers here, if this was 30 degrees, and we know that on this triangle this is 90 degrees right over here, we know that this triangle right over here is similar to that one there."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "This will be the first of our similarity postulates. We've called it angle-angle. If you could show that two corresponding angles are congruent, then we're dealing with similar triangles. For example, just to put some numbers here, if this was 30 degrees, and we know that on this triangle this is 90 degrees right over here, we know that this triangle right over here is similar to that one there. You can really just go to the third angle in a pretty straightforward way. You say, hey, this third angle is 60 degrees, so all three angles are the same. That's one of our constraints for similarity."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "For example, just to put some numbers here, if this was 30 degrees, and we know that on this triangle this is 90 degrees right over here, we know that this triangle right over here is similar to that one there. You can really just go to the third angle in a pretty straightforward way. You say, hey, this third angle is 60 degrees, so all three angles are the same. That's one of our constraints for similarity. The other thing we know about similarity is that the ratio between all of the sides are going to be the same. For example, if we have another triangle right over here, let me draw another triangle. I'll call this triangle x, y, and z."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "That's one of our constraints for similarity. The other thing we know about similarity is that the ratio between all of the sides are going to be the same. For example, if we have another triangle right over here, let me draw another triangle. I'll call this triangle x, y, and z. Let's say that we know that the ratio between AB and xy, we know that AB over xy, so the ratio between this side and this side, notice we're not saying that they're congruent, we're just saying that they're ratio. We're looking at the ratio now. We're saying AB over xy, let's say that that is equal to BC over yz, and that is equal to AC over xz."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "I'll call this triangle x, y, and z. Let's say that we know that the ratio between AB and xy, we know that AB over xy, so the ratio between this side and this side, notice we're not saying that they're congruent, we're just saying that they're ratio. We're looking at the ratio now. We're saying AB over xy, let's say that that is equal to BC over yz, and that is equal to AC over xz. Once again, this is one of the ways that we say, hey, this means similarity. If we have all three corresponding sides, the ratio between all three corresponding sides are the same, then we know we are dealing with similar triangles. This is what we call side-side-side similarity."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "We're saying AB over xy, let's say that that is equal to BC over yz, and that is equal to AC over xz. Once again, this is one of the ways that we say, hey, this means similarity. If we have all three corresponding sides, the ratio between all three corresponding sides are the same, then we know we are dealing with similar triangles. This is what we call side-side-side similarity. You don't want to get these confused with side-side-side congruence. These are all of our similarity postulates, or axioms, or things that we're going to assume and then we're going to build off of them to solve problems and prove other things. Side-side-side, when we're talking about congruence, means that the corresponding sides are congruent."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "This is what we call side-side-side similarity. You don't want to get these confused with side-side-side congruence. These are all of our similarity postulates, or axioms, or things that we're going to assume and then we're going to build off of them to solve problems and prove other things. Side-side-side, when we're talking about congruence, means that the corresponding sides are congruent. Side-side-side for similarity, we're saying that the ratio between corresponding sides are going to be the same. For example, if this right over here is 10, let's say this is 60, this right over here is 30, and this right over here is 30 square roots of 3. I just made those numbers right because we will soon learn what typical ratios are of the sides of 30, 60, 90 triangles."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "Side-side-side, when we're talking about congruence, means that the corresponding sides are congruent. Side-side-side for similarity, we're saying that the ratio between corresponding sides are going to be the same. For example, if this right over here is 10, let's say this is 60, this right over here is 30, and this right over here is 30 square roots of 3. I just made those numbers right because we will soon learn what typical ratios are of the sides of 30, 60, 90 triangles. Let's say this one over here is 6, 3, and 3 square roots of 3. Notice, AB over xy, 30 square roots of 3 over 3 square roots of 3, this will be 10. What is BC over xy?"}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "I just made those numbers right because we will soon learn what typical ratios are of the sides of 30, 60, 90 triangles. Let's say this one over here is 6, 3, and 3 square roots of 3. Notice, AB over xy, 30 square roots of 3 over 3 square roots of 3, this will be 10. What is BC over xy? 30 divided by 3 is 10. What is 60 divided by 6? AC over xz, that's going to be 10."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "What is BC over xy? 30 divided by 3 is 10. What is 60 divided by 6? AC over xz, that's going to be 10. In general, to go from the corresponding side here to the corresponding side there, we always multiply by 10 on every side. We're not saying the sides are the same for this side-side-side for similarity. We're saying that we're really just scaling them up by the same amount."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "AC over xz, that's going to be 10. In general, to go from the corresponding side here to the corresponding side there, we always multiply by 10 on every side. We're not saying the sides are the same for this side-side-side for similarity. We're saying that we're really just scaling them up by the same amount. Or, another way to think about it, the ratio between corresponding sides are the same. Now what about if we had, let's start another triangle right over here. Let me draw it like this."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "We're saying that we're really just scaling them up by the same amount. Or, another way to think about it, the ratio between corresponding sides are the same. Now what about if we had, let's start another triangle right over here. Let me draw it like this. Actually, I want to leave this here so we can have our list. Let me draw another triangle ABC. Let's draw another triangle ABC."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "Let me draw it like this. Actually, I want to leave this here so we can have our list. Let me draw another triangle ABC. Let's draw another triangle ABC. This is A, B, and C. Let's say that we know that this side, when we go to another triangle, we know that xy is AB multiplied by some constant. I can write it over here. xy is equal to some constant times AB."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "Let's draw another triangle ABC. This is A, B, and C. Let's say that we know that this side, when we go to another triangle, we know that xy is AB multiplied by some constant. I can write it over here. xy is equal to some constant times AB. Actually, let me make xy bigger. It doesn't have to be. That constant could be less than 1, in which case it would be a smaller value."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "xy is equal to some constant times AB. Actually, let me make xy bigger. It doesn't have to be. That constant could be less than 1, in which case it would be a smaller value. Let me just do it that way. Let me just make xy look a little bit bigger. Let's say that this is x and that is y."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "That constant could be less than 1, in which case it would be a smaller value. Let me just do it that way. Let me just make xy look a little bit bigger. Let's say that this is x and that is y. Let's say that we know that xy over AB is equal to some constant. Or, if you multiply both sides by AB, you would get xy is some scaled-up version of AB. Maybe AB is 5, xy is 10, then our constant would be 2."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "Let's say that this is x and that is y. Let's say that we know that xy over AB is equal to some constant. Or, if you multiply both sides by AB, you would get xy is some scaled-up version of AB. Maybe AB is 5, xy is 10, then our constant would be 2. We scaled it up by a factor of 2. Let's say we also know that angle ABC is congruent to angle XYZ. I'll add another point over here."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "Maybe AB is 5, xy is 10, then our constant would be 2. We scaled it up by a factor of 2. Let's say we also know that angle ABC is congruent to angle XYZ. I'll add another point over here. Let me draw another side right over here. This is Z. Let's say we also know that angle ABC is congruent to XYZ."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "I'll add another point over here. Let me draw another side right over here. This is Z. Let's say we also know that angle ABC is congruent to XYZ. Let's say we know that the ratio between BC and YZ is also this constant. The ratio between BC and YZ is also equal to the same constant. In the example where this is 5 and 10, maybe this is 3 and 6."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "Let's say we also know that angle ABC is congruent to XYZ. Let's say we know that the ratio between BC and YZ is also this constant. The ratio between BC and YZ is also equal to the same constant. In the example where this is 5 and 10, maybe this is 3 and 6. The constant, we're doubling the length of the side. Is this triangle, is triangle XYZ going to be similar? If you think about it, if you say that this is some multiple, if XY is the same multiple of AB as YZ is the multiple of BC, and the angle in between is congruent, there's only one triangle we can set up over here."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "In the example where this is 5 and 10, maybe this is 3 and 6. The constant, we're doubling the length of the side. Is this triangle, is triangle XYZ going to be similar? If you think about it, if you say that this is some multiple, if XY is the same multiple of AB as YZ is the multiple of BC, and the angle in between is congruent, there's only one triangle we can set up over here. We're only constrained to one triangle right over here. We're completely constraining the length of this side. The length of this side is going to have to be that same scale as that over there."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "If you think about it, if you say that this is some multiple, if XY is the same multiple of AB as YZ is the multiple of BC, and the angle in between is congruent, there's only one triangle we can set up over here. We're only constrained to one triangle right over here. We're completely constraining the length of this side. The length of this side is going to have to be that same scale as that over there. We call that side-angle-side similarity. Once again, we saw SSS and SAS in our congruence postulates, but we're saying something very different here. We're saying that in SAS, if the ratio between corresponding sides of the true triangle are the same, so AB and XY of one corresponding side, and then another corresponding side, that's that second side, so that's between BC and YZ, and the angle between them are congruent, then we're saying it's similar."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "The length of this side is going to have to be that same scale as that over there. We call that side-angle-side similarity. Once again, we saw SSS and SAS in our congruence postulates, but we're saying something very different here. We're saying that in SAS, if the ratio between corresponding sides of the true triangle are the same, so AB and XY of one corresponding side, and then another corresponding side, that's that second side, so that's between BC and YZ, and the angle between them are congruent, then we're saying it's similar. For SAS for congruency, we said that the sides actually had to be congruent. Here we're saying that the ratio between the corresponding sides just has to be the same. For example, SAS, just to apply it, if I have, let me just show some examples here."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "We're saying that in SAS, if the ratio between corresponding sides of the true triangle are the same, so AB and XY of one corresponding side, and then another corresponding side, that's that second side, so that's between BC and YZ, and the angle between them are congruent, then we're saying it's similar. For SAS for congruency, we said that the sides actually had to be congruent. Here we're saying that the ratio between the corresponding sides just has to be the same. For example, SAS, just to apply it, if I have, let me just show some examples here. Let's say I have a triangle here that is 3, 2, 4. Let's say we have another triangle here that has length 9, 6, and we also know that the angle in between are congruent, so that that angle is equal to that angle. What SAS in the similarity world tells you is that these triangles are definitely going to be similar triangles, that we're actually constraining, because there's actually only one triangle we can draw right over here."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "For example, SAS, just to apply it, if I have, let me just show some examples here. Let's say I have a triangle here that is 3, 2, 4. Let's say we have another triangle here that has length 9, 6, and we also know that the angle in between are congruent, so that that angle is equal to that angle. What SAS in the similarity world tells you is that these triangles are definitely going to be similar triangles, that we're actually constraining, because there's actually only one triangle we can draw right over here. It's a triangle where all of the sides are going to have to be scaled up by the same amount. There's only one long side right here that we can actually draw, and that's going to have to be scaled up by 3 as well. This is the only possible triangle."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "What SAS in the similarity world tells you is that these triangles are definitely going to be similar triangles, that we're actually constraining, because there's actually only one triangle we can draw right over here. It's a triangle where all of the sides are going to have to be scaled up by the same amount. There's only one long side right here that we can actually draw, and that's going to have to be scaled up by 3 as well. This is the only possible triangle. If you constrain this side, you're saying, look, this is 3 times that side, this is 3 times that side, and the angle between them is congruent. There's only one triangle we can make, and we know that there is a similar triangle there, where everything is scaled up by a factor of 3, so that one triangle we can draw has to be that one similar triangle. This is what we're talking about SAS."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "This is the only possible triangle. If you constrain this side, you're saying, look, this is 3 times that side, this is 3 times that side, and the angle between them is congruent. There's only one triangle we can make, and we know that there is a similar triangle there, where everything is scaled up by a factor of 3, so that one triangle we can draw has to be that one similar triangle. This is what we're talking about SAS. We're not saying that this side is congruent to that side, or that side is congruent to that side. We're saying that they're scaled up by the same factor. If we had another triangle that looked like this, so maybe this is 9, this is 4, and the angle between them were congruent, you couldn't say that they're similar, because this side is scaled up by a factor of 3, this side is only scaled up by a factor of 2, so this one right over there, you could not say that it is necessarily similar."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "This is what we're talking about SAS. We're not saying that this side is congruent to that side, or that side is congruent to that side. We're saying that they're scaled up by the same factor. If we had another triangle that looked like this, so maybe this is 9, this is 4, and the angle between them were congruent, you couldn't say that they're similar, because this side is scaled up by a factor of 3, this side is only scaled up by a factor of 2, so this one right over there, you could not say that it is necessarily similar. Likewise, if you had a triangle that had length 9 here and length 6 there, but you did not know that these two angles are the same, once again, you're not constraining this enough, and you would not know that those two triangles are necessarily similar, because you don't know that middle angle is the same. You might be saying, well, there were a few other postulates that we had. We had AAS when we dealt with congruency, but if you think about it, we've already shown that two angles by themselves are enough to show similarity, so why worry about an angle and angle and a side, or the ratio between the sides?"}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "If we had another triangle that looked like this, so maybe this is 9, this is 4, and the angle between them were congruent, you couldn't say that they're similar, because this side is scaled up by a factor of 3, this side is only scaled up by a factor of 2, so this one right over there, you could not say that it is necessarily similar. Likewise, if you had a triangle that had length 9 here and length 6 there, but you did not know that these two angles are the same, once again, you're not constraining this enough, and you would not know that those two triangles are necessarily similar, because you don't know that middle angle is the same. You might be saying, well, there were a few other postulates that we had. We had AAS when we dealt with congruency, but if you think about it, we've already shown that two angles by themselves are enough to show similarity, so why worry about an angle and angle and a side, or the ratio between the sides? Why even worry about that? We also had angle, side, angle, and congruence, but once again, we already know that two angles are enough, so we don't need to throw in this extra side, so we don't even need this right over here. So these are going to be our similarity postulates, and I want to remind you, side, side, side, this is different than the side, side, side for congruence."}, {"video_title": "Similarity postulates Similarity Geometry Khan Academy.mp3", "Sentence": "We had AAS when we dealt with congruency, but if you think about it, we've already shown that two angles by themselves are enough to show similarity, so why worry about an angle and angle and a side, or the ratio between the sides? Why even worry about that? We also had angle, side, angle, and congruence, but once again, we already know that two angles are enough, so we don't need to throw in this extra side, so we don't even need this right over here. So these are going to be our similarity postulates, and I want to remind you, side, side, side, this is different than the side, side, side for congruence. We're talking about the ratio between corresponding sides. We're not saying that they're actually congruent, and here, side, angle, side, it's different than the side, angle, side for congruence. It's kind of related, but here we're talking about the ratio between the sides, not the actual measures."}, {"video_title": "Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3", "Sentence": "Let's call this line L. And we see at point A is the point that the tangent line intersects with the circle. And then we've drawn a radius from the center of the circle to point A. Now what we want to do in this video is prove to ourselves that this radius and that this tangent line intersect at a right angle. We want to prove to ourselves that they intersect at a right angle. And the first step to doing that is we're going to feel good, we're going to prove to ourselves that point A is the closest point on line L to the center of our circle. So I want to prove that point A is closest point on L to point O. And I encourage you to pause the video and see if you can prove that to yourself."}, {"video_title": "Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3", "Sentence": "We want to prove to ourselves that they intersect at a right angle. And the first step to doing that is we're going to feel good, we're going to prove to ourselves that point A is the closest point on line L to the center of our circle. So I want to prove that point A is closest point on L to point O. And I encourage you to pause the video and see if you can prove that to yourself. Well, to think about that, just think about any other point on line L. Pick any other arbitrary point on line L. It could be this point right over here, it could be this point right over here, it could be this point right over here. And you immediately see that it sits outside of the circle. And if it's sitting outside of the circle, I'll pick this point here just so it'll become a little bit clearer on our diagram."}, {"video_title": "Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3", "Sentence": "And I encourage you to pause the video and see if you can prove that to yourself. Well, to think about that, just think about any other point on line L. Pick any other arbitrary point on line L. It could be this point right over here, it could be this point right over here, it could be this point right over here. And you immediately see that it sits outside of the circle. And if it's sitting outside of the circle, I'll pick this point here just so it'll become a little bit clearer on our diagram. If it's sitting outside of the circle, in order to get from point O to this point, we'll call this point B right over here, you have to go the length of the radius, and then you have to go some more. So this length, the length of segment OB, is clearly going to be longer than the length of the radius, because you have to go to the radius to get to the circle itself, and then you have to go a little bit further for any point that sits outside of the circle. So point A is the only point, by definition, this is a tangent line, it's the only point that sits on the circle."}, {"video_title": "Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3", "Sentence": "And if it's sitting outside of the circle, I'll pick this point here just so it'll become a little bit clearer on our diagram. If it's sitting outside of the circle, in order to get from point O to this point, we'll call this point B right over here, you have to go the length of the radius, and then you have to go some more. So this length, the length of segment OB, is clearly going to be longer than the length of the radius, because you have to go to the radius to get to the circle itself, and then you have to go a little bit further for any point that sits outside of the circle. So point A is the only point, by definition, this is a tangent line, it's the only point that sits on the circle. Every other point on line L sits outside of the circle. So it's going to be further. So point A, hopefully this makes you feel good, because if you pick any other point, it's going to sit outside of the circle, so you have to go to the radius and then some."}, {"video_title": "Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3", "Sentence": "So point A is the only point, by definition, this is a tangent line, it's the only point that sits on the circle. Every other point on line L sits outside of the circle. So it's going to be further. So point A, hopefully this makes you feel good, because if you pick any other point, it's going to sit outside of the circle, so you have to go to the radius and then some. So hopefully this makes you feel good that point A is the closest point on L to the center of the circle. Now, we're not done yet. We now have to convince ourselves that if we have a point and a line, that the segment connecting the point to the closest point on the line to that original point, that that's going to be perpendicular to the line."}, {"video_title": "Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3", "Sentence": "So point A, hopefully this makes you feel good, because if you pick any other point, it's going to sit outside of the circle, so you have to go to the radius and then some. So hopefully this makes you feel good that point A is the closest point on L to the center of the circle. Now, we're not done yet. We now have to convince ourselves that if we have a point and a line, that the segment connecting the point to the closest point on the line to that original point, that that's going to be perpendicular to the line. So let me give ourselves some space here. We want to prove that the segment connecting a point off the line and closest point on the line is perpendicular to the line. So what we want to do is we want to say, hey, if we have some line here, L, and if you were to take a point off the line, so let's say that's this point right over here, point O, and you want the segment connecting the point off the line to the closest point on the line."}, {"video_title": "Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3", "Sentence": "We now have to convince ourselves that if we have a point and a line, that the segment connecting the point to the closest point on the line to that original point, that that's going to be perpendicular to the line. So let me give ourselves some space here. We want to prove that the segment connecting a point off the line and closest point on the line is perpendicular to the line. So what we want to do is we want to say, hey, if we have some line here, L, and if you were to take a point off the line, so let's say that's this point right over here, point O, and you want the segment connecting the point off the line to the closest point on the line. So the closest point on the line, so let's say that this is the closest point on the line, we want to feel good that this segment connecting them, so let me do this in a new color, that the segment connecting them is going to be perpendicular to the line, that it goes just straight down like that, that it's going to be perpendicular. And I'm going to prove this by contradiction. I'm going to assume that it's not perpendicular."}, {"video_title": "Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3", "Sentence": "So what we want to do is we want to say, hey, if we have some line here, L, and if you were to take a point off the line, so let's say that's this point right over here, point O, and you want the segment connecting the point off the line to the closest point on the line. So the closest point on the line, so let's say that this is the closest point on the line, we want to feel good that this segment connecting them, so let me do this in a new color, that the segment connecting them is going to be perpendicular to the line, that it goes just straight down like that, that it's going to be perpendicular. And I'm going to prove this by contradiction. I'm going to assume that it's not perpendicular. So assume that the segment connecting a point off line and closest point to the line is not perpendicular to the line. So how can I visualize that? Well, I could draw my line right over here, so that's line L, and let's say I have my point O right over here, point O."}, {"video_title": "Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3", "Sentence": "I'm going to assume that it's not perpendicular. So assume that the segment connecting a point off line and closest point to the line is not perpendicular to the line. So how can I visualize that? Well, I could draw my line right over here, so that's line L, and let's say I have my point O right over here, point O. And let's say the closest point on line L, 2.0, let's say that it's not, so let's say it's over here, that if I were to connect these two points, that it's not perpendicular to line L. So this is the closest point, let's call this point A, and let's say that the segment connecting these two is not perpendicular to the line. So let me get, so let's assume this is not perpendicular. This angle right here is not 90 degrees."}, {"video_title": "Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, I could draw my line right over here, so that's line L, and let's say I have my point O right over here, point O. And let's say the closest point on line L, 2.0, let's say that it's not, so let's say it's over here, that if I were to connect these two points, that it's not perpendicular to line L. So this is the closest point, let's call this point A, and let's say that the segment connecting these two is not perpendicular to the line. So let me get, so let's assume this is not perpendicular. This angle right here is not 90 degrees. So if we assume that, the reason why this is going to be a proof by contradiction is I can show that this is not 90 degrees, that I can always find a point that is going to be closer, another point on line L that is going to be closer to point O, which contradicts the fact that this was supposed to be the closest point. A was supposed to be the closest point on line L to O. And how do I always find a closer point?"}, {"video_title": "Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3", "Sentence": "This angle right here is not 90 degrees. So if we assume that, the reason why this is going to be a proof by contradiction is I can show that this is not 90 degrees, that I can always find a point that is going to be closer, another point on line L that is going to be closer to point O, which contradicts the fact that this was supposed to be the closest point. A was supposed to be the closest point on line L to O. And how do I always find a closer point? Well, I construct a right triangle. I can construct a right triangle just like this. I can construct a right triangle like that."}, {"video_title": "Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3", "Sentence": "And how do I always find a closer point? Well, I construct a right triangle. I can construct a right triangle just like this. I can construct a right triangle like that. And we see that this distance, let's call this distance right over here A, and let's call the base of this triangle B, let me do this in a different color. So A, B, that's the base of the right triangle, and the hypotenuse is the distance from O to A. We could call that C. We know from the Pythagorean theorem that A squared plus B squared is going to be equal to C squared."}, {"video_title": "Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3", "Sentence": "I can construct a right triangle like that. And we see that this distance, let's call this distance right over here A, and let's call the base of this triangle B, let me do this in a different color. So A, B, that's the base of the right triangle, and the hypotenuse is the distance from O to A. We could call that C. We know from the Pythagorean theorem that A squared plus B squared is going to be equal to C squared. And so B squared, if we have a non-degenerate triangle right over here, this is going to be some positive value over here. And so A is going to be less than C. So this gets us to the conclusion. Because if this is some positive value here, and A and C are positive, everything at positive distances, then this tells us that A has to be less than C, that a non-hypotenuse side of a right triangle, of a non-degenerate right triangle, assuming it has some area, is going to be shorter than the hypotenuse."}, {"video_title": "Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3", "Sentence": "We could call that C. We know from the Pythagorean theorem that A squared plus B squared is going to be equal to C squared. And so B squared, if we have a non-degenerate triangle right over here, this is going to be some positive value over here. And so A is going to be less than C. So this gets us to the conclusion. Because if this is some positive value here, and A and C are positive, everything at positive distances, then this tells us that A has to be less than C, that a non-hypotenuse side of a right triangle, of a non-degenerate right triangle, assuming it has some area, is going to be shorter than the hypotenuse. The hypotenuse is the longest side. So A is going to be less than C, and that will tell us if A is less than C, then we've found another point. Let's call this point, I've used a lot of letters here, let's call that point D. D is going to be closer."}, {"video_title": "Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3", "Sentence": "Because if this is some positive value here, and A and C are positive, everything at positive distances, then this tells us that A has to be less than C, that a non-hypotenuse side of a right triangle, of a non-degenerate right triangle, assuming it has some area, is going to be shorter than the hypotenuse. The hypotenuse is the longest side. So A is going to be less than C, and that will tell us if A is less than C, then we've found another point. Let's call this point, I've used a lot of letters here, let's call that point D. D is going to be closer. So we've just set up a contradiction. We assumed A was the closest point on line L to point O, but we assumed that the segment connecting them is not at a 90 degree angle. If it's not at a 90 degree angle, then we can drop a perpendicular and find a closer point, which is a contradiction to the fact that A was supposed to be the closer point."}, {"video_title": "Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3", "Sentence": "Let's call this point, I've used a lot of letters here, let's call that point D. D is going to be closer. So we've just set up a contradiction. We assumed A was the closest point on line L to point O, but we assumed that the segment connecting them is not at a 90 degree angle. If it's not at a 90 degree angle, then we can drop a perpendicular and find a closer point, which is a contradiction to the fact that A was supposed to be the closer point. So this leads to a contradiction. Because you can actually find this is not the closest point. You can always find a closer point."}, {"video_title": "Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3", "Sentence": "If it's not at a 90 degree angle, then we can drop a perpendicular and find a closer point, which is a contradiction to the fact that A was supposed to be the closer point. So this leads to a contradiction. Because you can actually find this is not the closest point. You can always find a closer point. So therefore, the segment connecting a point off the line to the closest point to the line must be perpendicular. So the segment connecting a point off the line to a closest point on the line, that must be perpendicular to the line. Just like that, we hopefully feel good about the idea that if you have a radius, and the point at which it intersects a tangent line to the circle, that forms a 90 degree angle, the radius and the tangent line."}, {"video_title": "Constructing a perpendicular line using a compass and straightedge Geometry Khan Academy.mp3", "Sentence": "And actually, it'll be a perpendicular bisector of the segment formed by those two points. Now, they don't care whether we're bisecting anything, but they do care about it being perpendicular. So let's do this. I'm going to draw a circle with my compass. And so let's just pick that point right over there. I could adjust the radius if I like, but I might as well. I'll just leave it right over there."}, {"video_title": "Constructing a perpendicular line using a compass and straightedge Geometry Khan Academy.mp3", "Sentence": "I'm going to draw a circle with my compass. And so let's just pick that point right over there. I could adjust the radius if I like, but I might as well. I'll just leave it right over there. Now, let me draw another circle. And this time, I'm going to center it where the first circle intersects with my line. And then I'm going to adjust the radius to overlap with the first dot, I should say."}, {"video_title": "Constructing a perpendicular line using a compass and straightedge Geometry Khan Academy.mp3", "Sentence": "I'll just leave it right over there. Now, let me draw another circle. And this time, I'm going to center it where the first circle intersects with my line. And then I'm going to adjust the radius to overlap with the first dot, I should say. And now, where these two circles intersect, those are points that are equidistant from both of these centers that I just constructed. So let me draw a line that connects those two. And that line is going to be perpendicular to our original line."}, {"video_title": "Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So the first thing that we can think about, these aren't just diagonals. These are lines that are intersecting parallel lines. So you can also view them as transversals. And if we focus on DB right over here, we see that it intersects DC and AB. And since those we know are parallel, this is a parallelogram, we know that alternate interior angles must be congruent. So that angle must be equal to that angle there. And let me make a label here."}, {"video_title": "Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "And if we focus on DB right over here, we see that it intersects DC and AB. And since those we know are parallel, this is a parallelogram, we know that alternate interior angles must be congruent. So that angle must be equal to that angle there. And let me make a label here. Let me call that middle point E. So we know that angle ABE must be congruent to angle CDE by alternate interior angles of a transversal intersecting parallel lines. Alternate interior angles. Now if we look at diagonal AC, or we should call it transversal AC, we can make the same argument."}, {"video_title": "Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "And let me make a label here. Let me call that middle point E. So we know that angle ABE must be congruent to angle CDE by alternate interior angles of a transversal intersecting parallel lines. Alternate interior angles. Now if we look at diagonal AC, or we should call it transversal AC, we can make the same argument. It intersects here and here. These two lines are parallel. So alternate interior angles must be congruent."}, {"video_title": "Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "Now if we look at diagonal AC, or we should call it transversal AC, we can make the same argument. It intersects here and here. These two lines are parallel. So alternate interior angles must be congruent. So angle DEC must be congruent to angle BAE for the exact same reason. Now we have something interesting. If we look at this top triangle over here and this bottom triangle, we have one set of corresponding angles that are congruent."}, {"video_title": "Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So alternate interior angles must be congruent. So angle DEC must be congruent to angle BAE for the exact same reason. Now we have something interesting. If we look at this top triangle over here and this bottom triangle, we have one set of corresponding angles that are congruent. We have a side in between that's going to be congruent. Actually, let me write that down explicitly. We know, and we've proved this to ourselves in the previous video, that parallelograms not only are opposite sides parallel, they are also congruent."}, {"video_title": "Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "If we look at this top triangle over here and this bottom triangle, we have one set of corresponding angles that are congruent. We have a side in between that's going to be congruent. Actually, let me write that down explicitly. We know, and we've proved this to ourselves in the previous video, that parallelograms not only are opposite sides parallel, they are also congruent. So we know from the previous video that that side is equal to that side. So let me go back to what I was saying. We have two sets of corresponding angles that are congruent."}, {"video_title": "Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "We know, and we've proved this to ourselves in the previous video, that parallelograms not only are opposite sides parallel, they are also congruent. So we know from the previous video that that side is equal to that side. So let me go back to what I was saying. We have two sets of corresponding angles that are congruent. We have a side in between that's congruent. And then we have another set of corresponding angles that are congruent. So we know that this triangle is congruent to that triangle by angle side angle."}, {"video_title": "Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "We have two sets of corresponding angles that are congruent. We have a side in between that's congruent. And then we have another set of corresponding angles that are congruent. So we know that this triangle is congruent to that triangle by angle side angle. So we know that triangle, I'm going to go from the blue to the orange to the last one, triangle ABE is congruent to triangle blue, orange, and the last one, CDE, by angle side angle congruency. Now, what does that do for us? Well, we know if two triangles are congruent, all of their corresponding features, especially all of their corresponding sides, are congruent."}, {"video_title": "Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So we know that this triangle is congruent to that triangle by angle side angle. So we know that triangle, I'm going to go from the blue to the orange to the last one, triangle ABE is congruent to triangle blue, orange, and the last one, CDE, by angle side angle congruency. Now, what does that do for us? Well, we know if two triangles are congruent, all of their corresponding features, especially all of their corresponding sides, are congruent. So we know that side EC corresponds to side EA. Or I could say side AE corresponds to side CE. Their corresponding sides of congruent triangles, so their measures or their lengths must be the same."}, {"video_title": "Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "Well, we know if two triangles are congruent, all of their corresponding features, especially all of their corresponding sides, are congruent. So we know that side EC corresponds to side EA. Or I could say side AE corresponds to side CE. Their corresponding sides of congruent triangles, so their measures or their lengths must be the same. So AE must be equal to CE. Let me put two slashes since I already used one slash over here. Now, by the same exact logic, we know that DE, we know that, let me focus on this, we know that BE must be equal to DE."}, {"video_title": "Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "Their corresponding sides of congruent triangles, so their measures or their lengths must be the same. So AE must be equal to CE. Let me put two slashes since I already used one slash over here. Now, by the same exact logic, we know that DE, we know that, let me focus on this, we know that BE must be equal to DE. Once again, so their corresponding sides of two congruent triangles, so they must have the same length. So this is corresponding sides of congruent triangles. So BE is equal to DE."}, {"video_title": "Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "Now, by the same exact logic, we know that DE, we know that, let me focus on this, we know that BE must be equal to DE. Once again, so their corresponding sides of two congruent triangles, so they must have the same length. So this is corresponding sides of congruent triangles. So BE is equal to DE. And we've done our proof. We've shown that, look, diagonal DB is splitting AC into two segments of equal length and vice versa. AC is splitting DB into two segments of equal length."}, {"video_title": "Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So BE is equal to DE. And we've done our proof. We've shown that, look, diagonal DB is splitting AC into two segments of equal length and vice versa. AC is splitting DB into two segments of equal length. So they are bisecting each other. Now let's go the other way around. Let's prove to ourselves that if we have two diagonals of a quadrilateral that are bisecting each other, that we are dealing with a parallelogram."}, {"video_title": "Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "AC is splitting DB into two segments of equal length. So they are bisecting each other. Now let's go the other way around. Let's prove to ourselves that if we have two diagonals of a quadrilateral that are bisecting each other, that we are dealing with a parallelogram. So let me see. So we're going to assume that the two diagonals are bisecting each other. So we're assuming that that is equal to that and that that right over there is equal to that."}, {"video_title": "Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "Let's prove to ourselves that if we have two diagonals of a quadrilateral that are bisecting each other, that we are dealing with a parallelogram. So let me see. So we're going to assume that the two diagonals are bisecting each other. So we're assuming that that is equal to that and that that right over there is equal to that. Given that, we want to prove that this is a parallelogram. And to do that, we just have to remind ourselves that this angle is going to be equal to that angle. It's one of the first things we learn, because they are vertical angles."}, {"video_title": "Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So we're assuming that that is equal to that and that that right over there is equal to that. Given that, we want to prove that this is a parallelogram. And to do that, we just have to remind ourselves that this angle is going to be equal to that angle. It's one of the first things we learn, because they are vertical angles. So let me write this down. C, let me label this point, CED, angle CED, is going to be equal to or is congruent to angle BEA. And what is that?"}, {"video_title": "Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "It's one of the first things we learn, because they are vertical angles. So let me write this down. C, let me label this point, CED, angle CED, is going to be equal to or is congruent to angle BEA. And what is that? Well, that shows us that these two triangles are congruent, because we have a corresponding size that are congruent, an angle in between, and then another side. So we now know that the triangle, I'll keep this in yellow, triangle AEB is congruent to triangle DEC by side angle, side congruency. By SAS, congruent triangles."}, {"video_title": "Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "And what is that? Well, that shows us that these two triangles are congruent, because we have a corresponding size that are congruent, an angle in between, and then another side. So we now know that the triangle, I'll keep this in yellow, triangle AEB is congruent to triangle DEC by side angle, side congruency. By SAS, congruent triangles. Fair enough. Now, if we know that two triangles are congruent, we know that all of the corresponding sides and angles are congruent. So for example, we know that angle CDE is going to be congruent to angle BAE."}, {"video_title": "Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "By SAS, congruent triangles. Fair enough. Now, if we know that two triangles are congruent, we know that all of the corresponding sides and angles are congruent. So for example, we know that angle CDE is going to be congruent to angle BAE. BAE is going to be congruent to angle BAE, and this is just corresponding angles. And now we have this transversal of these two lines that could be parallel if the alternate interior angles are congruent, and we see that they are. These two are kind of candidate alternate interior angles, and they are congruent."}, {"video_title": "Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So for example, we know that angle CDE is going to be congruent to angle BAE. BAE is going to be congruent to angle BAE, and this is just corresponding angles. And now we have this transversal of these two lines that could be parallel if the alternate interior angles are congruent, and we see that they are. These two are kind of candidate alternate interior angles, and they are congruent. So AB must be parallel to CD. AB is parallel to CD by alternate interior angles congruent of parallel lines. I'm just writing in some shorthand to forgive the cryptic nature of it."}, {"video_title": "Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "These two are kind of candidate alternate interior angles, and they are congruent. So AB must be parallel to CD. AB is parallel to CD by alternate interior angles congruent of parallel lines. I'm just writing in some shorthand to forgive the cryptic nature of it. I'm saying it out. And so we can then do the exact same logic. We've just shown that these two sides are parallel."}, {"video_title": "Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "I'm just writing in some shorthand to forgive the cryptic nature of it. I'm saying it out. And so we can then do the exact same logic. We've just shown that these two sides are parallel. We can then do the exact same logic to show that these two sides are parallel. And I won't necessarily write it all out, but it's the exact same proof to show that these two. So first of all, we know that this angle is congruent to that angle right over there."}, {"video_title": "Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "We've just shown that these two sides are parallel. We can then do the exact same logic to show that these two sides are parallel. And I won't necessarily write it all out, but it's the exact same proof to show that these two. So first of all, we know that this angle is congruent to that angle right over there. And then we know, actually let me write it out. So we know that angle AEC is congruent to angle DEB. They are vertical angles."}, {"video_title": "Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So first of all, we know that this angle is congruent to that angle right over there. And then we know, actually let me write it out. So we know that angle AEC is congruent to angle DEB. They are vertical angles. And then we see that triangle AEC must be congruent to triangle DEB by side angle side. So then we have triangle AEC must be congruent to triangle DEB by SAS congruency. Then we know that corresponding angles must be congruent."}, {"video_title": "Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "They are vertical angles. And then we see that triangle AEC must be congruent to triangle DEB by side angle side. So then we have triangle AEC must be congruent to triangle DEB by SAS congruency. Then we know that corresponding angles must be congruent. So that we know that angle, so for example, angle CAE must be congruent to angle BDE. And this is their corresponding angles of congruent triangles. So CAE, let me do this in a new color, must be congruent to BDE."}, {"video_title": "Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "Then we know that corresponding angles must be congruent. So that we know that angle, so for example, angle CAE must be congruent to angle BDE. And this is their corresponding angles of congruent triangles. So CAE, let me do this in a new color, must be congruent to BDE. And now we have a transversal. So the alternate interior angles are congruent. So the two lines that the transversal is intersecting must be parallel."}, {"video_title": "Proof Diagonals of a parallelogram bisect each other Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So CAE, let me do this in a new color, must be congruent to BDE. And now we have a transversal. So the alternate interior angles are congruent. So the two lines that the transversal is intersecting must be parallel. So this must be parallel to that. So then we have AC must be parallel to BD by alternate interior angles. And we're done."}, {"video_title": "Angle bisector theorem examples Geometry Khan Academy.mp3", "Sentence": "So in this first triangle right over here, we're given that this side has length 3, this side has length 6. And this little dotted line here, this is clearly the angle bisector, because they're telling us that this angle is congruent to that angle right over there. And then they tell us that the length of just this part of this side right over here is 2. So from here to here is 2. And that this length is x. So let's figure out what x is. So the angle bisector theorem tells us that the ratio of 3 to 2 is going to be equal to 6 to x."}, {"video_title": "Angle bisector theorem examples Geometry Khan Academy.mp3", "Sentence": "So from here to here is 2. And that this length is x. So let's figure out what x is. So the angle bisector theorem tells us that the ratio of 3 to 2 is going to be equal to 6 to x. And we can just solve for x. So 3 to 2 is going to be equal to 6 to x. And then once again, you can just cross multiply, or you can multiply both sides by 2 and x."}, {"video_title": "Angle bisector theorem examples Geometry Khan Academy.mp3", "Sentence": "So the angle bisector theorem tells us that the ratio of 3 to 2 is going to be equal to 6 to x. And we can just solve for x. So 3 to 2 is going to be equal to 6 to x. And then once again, you can just cross multiply, or you can multiply both sides by 2 and x. It kind of gives you the same result. If you cross multiply, you get 3x. 3x is equal to 2 times 6 is 12. x is equal to, divide both sides by 3. x is equal to 4."}, {"video_title": "Angle bisector theorem examples Geometry Khan Academy.mp3", "Sentence": "And then once again, you can just cross multiply, or you can multiply both sides by 2 and x. It kind of gives you the same result. If you cross multiply, you get 3x. 3x is equal to 2 times 6 is 12. x is equal to, divide both sides by 3. x is equal to 4. So in this case, x is equal to 4. And this is kind of interesting, because we just realize now that this side, this entire side right over here, is going to be equal to 6. So even though it doesn't look that way based on how it's drawn, this is actually an isosceles triangle."}, {"video_title": "Angle bisector theorem examples Geometry Khan Academy.mp3", "Sentence": "3x is equal to 2 times 6 is 12. x is equal to, divide both sides by 3. x is equal to 4. So in this case, x is equal to 4. And this is kind of interesting, because we just realize now that this side, this entire side right over here, is going to be equal to 6. So even though it doesn't look that way based on how it's drawn, this is actually an isosceles triangle. It has a 6 and a 6. And then the base right over here is 3. It's kind of interesting."}, {"video_title": "Angle bisector theorem examples Geometry Khan Academy.mp3", "Sentence": "So even though it doesn't look that way based on how it's drawn, this is actually an isosceles triangle. It has a 6 and a 6. And then the base right over here is 3. It's kind of interesting. Over here, we're given that this length is 5. This length is 7. This entire side is 10."}, {"video_title": "Angle bisector theorem examples Geometry Khan Academy.mp3", "Sentence": "It's kind of interesting. Over here, we're given that this length is 5. This length is 7. This entire side is 10. And then we have this angle bisector right over there. And we need to figure out just this part of the triangle between this point, if we call this point A, and this point right over here. We need to find the length of AB right over here."}, {"video_title": "Angle bisector theorem examples Geometry Khan Academy.mp3", "Sentence": "This entire side is 10. And then we have this angle bisector right over there. And we need to figure out just this part of the triangle between this point, if we call this point A, and this point right over here. We need to find the length of AB right over here. So once again, angle bisector theorem, the ratio of 5 to this. Let me do this in a new color. The ratio of 5 to x is going to be equal to the ratio of 7 to this distance right over here."}, {"video_title": "Angle bisector theorem examples Geometry Khan Academy.mp3", "Sentence": "We need to find the length of AB right over here. So once again, angle bisector theorem, the ratio of 5 to this. Let me do this in a new color. The ratio of 5 to x is going to be equal to the ratio of 7 to this distance right over here. And what is that distance? Well, if the whole thing is 10, and this is x, then this distance right over here is going to be 10 minus x. So the ratio of 5 to x is equal to 7 over 10 minus x."}, {"video_title": "Angle bisector theorem examples Geometry Khan Academy.mp3", "Sentence": "The ratio of 5 to x is going to be equal to the ratio of 7 to this distance right over here. And what is that distance? Well, if the whole thing is 10, and this is x, then this distance right over here is going to be 10 minus x. So the ratio of 5 to x is equal to 7 over 10 minus x. And we can cross multiply. 5 times 10 minus x is 50 minus 5x. And then x times 7 is equal to 7x."}, {"video_title": "Angle bisector theorem examples Geometry Khan Academy.mp3", "Sentence": "So the ratio of 5 to x is equal to 7 over 10 minus x. And we can cross multiply. 5 times 10 minus x is 50 minus 5x. And then x times 7 is equal to 7x. Add 5x to both sides of this equation. You get 50 is equal to 12x. We can divide both sides by 12."}, {"video_title": "Angle bisector theorem examples Geometry Khan Academy.mp3", "Sentence": "And then x times 7 is equal to 7x. Add 5x to both sides of this equation. You get 50 is equal to 12x. We can divide both sides by 12. And we get 50 over 12 is equal to x. And we can reduce this. Let's see, if you divide the numerator and the denominator by 2, you get this is the same thing as 25 over 6, which is the same thing, if we want to write it as a mixed number, as 4."}, {"video_title": "Angle bisector theorem examples Geometry Khan Academy.mp3", "Sentence": "We can divide both sides by 12. And we get 50 over 12 is equal to x. And we can reduce this. Let's see, if you divide the numerator and the denominator by 2, you get this is the same thing as 25 over 6, which is the same thing, if we want to write it as a mixed number, as 4. 24 over 6 is 4. And then you have 1 6 left over. 4 and 1 6."}, {"video_title": "Angle bisector theorem examples Geometry Khan Academy.mp3", "Sentence": "Let's see, if you divide the numerator and the denominator by 2, you get this is the same thing as 25 over 6, which is the same thing, if we want to write it as a mixed number, as 4. 24 over 6 is 4. And then you have 1 6 left over. 4 and 1 6. So this length right over here is going to, oh sorry, this length right over here, x is 4 and 1 6. And then this length over here is going to be 10 minus 4 and 1 6. So what is that?"}, {"video_title": "Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "And what we need to do, we need to prove that the area of this rhombus is equal to 1 half times AC times BD. So we're essentially proving that the area of a rhombus is 1 half times the product of the lengths of its diagonals. So let's see what we can do over here. So there's a bunch of things we know about rhombi, and all rhombi are parallelograms, so there's tons of things that we know about parallelograms. First of all, if it's a rhombus, we know that all of the sides are congruent. So that side length is equal to that side length is equal to that side length is equal to that side length. Because it's a parallelogram, we know that diagonals bisect each other, so we know that this length, let me call this point over here B, let's call this E, we know that BE is going to be equal to ED."}, {"video_title": "Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So there's a bunch of things we know about rhombi, and all rhombi are parallelograms, so there's tons of things that we know about parallelograms. First of all, if it's a rhombus, we know that all of the sides are congruent. So that side length is equal to that side length is equal to that side length is equal to that side length. Because it's a parallelogram, we know that diagonals bisect each other, so we know that this length, let me call this point over here B, let's call this E, we know that BE is going to be equal to ED. So that's BE, we know that's going to be equal to ED. And we know that AE is equal to EC. We also know because this is a rhombus, and we proved this in the last video, that the diagonals, not only do they bisect each other, but they are also perpendicular."}, {"video_title": "Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "Because it's a parallelogram, we know that diagonals bisect each other, so we know that this length, let me call this point over here B, let's call this E, we know that BE is going to be equal to ED. So that's BE, we know that's going to be equal to ED. And we know that AE is equal to EC. We also know because this is a rhombus, and we proved this in the last video, that the diagonals, not only do they bisect each other, but they are also perpendicular. So we know that this is a right angle, this is a right angle, that is a right angle, and then this is a right angle. So the easiest way to think about it is, if we can show that this triangle ADC is congruent to triangle ABC, and if we can figure out the area of one of them, we can just double it. So the first part is pretty straightforward."}, {"video_title": "Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "We also know because this is a rhombus, and we proved this in the last video, that the diagonals, not only do they bisect each other, but they are also perpendicular. So we know that this is a right angle, this is a right angle, that is a right angle, and then this is a right angle. So the easiest way to think about it is, if we can show that this triangle ADC is congruent to triangle ABC, and if we can figure out the area of one of them, we can just double it. So the first part is pretty straightforward. So we can see that triangle ADC, we know that triangle ADC is going to be congruent to triangle ABC, and we know that by side-side-side congruency. This side is congruent to that side, this side is congruent to that side, and they both share AC right over here. So this is by side-side-side."}, {"video_title": "Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So the first part is pretty straightforward. So we can see that triangle ADC, we know that triangle ADC is going to be congruent to triangle ABC, and we know that by side-side-side congruency. This side is congruent to that side, this side is congruent to that side, and they both share AC right over here. So this is by side-side-side. And so we can say that the area, so because of that, we know that the area of ABCD is just going to be equal to 2 times the area of, we can pick either one of these, we can say 2 times the area of ABC. Because the area of ABCD, actually let me write it this way. The area of ABCD is equal to the area of ADC plus the area of ABC."}, {"video_title": "Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So this is by side-side-side. And so we can say that the area, so because of that, we know that the area of ABCD is just going to be equal to 2 times the area of, we can pick either one of these, we can say 2 times the area of ABC. Because the area of ABCD, actually let me write it this way. The area of ABCD is equal to the area of ADC plus the area of ABC. But since they're congruent, these two are going to be the same thing, so it's just going to be 2 times the area of ABC. Now what is the area of ABC? Well, area of a triangle is just 1 half base times height."}, {"video_title": "Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "The area of ABCD is equal to the area of ADC plus the area of ABC. But since they're congruent, these two are going to be the same thing, so it's just going to be 2 times the area of ABC. Now what is the area of ABC? Well, area of a triangle is just 1 half base times height. So area of ABC is just equal to 1 half times the base of that triangle times its height, which is equal to 1 half. What is the length of the base? Well, the length of the base is AC, so it's 1 half, I'll color code it."}, {"video_title": "Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "Well, area of a triangle is just 1 half base times height. So area of ABC is just equal to 1 half times the base of that triangle times its height, which is equal to 1 half. What is the length of the base? Well, the length of the base is AC, so it's 1 half, I'll color code it. The base is AC, and then what is the height? What is the height here? Well, we know that this diagonal right over here, that's a perpendicular bisector."}, {"video_title": "Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "Well, the length of the base is AC, so it's 1 half, I'll color code it. The base is AC, and then what is the height? What is the height here? Well, we know that this diagonal right over here, that's a perpendicular bisector. So the height is just the distance from BE. So it's AC times BE. That is the height."}, {"video_title": "Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "Well, we know that this diagonal right over here, that's a perpendicular bisector. So the height is just the distance from BE. So it's AC times BE. That is the height. This is an altitude. It intersects this base at a 90 degree angle. Or we could say BE is the same thing as 1 half times BD."}, {"video_title": "Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "That is the height. This is an altitude. It intersects this base at a 90 degree angle. Or we could say BE is the same thing as 1 half times BD. So this is equal to 1 half times AC. That's our base. And then our height is BE, which we're saying is the same thing as 1 half times BD."}, {"video_title": "Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "Or we could say BE is the same thing as 1 half times BD. So this is equal to 1 half times AC. That's our base. And then our height is BE, which we're saying is the same thing as 1 half times BD. Which is 1 half times BD. So that's the area of just ABC. That's just the area of this broader triangle right up there, or that larger triangle right up there, that half of the rhombus."}, {"video_title": "Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "And then our height is BE, which we're saying is the same thing as 1 half times BD. Which is 1 half times BD. So that's the area of just ABC. That's just the area of this broader triangle right up there, or that larger triangle right up there, that half of the rhombus. But we just said that the area of the whole thing is 2 times that. So if we go back, if we use both this information and this information right over here, we have the area of ABCD is going to be equal to 2 times the area of ABC. But the area of ABC is this thing right over here."}, {"video_title": "Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "That's just the area of this broader triangle right up there, or that larger triangle right up there, that half of the rhombus. But we just said that the area of the whole thing is 2 times that. So if we go back, if we use both this information and this information right over here, we have the area of ABCD is going to be equal to 2 times the area of ABC. But the area of ABC is this thing right over here. It is. So 2 times the area of ABC, area of ABC is that right over there. So 1 half times 1 half is 1 fourth times AC times BD."}, {"video_title": "Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "But the area of ABC is this thing right over here. It is. So 2 times the area of ABC, area of ABC is that right over there. So 1 half times 1 half is 1 fourth times AC times BD. And then you see where this is going. 2 times 1 fourth is 1 half times AC times BD. Fairly straightforward."}, {"video_title": "Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So 1 half times 1 half is 1 fourth times AC times BD. And then you see where this is going. 2 times 1 fourth is 1 half times AC times BD. Fairly straightforward. Which is a neat result. And actually I haven't done this in a video yet. I'll do it in the next video."}, {"video_title": "Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "Fairly straightforward. Which is a neat result. And actually I haven't done this in a video yet. I'll do it in the next video. There are other ways of finding the areas of parallelograms generally. It's essentially base times height. But for a rhombus, we could do that because it is a parallelogram."}, {"video_title": "Proof Rhombus area half product of diagonal length Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "I'll do it in the next video. There are other ways of finding the areas of parallelograms generally. It's essentially base times height. But for a rhombus, we could do that because it is a parallelogram. But we also have this other neat little result that we proved in this video. That if we know the lengths of the diagonals, the area of the rhombus is 1 half times the products of the lengths of the diagonals. Which is kind of a neat result."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Let's just do a ton of more examples just so that we make sure that we're getting this trig function thing down well. So let's construct ourselves some right triangles. Let's construct ourselves some right triangles. And I want to be very clear, the way I've defined it so far, this will only work in right triangles. So if you're trying to find the trig functions of angles that aren't part of right triangles, we're going to see that we're going to have to construct right triangles. But let's just focus on the right triangles for now. So let's say that I have a triangle where let's say this length down here is 7."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And I want to be very clear, the way I've defined it so far, this will only work in right triangles. So if you're trying to find the trig functions of angles that aren't part of right triangles, we're going to see that we're going to have to construct right triangles. But let's just focus on the right triangles for now. So let's say that I have a triangle where let's say this length down here is 7. And let's say the length of this side up here, let's say that that is 4. And let's figure out what the hypotenuse over here is going to be. So we know, let's call the hypotenuse h. We know that h squared is going to be equal to 7 squared plus 4 squared."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So let's say that I have a triangle where let's say this length down here is 7. And let's say the length of this side up here, let's say that that is 4. And let's figure out what the hypotenuse over here is going to be. So we know, let's call the hypotenuse h. We know that h squared is going to be equal to 7 squared plus 4 squared. We know that from the Pythagorean theorem. That the hypotenuse squared is equal to the sum of the squares of the other two sides. h squared is equal to 7 squared plus 4 squared."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So we know, let's call the hypotenuse h. We know that h squared is going to be equal to 7 squared plus 4 squared. We know that from the Pythagorean theorem. That the hypotenuse squared is equal to the sum of the squares of the other two sides. h squared is equal to 7 squared plus 4 squared. So this is equal to 49 plus 16. 49 plus 10 is 59. Plus 6 is 65."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "h squared is equal to 7 squared plus 4 squared. So this is equal to 49 plus 16. 49 plus 10 is 59. Plus 6 is 65. So this h squared, let me write h squared, it's a different shade of yellow. So we have h squared is equal to 65. Did I do that right?"}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Plus 6 is 65. So this h squared, let me write h squared, it's a different shade of yellow. So we have h squared is equal to 65. Did I do that right? 49 plus 10 is 59. Plus another 6 is 65. Or we could say that h is equal to, if we take the square root of both sides, square root of 65."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Did I do that right? 49 plus 10 is 59. Plus another 6 is 65. Or we could say that h is equal to, if we take the square root of both sides, square root of 65. And we really can't simplify this at all. This is 13, this is the same thing as 13 times 5. Both of those are not perfect squares and they're both primes."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Or we could say that h is equal to, if we take the square root of both sides, square root of 65. And we really can't simplify this at all. This is 13, this is the same thing as 13 times 5. Both of those are not perfect squares and they're both primes. You can't simplify this anymore. So this is equal to the square root of 65. Now let's find the trig functions for this angle up here."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Both of those are not perfect squares and they're both primes. You can't simplify this anymore. So this is equal to the square root of 65. Now let's find the trig functions for this angle up here. Let's call that angle up there theta. So whenever you do it, you always want to write down, at least for me it works out to write down, SOH CAH TOA. I have these vague memories of my trigonometry teacher."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Now let's find the trig functions for this angle up here. Let's call that angle up there theta. So whenever you do it, you always want to write down, at least for me it works out to write down, SOH CAH TOA. I have these vague memories of my trigonometry teacher. Maybe I read it in some book. I don't know about some type of Indian princess named SOH CAH TOA or whatever. But it's a very useful mnemonic."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "I have these vague memories of my trigonometry teacher. Maybe I read it in some book. I don't know about some type of Indian princess named SOH CAH TOA or whatever. But it's a very useful mnemonic. So we can apply SOH CAH TOA. Let's say we wanted to find the cosine. We want to find the cosine of our angle."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "But it's a very useful mnemonic. So we can apply SOH CAH TOA. Let's say we wanted to find the cosine. We want to find the cosine of our angle. You say SOH CAH TOA. So the CAH tells us what to do with cosine. The CAH part tells us that cosine is adjacent over hypotenuse."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "We want to find the cosine of our angle. You say SOH CAH TOA. So the CAH tells us what to do with cosine. The CAH part tells us that cosine is adjacent over hypotenuse. Cosine is equal to adjacent over hypotenuse. So let's look over here. To theta, what side is adjacent?"}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "The CAH part tells us that cosine is adjacent over hypotenuse. Cosine is equal to adjacent over hypotenuse. So let's look over here. To theta, what side is adjacent? Well, we know that the hypotenuse is this side over here. So it can't be that side. The only other side that's kind of adjacent to it, that isn't the hypotenuse, is this 4."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "To theta, what side is adjacent? Well, we know that the hypotenuse is this side over here. So it can't be that side. The only other side that's kind of adjacent to it, that isn't the hypotenuse, is this 4. So the adjacent side over here, that side is literally right next to the angle. It's one of the sides that kind of forms the angle. It's 4 over the hypotenuse."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "The only other side that's kind of adjacent to it, that isn't the hypotenuse, is this 4. So the adjacent side over here, that side is literally right next to the angle. It's one of the sides that kind of forms the angle. It's 4 over the hypotenuse. The hypotenuse, we already know, is square root of 65. So it's 4 over the square root of 65. And sometimes people want you to rationalize the denominator, which means they don't like to have an irrational number in the denominator, like the square root of 65."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "It's 4 over the hypotenuse. The hypotenuse, we already know, is square root of 65. So it's 4 over the square root of 65. And sometimes people want you to rationalize the denominator, which means they don't like to have an irrational number in the denominator, like the square root of 65. And if you want to rewrite this without an irrational number in the denominator, you can multiply the numerator and the denominator by square root of 65. This clearly will not change the number because we're multiplying it by something over itself. So we're multiplying the number by 1."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And sometimes people want you to rationalize the denominator, which means they don't like to have an irrational number in the denominator, like the square root of 65. And if you want to rewrite this without an irrational number in the denominator, you can multiply the numerator and the denominator by square root of 65. This clearly will not change the number because we're multiplying it by something over itself. So we're multiplying the number by 1. That won't change the number, but at least it gets rid of the irrational number in the denominator. So the numerator becomes 4 times the square root of 65. And the denominator, square root of 65 times square root of 65, is just going to be 65."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So we're multiplying the number by 1. That won't change the number, but at least it gets rid of the irrational number in the denominator. So the numerator becomes 4 times the square root of 65. And the denominator, square root of 65 times square root of 65, is just going to be 65. We didn't get rid of the irrational number. It's still there, but it's now in the numerator. Now let's do the other trig functions, or at least the other core trig functions."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And the denominator, square root of 65 times square root of 65, is just going to be 65. We didn't get rid of the irrational number. It's still there, but it's now in the numerator. Now let's do the other trig functions, or at least the other core trig functions. We'll learn in the future that there's actually a ton of them, but they're all derived from these. So let's think about what the sine of theta is. Once again, go to SOH CAH TOA."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Now let's do the other trig functions, or at least the other core trig functions. We'll learn in the future that there's actually a ton of them, but they're all derived from these. So let's think about what the sine of theta is. Once again, go to SOH CAH TOA. The SOH tells us what to do with sine. Sine is opposite over hypotenuse. So for this angle, what side is opposite?"}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Once again, go to SOH CAH TOA. The SOH tells us what to do with sine. Sine is opposite over hypotenuse. So for this angle, what side is opposite? Well, you just go opposite it. What it opens into, it's opposite the 7. So the opposite side is the 7."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So for this angle, what side is opposite? Well, you just go opposite it. What it opens into, it's opposite the 7. So the opposite side is the 7. This is right here. That is the opposite side. And then the hypotenuse, it's opposite over hypotenuse."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So the opposite side is the 7. This is right here. That is the opposite side. And then the hypotenuse, it's opposite over hypotenuse. The hypotenuse is the square root of 65. And once again, if we wanted to rationalize this, we could multiply it times the square root of 65 over the square root of 65. In the numerator, we'll get 7 square roots of 65."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And then the hypotenuse, it's opposite over hypotenuse. The hypotenuse is the square root of 65. And once again, if we wanted to rationalize this, we could multiply it times the square root of 65 over the square root of 65. In the numerator, we'll get 7 square roots of 65. And in the denominator, we will get just 65 again. Now let's do tangent. So if I asked you the tangent of theta, once again, go back to SOH CAH TOA."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "In the numerator, we'll get 7 square roots of 65. And in the denominator, we will get just 65 again. Now let's do tangent. So if I asked you the tangent of theta, once again, go back to SOH CAH TOA. The TOA part tells us what to do with tangent. It tells us that tangent is equal to opposite over adjacent. So for this angle, what is opposite?"}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So if I asked you the tangent of theta, once again, go back to SOH CAH TOA. The TOA part tells us what to do with tangent. It tells us that tangent is equal to opposite over adjacent. So for this angle, what is opposite? We've already figured it out. It's 7. It opens into the 7."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So for this angle, what is opposite? We've already figured it out. It's 7. It opens into the 7. It's opposite the 7. So it's 7 over what side is adjacent? Well, this 4 is adjacent."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "It opens into the 7. It's opposite the 7. So it's 7 over what side is adjacent? Well, this 4 is adjacent. So the adjacent side is 4. So it's 7 over 4. And we're done."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Well, this 4 is adjacent. So the adjacent side is 4. So it's 7 over 4. And we're done. We figured out all of the trig ratios for theta. Let's do another one. And I'll make it a little bit concrete."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And we're done. We figured out all of the trig ratios for theta. Let's do another one. And I'll make it a little bit concrete. Because right now we've been saying, oh, what's tangent of x? What's tangent of theta? Let's make it a little bit more concrete."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And I'll make it a little bit concrete. Because right now we've been saying, oh, what's tangent of x? What's tangent of theta? Let's make it a little bit more concrete. Let's say, let me draw another right triangle. Everything we're dealing with, these are going to be right triangles. Let's say the hypotenuse has length 4."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Let's make it a little bit more concrete. Let's say, let me draw another right triangle. Everything we're dealing with, these are going to be right triangles. Let's say the hypotenuse has length 4. Let's say that this side over here has length 2. And let's say that this length over here is going to be 2 times the square root of 3. We can verify that this works."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Let's say the hypotenuse has length 4. Let's say that this side over here has length 2. And let's say that this length over here is going to be 2 times the square root of 3. We can verify that this works. If you have this side squared, so you have, let me write it down, 2 times the square root of 3 squared plus 2 squared is equal to what? This is 2. This is going to be 4 times 3."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "We can verify that this works. If you have this side squared, so you have, let me write it down, 2 times the square root of 3 squared plus 2 squared is equal to what? This is 2. This is going to be 4 times 3. 4 times 3 plus 4. And this is going to be equal to 12 plus 4 is equal to 16. And 16 is indeed 4 squared."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "This is going to be 4 times 3. 4 times 3 plus 4. And this is going to be equal to 12 plus 4 is equal to 16. And 16 is indeed 4 squared. So this does equal 4 squared. It does equal 4 squared. It satisfies the Pythagorean theorem."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And 16 is indeed 4 squared. So this does equal 4 squared. It does equal 4 squared. It satisfies the Pythagorean theorem. And if you remember some of your work from 30-60-90 triangles that you might have learned in geometry, you might recognize that this is a 30-60-90 triangle. This right here is our right angle. I should have drawn it from the get-go to show that this is a right triangle."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "It satisfies the Pythagorean theorem. And if you remember some of your work from 30-60-90 triangles that you might have learned in geometry, you might recognize that this is a 30-60-90 triangle. This right here is our right angle. I should have drawn it from the get-go to show that this is a right triangle. This angle right over here is our 30-degree angle. And then this angle up here is a 60-degree angle. And it's a 30-60-90 because the side opposite the 30 degrees is half the hypotenuse."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "I should have drawn it from the get-go to show that this is a right triangle. This angle right over here is our 30-degree angle. And then this angle up here is a 60-degree angle. And it's a 30-60-90 because the side opposite the 30 degrees is half the hypotenuse. And then the side opposite the 60 degrees is the square root of 3 times the other side that's not the hypotenuse. So with that said, this isn't supposed to be a review of 30-60-90 triangles, although I just did it. Let's actually find the trig ratios for the different angles."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And it's a 30-60-90 because the side opposite the 30 degrees is half the hypotenuse. And then the side opposite the 60 degrees is the square root of 3 times the other side that's not the hypotenuse. So with that said, this isn't supposed to be a review of 30-60-90 triangles, although I just did it. Let's actually find the trig ratios for the different angles. So if I were to ask you, or if anyone were to ask you, what is the sine of 30 degrees? And remember, 30 degrees is one of the angles in this triangle, but it would apply whenever you have a 30-degree angle and you're dealing with a right triangle. We'll have broader definitions in the future."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Let's actually find the trig ratios for the different angles. So if I were to ask you, or if anyone were to ask you, what is the sine of 30 degrees? And remember, 30 degrees is one of the angles in this triangle, but it would apply whenever you have a 30-degree angle and you're dealing with a right triangle. We'll have broader definitions in the future. But if you say sine of 30 degrees, hey, this angle right over here is 30 degrees, so I can use this right triangle. And we just have to remember SOH CAH TOA. Let me rewrite it."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "We'll have broader definitions in the future. But if you say sine of 30 degrees, hey, this angle right over here is 30 degrees, so I can use this right triangle. And we just have to remember SOH CAH TOA. Let me rewrite it. SOH CAH TOA. SOH tells us what to do with sine. Sine is opposite over hypotenuse."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Let me rewrite it. SOH CAH TOA. SOH tells us what to do with sine. Sine is opposite over hypotenuse. Sine of 30 degrees is the opposite side. That is the opposite side, which is 2, over the hypotenuse. The hypotenuse here is 4."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Sine is opposite over hypotenuse. Sine of 30 degrees is the opposite side. That is the opposite side, which is 2, over the hypotenuse. The hypotenuse here is 4. It is 2 fourths, which is the same thing as 1 half. Sine of 30 degrees, you'll see, is always going to be equal to 1 half. Now, what is the cosine?"}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "The hypotenuse here is 4. It is 2 fourths, which is the same thing as 1 half. Sine of 30 degrees, you'll see, is always going to be equal to 1 half. Now, what is the cosine? What is the cosine of 30 degrees? Once again, go back to SOH CAH TOA. The CAH tells us what to do with cosine."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Now, what is the cosine? What is the cosine of 30 degrees? Once again, go back to SOH CAH TOA. The CAH tells us what to do with cosine. Cosine is adjacent over hypotenuse. So if we're looking at the 30-degree angle, it's the adjacent. This right over here is adjacent."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "The CAH tells us what to do with cosine. Cosine is adjacent over hypotenuse. So if we're looking at the 30-degree angle, it's the adjacent. This right over here is adjacent. It's right next to it. It's not the hypotenuse. It's the adjacent over the hypotenuse."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "This right over here is adjacent. It's right next to it. It's not the hypotenuse. It's the adjacent over the hypotenuse. So it's 2 square roots of 3 adjacent over the hypotenuse, over 4. Or if we simplify that, we divide the numerator and denominator by 2, it's the square root of 3 over 2. Finally, let's do the tangent."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "It's the adjacent over the hypotenuse. So it's 2 square roots of 3 adjacent over the hypotenuse, over 4. Or if we simplify that, we divide the numerator and denominator by 2, it's the square root of 3 over 2. Finally, let's do the tangent. The tangent of 30 degrees. We go back to SOH CAH TOA. TOA tells us what to do with tangent."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Finally, let's do the tangent. The tangent of 30 degrees. We go back to SOH CAH TOA. TOA tells us what to do with tangent. It's opposite over adjacent. We go to the 30-degree angle, because that's what we care about. Tangent of 30."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "TOA tells us what to do with tangent. It's opposite over adjacent. We go to the 30-degree angle, because that's what we care about. Tangent of 30. Opposite is 2, and the adjacent is 2 square roots of 3. It's right next to it. It's adjacent to it."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Tangent of 30. Opposite is 2, and the adjacent is 2 square roots of 3. It's right next to it. It's adjacent to it. Adjacent means next to. So 2 square roots of 3. So this is equal to, the 2's cancel out, 1 over the square root of 3."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "It's adjacent to it. Adjacent means next to. So 2 square roots of 3. So this is equal to, the 2's cancel out, 1 over the square root of 3. Or we can multiply the numerator and the denominator by the square root of 3. So we have square root of 3 over square root of 3. This is going to be equal to, the numerator is square root of 3, and then the denominator right over here is just going to be 3."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So this is equal to, the 2's cancel out, 1 over the square root of 3. Or we can multiply the numerator and the denominator by the square root of 3. So we have square root of 3 over square root of 3. This is going to be equal to, the numerator is square root of 3, and then the denominator right over here is just going to be 3. So we've rationalized it. Square root of 3 over 3. Fair enough."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "This is going to be equal to, the numerator is square root of 3, and then the denominator right over here is just going to be 3. So we've rationalized it. Square root of 3 over 3. Fair enough. Now let's use the same triangle to figure out the trig ratios for the 60 degrees, since we've already drawn it. So what is the sine of 60 degrees? I think you're hopefully getting the hang of it now."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Fair enough. Now let's use the same triangle to figure out the trig ratios for the 60 degrees, since we've already drawn it. So what is the sine of 60 degrees? I think you're hopefully getting the hang of it now. Sine is opposite over adjacent. So, from the Sohcahtoa, for the 60 degree angle, what side is opposite? Well, it opens out into the 2 square roots of 3."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "I think you're hopefully getting the hang of it now. Sine is opposite over adjacent. So, from the Sohcahtoa, for the 60 degree angle, what side is opposite? Well, it opens out into the 2 square roots of 3. So the opposite side is 2 square roots of 3. And for the 60 degree angle, the adjacent, or sorry, it's opposite over hypotenuse. Don't want to confuse you."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Well, it opens out into the 2 square roots of 3. So the opposite side is 2 square roots of 3. And for the 60 degree angle, the adjacent, or sorry, it's opposite over hypotenuse. Don't want to confuse you. So it's opposite over hypotenuse. So it's 2 square roots of 3 over 4. 4 is the hypotenuse."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Don't want to confuse you. So it's opposite over hypotenuse. So it's 2 square roots of 3 over 4. 4 is the hypotenuse. So it is equal to, this simplifies to square root of 3 over 2. What is the cosine of 60 degrees? So remember, Sohcahtoa, cosine is adjacent over hypotenuse."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "4 is the hypotenuse. So it is equal to, this simplifies to square root of 3 over 2. What is the cosine of 60 degrees? So remember, Sohcahtoa, cosine is adjacent over hypotenuse. Adjacent is the 2 side. It's right next to the 60 degree angle. So it's 2 over the hypotenuse, which is 4."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So remember, Sohcahtoa, cosine is adjacent over hypotenuse. Adjacent is the 2 side. It's right next to the 60 degree angle. So it's 2 over the hypotenuse, which is 4. So this is equal to 1 half. And then finally, what is the tangent? What is the tangent of 60 degrees?"}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So it's 2 over the hypotenuse, which is 4. So this is equal to 1 half. And then finally, what is the tangent? What is the tangent of 60 degrees? Well, tangent, Sohcahtoa, tangent is opposite over adjacent. Opposite the 60 degrees is 2 square roots of 3. And adjacent to that is 2."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "What is the tangent of 60 degrees? Well, tangent, Sohcahtoa, tangent is opposite over adjacent. Opposite the 60 degrees is 2 square roots of 3. And adjacent to that is 2. Adjacent to 60 degrees is 2. So it's opposite over adjacent. 2 square roots of 3 over 2, which is just equal to the square root of 3."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And adjacent to that is 2. Adjacent to 60 degrees is 2. So it's opposite over adjacent. 2 square roots of 3 over 2, which is just equal to the square root of 3. And I just want to, you know, look how these are related. The sine of 30 degrees is the same thing as the cosine of 60 degrees. The cosine of 30 degrees is the same thing as the sine of 60 degrees."}, {"video_title": "Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "2 square roots of 3 over 2, which is just equal to the square root of 3. And I just want to, you know, look how these are related. The sine of 30 degrees is the same thing as the cosine of 60 degrees. The cosine of 30 degrees is the same thing as the sine of 60 degrees. And then these guys are the inverse of each other. I think if you think a little bit about this triangle, it will start to make sense why. We'll keep extending this and give you a lot more practice in the next few videos."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "And I always do this before I have to convert between the two. If I do one revolution of a circle, how many radians is that going to be? Well, we know one revolution of a circle is 2 pi radians. And how many degrees is that? If I do one revolution around a circle, well, we know that that's 360. I can either write it with a little degree symbol right like that, or I could write it just like that. And this is really enough information for us to think about how to convert between radians and degrees."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "And how many degrees is that? If I do one revolution around a circle, well, we know that that's 360. I can either write it with a little degree symbol right like that, or I could write it just like that. And this is really enough information for us to think about how to convert between radians and degrees. If we want to simplify this a little bit, we can divide both sides by 2, and you could have pi radians are equal to 180 degrees, or another way to think about it, going halfway around a circle in radians is pi radians, or the arc that subtends that angle is pi radiuses. And that's also 180 degrees. And if you want to really think about, well, how many degrees are there per radian, you can divide both sides of this by pi."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "And this is really enough information for us to think about how to convert between radians and degrees. If we want to simplify this a little bit, we can divide both sides by 2, and you could have pi radians are equal to 180 degrees, or another way to think about it, going halfway around a circle in radians is pi radians, or the arc that subtends that angle is pi radiuses. And that's also 180 degrees. And if you want to really think about, well, how many degrees are there per radian, you can divide both sides of this by pi. So if you divide both sides of this by pi, you get one radian. I have to go from plural to singular. One radian is equal to 180 over pi degrees."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "And if you want to really think about, well, how many degrees are there per radian, you can divide both sides of this by pi. So if you divide both sides of this by pi, you get one radian. I have to go from plural to singular. One radian is equal to 180 over pi degrees. So all I did is I divided both sides by pi. And if you wanted to figure out how many radians are there per degree, you could divide both sides by 180. So you'd get pi over 180 radians is equal to 1 degree."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "One radian is equal to 180 over pi degrees. So all I did is I divided both sides by pi. And if you wanted to figure out how many radians are there per degree, you could divide both sides by 180. So you'd get pi over 180 radians is equal to 1 degree. So now I think we are ready to start converting. So let's convert 30 degrees to radians. So let's think about it."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "So you'd get pi over 180 radians is equal to 1 degree. So now I think we are ready to start converting. So let's convert 30 degrees to radians. So let's think about it. So I'm going to write it out. And actually, this might remind you of kind of unit analysis that you might do when you first did unit conversion, but it also works out here. So if I were to write 30 degrees, and this is how my brain likes to work with it, I like to write out the word degrees."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "So let's think about it. So I'm going to write it out. And actually, this might remind you of kind of unit analysis that you might do when you first did unit conversion, but it also works out here. So if I were to write 30 degrees, and this is how my brain likes to work with it, I like to write out the word degrees. And then I say, well, I want to convert to radians. So I really want to figure out how many radians are there per degree. So let me write this down."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "So if I were to write 30 degrees, and this is how my brain likes to work with it, I like to write out the word degrees. And then I say, well, I want to convert to radians. So I really want to figure out how many radians are there per degree. So let me write this down. I want to figure out how many radians do we have per degree. And I haven't filled out how many that is, but we see just the units will cancel out. If we have degrees times radians per degree, the degrees will cancel out and I'll be just left with radians."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "So let me write this down. I want to figure out how many radians do we have per degree. And I haven't filled out how many that is, but we see just the units will cancel out. If we have degrees times radians per degree, the degrees will cancel out and I'll be just left with radians. If I multiply the number of degrees I have times the number of radians per degree, we're going to get radians. And hopefully that makes intuitive sense as well. And here we just have to think about, well, if I have pi radians, how many degrees is that?"}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "If we have degrees times radians per degree, the degrees will cancel out and I'll be just left with radians. If I multiply the number of degrees I have times the number of radians per degree, we're going to get radians. And hopefully that makes intuitive sense as well. And here we just have to think about, well, if I have pi radians, how many degrees is that? Well, that's 180 degrees. It comes straight out of this right over here. Pi radians for every 180 degrees or pi over 180 radians per degree."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "And here we just have to think about, well, if I have pi radians, how many degrees is that? Well, that's 180 degrees. It comes straight out of this right over here. Pi radians for every 180 degrees or pi over 180 radians per degree. And this is going to get us to 30 times pi over 180, which we'll simplify to 30 over 180 is 1 over 6. So this is equal to pi over 6. Actually, let me write the units out."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "Pi radians for every 180 degrees or pi over 180 radians per degree. And this is going to get us to 30 times pi over 180, which we'll simplify to 30 over 180 is 1 over 6. So this is equal to pi over 6. Actually, let me write the units out. This is 30 radians, which is equal to pi over 6 radians. Now let's go the other way. Let's think about if we have pi over 3 radians, and I want to convert that to degrees."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "Actually, let me write the units out. This is 30 radians, which is equal to pi over 6 radians. Now let's go the other way. Let's think about if we have pi over 3 radians, and I want to convert that to degrees. So what am I going to get if I convert that to degrees? Well, here we're going to want to figure out how many degrees are there per radian. And one way to think about it is, well, think about the pi and the 180."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "Let's think about if we have pi over 3 radians, and I want to convert that to degrees. So what am I going to get if I convert that to degrees? Well, here we're going to want to figure out how many degrees are there per radian. And one way to think about it is, well, think about the pi and the 180. For every 180 degrees, you have pi radians. 180 degrees over pi radians, these are essentially the equivalent thing. Essentially, you're just multiplying this quantity by 1, but you're changing the units."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "And one way to think about it is, well, think about the pi and the 180. For every 180 degrees, you have pi radians. 180 degrees over pi radians, these are essentially the equivalent thing. Essentially, you're just multiplying this quantity by 1, but you're changing the units. The radians cancel out, and then the pi's cancel out, and you're left with 180 over 3 degrees. 180 over 3 is 60, and we could either write out the word degrees, or you can write degrees just like that. Now let's think about 45 degrees."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "Essentially, you're just multiplying this quantity by 1, but you're changing the units. The radians cancel out, and then the pi's cancel out, and you're left with 180 over 3 degrees. 180 over 3 is 60, and we could either write out the word degrees, or you can write degrees just like that. Now let's think about 45 degrees. So what about 45 degrees? And I'll write it like that just so you can figure it out as they're. Figure it out with that notation as well."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "Now let's think about 45 degrees. So what about 45 degrees? And I'll write it like that just so you can figure it out as they're. Figure it out with that notation as well. How many radians will this be equal to? Well, once again, we're going to want to think about how many radians do we have per degree. So we're going to multiply this times, well, we know we have pi radians for every 180 degrees, or we could even write it this way, pi radians for every 180 degrees."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "Figure it out with that notation as well. How many radians will this be equal to? Well, once again, we're going to want to think about how many radians do we have per degree. So we're going to multiply this times, well, we know we have pi radians for every 180 degrees, or we could even write it this way, pi radians for every 180 degrees. And here, this might be a little less intuitive, the degrees cancel out, and that's why I'd like to usually write out the word, and you're left with 45 pi over 180 radians. Actually, let me write this with the words written out. Maybe that's more intuitive when I'm thinking about it in terms of using the notation."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "So we're going to multiply this times, well, we know we have pi radians for every 180 degrees, or we could even write it this way, pi radians for every 180 degrees. And here, this might be a little less intuitive, the degrees cancel out, and that's why I'd like to usually write out the word, and you're left with 45 pi over 180 radians. Actually, let me write this with the words written out. Maybe that's more intuitive when I'm thinking about it in terms of using the notation. So 45 degrees times, we have pi radians for every 180 degrees. So we are left with, when you multiply, 45 times pi over 180, the degrees have canceled out, and you're just left with radians, which is equal to what? 45 is half of 90, which is half of 180, so this is 1 4th."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "Maybe that's more intuitive when I'm thinking about it in terms of using the notation. So 45 degrees times, we have pi radians for every 180 degrees. So we are left with, when you multiply, 45 times pi over 180, the degrees have canceled out, and you're just left with radians, which is equal to what? 45 is half of 90, which is half of 180, so this is 1 4th. This is equal to pi over 4 radians. Let's do one more over here. So let's say that we had negative pi over 2 radians."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "45 is half of 90, which is half of 180, so this is 1 4th. This is equal to pi over 4 radians. Let's do one more over here. So let's say that we had negative pi over 2 radians. What's that going to be in degrees? Well, once again, we have to figure out how many degrees are each of these radians. We know that there are 180 degrees for every pi radians, so we're going to get the radians cancel out, the pi's cancel out, and so you have negative 180 over 2."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "So in this problem here, we're told that the triangle ACE is isosceles. So that's this big triangle right here. It's isosceles, which means it has two equal sides. And we also know from isosceles triangles that the base angles must be equal. So these two base angles are going to be equal, and this side right over here is going to be equal in length to this side over here. We can say AC is going to be equal to CE. So we get all of that from this first statement right over there."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "And we also know from isosceles triangles that the base angles must be equal. So these two base angles are going to be equal, and this side right over here is going to be equal in length to this side over here. We can say AC is going to be equal to CE. So we get all of that from this first statement right over there. Then they give us some more clues or some more information. They say CG is equal to 24. So this is CG right over here."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "So we get all of that from this first statement right over there. Then they give us some more clues or some more information. They say CG is equal to 24. So this is CG right over here. It has length 24. They tell us BH is equal to DF. BH is equal to DF, so those two things are going to be congruent."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "So this is CG right over here. It has length 24. They tell us BH is equal to DF. BH is equal to DF, so those two things are going to be congruent. They're going to be the same length. Then they tell us that GF is equal to 12. So this is GF right over here."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "BH is equal to DF, so those two things are going to be congruent. They're going to be the same length. Then they tell us that GF is equal to 12. So this is GF right over here. So GF is equal to 12. That's that distance right over there. Then finally they tell us that FE is equal to 6."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "So this is GF right over here. So GF is equal to 12. That's that distance right over there. Then finally they tell us that FE is equal to 6. So this is FE. Then finally they ask us what is the area of CBHFD. So CBHFD."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "Then finally they tell us that FE is equal to 6. So this is FE. Then finally they ask us what is the area of CBHFD. So CBHFD. So they're asking us for the area of this part right over here. That part and that part right over there. That is CBHFD."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "So CBHFD. So they're asking us for the area of this part right over here. That part and that part right over there. That is CBHFD. So let's think about how we can do this. We can figure out the area of the larger triangle. Then from that we can subtract the areas of these little pieces at the end."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "That is CBHFD. So let's think about how we can do this. We can figure out the area of the larger triangle. Then from that we can subtract the areas of these little pieces at the end. Then we'll be able to figure out this middle area, this area that I've shaded. We don't have all the information yet to solve that. We know what the height or the altitude of this triangle is, but we don't know its base."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "Then from that we can subtract the areas of these little pieces at the end. Then we'll be able to figure out this middle area, this area that I've shaded. We don't have all the information yet to solve that. We know what the height or the altitude of this triangle is, but we don't know its base. If we knew its base, we'd say 1 half base times height, we'd get the area of this triangle. Then we'd have to subtract out these areas. We don't have full information there either."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "We know what the height or the altitude of this triangle is, but we don't know its base. If we knew its base, we'd say 1 half base times height, we'd get the area of this triangle. Then we'd have to subtract out these areas. We don't have full information there either. We don't know this height. Once we know that height, then we can figure out this height, but we also don't quite yet know what this length right over here is. Let's just take it piece by piece."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "We don't have full information there either. We don't know this height. Once we know that height, then we can figure out this height, but we also don't quite yet know what this length right over here is. Let's just take it piece by piece. The first thing we might want to do, and you might guess because we've been talking a lot about similarity, is making some type of argument about similarity here because there's a bunch of similar triangles. For example, triangle CGE shares this angle with triangle DFE. They both share this orange angle right here."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "Let's just take it piece by piece. The first thing we might want to do, and you might guess because we've been talking a lot about similarity, is making some type of argument about similarity here because there's a bunch of similar triangles. For example, triangle CGE shares this angle with triangle DFE. They both share this orange angle right here. They both have this right angle right over here. They have two angles in common. They are going to be similar by angle-angle."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "They both share this orange angle right here. They both have this right angle right over here. They have two angles in common. They are going to be similar by angle-angle. You can actually show that there's going to be a third angle in common because these two are parallel lines. We can write that triangle CGE is similar to triangle DFE. We know that by angle-angle."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "They are going to be similar by angle-angle. You can actually show that there's going to be a third angle in common because these two are parallel lines. We can write that triangle CGE is similar to triangle DFE. We know that by angle-angle. We have one set of corresponding angles congruent, and then this angle is in both triangles, so it is a set of corresponding congruent angles right over there. Then once we know that they are similar, we can set up the ratio between sides because we have some information about some of the sides. We know that the ratio between DF and this side right over here, which is a corresponding side, the ratio between DF and CG, which is 24, is going to be the same thing as the ratio between FE, which is 6, and GE, which is not 12."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "We know that by angle-angle. We have one set of corresponding angles congruent, and then this angle is in both triangles, so it is a set of corresponding congruent angles right over there. Then once we know that they are similar, we can set up the ratio between sides because we have some information about some of the sides. We know that the ratio between DF and this side right over here, which is a corresponding side, the ratio between DF and CG, which is 24, is going to be the same thing as the ratio between FE, which is 6, and GE, which is not 12. It's 12 plus 6. It is 18. Then let's see."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "We know that the ratio between DF and this side right over here, which is a corresponding side, the ratio between DF and CG, which is 24, is going to be the same thing as the ratio between FE, which is 6, and GE, which is not 12. It's 12 plus 6. It is 18. Then let's see. 6 over 18, this is just 1 over 3. You get 3DF. You get 3DF is equal to 24."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "Then let's see. 6 over 18, this is just 1 over 3. You get 3DF. You get 3DF is equal to 24. I just cross-multiplied, or you could multiply both sides by 24, multiply both sides by 3. You would get this. Actually, you could just multiply both sides times 24, and you'll get 24 times 1 third, but we'll just do it this way."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "You get 3DF is equal to 24. I just cross-multiplied, or you could multiply both sides by 24, multiply both sides by 3. You would get this. Actually, you could just multiply both sides times 24, and you'll get 24 times 1 third, but we'll just do it this way. Divide both sides by 3. You get DF is equal to 8. We found out that DF is equal to 8, that length right over there."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "Actually, you could just multiply both sides times 24, and you'll get 24 times 1 third, but we'll just do it this way. Divide both sides by 3. You get DF is equal to 8. We found out that DF is equal to 8, that length right over there. That's useful for us because we know that this length right over here is also equal to 8. Now what can we do? It seems like we can make another similarity argument because we have this angle right over here."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "We found out that DF is equal to 8, that length right over there. That's useful for us because we know that this length right over here is also equal to 8. Now what can we do? It seems like we can make another similarity argument because we have this angle right over here. It is congruent to that angle right over there. We also have this angle, which is going to be 90 degrees. We have a 90-degree angle there."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "It seems like we can make another similarity argument because we have this angle right over here. It is congruent to that angle right over there. We also have this angle, which is going to be 90 degrees. We have a 90-degree angle there. That by itself is actually enough to say that we have two similar triangles. We don't even have to show that they have a congruent side here. Actually, we're going to show that these are actually congruent triangles that we're dealing with right over here."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "We have a 90-degree angle there. That by itself is actually enough to say that we have two similar triangles. We don't even have to show that they have a congruent side here. Actually, we're going to show that these are actually congruent triangles that we're dealing with right over here. We have two angles. Actually, we could just go straight to that because when we talk about congruency, if you have an angle that's congruent to another angle, another angle that's congruent to another angle, and then a side that's congruent to another side, you're dealing with two congruent triangles. We can write triangle A, H, B is congruent to triangle E, F, D. We know that because we have angle, angle, side."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "Actually, we're going to show that these are actually congruent triangles that we're dealing with right over here. We have two angles. Actually, we could just go straight to that because when we talk about congruency, if you have an angle that's congruent to another angle, another angle that's congruent to another angle, and then a side that's congruent to another side, you're dealing with two congruent triangles. We can write triangle A, H, B is congruent to triangle E, F, D. We know that because we have angle, angle, side. Angle, angle, side postulate for congruency. If the two triangles are congruent, that makes things convenient. That means if this side is 8, that side is 8."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "We can write triangle A, H, B is congruent to triangle E, F, D. We know that because we have angle, angle, side. Angle, angle, side postulate for congruency. If the two triangles are congruent, that makes things convenient. That means if this side is 8, that side is 8. We already knew that. That's how we established our congruency. That means if this side has length 6, then the corresponding side on this triangle is also going to have length 6."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "That means if this side is 8, that side is 8. We already knew that. That's how we established our congruency. That means if this side has length 6, then the corresponding side on this triangle is also going to have length 6. We can write this length right over here is also going to be 6. I can imagine you can imagine where all of this is going to go, but we want to prove to ourselves. We want to know for sure what the area is."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "That means if this side has length 6, then the corresponding side on this triangle is also going to have length 6. We can write this length right over here is also going to be 6. I can imagine you can imagine where all of this is going to go, but we want to prove to ourselves. We want to know for sure what the area is. We don't want to say, hey, maybe this is the same thing as that. Let's just actually prove it to ourselves. How do we figure out?"}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "We want to know for sure what the area is. We don't want to say, hey, maybe this is the same thing as that. Let's just actually prove it to ourselves. How do we figure out? We've almost figured out the entire base of this triangle, but we still haven't figured out the length of HG. Now we can use a similarity argument again because we can see that triangle ABH is actually similar to triangle ACG. They both have this angle here, and then they both have a right angle."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "How do we figure out? We've almost figured out the entire base of this triangle, but we still haven't figured out the length of HG. Now we can use a similarity argument again because we can see that triangle ABH is actually similar to triangle ACG. They both have this angle here, and then they both have a right angle. They have one. ABH has a right angle there. ACG has a right angle right over there."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "They both have this angle here, and then they both have a right angle. They have one. ABH has a right angle there. ACG has a right angle right over there. So you have two angles, two corresponding angles are equal to each other. You're now dealing with similar triangles. We know that triangle ABH, I'll just write it as AHB since I already wrote it this way, AHB is similar to triangle AGC."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "ACG has a right angle right over there. So you have two angles, two corresponding angles are equal to each other. You're now dealing with similar triangles. We know that triangle ABH, I'll just write it as AHB since I already wrote it this way, AHB is similar to triangle AGC. You want to make sure you get the vertices in the right order. A is the orange angle, G is the right angle, and C is the unlabeled angle. This is similar to triangle AGC."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "We know that triangle ABH, I'll just write it as AHB since I already wrote it this way, AHB is similar to triangle AGC. You want to make sure you get the vertices in the right order. A is the orange angle, G is the right angle, and C is the unlabeled angle. This is similar to triangle AGC. What that does for us is now we can use the ratios to figure out what HG is equal to. What can we say over here? We can say that 8 over 24, BH over its corresponding side of the larger triangle, so we say 8 over 24 is equal to 6 over not HG but over AG."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "This is similar to triangle AGC. What that does for us is now we can use the ratios to figure out what HG is equal to. What can we say over here? We can say that 8 over 24, BH over its corresponding side of the larger triangle, so we say 8 over 24 is equal to 6 over not HG but over AG. I think you can see where this is going. You have 1 third is equal to 6 over AG, or we can cross multiply here and we can get AG is equal to 18. This entire length right over here is 18."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "We can say that 8 over 24, BH over its corresponding side of the larger triangle, so we say 8 over 24 is equal to 6 over not HG but over AG. I think you can see where this is going. You have 1 third is equal to 6 over AG, or we can cross multiply here and we can get AG is equal to 18. This entire length right over here is 18. If AG is 18 and AH is 6, then HG is 12. This is what you might have guessed if you were just trying to guess the answer right over here. Now we have proven to ourselves that this base has length of 18 here and then we have another 18 here."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "This entire length right over here is 18. If AG is 18 and AH is 6, then HG is 12. This is what you might have guessed if you were just trying to guess the answer right over here. Now we have proven to ourselves that this base has length of 18 here and then we have another 18 here. It has a length of 36. The entire base here is 36. Now we can figure out the area of this larger, of the entire isosceles triangle."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "Now we have proven to ourselves that this base has length of 18 here and then we have another 18 here. It has a length of 36. The entire base here is 36. Now we can figure out the area of this larger, of the entire isosceles triangle. The area of ACE is going to be equal to 1 half times the base, which is 36, times 24. This is going to be the same thing as 1 half times 36 is 18 times 24. I'll just do that over here on the top."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "Now we can figure out the area of this larger, of the entire isosceles triangle. The area of ACE is going to be equal to 1 half times the base, which is 36, times 24. This is going to be the same thing as 1 half times 36 is 18 times 24. I'll just do that over here on the top. 18 times 24, 8 times 4 is 32, 1 times 4 is 4 plus 3 is 7. We put a zero here because we're not dealing with 2 but 20. You have 2 times 8 is 16, 2 times 1 is 2 plus 1, so it's 360."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "I'll just do that over here on the top. 18 times 24, 8 times 4 is 32, 1 times 4 is 4 plus 3 is 7. We put a zero here because we're not dealing with 2 but 20. You have 2 times 8 is 16, 2 times 1 is 2 plus 1, so it's 360. Then you have the 2, 7 plus 6 is 13, 1 plus 3 is 4. The area of ACE is equal to 432, but we're not done yet. This area that we care about is the area of the entire triangle minus this area and minus this area right over here."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "You have 2 times 8 is 16, 2 times 1 is 2 plus 1, so it's 360. Then you have the 2, 7 plus 6 is 13, 1 plus 3 is 4. The area of ACE is equal to 432, but we're not done yet. This area that we care about is the area of the entire triangle minus this area and minus this area right over here. What is the area of each of these little wedges right over here? It's going to be 1 half times 8 times 6. 1 half times 8 is 4 times 6, so this is going to be 24 right over there."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "This area that we care about is the area of the entire triangle minus this area and minus this area right over here. What is the area of each of these little wedges right over here? It's going to be 1 half times 8 times 6. 1 half times 8 is 4 times 6, so this is going to be 24 right over there. This is going to be another 24 right over there. This is going to be equal to 432 minus 24 minus 24 or minus 48, which is equal to, and we could try this to do this in our head, if we subtract 32, we're going to get to 400. Then we're going to have to subtract another 16."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "1 half times 8 is 4 times 6, so this is going to be 24 right over there. This is going to be another 24 right over there. This is going to be equal to 432 minus 24 minus 24 or minus 48, which is equal to, and we could try this to do this in our head, if we subtract 32, we're going to get to 400. Then we're going to have to subtract another 16. If you subtract 10 from 400, you get 390, so you get to 384, whatever the units were. If these were in meters, then this would be meters squared. If this was centimeters, this would be centimeters squared."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3", "Sentence": "Then we're going to have to subtract another 16. If you subtract 10 from 400, you get 390, so you get to 384, whatever the units were. If these were in meters, then this would be meters squared. If this was centimeters, this would be centimeters squared. Did I do that right? Let me go the other way. If I add 40 to this, 24 plus another 8 gets me to 432."}, {"video_title": "Properties perserved after rigid transformations.mp3", "Sentence": "In particular, we're gonna think about rotations and reflections in this video. And both of those are rigid transformations, which means that the length between corresponding points do not change. So for example, let's say we take this circle A, it's centered at point A, and we were to rotate it around point P. Point P is the center of rotation. And just say, for the sake of argument, we rotate it clockwise a certain angle. So let's say we end up right over, so we're gonna rotate that way. And let's say our center ends up right over here. So our new circle, the image after the rotation, might look something like this."}, {"video_title": "Properties perserved after rigid transformations.mp3", "Sentence": "And just say, for the sake of argument, we rotate it clockwise a certain angle. So let's say we end up right over, so we're gonna rotate that way. And let's say our center ends up right over here. So our new circle, the image after the rotation, might look something like this. And I'm hand drawing it, so you gotta forgive that it's not that well hand drawn of a circle. But the circle might look something like this. And so the clear things that are preserved, or maybe it's not so clear, and we're gonna hopefully make them clear right now."}, {"video_title": "Properties perserved after rigid transformations.mp3", "Sentence": "So our new circle, the image after the rotation, might look something like this. And I'm hand drawing it, so you gotta forgive that it's not that well hand drawn of a circle. But the circle might look something like this. And so the clear things that are preserved, or maybe it's not so clear, and we're gonna hopefully make them clear right now. Things that are preserved under a rigid transformation like this rotation right over here, this is clearly a rotation. Things that are preserved, well, you have things like the radius of the circle, the radius length, I could say, to be more particular. The radius here is two."}, {"video_title": "Properties perserved after rigid transformations.mp3", "Sentence": "And so the clear things that are preserved, or maybe it's not so clear, and we're gonna hopefully make them clear right now. Things that are preserved under a rigid transformation like this rotation right over here, this is clearly a rotation. Things that are preserved, well, you have things like the radius of the circle, the radius length, I could say, to be more particular. The radius here is two. The radius here is also two, right over there. You have things like the perimeter. Well, if the radius is preserved, the perimeter of a circle, which we call a circumference, well, that's just a function of the radius."}, {"video_title": "Properties perserved after rigid transformations.mp3", "Sentence": "The radius here is two. The radius here is also two, right over there. You have things like the perimeter. Well, if the radius is preserved, the perimeter of a circle, which we call a circumference, well, that's just a function of the radius. We're talking about two times pi times the radius. So the perimeter, of course, is going to be preserved. In fact, that follows from the fact that the length of the radius is preserved."}, {"video_title": "Properties perserved after rigid transformations.mp3", "Sentence": "Well, if the radius is preserved, the perimeter of a circle, which we call a circumference, well, that's just a function of the radius. We're talking about two times pi times the radius. So the perimeter, of course, is going to be preserved. In fact, that follows from the fact that the length of the radius is preserved. And of course, if the radius is preserved, then the area is also going to be preserved. The area is just pi times the radius squared. So if they have the same radius, they're gonna have all of these in common."}, {"video_title": "Properties perserved after rigid transformations.mp3", "Sentence": "In fact, that follows from the fact that the length of the radius is preserved. And of course, if the radius is preserved, then the area is also going to be preserved. The area is just pi times the radius squared. So if they have the same radius, they're gonna have all of these in common. And you can also, that feels intuitively right. So what is not preserved? Not preserved."}, {"video_title": "Properties perserved after rigid transformations.mp3", "Sentence": "So if they have the same radius, they're gonna have all of these in common. And you can also, that feels intuitively right. So what is not preserved? Not preserved. And this is in general true of rigid transformations, is that they will preserve the distance between corresponding points. If we're transforming a shape, they'll preserve things like perimeter and area. In this case, instead of perimeter, I could say circumference, circumference."}, {"video_title": "Properties perserved after rigid transformations.mp3", "Sentence": "Not preserved. And this is in general true of rigid transformations, is that they will preserve the distance between corresponding points. If we're transforming a shape, they'll preserve things like perimeter and area. In this case, instead of perimeter, I could say circumference, circumference. So they'll preserve things like that. They'll preserve angles. We don't have clear angles in this picture, but they'll preserve things like angles."}, {"video_title": "Properties perserved after rigid transformations.mp3", "Sentence": "In this case, instead of perimeter, I could say circumference, circumference. So they'll preserve things like that. They'll preserve angles. We don't have clear angles in this picture, but they'll preserve things like angles. But what they won't preserve is the coordinates, coordinates of corresponding points. They might sometimes, but not always. So for example, the coordinate of the center here is for sure going to change."}, {"video_title": "Properties perserved after rigid transformations.mp3", "Sentence": "We don't have clear angles in this picture, but they'll preserve things like angles. But what they won't preserve is the coordinates, coordinates of corresponding points. They might sometimes, but not always. So for example, the coordinate of the center here is for sure going to change. We go from the coordinate negative three comma zero to here we went to the coordinate negative one comma two. So the coordinates are not preserved, coordinates of the center. Let's do another example with a non-circular shape, and we'll do a different type of transformation."}, {"video_title": "Properties perserved after rigid transformations.mp3", "Sentence": "So for example, the coordinate of the center here is for sure going to change. We go from the coordinate negative three comma zero to here we went to the coordinate negative one comma two. So the coordinates are not preserved, coordinates of the center. Let's do another example with a non-circular shape, and we'll do a different type of transformation. In this situation, let us do a reflection. So we have a quadrilateral here, quadrilateral ABCD. And we wanna think about what is preserved or not preserved as we do a reflection across the line L. So let me write that down."}, {"video_title": "Properties perserved after rigid transformations.mp3", "Sentence": "Let's do another example with a non-circular shape, and we'll do a different type of transformation. In this situation, let us do a reflection. So we have a quadrilateral here, quadrilateral ABCD. And we wanna think about what is preserved or not preserved as we do a reflection across the line L. So let me write that down. We're gonna have a reflection in this situation. And we could even think about this without even doing the reflection ourselves, but let's just do the reflection really fast. So we're reflecting across the line X, Y is equal to X."}, {"video_title": "Properties perserved after rigid transformations.mp3", "Sentence": "And we wanna think about what is preserved or not preserved as we do a reflection across the line L. So let me write that down. We're gonna have a reflection in this situation. And we could even think about this without even doing the reflection ourselves, but let's just do the reflection really fast. So we're reflecting across the line X, Y is equal to X. So what it essentially does to the coordinates is it swaps the X and Y coordinates, but you don't have to know that for the sake of this video. So B prime would be right over here, A prime would be right over there, D prime would be right over here. And since C is right on the line L, its image, C prime, won't change."}, {"video_title": "Properties perserved after rigid transformations.mp3", "Sentence": "So we're reflecting across the line X, Y is equal to X. So what it essentially does to the coordinates is it swaps the X and Y coordinates, but you don't have to know that for the sake of this video. So B prime would be right over here, A prime would be right over there, D prime would be right over here. And since C is right on the line L, its image, C prime, won't change. And so our new, when we reflect over the line L, and you don't have to know for the sake of this video exactly how I did that fairly quickly. I really just want you to see what the reflection looks like. The real appreciation here is to think about, well, what happens with rigid transformations?"}, {"video_title": "Properties perserved after rigid transformations.mp3", "Sentence": "And since C is right on the line L, its image, C prime, won't change. And so our new, when we reflect over the line L, and you don't have to know for the sake of this video exactly how I did that fairly quickly. I really just want you to see what the reflection looks like. The real appreciation here is to think about, well, what happens with rigid transformations? So it's gonna look something like this. The reflection, the reflection looks something like this. So what's preserved?"}, {"video_title": "Properties perserved after rigid transformations.mp3", "Sentence": "The real appreciation here is to think about, well, what happens with rigid transformations? So it's gonna look something like this. The reflection, the reflection looks something like this. So what's preserved? And in general, this is good to know for any rigid transformation, what's preserved? Well, side lengths. That's actually one way that we even use to define what a rigid transformation is, a transformation that preserves the lengths between corresponding points."}, {"video_title": "Properties perserved after rigid transformations.mp3", "Sentence": "So what's preserved? And in general, this is good to know for any rigid transformation, what's preserved? Well, side lengths. That's actually one way that we even use to define what a rigid transformation is, a transformation that preserves the lengths between corresponding points. Angle measures, angle measures. So for example, this angle here, the angle A, is gonna be the same as the angle A prime over here. Side lengths, the distance between A and B is gonna be the same as the distance between A prime and B prime."}, {"video_title": "Properties perserved after rigid transformations.mp3", "Sentence": "That's actually one way that we even use to define what a rigid transformation is, a transformation that preserves the lengths between corresponding points. Angle measures, angle measures. So for example, this angle here, the angle A, is gonna be the same as the angle A prime over here. Side lengths, the distance between A and B is gonna be the same as the distance between A prime and B prime. Perimeter, if you have the same side lengths and the same angles, then perimeter and area are also going to be preserved, just like we saw with the rotation example. These are rigid transformations. These are the types of things that are preserved."}, {"video_title": "Properties perserved after rigid transformations.mp3", "Sentence": "Side lengths, the distance between A and B is gonna be the same as the distance between A prime and B prime. Perimeter, if you have the same side lengths and the same angles, then perimeter and area are also going to be preserved, just like we saw with the rotation example. These are rigid transformations. These are the types of things that are preserved. Well, what is not preserved? Not preserved, and this just goes back to the example we just looked at. Well, coordinates are not preserved."}, {"video_title": "Properties perserved after rigid transformations.mp3", "Sentence": "These are the types of things that are preserved. Well, what is not preserved? Not preserved, and this just goes back to the example we just looked at. Well, coordinates are not preserved. So as we see, the image of A, A prime, has different coordinates than A. B prime has different coordinates than B. C prime, in this case, happens to have the same coordinates as C because C happened to sit on the line that we're reflecting over."}, {"video_title": "Properties perserved after rigid transformations.mp3", "Sentence": "Well, coordinates are not preserved. So as we see, the image of A, A prime, has different coordinates than A. B prime has different coordinates than B. C prime, in this case, happens to have the same coordinates as C because C happened to sit on the line that we're reflecting over. But D prime definitely does not have the same coordinates as D. So most of, or let me say, coordinates of A, B, or A, B, D, coordinates of A, B, D, not preserved after transformation, or their images, they don't have the same coordinates, after transformation. The one coordinate that happened to be preserved here is C's coordinates, because it was right on the line of reflection. And you could also look at other properties of how it might relate, how different segments might relate to lines that were not being transformed."}, {"video_title": "Properties perserved after rigid transformations.mp3", "Sentence": "C prime, in this case, happens to have the same coordinates as C because C happened to sit on the line that we're reflecting over. But D prime definitely does not have the same coordinates as D. So most of, or let me say, coordinates of A, B, or A, B, D, coordinates of A, B, D, not preserved after transformation, or their images, they don't have the same coordinates, after transformation. The one coordinate that happened to be preserved here is C's coordinates, because it was right on the line of reflection. And you could also look at other properties of how it might relate, how different segments might relate to lines that were not being transformed. So for example, right over here, before transformation, C, D is parallel to the Y axis. You see that right over there. But after the transformation, C prime, D prime, so this could be C prime, D prime, is no longer parallel to the Y axis."}, {"video_title": "Inscribed quadrilaterals proof Mathematics II High School Math Khan Academy.mp3", "Sentence": "So I have a arbitrary inscribed quadrilateral in this circle. And what I want to prove is that for any inscribed quadrilateral, that opposite angles are supplementary. So when I say they're supplementary, the measure of this angle plus the measure of this angle need to be 180 degrees. The measure of this angle plus the measure of this angle need to be 180 degrees. And the way I'm gonna prove it is we're gonna assume that this, the measure of this angle right over here, that this is x degrees. And so from that, if we can prove that the measure of this opposite angle is 180 minus x degrees, then we've proven that opposite angles for an arbitrary quadrilateral that's inscribed in a circle are supplementary, because if this is 180 minus x, 180 minus x plus x is going to be 180 degrees. So I encourage you to pause the video and see if you can do that proof."}, {"video_title": "Inscribed quadrilaterals proof Mathematics II High School Math Khan Academy.mp3", "Sentence": "The measure of this angle plus the measure of this angle need to be 180 degrees. And the way I'm gonna prove it is we're gonna assume that this, the measure of this angle right over here, that this is x degrees. And so from that, if we can prove that the measure of this opposite angle is 180 minus x degrees, then we've proven that opposite angles for an arbitrary quadrilateral that's inscribed in a circle are supplementary, because if this is 180 minus x, 180 minus x plus x is going to be 180 degrees. So I encourage you to pause the video and see if you can do that proof. And I'll give you a little bit of a hint. It's going to involve the measure of the arcs that the various angles intercept. So let's think about it a little bit."}, {"video_title": "Inscribed quadrilaterals proof Mathematics II High School Math Khan Academy.mp3", "Sentence": "So I encourage you to pause the video and see if you can do that proof. And I'll give you a little bit of a hint. It's going to involve the measure of the arcs that the various angles intercept. So let's think about it a little bit. This angle that has a measure of x degrees, it intercepts this arc. So we see one side of the angle goes and intercepts the circle there. The other side right over there."}, {"video_title": "Inscribed quadrilaterals proof Mathematics II High School Math Khan Academy.mp3", "Sentence": "So let's think about it a little bit. This angle that has a measure of x degrees, it intercepts this arc. So we see one side of the angle goes and intercepts the circle there. The other side right over there. And so the arc that it intercepts, I am highlighting in yellow. I am highlighting it in yellow. Trying to color it in."}, {"video_title": "Inscribed quadrilaterals proof Mathematics II High School Math Khan Academy.mp3", "Sentence": "The other side right over there. And so the arc that it intercepts, I am highlighting in yellow. I am highlighting it in yellow. Trying to color it in. So there you go. Not a great job at coloring it in, but you get the point. That's the arc that it intercepts."}, {"video_title": "Inscribed quadrilaterals proof Mathematics II High School Math Khan Academy.mp3", "Sentence": "Trying to color it in. So there you go. Not a great job at coloring it in, but you get the point. That's the arc that it intercepts. And we've already learned in previous videos that the relationship between an inscribed angle, the vertex of this angle sits on the circle, the relationship between an inscribed angle and the measure of the arc that it intercepts is that the measure of the inscribed angle is half the measure of the arc that it intercepts. So if this angle measures x degrees, then the measure of this arc is going to be 2x. 2x degrees."}, {"video_title": "Inscribed quadrilaterals proof Mathematics II High School Math Khan Academy.mp3", "Sentence": "That's the arc that it intercepts. And we've already learned in previous videos that the relationship between an inscribed angle, the vertex of this angle sits on the circle, the relationship between an inscribed angle and the measure of the arc that it intercepts is that the measure of the inscribed angle is half the measure of the arc that it intercepts. So if this angle measures x degrees, then the measure of this arc is going to be 2x. 2x degrees. All right, well that's kind of interesting, but let's keep going. If the measure of that arc is 2x degrees, what is the measure of this arc right over here, the arc that completes the circle? Well if you go all the way around the circle, that's 360 degrees."}, {"video_title": "Inscribed quadrilaterals proof Mathematics II High School Math Khan Academy.mp3", "Sentence": "2x degrees. All right, well that's kind of interesting, but let's keep going. If the measure of that arc is 2x degrees, what is the measure of this arc right over here, the arc that completes the circle? Well if you go all the way around the circle, that's 360 degrees. So this blue arc that I'm showing you right now, that's going to have a measure of 360 minus 2x, minus 2x, minus 2x degrees. 360 is all the way around. The blue one is all the way around minus the yellow arc."}, {"video_title": "Inscribed quadrilaterals proof Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well if you go all the way around the circle, that's 360 degrees. So this blue arc that I'm showing you right now, that's going to have a measure of 360 minus 2x, minus 2x, minus 2x degrees. 360 is all the way around. The blue one is all the way around minus the yellow arc. What you have left over if you subtract out the yellow arc is you have this blue arc. Now what's the angle that intercepts this blue arc? What's the inscribed angle that intercepts this blue arc right over here?"}, {"video_title": "Inscribed quadrilaterals proof Mathematics II High School Math Khan Academy.mp3", "Sentence": "The blue one is all the way around minus the yellow arc. What you have left over if you subtract out the yellow arc is you have this blue arc. Now what's the angle that intercepts this blue arc? What's the inscribed angle that intercepts this blue arc right over here? What's this angle? It's the angle that we wanted to figure out in terms of x. It is, well I'm having trouble changing colors."}, {"video_title": "Inscribed quadrilaterals proof Mathematics II High School Math Khan Academy.mp3", "Sentence": "What's the inscribed angle that intercepts this blue arc right over here? What's this angle? It's the angle that we wanted to figure out in terms of x. It is, well I'm having trouble changing colors. It is that angle right over there. Notice the two sides of this angle, they intercept, this angle intercepts that arc. So once again, the measure of an inscribed angle is going to be half the measure of the arc that it intercepts."}, {"video_title": "Inscribed quadrilaterals proof Mathematics II High School Math Khan Academy.mp3", "Sentence": "It is, well I'm having trouble changing colors. It is that angle right over there. Notice the two sides of this angle, they intercept, this angle intercepts that arc. So once again, the measure of an inscribed angle is going to be half the measure of the arc that it intercepts. So what's 1 half? What is 1 half times 360 minus 2x? Well 1 half times 360 is 180."}, {"video_title": "Inscribed quadrilaterals proof Mathematics II High School Math Khan Academy.mp3", "Sentence": "So once again, the measure of an inscribed angle is going to be half the measure of the arc that it intercepts. So what's 1 half? What is 1 half times 360 minus 2x? Well 1 half times 360 is 180. 1 half times 2x is x. So the measure of this angle is going to be 180 minus x degrees. 180 minus x degrees."}, {"video_title": "Inscribed quadrilaterals proof Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well 1 half times 360 is 180. 1 half times 2x is x. So the measure of this angle is going to be 180 minus x degrees. 180 minus x degrees. And just like that, we've proven that these opposite sides for this arbitrary inscribed quadrilateral, that they are supplementary. You add these together, x plus 180 minus x, you're going to get 180 degrees. So they are supplementary."}, {"video_title": "Constructing an angle bisector using a compass and straightedge Geometry Khan Academy.mp3", "Sentence": "And they want us to make a line that goes right in between that angle, that divides that angle into two angles that have equal measure, that have half the measure of the first angle. So let's first find two points that are equidistant from this point right over here on each of these rays. So to do that, let's draw one circle here. And I could make this of any radius. Wherever this intersects with the rays, that's where I'm going to put a point. So let's say here and here. Notice both of these points, since they're both on this circle, are going to be equidistant from this point, which is the center of the circle."}, {"video_title": "Constructing an angle bisector using a compass and straightedge Geometry Khan Academy.mp3", "Sentence": "And I could make this of any radius. Wherever this intersects with the rays, that's where I'm going to put a point. So let's say here and here. Notice both of these points, since they're both on this circle, are going to be equidistant from this point, which is the center of the circle. Now, what I want to do is construct a line that is equidistant from both of these points. And we've done that already when we looked at perpendicular bisectors for lines in this construction module. So let's do that."}, {"video_title": "Constructing an angle bisector using a compass and straightedge Geometry Khan Academy.mp3", "Sentence": "Notice both of these points, since they're both on this circle, are going to be equidistant from this point, which is the center of the circle. Now, what I want to do is construct a line that is equidistant from both of these points. And we've done that already when we looked at perpendicular bisectors for lines in this construction module. So let's do that. So let's add a compass. And so what I want to do, this circle is centered at this point. And it has a radius equal to the distance between this point and that point."}, {"video_title": "Constructing an angle bisector using a compass and straightedge Geometry Khan Academy.mp3", "Sentence": "So let's do that. So let's add a compass. And so what I want to do, this circle is centered at this point. And it has a radius equal to the distance between this point and that point. And then I do that again. So this circle is centered at this point and has a radius equal to the distance between that point and that point. And then the two places where they intersect are equidistant to both of these points."}, {"video_title": "Constructing an angle bisector using a compass and straightedge Geometry Khan Academy.mp3", "Sentence": "And it has a radius equal to the distance between this point and that point. And then I do that again. So this circle is centered at this point and has a radius equal to the distance between that point and that point. And then the two places where they intersect are equidistant to both of these points. They're equidistant to both of these points. And so we can now draw our angle bisector just like that. And you might say, well, how do we really know that this angle is equal to this angle?"}, {"video_title": "Constructing an angle bisector using a compass and straightedge Geometry Khan Academy.mp3", "Sentence": "And then the two places where they intersect are equidistant to both of these points. They're equidistant to both of these points. And so we can now draw our angle bisector just like that. And you might say, well, how do we really know that this angle is equal to this angle? Well, there's a couple of ways we can tell. We know this distance right over here is equal to this distance right over there. We know that this distance over here is equal to this distance over here."}, {"video_title": "Constructing an angle bisector using a compass and straightedge Geometry Khan Academy.mp3", "Sentence": "And you might say, well, how do we really know that this angle is equal to this angle? Well, there's a couple of ways we can tell. We know this distance right over here is equal to this distance right over there. We know that this distance over here is equal to this distance over here. And both of these triangles share this line. So essentially, if you look at this point, this point, and this point, that forms a triangle. And if you look at this point, this point, and this point, that forms a triangle."}, {"video_title": "Constructing an angle bisector using a compass and straightedge Geometry Khan Academy.mp3", "Sentence": "We know that this distance over here is equal to this distance over here. And both of these triangles share this line. So essentially, if you look at this point, this point, and this point, that forms a triangle. And if you look at this point, this point, and this point, that forms a triangle. We know those two triangles are congruent. So this angle must be equal to this angle. These are the corresponding angles."}, {"video_title": "Shape properties after a sequence of transformations.mp3", "Sentence": "In past videos, we've thought about whether segment lengths or angle measures are preserved with a transformation. What we're now going to think about is what is preserved with a sequence of transformations. And in particular, we're gonna think about angle measure, angle measure, and segment lengths. So if you're transforming some type of a shape, segment, segment lengths. So let's look at this first example. We say a sequence of transformations is described below. So we first do a translation, then we do a reflection over a horizontal line, PQ, then we do a vertical stretch about PQ."}, {"video_title": "Shape properties after a sequence of transformations.mp3", "Sentence": "So if you're transforming some type of a shape, segment, segment lengths. So let's look at this first example. We say a sequence of transformations is described below. So we first do a translation, then we do a reflection over a horizontal line, PQ, then we do a vertical stretch about PQ. What is this going to do? Is this going to preserve angle measures, and is this going to preserve segment lengths? Well, a translation is a rigid transformation, and so that will preserve both angle measures and segment lengths."}, {"video_title": "Shape properties after a sequence of transformations.mp3", "Sentence": "So we first do a translation, then we do a reflection over a horizontal line, PQ, then we do a vertical stretch about PQ. What is this going to do? Is this going to preserve angle measures, and is this going to preserve segment lengths? Well, a translation is a rigid transformation, and so that will preserve both angle measures and segment lengths. So after that, angle measures and segment lengths are still going to be the same. A reflection over a horizontal line, PQ. Well, a reflection is also a rigid transformation, and so we will continue to preserve angle measure and segment lengths."}, {"video_title": "Shape properties after a sequence of transformations.mp3", "Sentence": "Well, a translation is a rigid transformation, and so that will preserve both angle measures and segment lengths. So after that, angle measures and segment lengths are still going to be the same. A reflection over a horizontal line, PQ. Well, a reflection is also a rigid transformation, and so we will continue to preserve angle measure and segment lengths. Then they say a vertical stretch about PQ. Well, let's just think about what a vertical stretch does. So if I have some triangle right over here, if I have some triangle that looks like this, this is triangle ABC, and if you were to do a vertical stretch, what's going to happen?"}, {"video_title": "Shape properties after a sequence of transformations.mp3", "Sentence": "Well, a reflection is also a rigid transformation, and so we will continue to preserve angle measure and segment lengths. Then they say a vertical stretch about PQ. Well, let's just think about what a vertical stretch does. So if I have some triangle right over here, if I have some triangle that looks like this, this is triangle ABC, and if you were to do a vertical stretch, what's going to happen? Well, let's just imagine that we take these sides and we stretch them out so that we now have A is over here, or A prime, I should say, is over there. Let's say that B prime is now over here. This isn't going to be exact."}, {"video_title": "Shape properties after a sequence of transformations.mp3", "Sentence": "So if I have some triangle right over here, if I have some triangle that looks like this, this is triangle ABC, and if you were to do a vertical stretch, what's going to happen? Well, let's just imagine that we take these sides and we stretch them out so that we now have A is over here, or A prime, I should say, is over there. Let's say that B prime is now over here. This isn't going to be exact. Well, what just happened to my triangle? Well, the measure of angle C is for sure going to be different now, and my segment lengths are for sure going to be different now. A prime, C prime is going to be different than AC in terms of segment length."}, {"video_title": "Shape properties after a sequence of transformations.mp3", "Sentence": "This isn't going to be exact. Well, what just happened to my triangle? Well, the measure of angle C is for sure going to be different now, and my segment lengths are for sure going to be different now. A prime, C prime is going to be different than AC in terms of segment length. So a vertical stretch, if we're talking about a stretch in general, this is going to preserve neither. So neither preserved. Neither preserved."}, {"video_title": "Shape properties after a sequence of transformations.mp3", "Sentence": "A prime, C prime is going to be different than AC in terms of segment length. So a vertical stretch, if we're talking about a stretch in general, this is going to preserve neither. So neither preserved. Neither preserved. So in general, if you're doing rigid transformation after rigid transformation, you're gonna preserve both angles and segment lengths, but if you throw a stretch in there, then all bets are off. You're not going to preserve either of them. Let's do another example."}, {"video_title": "Shape properties after a sequence of transformations.mp3", "Sentence": "Neither preserved. So in general, if you're doing rigid transformation after rigid transformation, you're gonna preserve both angles and segment lengths, but if you throw a stretch in there, then all bets are off. You're not going to preserve either of them. Let's do another example. A sequence of transformations is described below, and so they give three transformations. So pause this video and think about whether angle measures, segment lengths, or will either both or neither or only one of them be preserved? All right, so first we have a rotation about a point P. That's a rigid transformation."}, {"video_title": "Shape properties after a sequence of transformations.mp3", "Sentence": "Let's do another example. A sequence of transformations is described below, and so they give three transformations. So pause this video and think about whether angle measures, segment lengths, or will either both or neither or only one of them be preserved? All right, so first we have a rotation about a point P. That's a rigid transformation. It would preserve both segment lengths and angle measures. Then you have a translation, which is also a rigid transformation, and so that would preserve both again. Then we have a rotation about point P. So once again, another rigid transformation."}, {"video_title": "Shape properties after a sequence of transformations.mp3", "Sentence": "All right, so first we have a rotation about a point P. That's a rigid transformation. It would preserve both segment lengths and angle measures. Then you have a translation, which is also a rigid transformation, and so that would preserve both again. Then we have a rotation about point P. So once again, another rigid transformation. So in this situation, everything is going to be preserved. So both angle measure, angle measure, and segment length are going to be preserved in this example. Let's do one more example."}, {"video_title": "Shape properties after a sequence of transformations.mp3", "Sentence": "Then we have a rotation about point P. So once again, another rigid transformation. So in this situation, everything is going to be preserved. So both angle measure, angle measure, and segment length are going to be preserved in this example. Let's do one more example. So here, once again, we have a sequence of transformations, and so pause this video again and see if you can figure out whether angle measures, segment lengths, both or neither are going to be preserved. So the first transformation is a dilation. So a dilation is a non-rigid transformation."}, {"video_title": "Shape properties after a sequence of transformations.mp3", "Sentence": "Let's do one more example. So here, once again, we have a sequence of transformations, and so pause this video again and see if you can figure out whether angle measures, segment lengths, both or neither are going to be preserved. So the first transformation is a dilation. So a dilation is a non-rigid transformation. So segment lengths not preserved. Segment lengths not preserved. And we've seen this in multiple videos already."}, {"video_title": "Shape properties after a sequence of transformations.mp3", "Sentence": "So a dilation is a non-rigid transformation. So segment lengths not preserved. Segment lengths not preserved. And we've seen this in multiple videos already. But in a dilation, angles are preserved. Angles preserved. So already we've lost our segment lengths, but we still got our angles."}, {"video_title": "Shape properties after a sequence of transformations.mp3", "Sentence": "And we've seen this in multiple videos already. But in a dilation, angles are preserved. Angles preserved. So already we've lost our segment lengths, but we still got our angles. Then we have a rotation about another point Q. So this is a rigid transformation. It would preserve both, but we've already lost our segment lengths, but angles are going to continue to be preserved."}, {"video_title": "Constructing circumscribing circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "Construct a circle circumscribing the triangle. So that would be a circle that touches the vertices, the three vertices of this triangle. So we can construct it using a compass and a straight edge, or a virtual compass and a virtual straight edge. So what we want to do is center the circle at the perpendicular bisectors of the sides, or sometimes that's called the circumcenter of this triangle. So let's do that. And so let's think about, let's try to construct where the perpendicular bisectors of the sides are. So let me put a circle right over here whose radius is longer than this side right over here."}, {"video_title": "Constructing circumscribing circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "So what we want to do is center the circle at the perpendicular bisectors of the sides, or sometimes that's called the circumcenter of this triangle. So let's do that. And so let's think about, let's try to construct where the perpendicular bisectors of the sides are. So let me put a circle right over here whose radius is longer than this side right over here. Now let me get one that has the same size. So let me make it the same size as the one I just did. And let me put it right over here."}, {"video_title": "Constructing circumscribing circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "So let me put a circle right over here whose radius is longer than this side right over here. Now let me get one that has the same size. So let me make it the same size as the one I just did. And let me put it right over here. And this allows us to construct a perpendicular bisector. So if I go through that point and this point right over here, this bisects this side over here, and it's at a right angle. So now let's do that for the other sides."}, {"video_title": "Constructing circumscribing circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "And let me put it right over here. And this allows us to construct a perpendicular bisector. So if I go through that point and this point right over here, this bisects this side over here, and it's at a right angle. So now let's do that for the other sides. So if I move this over here, and I really just have to do it for one of the other sides because the intersection of two lines is gonna give me a point. So I can do it for this side right over here. Let me scroll down so you can see a little bit clearer."}, {"video_title": "Constructing circumscribing circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "So now let's do that for the other sides. So if I move this over here, and I really just have to do it for one of the other sides because the intersection of two lines is gonna give me a point. So I can do it for this side right over here. Let me scroll down so you can see a little bit clearer. So let me add another straight edge right over here. So I'm gonna go through that point, and I'm gonna go through this point. So that's the perpendicular bisector of this side right over here."}, {"video_title": "Constructing circumscribing circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "Let me scroll down so you can see a little bit clearer. So let me add another straight edge right over here. So I'm gonna go through that point, and I'm gonna go through this point. So that's the perpendicular bisector of this side right over here. Now I could do the third side, and it should intersect at that point. I'm not ultra, ultra precise, but I'm close enough. And now I just have to center one of these circles."}, {"video_title": "Constructing circumscribing circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "So that's the perpendicular bisector of this side right over here. Now I could do the third side, and it should intersect at that point. I'm not ultra, ultra precise, but I'm close enough. And now I just have to center one of these circles. Let me move one of these away. So let me just get rid of this one. And I just have to move this circle to the circumcenter and adjust its radius so that it gets pretty close to touching the three sides, the three vertices of this triangle."}, {"video_title": "Constructing circumscribing circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "And now I just have to center one of these circles. Let me move one of these away. So let me just get rid of this one. And I just have to move this circle to the circumcenter and adjust its radius so that it gets pretty close to touching the three sides, the three vertices of this triangle. It doesn't have to be perfect. I think this exercise has some margin for error. But they really want to see that you've made an attempt at drawing the perpendicular bisectors of the sides to find the circumcenter, and then you put a circle right over there."}, {"video_title": "Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3", "Sentence": "And a compass is something that allows you to draw circles centered where you want them to be centered of different radii. And your typical compass would be like a metal thing. It has a pin on one end and it's kind of shaped like an angle and then you have a pencil at the other end. Now I don't have real physical rulers and pencils in front of me, but I have the virtual equivalent. I can say add a compass and now I can draw a circle. I can pick where I want to center it and I can change the radius. And I can draw a straight line segment."}, {"video_title": "Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3", "Sentence": "Now I don't have real physical rulers and pencils in front of me, but I have the virtual equivalent. I can say add a compass and now I can draw a circle. I can pick where I want to center it and I can change the radius. And I can draw a straight line segment. So I can move it around and I can draw, this is equivalent to having a straight edge. And so using these tools, I want to construct a line going through P that is tangent to the circle. Now it might be tempting, oh, I could just, let me just try to draw a line."}, {"video_title": "Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3", "Sentence": "And I can draw a straight line segment. So I can move it around and I can draw, this is equivalent to having a straight edge. And so using these tools, I want to construct a line going through P that is tangent to the circle. Now it might be tempting, oh, I could just, let me just try to draw a line. Maybe it looks something like this. Do you remember a tangent? A tangent line will touch the circle at exactly one point and that point, since it's going through P, should be the point P. Another way to think about a tangent line is it's going to be perpendicular to the radius between that point and the center."}, {"video_title": "Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3", "Sentence": "Now it might be tempting, oh, I could just, let me just try to draw a line. Maybe it looks something like this. Do you remember a tangent? A tangent line will touch the circle at exactly one point and that point, since it's going through P, should be the point P. Another way to think about a tangent line is it's going to be perpendicular to the radius between that point and the center. Now what I just drew actually looks pretty good, but it's not so precise. I don't know if it's exactly perpendicular to the radius. I don't know if it's touching at exactly one point right over there."}, {"video_title": "Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3", "Sentence": "A tangent line will touch the circle at exactly one point and that point, since it's going through P, should be the point P. Another way to think about a tangent line is it's going to be perpendicular to the radius between that point and the center. Now what I just drew actually looks pretty good, but it's not so precise. I don't know if it's exactly perpendicular to the radius. I don't know if it's touching at exactly one point right over there. So what we're going to do is use our virtual compass and our virtual straight edge to try to do a more precise drawing. So let's do that. So the first thing I'm going to do is I'm going to set up P as the midpoint of a line where the center of the circle is one other end of the line."}, {"video_title": "Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3", "Sentence": "I don't know if it's touching at exactly one point right over there. So what we're going to do is use our virtual compass and our virtual straight edge to try to do a more precise drawing. So let's do that. So the first thing I'm going to do is I'm going to set up P as the midpoint of a line where the center of the circle is one other end of the line. And the way I can do that, let me add a compass here, and let me construct a circle that has the same, okay, so I've set up my compass to have the same radius as my original circle, but now let me move it over right over here. So now it's centered at P. Well, why is this useful? Well, now a diameter of this new circle is going to be a segment that is centered at P. So I'm going to have a segment where P is the midpoint, and then my center of my original circle is going to be one of the endpoints."}, {"video_title": "Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3", "Sentence": "So the first thing I'm going to do is I'm going to set up P as the midpoint of a line where the center of the circle is one other end of the line. And the way I can do that, let me add a compass here, and let me construct a circle that has the same, okay, so I've set up my compass to have the same radius as my original circle, but now let me move it over right over here. So now it's centered at P. Well, why is this useful? Well, now a diameter of this new circle is going to be a segment that is centered at P. So I'm going to have a segment where P is the midpoint, and then my center of my original circle is going to be one of the endpoints. So let's do that. So I'm going to add a straight edge. So I'll make that one of the endpoints."}, {"video_title": "Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3", "Sentence": "Well, now a diameter of this new circle is going to be a segment that is centered at P. So I'm going to have a segment where P is the midpoint, and then my center of my original circle is going to be one of the endpoints. So let's do that. So I'm going to add a straight edge. So I'll make that one of the endpoints. And I'm going to go through P all the way to the other side of my new circle. Now, what's the whole point of me, let me try to do it as well as I can, what's the whole point of me doing that? Well, now I've made P the midpoint of a segment."}, {"video_title": "Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3", "Sentence": "So I'll make that one of the endpoints. And I'm going to go through P all the way to the other side of my new circle. Now, what's the whole point of me, let me try to do it as well as I can, what's the whole point of me doing that? Well, now I've made P the midpoint of a segment. So if I can construct a perpendicular bisector of the segment, it will go through P because P is the midpoint, and then that thing, it's going to be exactly perpendicular to the radius because the radius, the original radius, let me be careful here, the original radius is part of this segment. So let's see how I can do this. So what I could do is, I'm going to draw another circle."}, {"video_title": "Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3", "Sentence": "Well, now I've made P the midpoint of a segment. So if I can construct a perpendicular bisector of the segment, it will go through P because P is the midpoint, and then that thing, it's going to be exactly perpendicular to the radius because the radius, the original radius, let me be careful here, the original radius is part of this segment. So let's see how I can do this. So what I could do is, I'm going to draw another circle. I'm going to center this one at the original circle. And I'm going to make it have a different radius, maybe a radius something like that. And now I'm going to construct another circle of this larger size, but I'm going to center it at this point right over here."}, {"video_title": "Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3", "Sentence": "So what I could do is, I'm going to draw another circle. I'm going to center this one at the original circle. And I'm going to make it have a different radius, maybe a radius something like that. And now I'm going to construct another circle of this larger size, but I'm going to center it at this point right over here. And I think you'll see quickly what this will accomplish. So I'm going to construct another circle of that same larger radius, of that same larger radius, just like that. And now I'm going to move it over here."}, {"video_title": "Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3", "Sentence": "And now I'm going to construct another circle of this larger size, but I'm going to center it at this point right over here. And I think you'll see quickly what this will accomplish. So I'm going to construct another circle of that same larger radius, of that same larger radius, just like that. And now I'm going to move it over here. Over here. So what's interesting about the intersection of these two larger circles? Well this point right over here is equidistant to this end of the segment and to this end of the segment."}, {"video_title": "Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3", "Sentence": "And now I'm going to move it over here. Over here. So what's interesting about the intersection of these two larger circles? Well this point right over here is equidistant to this end of the segment and to this end of the segment. Because remember, these two larger circles have the same radius. So if I'm sitting on both of them, then I am that distance away from this point, and I am sitting that distance away from this point. So something that is equidistant from the two end points of a segment, it's going to sit on the perpendicular bisector."}, {"video_title": "Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3", "Sentence": "Well this point right over here is equidistant to this end of the segment and to this end of the segment. Because remember, these two larger circles have the same radius. So if I'm sitting on both of them, then I am that distance away from this point, and I am sitting that distance away from this point. So something that is equidistant from the two end points of a segment, it's going to sit on the perpendicular bisector. So this point is going to sit on the perpendicular bisector, and this point is going to be sit on the perpendicular bisector. So now we can draw a perpendicular bisector. We can go from this point right over here, the intersection of our two larger circles, this point that is equidistant from my two centers of the large circle, to this point that is equidistant to the two centers of the large circle."}, {"video_title": "Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3", "Sentence": "So something that is equidistant from the two end points of a segment, it's going to sit on the perpendicular bisector. So this point is going to sit on the perpendicular bisector, and this point is going to be sit on the perpendicular bisector. So now we can draw a perpendicular bisector. We can go from this point right over here, the intersection of our two larger circles, this point that is equidistant from my two centers of the large circle, to this point that is equidistant to the two centers of the large circle. And once again, it's equidistant to the two centers of the large circle, but those points are also the end points of this segment. So these two points are on the perpendicular bisector. You just need two points for a line."}, {"video_title": "Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3", "Sentence": "We can go from this point right over here, the intersection of our two larger circles, this point that is equidistant from my two centers of the large circle, to this point that is equidistant to the two centers of the large circle. And once again, it's equidistant to the two centers of the large circle, but those points are also the end points of this segment. So these two points are on the perpendicular bisector. You just need two points for a line. So I've just constructed a perpendicular bisector to P. And it's perpendicular, once again, to the radius from the center to P of our original circle. Now, that is going to be a tangent line. Because if we go only through P, or we go through P, and we are exactly perpendicular to the radius from P to the center, then this line that we've just constructed is actually tangent."}, {"video_title": "Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3", "Sentence": "You just need two points for a line. So I've just constructed a perpendicular bisector to P. And it's perpendicular, once again, to the radius from the center to P of our original circle. Now, that is going to be a tangent line. Because if we go only through P, or we go through P, and we are exactly perpendicular to the radius from P to the center, then this line that we've just constructed is actually tangent. So anyway, it might seem like a lot of work to do all of this. You know, I could have started just eyeballing it. But when we do it like this, we can feel really, really, really, really, really good that we're being precise."}, {"video_title": "Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3", "Sentence": "Because if we go only through P, or we go through P, and we are exactly perpendicular to the radius from P to the center, then this line that we've just constructed is actually tangent. So anyway, it might seem like a lot of work to do all of this. You know, I could have started just eyeballing it. But when we do it like this, we can feel really, really, really, really, really good that we're being precise. Imagine if you were trying to do this in a larger scale, or if you're trying to engineer some very precise instrument. You would want to do it this way. You want to draw a very precise drawing, maybe an architectural drawing."}, {"video_title": "Constructing square inscribed in circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "Construct a square inscribed inside the circle. In order to do this, we just have to remember that a square, well, we know all squares, all four sides are congruent and they intersect at right angles, and we also have to remember that the two diagonals of the square are going to be perpendicular bisectors of each other. So let's see if we can construct two lines that are perpendicular bisectors of each other, and actually where those two lines intersect our bigger circle, those are going to be the vertices of our square. So let's throw a straight edge right over here, and let's make a diameter. So that's a diameter right over here. It just goes through the circle, goes through the center of the circle, and two sides of the circle. And now let's think about how we can construct a perpendicular bisector of this, and other compass construction or construction videos."}, {"video_title": "Constructing square inscribed in circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "So let's throw a straight edge right over here, and let's make a diameter. So that's a diameter right over here. It just goes through the circle, goes through the center of the circle, and two sides of the circle. And now let's think about how we can construct a perpendicular bisector of this, and other compass construction or construction videos. But what we can do is we can put a circle, let's throw a circle right over here. We've got to make its radius bigger than the center. What we're going to do is we're going to reuse this, we're going to make another circle that's the exact same size, put it there, and where they intersect is going to be exactly along, those two points of intersection are going to be along a perpendicular bisector."}, {"video_title": "Constructing square inscribed in circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "And now let's think about how we can construct a perpendicular bisector of this, and other compass construction or construction videos. But what we can do is we can put a circle, let's throw a circle right over here. We've got to make its radius bigger than the center. What we're going to do is we're going to reuse this, we're going to make another circle that's the exact same size, put it there, and where they intersect is going to be exactly along, those two points of intersection are going to be along a perpendicular bisector. So that's one of them. Let's do another one. So I'll draw a circle of the exact same dimensions, so I'll center it at the same place."}, {"video_title": "Constructing square inscribed in circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "What we're going to do is we're going to reuse this, we're going to make another circle that's the exact same size, put it there, and where they intersect is going to be exactly along, those two points of intersection are going to be along a perpendicular bisector. So that's one of them. Let's do another one. So I'll draw a circle of the exact same dimensions, so I'll center it at the same place. I'll drag it out there. That looks pretty good. I'll move it on to this side, the other side of my diameter."}, {"video_title": "Constructing square inscribed in circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "So I'll draw a circle of the exact same dimensions, so I'll center it at the same place. I'll drag it out there. That looks pretty good. I'll move it on to this side, the other side of my diameter. So that looks pretty good. And notice, if I connect that point to that point, I have now, I will have constructed a perpendicular bisector of this original segment. So let's do that."}, {"video_title": "Constructing square inscribed in circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "I'll move it on to this side, the other side of my diameter. So that looks pretty good. And notice, if I connect that point to that point, I have now, I will have constructed a perpendicular bisector of this original segment. So let's do that. Let's connect those two points. So that point and that point, and then we could just keep going all the way to, so keep going all the way to the end of the circle. Go all the way over, all the way over there."}, {"video_title": "Constructing square inscribed in circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "So let's do that. Let's connect those two points. So that point and that point, and then we could just keep going all the way to, so keep going all the way to the end of the circle. Go all the way over, all the way over there. That looks pretty good. And now we just have to connect these four points to have a square. So let's do that."}, {"video_title": "Constructing square inscribed in circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "Go all the way over, all the way over there. That looks pretty good. And now we just have to connect these four points to have a square. So let's do that. So I'll connect that and that, and then I will connect, draw another straight edge there, I will connect that with that, and then two more to go. I'll connect this with that, and then one more. I can connect this with that, and there you go."}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "What we're going to do in this video is study a proof of the Pythagorean Theorem that was first discovered, first discovered, or as far as we know, first discovered by James Garfield in 1876. And what's exciting about this is he was not a professional mathematician. You might know James Garfield as the 20th president of the United States. He was elected president, he was elected four years after, in 1880, and then he became president in 1881. And he did this proof while he was a sitting member of the United States House of Representatives. And what's exciting about that is it shows that Abraham Lincoln was not the only U.S. politician, or not the only U.S. president who was really into geometry. And what Garfield realized is if you construct a right triangle, so I'm going to do my best attempt to construct one."}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "He was elected president, he was elected four years after, in 1880, and then he became president in 1881. And he did this proof while he was a sitting member of the United States House of Representatives. And what's exciting about that is it shows that Abraham Lincoln was not the only U.S. politician, or not the only U.S. president who was really into geometry. And what Garfield realized is if you construct a right triangle, so I'm going to do my best attempt to construct one. So let me construct one right here. So let's say that this side right over here is length b. Let's say this side is length a."}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "And what Garfield realized is if you construct a right triangle, so I'm going to do my best attempt to construct one. So let me construct one right here. So let's say that this side right over here is length b. Let's say this side is length a. And let's say that this side, the hypotenuse of my right triangle, has length c. So I've just constructed a right triangle, and let me make it clear, it is a right triangle. He essentially flipped and rotated this right triangle to construct another one that is congruent to the first one. So let me construct that."}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "Let's say this side is length a. And let's say that this side, the hypotenuse of my right triangle, has length c. So I've just constructed a right triangle, and let me make it clear, it is a right triangle. He essentially flipped and rotated this right triangle to construct another one that is congruent to the first one. So let me construct that. So we're going to have length b, and it's collinear with length a. It's along the same line, I should say. They don't overlap with each other."}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "So let me construct that. So we're going to have length b, and it's collinear with length a. It's along the same line, I should say. They don't overlap with each other. So this is side of length b. And then you have a side of length, let me draw it, this should be a little bit taller, side of length b. Then you have your side of length a at a right angle."}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "They don't overlap with each other. So this is side of length b. And then you have a side of length, let me draw it, this should be a little bit taller, side of length b. Then you have your side of length a at a right angle. Your side of length a comes in at a right angle. And then you have your side of length c. You have your side of length c. So the first thing we need to think about is what's the angle between these two sides? What's this mystery angle?"}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "Then you have your side of length a at a right angle. Your side of length a comes in at a right angle. And then you have your side of length c. You have your side of length c. So the first thing we need to think about is what's the angle between these two sides? What's this mystery angle? What's that mystery angle going to be? Well, it looks like something, but let's see if we can prove to ourselves it really is what we think it looks like. If we look at this original triangle, and we call this angle theta, what's this angle over here, the angle that's between sides of length a and length c, what's the measure of this angle going to be?"}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "What's this mystery angle? What's that mystery angle going to be? Well, it looks like something, but let's see if we can prove to ourselves it really is what we think it looks like. If we look at this original triangle, and we call this angle theta, what's this angle over here, the angle that's between sides of length a and length c, what's the measure of this angle going to be? Well, theta plus this angle have to add up to 90, because you add those two together, they add up to 90, and then you have another 90, it's going to get 180 degrees for the interior angles of this triangle. So these two have to add up to 90, this angle's going to be 90 minus theta. Well, if this triangle up here is congruent, and we've constructed it so it is congruent, the corresponding angle to this one is this angle right over here, so this is also going to be theta, and this right over here is going to be 90 minus theta."}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "If we look at this original triangle, and we call this angle theta, what's this angle over here, the angle that's between sides of length a and length c, what's the measure of this angle going to be? Well, theta plus this angle have to add up to 90, because you add those two together, they add up to 90, and then you have another 90, it's going to get 180 degrees for the interior angles of this triangle. So these two have to add up to 90, this angle's going to be 90 minus theta. Well, if this triangle up here is congruent, and we've constructed it so it is congruent, the corresponding angle to this one is this angle right over here, so this is also going to be theta, and this right over here is going to be 90 minus theta. So given that this is theta, this is 90 minus theta, what is our angle going to be? Well, they all collectively kind of go 180 degrees, so you have theta plus 90 minus theta, plus our mystery angle is going to be equal to 180 degrees, the thetas cancel out, theta minus theta, and you have 90 plus our mystery angle is 180 degrees, subtract 90 from both sides, and you are left with your mystery angle equaling 90 degrees. So that all worked out well, so let me make that clear."}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "Well, if this triangle up here is congruent, and we've constructed it so it is congruent, the corresponding angle to this one is this angle right over here, so this is also going to be theta, and this right over here is going to be 90 minus theta. So given that this is theta, this is 90 minus theta, what is our angle going to be? Well, they all collectively kind of go 180 degrees, so you have theta plus 90 minus theta, plus our mystery angle is going to be equal to 180 degrees, the thetas cancel out, theta minus theta, and you have 90 plus our mystery angle is 180 degrees, subtract 90 from both sides, and you are left with your mystery angle equaling 90 degrees. So that all worked out well, so let me make that clear. That's going to be useful for us in a second. So we can now say definitively that this is 90 degrees, this is a right angle. Now what we are going to do is we are going to construct a trapezoid."}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "So that all worked out well, so let me make that clear. That's going to be useful for us in a second. So we can now say definitively that this is 90 degrees, this is a right angle. Now what we are going to do is we are going to construct a trapezoid. This side A is parallel to side B down here, the way it's been constructed, this is just one side right over here, this goes straight up, and now let's just connect these two sides right over there. So there's a couple of ways to think about the area of this trapezoid. One is we can just think of it as a trapezoid and come up with its area, and then we can think about it as the sum of the areas of its components."}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "Now what we are going to do is we are going to construct a trapezoid. This side A is parallel to side B down here, the way it's been constructed, this is just one side right over here, this goes straight up, and now let's just connect these two sides right over there. So there's a couple of ways to think about the area of this trapezoid. One is we can just think of it as a trapezoid and come up with its area, and then we can think about it as the sum of the areas of its components. So let's just first think of it as a trapezoid. So what do we know about the area of a trapezoid? Well the area of a trapezoid is going to be the height of the trapezoid, which is A plus B, A plus B, that's the height of the trapezoid, times the mean of the top and the bottom, or the average of the top and the bottom."}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "One is we can just think of it as a trapezoid and come up with its area, and then we can think about it as the sum of the areas of its components. So let's just first think of it as a trapezoid. So what do we know about the area of a trapezoid? Well the area of a trapezoid is going to be the height of the trapezoid, which is A plus B, A plus B, that's the height of the trapezoid, times the mean of the top and the bottom, or the average of the top and the bottom. So this times 1 half times A plus B. And the intuition there, you're taking the height times the average of this bottom and the top. The average of the bottom and the top gives you the area of the trapezoid."}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "Well the area of a trapezoid is going to be the height of the trapezoid, which is A plus B, A plus B, that's the height of the trapezoid, times the mean of the top and the bottom, or the average of the top and the bottom. So this times 1 half times A plus B. And the intuition there, you're taking the height times the average of this bottom and the top. The average of the bottom and the top gives you the area of the trapezoid. Now, how can we also figure out the area with its component parts? So this should be, regardless of how we calculate the area, as long as we do correct things, we should come up with the same result. So how else can we come up with this area?"}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "The average of the bottom and the top gives you the area of the trapezoid. Now, how can we also figure out the area with its component parts? So this should be, regardless of how we calculate the area, as long as we do correct things, we should come up with the same result. So how else can we come up with this area? Well, we could say it's the area of the two right triangles. The area of each of them is 1 half A times B. But there's two of them."}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "So how else can we come up with this area? Well, we could say it's the area of the two right triangles. The area of each of them is 1 half A times B. But there's two of them. Let me do that B in that same blue color. But there's two of these right triangles, so let's multiply by two. So this 2 times 1 half AB, that takes into consideration this bottom right triangle and this top one."}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "But there's two of them. Let me do that B in that same blue color. But there's two of these right triangles, so let's multiply by two. So this 2 times 1 half AB, that takes into consideration this bottom right triangle and this top one. And what's the area of this large one that I will color in in green? What's the area of this large one? Well, that's pretty straightforward."}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "So this 2 times 1 half AB, that takes into consideration this bottom right triangle and this top one. And what's the area of this large one that I will color in in green? What's the area of this large one? Well, that's pretty straightforward. It's just 1 half C times C. So plus 1 half C times C, which is 1 half C squared. Now, let's simplify this thing and see what we come up with. And you might guess where all of this is going."}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "Well, that's pretty straightforward. It's just 1 half C times C. So plus 1 half C times C, which is 1 half C squared. Now, let's simplify this thing and see what we come up with. And you might guess where all of this is going. So let's see what we get. So we can rearrange this. So this is 1 half times A plus B squared is going to be equal to 2 times 1 half."}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "And you might guess where all of this is going. So let's see what we get. So we can rearrange this. So this is 1 half times A plus B squared is going to be equal to 2 times 1 half. Well, that's just going to be 1. So it's going to be equal to A times B plus 1 half C squared. Well, I don't like these 1 halves laying around, so let's multiply both sides of this equation by 2."}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "So this is 1 half times A plus B squared is going to be equal to 2 times 1 half. Well, that's just going to be 1. So it's going to be equal to A times B plus 1 half C squared. Well, I don't like these 1 halves laying around, so let's multiply both sides of this equation by 2. I'm just going to multiply both sides of this equation by 2. On the left-hand side, I'm just left with A plus B squared. So let me write that."}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "Well, I don't like these 1 halves laying around, so let's multiply both sides of this equation by 2. I'm just going to multiply both sides of this equation by 2. On the left-hand side, I'm just left with A plus B squared. So let me write that. A plus B squared. And on the right-hand side, I am left with 2AB. Trying to keep the color coding right."}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "So let me write that. A plus B squared. And on the right-hand side, I am left with 2AB. Trying to keep the color coding right. And then 2 times 1 half C squared, that's just going to be C squared. Plus C squared. Well, what happens if you multiply out A plus B times A plus B?"}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "Trying to keep the color coding right. And then 2 times 1 half C squared, that's just going to be C squared. Plus C squared. Well, what happens if you multiply out A plus B times A plus B? What is A plus B squared? Well, it's going to be A squared plus 2AB plus B squared. And then our right-hand side is still going to be equal to all of this business."}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "Well, what happens if you multiply out A plus B times A plus B? What is A plus B squared? Well, it's going to be A squared plus 2AB plus B squared. And then our right-hand side is still going to be equal to all of this business. And changing all of the colors is difficult for me. So let me copy and let me paste it. So it's still going to be equal to the right-hand side."}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "And then our right-hand side is still going to be equal to all of this business. And changing all of the colors is difficult for me. So let me copy and let me paste it. So it's still going to be equal to the right-hand side. Well, this is interesting. How can we simplify this? Is there anything that we can subtract from both sides?"}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "So it's still going to be equal to the right-hand side. Well, this is interesting. How can we simplify this? Is there anything that we can subtract from both sides? Well, sure there is. You have a 2AB on the left-hand side. You have a 2AB on the right-hand side."}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "Is there anything that we can subtract from both sides? Well, sure there is. You have a 2AB on the left-hand side. You have a 2AB on the right-hand side. Let's subtract 2AB from both sides. If you subtract 2AB from both sides, what are you left with? You are left with the Pythagorean theorem."}, {"video_title": "Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "You have a 2AB on the right-hand side. Let's subtract 2AB from both sides. If you subtract 2AB from both sides, what are you left with? You are left with the Pythagorean theorem. So you're left with A squared plus B squared is equal to C squared. Very, very exciting. And for that, we have to thank the 20th president of the United States, James Garfield."}, {"video_title": "Possible transformations example Transformations Geometry Khan Academy.mp3", "Sentence": "Transformation C maps negative two, three, to four, negative one. So let me do negative two, comma, three. And it maps that to four, negative one. To four, negative one. And point negative five, comma, five. Negative five, comma, five. Negative five, comma, five."}, {"video_title": "Possible transformations example Transformations Geometry Khan Academy.mp3", "Sentence": "To four, negative one. And point negative five, comma, five. Negative five, comma, five. Negative five, comma, five. It maps that to seven, negative three. To seven, negative three. So seven, negative three."}, {"video_title": "Possible transformations example Transformations Geometry Khan Academy.mp3", "Sentence": "Negative five, comma, five. It maps that to seven, negative three. To seven, negative three. So seven, negative three. And so let's think about this a little bit. How could we get from this point to this point, and that point to that point? Now it's tempting to view this that maybe a translation is possible."}, {"video_title": "Possible transformations example Transformations Geometry Khan Academy.mp3", "Sentence": "So seven, negative three. And so let's think about this a little bit. How could we get from this point to this point, and that point to that point? Now it's tempting to view this that maybe a translation is possible. Because if you imagined a line like that, you could say, hey, let's just shift this whole thing down. And then to the right, these two things happen to have the same slope. They both have a slope of negative 2 3rds."}, {"video_title": "Possible transformations example Transformations Geometry Khan Academy.mp3", "Sentence": "Now it's tempting to view this that maybe a translation is possible. Because if you imagined a line like that, you could say, hey, let's just shift this whole thing down. And then to the right, these two things happen to have the same slope. They both have a slope of negative 2 3rds. And so this point would map to this point, and that point would map to that point. But that's not what we want. We don't want negative two, three to map to seven, negative three."}, {"video_title": "Possible transformations example Transformations Geometry Khan Academy.mp3", "Sentence": "They both have a slope of negative 2 3rds. And so this point would map to this point, and that point would map to that point. But that's not what we want. We don't want negative two, three to map to seven, negative three. We want negative two, three to map to four, negative one. So you could get this line over this line, but we won't map the points that we want to map. So this can't be, at least I can't think of a way, that this could actually be a translation."}, {"video_title": "Possible transformations example Transformations Geometry Khan Academy.mp3", "Sentence": "We don't want negative two, three to map to seven, negative three. We want negative two, three to map to four, negative one. So you could get this line over this line, but we won't map the points that we want to map. So this can't be, at least I can't think of a way, that this could actually be a translation. Now let's think about whether our transformation could be a reflection. Well, if we imagine a line that has, let's see, these both have a slope of negative 3. These both have a slope of negative 2 3rds."}, {"video_title": "Possible transformations example Transformations Geometry Khan Academy.mp3", "Sentence": "So this can't be, at least I can't think of a way, that this could actually be a translation. Now let's think about whether our transformation could be a reflection. Well, if we imagine a line that has, let's see, these both have a slope of negative 3. These both have a slope of negative 2 3rds. So if you imagine a line that had a slope of positive 3 halves, that was equidistant from both. And I don't know if this is, let's see, is this equidistant? Is this equidistant from both of them?"}, {"video_title": "Possible transformations example Transformations Geometry Khan Academy.mp3", "Sentence": "These both have a slope of negative 2 3rds. So if you imagine a line that had a slope of positive 3 halves, that was equidistant from both. And I don't know if this is, let's see, is this equidistant? Is this equidistant from both of them? It's either going to be that line or this line right over, or that line. Actually, that line looks better. So that one."}, {"video_title": "Possible transformations example Transformations Geometry Khan Academy.mp3", "Sentence": "Is this equidistant from both of them? It's either going to be that line or this line right over, or that line. Actually, that line looks better. So that one. And once again, I'm just eyeballing it. So a line that has slope of positive 3 halves. So this one looks right in between the two."}, {"video_title": "Possible transformations example Transformations Geometry Khan Academy.mp3", "Sentence": "So that one. And once again, I'm just eyeballing it. So a line that has slope of positive 3 halves. So this one looks right in between the two. Or actually, it could be someplace in between. But either way, we just have to think about it qualitatively. So imagine a line that looked something like that."}, {"video_title": "Possible transformations example Transformations Geometry Khan Academy.mp3", "Sentence": "So this one looks right in between the two. Or actually, it could be someplace in between. But either way, we just have to think about it qualitatively. So imagine a line that looked something like that. And if you were to reflect over this line, then this point would map to this point, which is what we want. And this purple point, negative 5 comma 5, would map to that point. It would be reflected over."}, {"video_title": "Possible transformations example Transformations Geometry Khan Academy.mp3", "Sentence": "So imagine a line that looked something like that. And if you were to reflect over this line, then this point would map to this point, which is what we want. And this purple point, negative 5 comma 5, would map to that point. It would be reflected over. So it's pretty clear that this could be a reflection. Now, rotation actually makes even more sense, or at least in my brain, makes a little more sense. If you were to rotate around this point right over here, this point would map to that point."}, {"video_title": "Possible transformations example Transformations Geometry Khan Academy.mp3", "Sentence": "It would be reflected over. So it's pretty clear that this could be a reflection. Now, rotation actually makes even more sense, or at least in my brain, makes a little more sense. If you were to rotate around this point right over here, this point would map to that point. And that point would map to that point. So a rotation also seems like a possibility for transformation C. Now let's think about transformation D. We are going from 4, negative 1 to 7, negative 3. Actually, maybe I'll put that in magenta as well, to 7, negative 3, just like that."}, {"video_title": "Possible transformations example Transformations Geometry Khan Academy.mp3", "Sentence": "If you were to rotate around this point right over here, this point would map to that point. And that point would map to that point. So a rotation also seems like a possibility for transformation C. Now let's think about transformation D. We are going from 4, negative 1 to 7, negative 3. Actually, maybe I'll put that in magenta as well, to 7, negative 3, just like that. And we want to go from negative 5, 5 to negative 2, 3. So I could definitely imagine a translation right over here. This point went 3 to the right and 2 down."}, {"video_title": "Possible transformations example Transformations Geometry Khan Academy.mp3", "Sentence": "Actually, maybe I'll put that in magenta as well, to 7, negative 3, just like that. And we want to go from negative 5, 5 to negative 2, 3. So I could definitely imagine a translation right over here. This point went 3 to the right and 2 down. This point went 3 to the right and 2 down. So a translation definitely makes sense. Now let's think about a reflection."}, {"video_title": "Possible transformations example Transformations Geometry Khan Academy.mp3", "Sentence": "This point went 3 to the right and 2 down. This point went 3 to the right and 2 down. So a translation definitely makes sense. Now let's think about a reflection. So it would be tempting to get from this point to this point. I could reflect around that. But that won't help this one over here."}, {"video_title": "Possible transformations example Transformations Geometry Khan Academy.mp3", "Sentence": "Now let's think about a reflection. So it would be tempting to get from this point to this point. I could reflect around that. But that won't help this one over here. And to get from that point to that point, I could reflect around that. But once again, that's not going to help that point over there. So a reflection really doesn't seem to do the trick."}, {"video_title": "Possible transformations example Transformations Geometry Khan Academy.mp3", "Sentence": "But that won't help this one over here. And to get from that point to that point, I could reflect around that. But once again, that's not going to help that point over there. So a reflection really doesn't seem to do the trick. And what about a rotation? Well, to go from this point to this point, we could rotate around this point. We could go there."}, {"video_title": "Possible transformations example Transformations Geometry Khan Academy.mp3", "Sentence": "So a reflection really doesn't seem to do the trick. And what about a rotation? Well, to go from this point to this point, we could rotate around this point. We could go there. But that won't help this point right over here. While this is rotating there, this point is going to rotate around like that. And it's going to end up someplace out here."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "A lot of what geometry is about is proving things about the world. In order to really prove things about the world, we have to be very careful, very precise, very exact with our language so that we know what we're proving and we know what we're assuming and what type of deductions we are making as we prove things. And so to get some practice being precise and exact with our language, I'm going to go through some exercises from the geometric definitions exercise on Khan Academy. So this first one says, three students attempt to define what it means for lines L and M to be perpendicular. Can you match the teacher's comments to the definitions? All right, so it looks like three different students attempt definitions of what it means to be perpendicular, and then there's these teacher's comments that we can move around. So we're going to, I guess, pretend that we're the teacher."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So this first one says, three students attempt to define what it means for lines L and M to be perpendicular. Can you match the teacher's comments to the definitions? All right, so it looks like three different students attempt definitions of what it means to be perpendicular, and then there's these teacher's comments that we can move around. So we're going to, I guess, pretend that we're the teacher. So Ruby's definition for being perpendicular, L and M, lines L and M, are perpendicular if they never meet. Well, that's not true. In fact, perpendicular lines, for sure, will intersect."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So we're going to, I guess, pretend that we're the teacher. So Ruby's definition for being perpendicular, L and M, lines L and M, are perpendicular if they never meet. Well, that's not true. In fact, perpendicular lines, for sure, will intersect. So in fact, they intersect at right angles. So that is not going to be correct. And so, actually, this looks right."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "In fact, perpendicular lines, for sure, will intersect. So in fact, they intersect at right angles. So that is not going to be correct. And so, actually, this looks right. Were you thinking of parallel lines? Because that looks like what she was trying to define. If things are on the same plane and they never intersect, then you are talking about parallel lines."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And so, actually, this looks right. Were you thinking of parallel lines? Because that looks like what she was trying to define. If things are on the same plane and they never intersect, then you are talking about parallel lines. I also get Shreya's definition. L and M are perpendicular if they meet at one point and one of the angles at their point of intersection is a right angle. Well, that seems spot on."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "If things are on the same plane and they never intersect, then you are talking about parallel lines. I also get Shreya's definition. L and M are perpendicular if they meet at one point and one of the angles at their point of intersection is a right angle. Well, that seems spot on. So let's see. I would say, woo-hoo, nice work. I couldn't have said it better myself."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Well, that seems spot on. So let's see. I would say, woo-hoo, nice work. I couldn't have said it better myself. Now let's just make sure this comment matches for this definition. Abhishek says, L and M are perpendicular if they meet at a single point such that the two lines make a T. Well, that's, in a hand-wavy way, kind of right. When you imagine perpendicular lines, you could imagine them kind of forming a cross or I guess a T would be part of it."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "I couldn't have said it better myself. Now let's just make sure this comment matches for this definition. Abhishek says, L and M are perpendicular if they meet at a single point such that the two lines make a T. Well, that's, in a hand-wavy way, kind of right. When you imagine perpendicular lines, you could imagine them kind of forming a cross or I guess a T would be part of it. But I think this comment is spot on, the teacher's comment. Your definition is kind of correct, but it lacks mathematical precision. What does he mean by a T?"}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "When you imagine perpendicular lines, you could imagine them kind of forming a cross or I guess a T would be part of it. But I think this comment is spot on, the teacher's comment. Your definition is kind of correct, but it lacks mathematical precision. What does he mean by a T? What does it mean to make a T? Shreya's definition is much more precise. They're perpendicular if they meet at one point and one of their angles at their point of intersection is a right angle, is a 90-degree angle."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "What does he mean by a T? What does it mean to make a T? Shreya's definition is much more precise. They're perpendicular if they meet at one point and one of their angles at their point of intersection is a right angle, is a 90-degree angle. Let's check our answer. So let's do a few more of these. This is actually a lot of fun."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "They're perpendicular if they meet at one point and one of their angles at their point of intersection is a right angle, is a 90-degree angle. Let's check our answer. So let's do a few more of these. This is actually a lot of fun. So once again, we're gonna have three students attempting to define, but now they're going to define an object called an angle. Can you match the teacher's comments to the definitions? So Ruby, well, these same three students, Ruby says the amount of turn between two straight lines that have a common vertex."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "This is actually a lot of fun. So once again, we're gonna have three students attempting to define, but now they're going to define an object called an angle. Can you match the teacher's comments to the definitions? So Ruby, well, these same three students, Ruby says the amount of turn between two straight lines that have a common vertex. Well, this is kind of getting there. The definition of an angle, we typically talk about two rays with a common vertex. She's talking about two lines with a common vertex."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So Ruby, well, these same three students, Ruby says the amount of turn between two straight lines that have a common vertex. Well, this is kind of getting there. The definition of an angle, we typically talk about two rays with a common vertex. She's talking about two lines with a common vertex. And she's talking about the amount of turn. So she's really talking about more of kind of the measure of an angle. So let's see what comment here."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "She's talking about two lines with a common vertex. And she's talking about the amount of turn. So she's really talking about more of kind of the measure of an angle. So let's see what comment here. So you seem to be getting at the idea of a measure of an angle and not the definition of an angle itself. So this is actually right. I would put this one right here."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So let's see what comment here. So you seem to be getting at the idea of a measure of an angle and not the definition of an angle itself. So this is actually right. I would put this one right here. We just got lucky this was already aligned. So Shreya's definition, two lines that come together. So once again, this is kind of, the definition of an angle is two rays with a common vertex."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "I would put this one right here. We just got lucky this was already aligned. So Shreya's definition, two lines that come together. So once again, this is kind of, the definition of an angle is two rays with a common vertex. So two lines that come together, this is just intersecting lines. Now when that happens, you might be forming some angles, but I would just say, were you thinking of intersecting lines? And let's see what Abhishek says."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So once again, this is kind of, the definition of an angle is two rays with a common vertex. So two lines that come together, this is just intersecting lines. Now when that happens, you might be forming some angles, but I would just say, were you thinking of intersecting lines? And let's see what Abhishek says. A figure composed of two rays sharing a common endpoint. The common endpoint is known as the vertex. Yep, that's a good definition of an angle."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And let's see what Abhishek says. A figure composed of two rays sharing a common endpoint. The common endpoint is known as the vertex. Yep, that's a good definition of an angle. So Abhishek got it this time. Let's do another one. So three students are now attempting to define what it means for two lines to be parallel."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Yep, that's a good definition of an angle. So Abhishek got it this time. Let's do another one. So three students are now attempting to define what it means for two lines to be parallel. So now let's match the teacher's comments. So Daniela says two lines are parallel if they are distinct and one can be translated on top of the other. All right, so that actually seems pretty interesting."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So three students are now attempting to define what it means for two lines to be parallel. So now let's match the teacher's comments. So Daniela says two lines are parallel if they are distinct and one can be translated on top of the other. All right, so that actually seems pretty interesting. That's actually not the first way that I would have defined parallel lines. I would have said, hey, if they're on the same plane and they don't intersect, then they are parallel. But this seems pretty good because if you're translating something, you're not, you aren't going to rotate it."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "All right, so that actually seems pretty interesting. That's actually not the first way that I would have defined parallel lines. I would have said, hey, if they're on the same plane and they don't intersect, then they are parallel. But this seems pretty good because if you're translating something, you're not, you aren't going to rotate it. You're not going to change its direction, I guess one way to think about it. And so if you're translating one, if you can, if they're two different lines, but you can shift them without changing their direction, which is what translation is all about, on top of each other, that actually feels pretty good. So I'll put that right over there."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "But this seems pretty good because if you're translating something, you're not, you aren't going to rotate it. You're not going to change its direction, I guess one way to think about it. And so if you're translating one, if you can, if they're two different lines, but you can shift them without changing their direction, which is what translation is all about, on top of each other, that actually feels pretty good. So I'll put that right over there. So Ori says two lines are parallel if they are close together but don't intersect. So if you're trying to define parallel lines, parallel lines, it doesn't matter if they're close together or not. They just have to be in the same plane and not intersect."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So I'll put that right over there. So Ori says two lines are parallel if they are close together but don't intersect. So if you're trying to define parallel lines, parallel lines, it doesn't matter if they're close together or not. They just have to be in the same plane and not intersect. They could be very far apart and they could still be parallel. So this isn't an incorrect statement. You could have two lines that are close together and don't intersect on the same plane and they are going to be parallel."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "They just have to be in the same plane and not intersect. They could be very far apart and they could still be parallel. So this isn't an incorrect statement. You could have two lines that are close together and don't intersect on the same plane and they are going to be parallel. But this isn't a good definition because you can also have parallel lines that are far apart. And so, actually I'd go with this statement right here. Part of your definition is correct, but the other part is not."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "You could have two lines that are close together and don't intersect on the same plane and they are going to be parallel. But this isn't a good definition because you can also have parallel lines that are far apart. And so, actually I'd go with this statement right here. Part of your definition is correct, but the other part is not. Parallel lines don't have to be close together. So this isn't a good definition of parallel lines. And let's see, Kaori."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Part of your definition is correct, but the other part is not. Parallel lines don't have to be close together. So this isn't a good definition of parallel lines. And let's see, Kaori. Two lines are parallel as long as they aren't perpendicular. Well, that's just not true because you can intersect. You can have two lines that intersect at non-right angles and they're not parallel and they're also not perpendicular."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And let's see, Kaori. Two lines are parallel as long as they aren't perpendicular. Well, that's just not true because you can intersect. You can have two lines that intersect at non-right angles and they're not parallel and they're also not perpendicular. So this is, you know, sorry, your definition is incorrect. This is actually a lot of fun pretending to be the teacher. Let's do another one."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "You can have two lines that intersect at non-right angles and they're not parallel and they're also not perpendicular. So this is, you know, sorry, your definition is incorrect. This is actually a lot of fun pretending to be the teacher. Let's do another one. All right. So three students attempt to define what a line segment is. And we have a depiction of a line segment right over here."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Let's do another one. All right. So three students attempt to define what a line segment is. And we have a depiction of a line segment right over here. We have point P, point Q, and the line segment is all the points in between P and Q. So let's match the teacher's comments to the definitions. Ivy's definition."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And we have a depiction of a line segment right over here. We have point P, point Q, and the line segment is all the points in between P and Q. So let's match the teacher's comments to the definitions. Ivy's definition. All of the points in line with P and Q extending infinitely in both directions. Well, that would be the definition of a line. That would be the line PQ."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Ivy's definition. All of the points in line with P and Q extending infinitely in both directions. Well, that would be the definition of a line. That would be the line PQ. That would be if you're extending infinitely in both directions. So I would say are you thinking of a line instead of a line segment? Ethan's definition."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "That would be the line PQ. That would be if you're extending infinitely in both directions. So I would say are you thinking of a line instead of a line segment? Ethan's definition. The exact distance from P to Q. Well, that's just the length of a line segment. That's not exactly what a line segment is."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Ethan's definition. The exact distance from P to Q. Well, that's just the length of a line segment. That's not exactly what a line segment is. Let's see, Ibuka's definition. The points P and Q, which are called endpoints, and all of the points in a straight line between points P and Q. Yep, that looks like a good definition for a line segment. So we can just check our answer."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "That's not exactly what a line segment is. Let's see, Ibuka's definition. The points P and Q, which are called endpoints, and all of the points in a straight line between points P and Q. Yep, that looks like a good definition for a line segment. So we can just check our answer. So looking good. Let's do one more of this. I'm just really enjoying pretending to be a teacher."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So we can just check our answer. So looking good. Let's do one more of this. I'm just really enjoying pretending to be a teacher. All right, three students attempted to find what a circle is. Define what a circle is. Can you match the teacher's comments to the definitions?"}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "I'm just really enjoying pretending to be a teacher. All right, three students attempted to find what a circle is. Define what a circle is. Can you match the teacher's comments to the definitions? Duru, the set of all points in a plane that are the same distance away from some given point, which we call the center. That actually seems like a pretty good definition of a circle. So, stupendous, well done."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Can you match the teacher's comments to the definitions? Duru, the set of all points in a plane that are the same distance away from some given point, which we call the center. That actually seems like a pretty good definition of a circle. So, stupendous, well done. Oliver's definition. The set of all points in 3D space that are the same distance from a center point. Well, if we're talking about 3D space and the set of all points that are equidistant from that point in 3D space, now we're talking about a sphere, not a circle."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So, stupendous, well done. Oliver's definition. The set of all points in 3D space that are the same distance from a center point. Well, if we're talking about 3D space and the set of all points that are equidistant from that point in 3D space, now we're talking about a sphere, not a circle. And so, you seem to be confusing a circle with a sphere. And then finally, a perfectly round shape. Well, that's kinda true, but if you're talking about three dimensions, you could be talking about a sphere."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Well, if we're talking about 3D space and the set of all points that are equidistant from that point in 3D space, now we're talking about a sphere, not a circle. And so, you seem to be confusing a circle with a sphere. And then finally, a perfectly round shape. Well, that's kinda true, but if you're talking about three dimensions, you could be talking about a sphere. If you go beyond three dimensions, hypersphere, whatever else. In two dimensions, yeah, a perfectly round shape. Most people would call it a circle, but that doesn't have a lot of precision to it."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Well, that's kinda true, but if you're talking about three dimensions, you could be talking about a sphere. If you go beyond three dimensions, hypersphere, whatever else. In two dimensions, yeah, a perfectly round shape. Most people would call it a circle, but that doesn't have a lot of precision to it. It doesn't give us a lot that we can work with from a mathematical point of view. So, I would say, actually, what the teacher's saying. Your definition needs to be much more precise."}, {"video_title": "Geometric precision practice Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Most people would call it a circle, but that doesn't have a lot of precision to it. It doesn't give us a lot that we can work with from a mathematical point of view. So, I would say, actually, what the teacher's saying. Your definition needs to be much more precise. Duru's definition is much, much more precise. The set of all points that are equidistant in a plane that are equidistant away from a given point, which we call the center. So, yep, Carlos could use a little bit more precision."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "So I just have an arbitrary triangle right over here, triangle ABC. What I'm going to do is I'm going to draw an angle bisector for this angle up here. We could have done it with any of the three angles, but I'll just do this one. It'll make our proof a little bit easier. So I'm just going to bisect this angle, angle ABC. So let's just say that that's the angle bisector angle ABC, and so this angle right over here is equal to this angle right over here. And let me call this point down here, let me call it point D. The angle bisector theorem tells us that the ratio between the sides that aren't this bisector, so when I put this angle bisector here, it created two smaller triangles out of that larger one."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "It'll make our proof a little bit easier. So I'm just going to bisect this angle, angle ABC. So let's just say that that's the angle bisector angle ABC, and so this angle right over here is equal to this angle right over here. And let me call this point down here, let me call it point D. The angle bisector theorem tells us that the ratio between the sides that aren't this bisector, so when I put this angle bisector here, it created two smaller triangles out of that larger one. The angle bisector theorem tells us the ratios between the other sides of these two triangles that we've now created are going to be the same. So it tells us that the ratio of AB to AD is going to be equal to the ratio of BC to, you could say, CD. So the ratio of, I'll color code it, the ratio of that, which is this, to this, is going to be equal to the ratio of this, which is that, to this right over here, to CD, which is that over here."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "And let me call this point down here, let me call it point D. The angle bisector theorem tells us that the ratio between the sides that aren't this bisector, so when I put this angle bisector here, it created two smaller triangles out of that larger one. The angle bisector theorem tells us the ratios between the other sides of these two triangles that we've now created are going to be the same. So it tells us that the ratio of AB to AD is going to be equal to the ratio of BC to, you could say, CD. So the ratio of, I'll color code it, the ratio of that, which is this, to this, is going to be equal to the ratio of this, which is that, to this right over here, to CD, which is that over here. So once you see the ratio of that to that, it's going to be the same as the ratio of that to that. So that's kind of a cool result, but you can't just left it on faith because it's a cool result. You want to prove it to ourselves."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "So the ratio of, I'll color code it, the ratio of that, which is this, to this, is going to be equal to the ratio of this, which is that, to this right over here, to CD, which is that over here. So once you see the ratio of that to that, it's going to be the same as the ratio of that to that. So that's kind of a cool result, but you can't just left it on faith because it's a cool result. You want to prove it to ourselves. And so you could imagine right over here we have some ratios set up. So we're going to prove it using similar triangles. And unfortunate for us, these two triangles right here aren't necessarily similar."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "You want to prove it to ourselves. And so you could imagine right over here we have some ratios set up. So we're going to prove it using similar triangles. And unfortunate for us, these two triangles right here aren't necessarily similar. We know that these two angles are congruent to each other, but we don't know whether this angle is equal to that angle or that angle. We don't know. We can't make any statements like that."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "And unfortunate for us, these two triangles right here aren't necessarily similar. We know that these two angles are congruent to each other, but we don't know whether this angle is equal to that angle or that angle. We don't know. We can't make any statements like that. So in order to actually set up this type of a statement, we'll have to construct maybe another triangle that will be similar to one of these right over here. And one way to do it would be to draw another line. And this is a bit of this proof wasn't obvious to me the first time that I thought about it, so don't worry if it's not obvious to you."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "We can't make any statements like that. So in order to actually set up this type of a statement, we'll have to construct maybe another triangle that will be similar to one of these right over here. And one way to do it would be to draw another line. And this is a bit of this proof wasn't obvious to me the first time that I thought about it, so don't worry if it's not obvious to you. Is that what happens is if we can continue this bisector, this angle bisector right over here. So let's just continue it. It just keeps going on and on and on."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "And this is a bit of this proof wasn't obvious to me the first time that I thought about it, so don't worry if it's not obvious to you. Is that what happens is if we can continue this bisector, this angle bisector right over here. So let's just continue it. It just keeps going on and on and on. And let's also, maybe we can construct a similar triangle to this triangle over here if we draw a line that's parallel to AB down here. So let's try to do that. So I'm just going to say, if C is not on AB, you can always find a point that goes through or a line that goes through C that is parallel to AB."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "It just keeps going on and on and on. And let's also, maybe we can construct a similar triangle to this triangle over here if we draw a line that's parallel to AB down here. So let's try to do that. So I'm just going to say, if C is not on AB, you can always find a point that goes through or a line that goes through C that is parallel to AB. So let's just by definition, let's just create another line right over here. And let's say, and let's call this point right over here F. And let's just pick this line in such a way that FC is parallel to AB. So this is parallel to that right over there."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "So I'm just going to say, if C is not on AB, you can always find a point that goes through or a line that goes through C that is parallel to AB. So let's just by definition, let's just create another line right over here. And let's say, and let's call this point right over here F. And let's just pick this line in such a way that FC is parallel to AB. So this is parallel to that right over there. So FC is parallel to AB. And we can just construct it that way. And now we have some interesting things."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "So this is parallel to that right over there. So FC is parallel to AB. And we can just construct it that way. And now we have some interesting things. And we did it that way so that we can make these two triangles be similar to each other. So let's see that. Let's see what happens."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "And now we have some interesting things. And we did it that way so that we can make these two triangles be similar to each other. So let's see that. Let's see what happens. So before we even think about similarity, let's think about what some of the angles or what we know about some of the angles here. We know that we have alternate interior angles. So just think about these two parallel lines."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "Let's see what happens. So before we even think about similarity, let's think about what some of the angles or what we know about some of the angles here. We know that we have alternate interior angles. So just think about these two parallel lines. So I could imagine AB keeps going like that. FC keeps going like that. And line BD right here is a transversal."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "So just think about these two parallel lines. So I could imagine AB keeps going like that. FC keeps going like that. And line BD right here is a transversal. Then whatever this angle is, this angle is going to be as well from alternate interior angles, which we've talked a lot about when we first talked about angles with transversals and all of that. So these two angles are going to be the same. But this angle and this angle are also going to be the same."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "And line BD right here is a transversal. Then whatever this angle is, this angle is going to be as well from alternate interior angles, which we've talked a lot about when we first talked about angles with transversals and all of that. So these two angles are going to be the same. But this angle and this angle are also going to be the same. Because this angle and that angle are the same. This is a bisector. So because this is a bisector, we know that angle ABD is the same as angle DBC."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "But this angle and this angle are also going to be the same. Because this angle and that angle are the same. This is a bisector. So because this is a bisector, we know that angle ABD is the same as angle DBC. So whatever this angle is, that angle is, and so is this angle. And that gives us some kind of an interesting result. Because here we have a situation where if you look at this larger triangle, BFC, we have two base angles that are the same, which means this must be an isosceles triangle."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "So because this is a bisector, we know that angle ABD is the same as angle DBC. So whatever this angle is, that angle is, and so is this angle. And that gives us some kind of an interesting result. Because here we have a situation where if you look at this larger triangle, BFC, we have two base angles that are the same, which means this must be an isosceles triangle. So BC must be the same as FC. So that was kind of cool. We just used the transversal and the alternate interior angles to show that these are isosceles and that BC and FC are the same thing."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "Because here we have a situation where if you look at this larger triangle, BFC, we have two base angles that are the same, which means this must be an isosceles triangle. So BC must be the same as FC. So that was kind of cool. We just used the transversal and the alternate interior angles to show that these are isosceles and that BC and FC are the same thing. And that could be useful. Because we know that we have a feeling that this triangle and this triangle are going to be similar. We haven't proven it yet."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "We just used the transversal and the alternate interior angles to show that these are isosceles and that BC and FC are the same thing. And that could be useful. Because we know that we have a feeling that this triangle and this triangle are going to be similar. We haven't proven it yet. But how will that help us get something about BC up here? But we just showed that BC and FC are the same thing. So this is going to be the same thing."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "We haven't proven it yet. But how will that help us get something about BC up here? But we just showed that BC and FC are the same thing. So this is going to be the same thing. If we want to prove it, if we can prove that FC, the ratio of AB to AD is the same thing as the ratio of FC to CD, we're going to be there. Because BC, we just showed, is equal to FC. But let's not start with the theorem."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "So this is going to be the same thing. If we want to prove it, if we can prove that FC, the ratio of AB to AD is the same thing as the ratio of FC to CD, we're going to be there. Because BC, we just showed, is equal to FC. But let's not start with the theorem. Let's actually get to the theorem. So FC is parallel to AB, able to set up this one isosceles triangle, show these sides are congruent. Now let's look at some of the other angles here and make ourselves feel good about it."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "But let's not start with the theorem. Let's actually get to the theorem. So FC is parallel to AB, able to set up this one isosceles triangle, show these sides are congruent. Now let's look at some of the other angles here and make ourselves feel good about it. Well, if we look at triangle ABD, so this triangle right over here, and triangle FDC, we already established that they have one set of angles that are the same. And then they also both, ABD has this angle right over here, which is a vertical angle with this one over here. So they're congruent."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "Now let's look at some of the other angles here and make ourselves feel good about it. Well, if we look at triangle ABD, so this triangle right over here, and triangle FDC, we already established that they have one set of angles that are the same. And then they also both, ABD has this angle right over here, which is a vertical angle with this one over here. So they're congruent. And we know if two triangles have two angles that are the same, actually, the third one's going to be the same as well. Or you could say by the angle-angle similarity postulate, these two triangles are similar. So let me write that down."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "So they're congruent. And we know if two triangles have two angles that are the same, actually, the third one's going to be the same as well. Or you could say by the angle-angle similarity postulate, these two triangles are similar. So let me write that down. You want to make sure you get the corresponding sides right. We now know by angle-angle, and I'm going to start at the green angle, that triangle B, and then the blue angle, BDA is similar to triangle, so once again, let's start with the green angle F, then you go to the blue angle FDC. And here, we want to eventually get to the angle bisector theorem, so we want to look at the ratio between AB and AD."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "So let me write that down. You want to make sure you get the corresponding sides right. We now know by angle-angle, and I'm going to start at the green angle, that triangle B, and then the blue angle, BDA is similar to triangle, so once again, let's start with the green angle F, then you go to the blue angle FDC. And here, we want to eventually get to the angle bisector theorem, so we want to look at the ratio between AB and AD. Similar triangles, either you could find the ratio between corresponding sides are going to be similar triangles, or you could find the ratio between two sides of a similar triangle and compare them to the ratio of the same two corresponding sides on the other similar triangle, and they should be the same. So by similar triangles, we know that the ratio of AB, and this, by the way, was by angle-angle similarity. Want to write that down."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "And here, we want to eventually get to the angle bisector theorem, so we want to look at the ratio between AB and AD. Similar triangles, either you could find the ratio between corresponding sides are going to be similar triangles, or you could find the ratio between two sides of a similar triangle and compare them to the ratio of the same two corresponding sides on the other similar triangle, and they should be the same. So by similar triangles, we know that the ratio of AB, and this, by the way, was by angle-angle similarity. Want to write that down. So now that we know they're similar, we know that the ratio of AB to AD is going to be equal to, and we could even look here for the corresponding sides, the ratio of AB, the corresponding side is going to be CF. It's going to equal CF over AD. AD is the same thing as CD."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "Want to write that down. So now that we know they're similar, we know that the ratio of AB to AD is going to be equal to, and we could even look here for the corresponding sides, the ratio of AB, the corresponding side is going to be CF. It's going to equal CF over AD. AD is the same thing as CD. And so we know the ratio of AB to AD is equal to CF over CD. But we just proved to ourselves, because this is an isosceles triangle, that CF is the same thing as BC right over here. So CF is the same thing as BC."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "AD is the same thing as CD. And so we know the ratio of AB to AD is equal to CF over CD. But we just proved to ourselves, because this is an isosceles triangle, that CF is the same thing as BC right over here. So CF is the same thing as BC. And we're done. We've just proven AB over AD is equal to BC over CD. So there's kind of two things we had to do here."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "So CF is the same thing as BC. And we're done. We've just proven AB over AD is equal to BC over CD. So there's kind of two things we had to do here. One, construct this other triangle that allowed us, assuming this was parallel, that gave us two things. That gave us another angle to show that they're similar, and also allowed us to establish, sorry, I have something stuck in my throat. Just coughed off camera."}, {"video_title": "Angle bisector theorem proof Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "So there's kind of two things we had to do here. One, construct this other triangle that allowed us, assuming this was parallel, that gave us two things. That gave us another angle to show that they're similar, and also allowed us to establish, sorry, I have something stuck in my throat. Just coughed off camera. So I should go get a drink of water after this. So we were able to, using constructing this triangle here, we were able to both show it's similar and to construct this larger isosceles triangle to show, look, if we can find a ratio between two sides of this triangle and this one, then that's going to be the ratio of this. If we could find the ratio of this side to this side, it's the same as the ratio of this side to this side."}, {"video_title": "Intro to arc measure Mathematics II High School Math Khan Academy.mp3", "Sentence": "What I want to talk about in this video is the notion of arc measure when we're dealing with circles. And as we'll see, sometimes when you see something like arc measure, you might think it's the length of an arc, but arc length is actually a different idea, so we will compare these two things. Arc length to arc measure. So arc measure, all that is, it's just a fancy way of saying, if I have a circle right over here, this is my best attempt at drawing a circle. I have a circle here. The center of the circle, let's call that point O. And let me put some other points over here."}, {"video_title": "Intro to arc measure Mathematics II High School Math Khan Academy.mp3", "Sentence": "So arc measure, all that is, it's just a fancy way of saying, if I have a circle right over here, this is my best attempt at drawing a circle. I have a circle here. The center of the circle, let's call that point O. And let me put some other points over here. So let's say that this is point A. Let's say this is point B. And let's say this is point C right over here."}, {"video_title": "Intro to arc measure Mathematics II High School Math Khan Academy.mp3", "Sentence": "And let me put some other points over here. So let's say that this is point A. Let's say this is point B. And let's say this is point C right over here. And let's say that I have, let's say the central angle right over here, because it includes the center of the circle. So the central angle, angle AOB. Let's say it has a measure of, let's say it has a measure of 120 degrees."}, {"video_title": "Intro to arc measure Mathematics II High School Math Khan Academy.mp3", "Sentence": "And let's say this is point C right over here. And let's say that I have, let's say the central angle right over here, because it includes the center of the circle. So the central angle, angle AOB. Let's say it has a measure of, let's say it has a measure of 120 degrees. And if someone were to say, what is the measure of arc AB? So let me write that down. The measure, so if someone were to say, what is the measure of arc AB?"}, {"video_title": "Intro to arc measure Mathematics II High School Math Khan Academy.mp3", "Sentence": "Let's say it has a measure of, let's say it has a measure of 120 degrees. And if someone were to say, what is the measure of arc AB? So let me write that down. The measure, so if someone were to say, what is the measure of arc AB? And they'd write it like this. So that's referring to arc AB right over here. It's the minor arc."}, {"video_title": "Intro to arc measure Mathematics II High School Math Khan Academy.mp3", "Sentence": "The measure, so if someone were to say, what is the measure of arc AB? And they'd write it like this. So that's referring to arc AB right over here. It's the minor arc. So there's two ways to connect AB. You could connect it right over here. This is the shorter distance."}, {"video_title": "Intro to arc measure Mathematics II High School Math Khan Academy.mp3", "Sentence": "It's the minor arc. So there's two ways to connect AB. You could connect it right over here. This is the shorter distance. Or you could go the other way around, which would be, which is considered the major arc. Now if someone's referring to the major arc, they would say mark ACB. So when you're given just two letters, you assume it's the shortest distance between the two."}, {"video_title": "Intro to arc measure Mathematics II High School Math Khan Academy.mp3", "Sentence": "This is the shorter distance. Or you could go the other way around, which would be, which is considered the major arc. Now if someone's referring to the major arc, they would say mark ACB. So when you're given just two letters, you assume it's the shortest distance between the two. You assume that it is the minor arc. In order to specify the major arc, you would give the third letter to go the long way around. So the measure of arc AB, and sometimes you'll see it with the parentheses right over here, all this is, this is the same thing as the measure of the central angle that intercepts that arc."}, {"video_title": "Intro to arc measure Mathematics II High School Math Khan Academy.mp3", "Sentence": "So when you're given just two letters, you assume it's the shortest distance between the two. You assume that it is the minor arc. In order to specify the major arc, you would give the third letter to go the long way around. So the measure of arc AB, and sometimes you'll see it with the parentheses right over here, all this is, this is the same thing as the measure of the central angle that intercepts that arc. Well, the central angle that intercepts that arc has a measure of 120 degrees. So this is just going to be 120 degrees. Now some of y'all might be saying, well, what about the major arc?"}, {"video_title": "Intro to arc measure Mathematics II High School Math Khan Academy.mp3", "Sentence": "So the measure of arc AB, and sometimes you'll see it with the parentheses right over here, all this is, this is the same thing as the measure of the central angle that intercepts that arc. Well, the central angle that intercepts that arc has a measure of 120 degrees. So this is just going to be 120 degrees. Now some of y'all might be saying, well, what about the major arc? Well, let's write that. So if we're talking about arc ACB, so we're going the other way around. So this is the major arc."}, {"video_title": "Intro to arc measure Mathematics II High School Math Khan Academy.mp3", "Sentence": "Now some of y'all might be saying, well, what about the major arc? Well, let's write that. So if we're talking about arc ACB, so we're going the other way around. So this is the major arc. So what is the measure of arc ACB? Once again, we're using three letters so that we're specifying the major arc. Well, this angle, this central angle right over here, to go all the way around the circle is 360 degrees."}, {"video_title": "Intro to arc measure Mathematics II High School Math Khan Academy.mp3", "Sentence": "So this is the major arc. So what is the measure of arc ACB? Once again, we're using three letters so that we're specifying the major arc. Well, this angle, this central angle right over here, to go all the way around the circle is 360 degrees. So this is going to be the 360 minus the 120 that we're not including. So 360 degrees minus 120 is going to be 240 degrees. So the measure of this angle right over here is 240 degrees."}, {"video_title": "Intro to arc measure Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, this angle, this central angle right over here, to go all the way around the circle is 360 degrees. So this is going to be the 360 minus the 120 that we're not including. So 360 degrees minus 120 is going to be 240 degrees. So the measure of this angle right over here is 240 degrees. So the arc, the measure of this arc, have to be careful not to say length of that arc, the measure of this arc is going to be the same as the measure of the central angle. So it's going to be 240 degrees. Now, this is going to be, these arc measures are going to be the case regardless of the size of the circle."}, {"video_title": "Intro to arc measure Mathematics II High School Math Khan Academy.mp3", "Sentence": "So the measure of this angle right over here is 240 degrees. So the arc, the measure of this arc, have to be careful not to say length of that arc, the measure of this arc is going to be the same as the measure of the central angle. So it's going to be 240 degrees. Now, this is going to be, these arc measures are going to be the case regardless of the size of the circle. And that's where the difference starts to be from arc measure to arc length. So I could have two circles. So this circle right over here and that circle right over here."}, {"video_title": "Intro to arc measure Mathematics II High School Math Khan Academy.mp3", "Sentence": "Now, this is going to be, these arc measures are going to be the case regardless of the size of the circle. And that's where the difference starts to be from arc measure to arc length. So I could have two circles. So this circle right over here and that circle right over here. And as long as the central angle that intercepts the arc has the same degree measure, so let's say that that degree measure is the same as, these are central angles, so we're assuming the vertex of the angle is the center of the circle. As long as these two are the same, these two central angles have the same degree measure, then the arc measures, then the corresponding arc measures are going to be the same. But clearly these two arc lengths are different."}, {"video_title": "Intro to arc measure Mathematics II High School Math Khan Academy.mp3", "Sentence": "So this circle right over here and that circle right over here. And as long as the central angle that intercepts the arc has the same degree measure, so let's say that that degree measure is the same as, these are central angles, so we're assuming the vertex of the angle is the center of the circle. As long as these two are the same, these two central angles have the same degree measure, then the arc measures, then the corresponding arc measures are going to be the same. But clearly these two arc lengths are different. The arc length is not going to depend only on the measure of the central angle, the arc length is going to depend on the size of the actual circle. Arc measure is only dependent on the measure of the central angle that intercepts that arc. So your maximum arc measure is going to be 360 degrees."}, {"video_title": "Intro to arc measure Mathematics II High School Math Khan Academy.mp3", "Sentence": "But clearly these two arc lengths are different. The arc length is not going to depend only on the measure of the central angle, the arc length is going to depend on the size of the actual circle. Arc measure is only dependent on the measure of the central angle that intercepts that arc. So your maximum arc measure is going to be 360 degrees. Your minimum arc measure is going to be zero degrees. It's measured in degrees, not in units of length that arc length would be measured in. So let me write this down."}, {"video_title": "Intro to arc measure Mathematics II High School Math Khan Academy.mp3", "Sentence": "So your maximum arc measure is going to be 360 degrees. Your minimum arc measure is going to be zero degrees. It's measured in degrees, not in units of length that arc length would be measured in. So let me write this down. This only depends, so this is, what's going to drive this is the measure, measure of central angle, central angle that intercepts the arc, that intercepts, intercepts the arc. When you talk about arc length, yes, it's going to be dependent on the angle, but it's also dependent, it's going to be dependent on the measure of that central angle plus the size of the circle. Size of the circle."}, {"video_title": "Solving inscribed quadrilaterals Mathematics II High School Math Khan Academy.mp3", "Sentence": "What I want to do in this video, see if we can find the measure of angle D. If we could find the measure of angle D. And like always, pause this video and see if you can figure it out. And I'll give you a little bit of a hint. It'll involve thinking about how an inscribed angle relates to the corresponding, to the measure of the arc that it intercepts. So think about it like that. All right, so let's work on this a little bit. So what do we know? What do we know?"}, {"video_title": "Solving inscribed quadrilaterals Mathematics II High School Math Khan Academy.mp3", "Sentence": "So think about it like that. All right, so let's work on this a little bit. So what do we know? What do we know? Well, angle D, angle D intercepts an arc. It intercepts this fairly large arc that I'm going to highlight right now in this purple color. So it intercepts that arc."}, {"video_title": "Solving inscribed quadrilaterals Mathematics II High School Math Khan Academy.mp3", "Sentence": "What do we know? Well, angle D, angle D intercepts an arc. It intercepts this fairly large arc that I'm going to highlight right now in this purple color. So it intercepts that arc. We don't know the measure of that arc, or at least we don't know the measure of that arc yet. If we did know the measure of this arc that I'm highlighting, then we know that the measure of angle D would just be half that, because the measure of an inscribed angle is half the measure of the arc that it intercepts. We've seen that multiple times."}, {"video_title": "Solving inscribed quadrilaterals Mathematics II High School Math Khan Academy.mp3", "Sentence": "So it intercepts that arc. We don't know the measure of that arc, or at least we don't know the measure of that arc yet. If we did know the measure of this arc that I'm highlighting, then we know that the measure of angle D would just be half that, because the measure of an inscribed angle is half the measure of the arc that it intercepts. We've seen that multiple times. So if we knew the measure of this arc, we would be able to figure out what the measure of angle D is. But we do know, we don't know the measure of that arc, but we do know the measure of another arc. We do know the measure of the arc that completes the circle."}, {"video_title": "Solving inscribed quadrilaterals Mathematics II High School Math Khan Academy.mp3", "Sentence": "We've seen that multiple times. So if we knew the measure of this arc, we would be able to figure out what the measure of angle D is. But we do know, we don't know the measure of that arc, but we do know the measure of another arc. We do know the measure of the arc that completes the circle. So we do know the measure of this arc. You might be saying, hey, wait, how do we know that measure? It's not labeled."}, {"video_title": "Solving inscribed quadrilaterals Mathematics II High School Math Khan Academy.mp3", "Sentence": "We do know the measure of the arc that completes the circle. So we do know the measure of this arc. You might be saying, hey, wait, how do we know that measure? It's not labeled. Well, the reason why we know the measure of this arc that I've just highlighted in this teal color is because the inscribed angle that intercepts it, they gave us the information. They said this is a 45 degree angle. So if this is a 45 degree angle, then this over here is a 90 degree arc."}, {"video_title": "Solving inscribed quadrilaterals Mathematics II High School Math Khan Academy.mp3", "Sentence": "It's not labeled. Well, the reason why we know the measure of this arc that I've just highlighted in this teal color is because the inscribed angle that intercepts it, they gave us the information. They said this is a 45 degree angle. So if this is a 45 degree angle, then this over here is a 90 degree arc. The measure of this arc is 90 degrees. The measure of arc, I guess you could say this is the measure of arc, let me write it this way. The measure of arc WL, WL is equal to 90 degrees."}, {"video_title": "Solving inscribed quadrilaterals Mathematics II High School Math Khan Academy.mp3", "Sentence": "So if this is a 45 degree angle, then this over here is a 90 degree arc. The measure of this arc is 90 degrees. The measure of arc, I guess you could say this is the measure of arc, let me write it this way. The measure of arc WL, WL is equal to 90 degrees. It's twice that, the inscribed angle that intercepts it. Now why is that helpful? Well, if you go all the way around the circle, you're 360 degrees."}, {"video_title": "Solving inscribed quadrilaterals Mathematics II High School Math Khan Academy.mp3", "Sentence": "The measure of arc WL, WL is equal to 90 degrees. It's twice that, the inscribed angle that intercepts it. Now why is that helpful? Well, if you go all the way around the circle, you're 360 degrees. So this purple arc that we cared about, that we said, hey, if we could figure out the measure of that, we're gonna be able to figure out the measure of angle D, that plus arc WL, they are going to add up to 360 degrees. Let me write that down. So the measure of arc, let's see, and this is going to be, this is going to be a major arc right over here."}, {"video_title": "Solving inscribed quadrilaterals Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, if you go all the way around the circle, you're 360 degrees. So this purple arc that we cared about, that we said, hey, if we could figure out the measure of that, we're gonna be able to figure out the measure of angle D, that plus arc WL, they are going to add up to 360 degrees. Let me write that down. So the measure of arc, let's see, and this is going to be, this is going to be a major arc right over here. This is so LIW, the measure of arc LIW plus the measure of arc WL, plus the measure of arc WL, WL plus this right over here, that's going to be equal to 360 degrees. This is going to be equal to 360 degrees. Now we already know that this is 90 degrees."}, {"video_title": "Solving inscribed quadrilaterals Mathematics II High School Math Khan Academy.mp3", "Sentence": "So the measure of arc, let's see, and this is going to be, this is going to be a major arc right over here. This is so LIW, the measure of arc LIW plus the measure of arc WL, plus the measure of arc WL, WL plus this right over here, that's going to be equal to 360 degrees. This is going to be equal to 360 degrees. Now we already know that this is 90 degrees. We already know WL is 90 degrees. So if you subtract 90 degrees from both sides, you get that the measure of this large arc right over here, measure of arc LIW is going to be equal to 270 degrees. 270 degrees."}, {"video_title": "Solving inscribed quadrilaterals Mathematics II High School Math Khan Academy.mp3", "Sentence": "Now we already know that this is 90 degrees. We already know WL is 90 degrees. So if you subtract 90 degrees from both sides, you get that the measure of this large arc right over here, measure of arc LIW is going to be equal to 270 degrees. 270 degrees. I just took 300, I went all the way around the circle, I subtracted out this 90 degrees, and I'm left with 270 degrees. So let me write that down. This is the measure of this arc in purple is 270 degrees."}, {"video_title": "Solving inscribed quadrilaterals Mathematics II High School Math Khan Academy.mp3", "Sentence": "270 degrees. I just took 300, I went all the way around the circle, I subtracted out this 90 degrees, and I'm left with 270 degrees. So let me write that down. This is the measure of this arc in purple is 270 degrees. And now we can figure out the measure of angle D. It's an inscribed angle that intercepts that arc, so it's going to have half the measure. The angle's going to have half the measure. So half of 270 is 135 degrees."}, {"video_title": "Solving inscribed quadrilaterals Mathematics II High School Math Khan Academy.mp3", "Sentence": "This is the measure of this arc in purple is 270 degrees. And now we can figure out the measure of angle D. It's an inscribed angle that intercepts that arc, so it's going to have half the measure. The angle's going to have half the measure. So half of 270 is 135 degrees. And we're done. You might notice something interesting, that if you add 135 degrees plus 45 degrees, that they add up to 180 degrees. So it looks like at least for this case, that these angles, these opposite angles of this inscribed quadrilateral, it looks like they are supplementary."}, {"video_title": "Solving inscribed quadrilaterals Mathematics II High School Math Khan Academy.mp3", "Sentence": "So half of 270 is 135 degrees. And we're done. You might notice something interesting, that if you add 135 degrees plus 45 degrees, that they add up to 180 degrees. So it looks like at least for this case, that these angles, these opposite angles of this inscribed quadrilateral, it looks like they are supplementary. So an interesting question is, are they always going to be supplementary? If you have a quadrilateral, an arbitrary quadrilateral, inscribed in a circle, so each of the vertices of the quadrilateral sit on the circle, if you have that, are opposite angles of that quadrilateral, are they always supplementary? Do they always add up to 180 degrees?"}, {"video_title": "Solving inscribed quadrilaterals Mathematics II High School Math Khan Academy.mp3", "Sentence": "So it looks like at least for this case, that these angles, these opposite angles of this inscribed quadrilateral, it looks like they are supplementary. So an interesting question is, are they always going to be supplementary? If you have a quadrilateral, an arbitrary quadrilateral, inscribed in a circle, so each of the vertices of the quadrilateral sit on the circle, if you have that, are opposite angles of that quadrilateral, are they always supplementary? Do they always add up to 180 degrees? So I encourage you to think about that, and even prove it if you get a chance. And the proof is very close to what we just did here. In order to prove it, you would just have to do it with more general numbers, like instead of saying 45 degrees, you could call this x."}, {"video_title": "Solving inscribed quadrilaterals Mathematics II High School Math Khan Academy.mp3", "Sentence": "Do they always add up to 180 degrees? So I encourage you to think about that, and even prove it if you get a chance. And the proof is very close to what we just did here. In order to prove it, you would just have to do it with more general numbers, like instead of saying 45 degrees, you could call this x. And then you would want to prove that this right over here would have to be 180 minus x. So I encourage you to do that on your own. But I'm going to do it in a video as well, so you can check if our reasoning is similar."}, {"video_title": "Multi-step word problem with Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "There are 10 lights per meter of the string. How many total lights are on the string? So pause this video and see if you can work this out. All right, now let's work through this together. And I think this one warrants some type of a diagram. So let me draw this doorframe that looks like this. And this doorframe is 2.5 meters tall."}, {"video_title": "Multi-step word problem with Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "All right, now let's work through this together. And I think this one warrants some type of a diagram. So let me draw this doorframe that looks like this. And this doorframe is 2.5 meters tall. So that's its height right over there. And what they're going to do, what Laney's doing is she is stringing up these lights. So that's this yellow right over here."}, {"video_title": "Multi-step word problem with Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "And this doorframe is 2.5 meters tall. So that's its height right over there. And what they're going to do, what Laney's doing is she is stringing up these lights. So that's this yellow right over here. So it goes up to the top of that doorframe. And then they run the rest of the light in a straight line to a point on the ground that is six meters from the base of the doorframe. So let me show a point that is six meters from the base of the doorframe."}, {"video_title": "Multi-step word problem with Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "So that's this yellow right over here. So it goes up to the top of that doorframe. And then they run the rest of the light in a straight line to a point on the ground that is six meters from the base of the doorframe. So let me show a point that is six meters from the base of the doorframe. So it would look something like, maybe like that. So this distance right over here is six meters. So they run the rest of the light from the top of the doorframe to that point that's six meters away."}, {"video_title": "Multi-step word problem with Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "So let me show a point that is six meters from the base of the doorframe. So it would look something like, maybe like that. So this distance right over here is six meters. So they run the rest of the light from the top of the doorframe to that point that's six meters away. So the yellow right over here, that is the light. And so we need to figure out how many total lights are on the string. So the way I would tackle it is, first of all, I wanna figure out how long is the total string."}, {"video_title": "Multi-step word problem with Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "So they run the rest of the light from the top of the doorframe to that point that's six meters away. So the yellow right over here, that is the light. And so we need to figure out how many total lights are on the string. So the way I would tackle it is, first of all, I wanna figure out how long is the total string. And to figure that out, I just need to figure out, okay, it's going to be 2.5 plus whatever the hypotenuse is of this right triangle. I think it's safe to assume that this is a standard house where doorframes are at a right angle to the floor. And so we have to figure out the length of this hypotenuse."}, {"video_title": "Multi-step word problem with Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "So the way I would tackle it is, first of all, I wanna figure out how long is the total string. And to figure that out, I just need to figure out, okay, it's going to be 2.5 plus whatever the hypotenuse is of this right triangle. I think it's safe to assume that this is a standard house where doorframes are at a right angle to the floor. And so we have to figure out the length of this hypotenuse. And if we know that plus this 2.5 meters, then we know how long the entire string of lights are. And then we just have to really multiply it by 10 because there's 10 lights per meter of the string. So let's do that."}, {"video_title": "Multi-step word problem with Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "And so we have to figure out the length of this hypotenuse. And if we know that plus this 2.5 meters, then we know how long the entire string of lights are. And then we just have to really multiply it by 10 because there's 10 lights per meter of the string. So let's do that. So how do we figure out the hypotenuse here? Well, of course we would use Pythagorean theorem. So let's call this, let's call this H for hypotenuse."}, {"video_title": "Multi-step word problem with Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "So let's do that. So how do we figure out the hypotenuse here? Well, of course we would use Pythagorean theorem. So let's call this, let's call this H for hypotenuse. We know that the hypotenuse squared is equal to 2.5 squared plus six squared. So this is going to be equal to 6.25 plus 36, which is equal to 42.25. Or we could say that the hypotenuse is going to be equal to the square root of 42.25."}, {"video_title": "Multi-step word problem with Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "So let's call this, let's call this H for hypotenuse. We know that the hypotenuse squared is equal to 2.5 squared plus six squared. So this is going to be equal to 6.25 plus 36, which is equal to 42.25. Or we could say that the hypotenuse is going to be equal to the square root of 42.25. And I could get my calculator out at this point, but I'll actually just keep using this expression to figure out the total number of lights. So what's the total length of the string? We have to be careful here."}, {"video_title": "Multi-step word problem with Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "Or we could say that the hypotenuse is going to be equal to the square root of 42.25. And I could get my calculator out at this point, but I'll actually just keep using this expression to figure out the total number of lights. So what's the total length of the string? We have to be careful here. A lot of folks would say, oh, I figured out the hypotenuse. Let me just multiply that by 10. In fact, my brain almost did that just now."}, {"video_title": "Multi-step word problem with Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "We have to be careful here. A lot of folks would say, oh, I figured out the hypotenuse. Let me just multiply that by 10. In fact, my brain almost did that just now. But we gotta realize that the entire string is the hypotenuse plus this 2.5. So the whole string length, let me write it this way, string length is equal to 2.5 plus the square root of 42.25. And then we would just multiply that times 10 to get the total number of lights."}, {"video_title": "Multi-step word problem with Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "In fact, my brain almost did that just now. But we gotta realize that the entire string is the hypotenuse plus this 2.5. So the whole string length, let me write it this way, string length is equal to 2.5 plus the square root of 42.25. And then we would just multiply that times 10 to get the total number of lights. So now let's actually get the calculator out. So we have 42.25, and then we were to take the square root of that, gets us to 6.5, and then we add the other 2.5 plus 2.5 equals that. That's the total length of string."}, {"video_title": "Multi-step word problem with Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "And then we would just multiply that times 10 to get the total number of lights. So now let's actually get the calculator out. So we have 42.25, and then we were to take the square root of that, gets us to 6.5, and then we add the other 2.5 plus 2.5 equals that. That's the total length of string. So the total length of string is nine meters, and there are 10 lights per meter. So the number of lights, number of lights, is equal to nine total meters of string times 10 lights per meter, which would give us 90 lights. Now, some of you might debate, if you think really deeply about it, is if you have a light right at the beginning, if these were kind of set up like fence posts, that maybe you could argue that there's one extra light in there."}, {"video_title": "Dilation scale factor examples.mp3", "Sentence": "We are told that pentagon A prime, B prime, C prime, D prime, E prime, which is in red right over here, is the image of pentagon ABCDE under a dilation. So that's ABCDE. What is the scale factor of the dilation? So they don't even tell us the center of the dilation, but in order to figure out the scale factor, you just have to realize when you do a dilation, the distance between corresponding points will change according to the scale factor. So for example, we could look at the distance between point A and point B right over here. What is our change in y? Our change in, or even what is our distance?"}, {"video_title": "Dilation scale factor examples.mp3", "Sentence": "So they don't even tell us the center of the dilation, but in order to figure out the scale factor, you just have to realize when you do a dilation, the distance between corresponding points will change according to the scale factor. So for example, we could look at the distance between point A and point B right over here. What is our change in y? Our change in, or even what is our distance? Our change in y is our distance because we don't have a change in x. Well, this is one, two, three, four, five, six. So this length right over here is equal to six."}, {"video_title": "Dilation scale factor examples.mp3", "Sentence": "Our change in, or even what is our distance? Our change in y is our distance because we don't have a change in x. Well, this is one, two, three, four, five, six. So this length right over here is equal to six. Now what about the corresponding side from A prime to B prime? Well, this length right over here is equal to two. And so you could see we went from having a length of six to a length of two, so you would have to multiply by 1 3rd."}, {"video_title": "Dilation scale factor examples.mp3", "Sentence": "So this length right over here is equal to six. Now what about the corresponding side from A prime to B prime? Well, this length right over here is equal to two. And so you could see we went from having a length of six to a length of two, so you would have to multiply by 1 3rd. So our scale factor right over here is 1 3rd. Now you might be saying, okay, that was pretty straightforward because we had a very clear, you could just see the distance between A and B. How would you do it if you didn't have a vertical or a horizontal line?"}, {"video_title": "Dilation scale factor examples.mp3", "Sentence": "And so you could see we went from having a length of six to a length of two, so you would have to multiply by 1 3rd. So our scale factor right over here is 1 3rd. Now you might be saying, okay, that was pretty straightforward because we had a very clear, you could just see the distance between A and B. How would you do it if you didn't have a vertical or a horizontal line? Well, one way to think about it is the changes in y and the changes in x would scale accordingly. So if you looked at the distance between point A and point E, our change in y is negative three right over here and our change in x is positive three right over here. And you could see over here between A prime and E prime, our change in y is negative one, which is 1 3rd of negative three, and our change in x is one, which is 1 3rd of three."}, {"video_title": "Dilation scale factor examples.mp3", "Sentence": "How would you do it if you didn't have a vertical or a horizontal line? Well, one way to think about it is the changes in y and the changes in x would scale accordingly. So if you looked at the distance between point A and point E, our change in y is negative three right over here and our change in x is positive three right over here. And you could see over here between A prime and E prime, our change in y is negative one, which is 1 3rd of negative three, and our change in x is one, which is 1 3rd of three. So once again, you see our scale factor being 1 3rd. Let's do another example. So we are told that pentagon A prime, B prime, C prime, D prime, E prime is the image, and they haven't drawn that here, is the image of pentagon A, B, C, D, E under a dilation with a scale factor of 5 1st."}, {"video_title": "Dilation scale factor examples.mp3", "Sentence": "And you could see over here between A prime and E prime, our change in y is negative one, which is 1 3rd of negative three, and our change in x is one, which is 1 3rd of three. So once again, you see our scale factor being 1 3rd. Let's do another example. So we are told that pentagon A prime, B prime, C prime, D prime, E prime is the image, and they haven't drawn that here, is the image of pentagon A, B, C, D, E under a dilation with a scale factor of 5 1st. So they're giving us our scale factor. What is the length of segment A prime, E prime? So as I was mentioning while I read it, they didn't actually draw this one out."}, {"video_title": "Dilation scale factor examples.mp3", "Sentence": "So we are told that pentagon A prime, B prime, C prime, D prime, E prime is the image, and they haven't drawn that here, is the image of pentagon A, B, C, D, E under a dilation with a scale factor of 5 1st. So they're giving us our scale factor. What is the length of segment A prime, E prime? So as I was mentioning while I read it, they didn't actually draw this one out. So how do we figure out the length of a segment? Well, I encourage you to pause the video and try to think about it. Well, they give us the scale factor, and so what it tells us, if the scale factor is 5 1st, that means that the corresponding lengths will change by a factor of 5 1st."}, {"video_title": "Dilation scale factor examples.mp3", "Sentence": "So as I was mentioning while I read it, they didn't actually draw this one out. So how do we figure out the length of a segment? Well, I encourage you to pause the video and try to think about it. Well, they give us the scale factor, and so what it tells us, if the scale factor is 5 1st, that means that the corresponding lengths will change by a factor of 5 1st. So to figure out the length of segment A prime, E prime, this is going to be, you could think of it as the image of segment AE, and so you could see that the length of AE is equal to two, and so the length of A prime, E prime is going to be equal to AE, which is two, times the scale factor, times 5 1st. This is our scale factor right over here, and of course, what's two times 5 1st? Well, it is going to be equal to five, five of these units right over here."}, {"video_title": "Dilation scale factor examples.mp3", "Sentence": "Well, they give us the scale factor, and so what it tells us, if the scale factor is 5 1st, that means that the corresponding lengths will change by a factor of 5 1st. So to figure out the length of segment A prime, E prime, this is going to be, you could think of it as the image of segment AE, and so you could see that the length of AE is equal to two, and so the length of A prime, E prime is going to be equal to AE, which is two, times the scale factor, times 5 1st. This is our scale factor right over here, and of course, what's two times 5 1st? Well, it is going to be equal to five, five of these units right over here. So in this case, we didn't even have to draw A prime, B prime, C prime, D prime, E prime. In fact, they haven't even given us enough information. I could draw the scale of that, but I actually don't know where to put it because they didn't even give us our center of dilation, but we know that corresponding sides or the lengths between corresponding points are going to be scaled by the scale factor."}, {"video_title": "Dilation scale factor examples.mp3", "Sentence": "Well, it is going to be equal to five, five of these units right over here. So in this case, we didn't even have to draw A prime, B prime, C prime, D prime, E prime. In fact, they haven't even given us enough information. I could draw the scale of that, but I actually don't know where to put it because they didn't even give us our center of dilation, but we know that corresponding sides or the lengths between corresponding points are going to be scaled by the scale factor. Now, with that in mind, let's do another example. So we are told that triangle A prime, B prime, C prime, which they depicted right over here, is the image of triangle ABC, which they did not depict under a dilation with a scale factor of two. What is the length of segment AB?"}, {"video_title": "Dilation scale factor examples.mp3", "Sentence": "I could draw the scale of that, but I actually don't know where to put it because they didn't even give us our center of dilation, but we know that corresponding sides or the lengths between corresponding points are going to be scaled by the scale factor. Now, with that in mind, let's do another example. So we are told that triangle A prime, B prime, C prime, which they depicted right over here, is the image of triangle ABC, which they did not depict under a dilation with a scale factor of two. What is the length of segment AB? Once again, they haven't drawn AB here. How do we figure it out? Well, it's gonna be a similar way as the last example, but here, they've given us the image and they didn't give us the original, so how do we do it?"}, {"video_title": "Dilation scale factor examples.mp3", "Sentence": "What is the length of segment AB? Once again, they haven't drawn AB here. How do we figure it out? Well, it's gonna be a similar way as the last example, but here, they've given us the image and they didn't give us the original, so how do we do it? Well, the key, and pause the video again and try to do it on your own. Well, the key realization here is that if you take the length of segment AB and you were to multiply by the scale factor, so you multiply it by two, then you're going to get the length of segment A prime, B prime. The image's length is equal to the scale factor times the corresponding length on our original triangle."}, {"video_title": "Dilation scale factor examples.mp3", "Sentence": "Well, it's gonna be a similar way as the last example, but here, they've given us the image and they didn't give us the original, so how do we do it? Well, the key, and pause the video again and try to do it on your own. Well, the key realization here is that if you take the length of segment AB and you were to multiply by the scale factor, so you multiply it by two, then you're going to get the length of segment A prime, B prime. The image's length is equal to the scale factor times the corresponding length on our original triangle. So what is the length of A prime, B prime? Well, this is straightforward to figure out. It is one, two, three, four, five, six, seven, eight."}, {"video_title": "Constructing a perpendicular bisector using a compass and straightedge Geometry Khan Academy.mp3", "Sentence": "So the fact that it's perpendicular means that this line will make a 90 degree angle where it intersects with AB. And it's going to bisect it, so it's going to go halfway in between. And I have at my disposal some tools. I can put out, I can draw things with a compass, and I can add a straight edge. So let's try this out. So let me add a compass. And so this is kind of a virtual compass."}, {"video_title": "Constructing a perpendicular bisector using a compass and straightedge Geometry Khan Academy.mp3", "Sentence": "I can put out, I can draw things with a compass, and I can add a straight edge. So let's try this out. So let me add a compass. And so this is kind of a virtual compass. So in a real compass, it's one of those little metal things where you can pivot it on one point, and you can draw a circle of any radius. And so here I'm going to center it at A. And I'm going to make the radius equal to the length of AB."}, {"video_title": "Constructing a perpendicular bisector using a compass and straightedge Geometry Khan Academy.mp3", "Sentence": "And so this is kind of a virtual compass. So in a real compass, it's one of those little metal things where you can pivot it on one point, and you can draw a circle of any radius. And so here I'm going to center it at A. And I'm going to make the radius equal to the length of AB. Now I'm going to add another circle with my compass. And now I'm going to center it at B and make the radius equal to AB. And now this gives me two points that I can actually use to draw my perpendicular bisector."}, {"video_title": "Constructing a perpendicular bisector using a compass and straightedge Geometry Khan Academy.mp3", "Sentence": "And I'm going to make the radius equal to the length of AB. Now I'm going to add another circle with my compass. And now I'm going to center it at B and make the radius equal to AB. And now this gives me two points that I can actually use to draw my perpendicular bisector. If I connect this point and this point, it is going to bisect AB, and it's also going to be perpendicular. So let's add a straight edge here. So this is to draw a line."}, {"video_title": "Constructing a perpendicular bisector using a compass and straightedge Geometry Khan Academy.mp3", "Sentence": "And now this gives me two points that I can actually use to draw my perpendicular bisector. If I connect this point and this point, it is going to bisect AB, and it's also going to be perpendicular. So let's add a straight edge here. So this is to draw a line. So I'm going to draw a line between that point and that point right over there. Let me scroll down so you can look at it a little bit clearer. So there you go."}, {"video_title": "Constructing a perpendicular bisector using a compass and straightedge Geometry Khan Academy.mp3", "Sentence": "So this is to draw a line. So I'm going to draw a line between that point and that point right over there. Let me scroll down so you can look at it a little bit clearer. So there you go. That's my construction. I've made a perpendicular bisector for segment AB. Check my answer."}, {"video_title": "Geometric constructions perpendicular line through a point off the line Geometry Khan Academy.mp3", "Sentence": "And my goal is to draw a new line that goes through this point and is perpendicular to my original line. How do I do that? Well, you might imagine that our compass will come in handy. It's been handy before. And so what I will do is, I'll pick an arbitrary point on our original line, let's say this point right over here, and then I'll adjust my compass. So the distance between the pivot point and my pencil tip is the same as the distance between those two points. And then I can now use my compass to trace out an arc of that radius."}, {"video_title": "Geometric constructions perpendicular line through a point off the line Geometry Khan Academy.mp3", "Sentence": "It's been handy before. And so what I will do is, I'll pick an arbitrary point on our original line, let's say this point right over here, and then I'll adjust my compass. So the distance between the pivot point and my pencil tip is the same as the distance between those two points. And then I can now use my compass to trace out an arc of that radius. So there you go. Now my next step is to find another point on my original line that has the same distance from that point that is off the line. And I can do that by centering my compass on that offline point, and then drawing another arc."}, {"video_title": "Geometric constructions perpendicular line through a point off the line Geometry Khan Academy.mp3", "Sentence": "And then I can now use my compass to trace out an arc of that radius. So there you go. Now my next step is to find another point on my original line that has the same distance from that point that is off the line. And I can do that by centering my compass on that offline point, and then drawing another arc. And I can see very clearly that this point also has the same distance from this point up here. And then I can center my compass on that point. And notice I haven't changed the radius of my compass to draw another arc like this, to draw another arc like this."}, {"video_title": "Geometric constructions perpendicular line through a point off the line Geometry Khan Academy.mp3", "Sentence": "And I can do that by centering my compass on that offline point, and then drawing another arc. And I can see very clearly that this point also has the same distance from this point up here. And then I can center my compass on that point. And notice I haven't changed the radius of my compass to draw another arc like this, to draw another arc like this. And then what I can do is connect this point and that point, and it at least looks perpendicular, but we're going to prove to ourselves that it is indeed perpendicular to our original line. So let me just draw it so you have that like that. So how do we feel good that this new line that I just drew is perpendicular to our original one?"}, {"video_title": "Geometric constructions perpendicular line through a point off the line Geometry Khan Academy.mp3", "Sentence": "And notice I haven't changed the radius of my compass to draw another arc like this, to draw another arc like this. And then what I can do is connect this point and that point, and it at least looks perpendicular, but we're going to prove to ourselves that it is indeed perpendicular to our original line. So let me just draw it so you have that like that. So how do we feel good that this new line that I just drew is perpendicular to our original one? Well, let's connect the dots that we made. So if we connect all the dots, we're going to get a rhombus. We know that this distance, this distance is the same as this distance, the same as this one right over here, which is the same as this distance."}, {"video_title": "Geometric constructions perpendicular line through a point off the line Geometry Khan Academy.mp3", "Sentence": "So how do we feel good that this new line that I just drew is perpendicular to our original one? Well, let's connect the dots that we made. So if we connect all the dots, we're going to get a rhombus. We know that this distance, this distance is the same as this distance, the same as this one right over here, which is the same as this distance. Let me make sure I get my straight edge right. Same as that distance, which is the same as this distance, same as that distance. And then, so this is a rhombus, and we know that the diagonals of a rhombus intersect at right angles."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "We tend to be told in algebra class that if we have a line, our line will have a constant rate of change of y with respect to x. Or another way of thinking about it, that a line will have a constant inclination, or that our line will have a constant slope. And our slope is literally defined as your change in y. This triangle is the Greek letter delta. It's a shorthand for change in. It means change in y. Delta y means change in y. Over change in x."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "This triangle is the Greek letter delta. It's a shorthand for change in. It means change in y. Delta y means change in y. Over change in x. And if you're dealing with a line, this right over here is constant. Constant for a line. What I want to do in this video is to actually prove that using similar triangles from geometry."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "Over change in x. And if you're dealing with a line, this right over here is constant. Constant for a line. What I want to do in this video is to actually prove that using similar triangles from geometry. So let's think about two sets of two points. So let's say that's a point there. Let me do it in a different color."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "What I want to do in this video is to actually prove that using similar triangles from geometry. So let's think about two sets of two points. So let's say that's a point there. Let me do it in a different color. Let me start at this point. And let me end up at that point. So what is our change in x between these two points?"}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "Let me do it in a different color. Let me start at this point. And let me end up at that point. So what is our change in x between these two points? So this point's x value is right over here. This point's x value is right over here. So our change in x is going to be that right over there."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "So what is our change in x between these two points? So this point's x value is right over here. This point's x value is right over here. So our change in x is going to be that right over there. And what's our change in y? Well, this point's y value is right over here. This point's y value is right over here."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "So our change in x is going to be that right over there. And what's our change in y? Well, this point's y value is right over here. This point's y value is right over here. So this height or this height is our change in y. So that is our change in y. Now, let's look at two other points."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "This point's y value is right over here. So this height or this height is our change in y. So that is our change in y. Now, let's look at two other points. Let's say I have this point and this point right over here. And let's do the same exercise. What's the change in x?"}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "Now, let's look at two other points. Let's say I have this point and this point right over here. And let's do the same exercise. What's the change in x? Well, let's see. If we're going this point's x value is here. This point's x value is here."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "What's the change in x? Well, let's see. If we're going this point's x value is here. This point's x value is here. So if we start here and we go this far, this would be the change in x between this point and this point. And this is going to be the change. Let me do that in the same green color."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "This point's x value is here. So if we start here and we go this far, this would be the change in x between this point and this point. And this is going to be the change. Let me do that in the same green color. So this is going to be the change in x between those two points. And our change in y, well, this y value is here. This y value is up here."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "Let me do that in the same green color. So this is going to be the change in x between those two points. And our change in y, well, this y value is here. This y value is up here. So our change in y is going to be that right over here. So what I need to show, I'm just picking two arbitrary points. I need to show that the ratio of this change in y to this change of x is going to be the same as the ratio of this change in y to this change of x."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "This y value is up here. So our change in y is going to be that right over here. So what I need to show, I'm just picking two arbitrary points. I need to show that the ratio of this change in y to this change of x is going to be the same as the ratio of this change in y to this change of x. Or the ratio of this purple side to this green side is going to be the same as the ratio of this purple side to this green side. Remember, I'm just picking two sets of arbitrary points here. And the way that I will show it is through similarity."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "I need to show that the ratio of this change in y to this change of x is going to be the same as the ratio of this change in y to this change of x. Or the ratio of this purple side to this green side is going to be the same as the ratio of this purple side to this green side. Remember, I'm just picking two sets of arbitrary points here. And the way that I will show it is through similarity. If I can show that this triangle is similar to this triangle, then we are all set up. And just as a reminder of what similarity is, two triangles are similar. And there's multiple ways of thinking about it."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "And the way that I will show it is through similarity. If I can show that this triangle is similar to this triangle, then we are all set up. And just as a reminder of what similarity is, two triangles are similar. And there's multiple ways of thinking about it. So you're similar if and only if all corresponding, or I should say all three angles, are the same or are congruent. So all three, and let me be careful here. They don't have to be the same exact angle."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "And there's multiple ways of thinking about it. So you're similar if and only if all corresponding, or I should say all three angles, are the same or are congruent. So all three, and let me be careful here. They don't have to be the same exact angle. The corresponding angles have to be the same. So corresponding, I always misspell it. Corresponding angles are going to be equal."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "They don't have to be the same exact angle. The corresponding angles have to be the same. So corresponding, I always misspell it. Corresponding angles are going to be equal. Or we could say they are congruent. So for example, if I have this triangle right over here, and this is 30, this is 60, and this is 90. And then I have this triangle right over here."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "Corresponding angles are going to be equal. Or we could say they are congruent. So for example, if I have this triangle right over here, and this is 30, this is 60, and this is 90. And then I have this triangle right over here. I'll try to draw it. So I have this triangle where this is 30 degrees, this is 60 degrees, and this is 90 degrees. Even though their side lengths are different, these are going to be similar triangles."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "And then I have this triangle right over here. I'll try to draw it. So I have this triangle where this is 30 degrees, this is 60 degrees, and this is 90 degrees. Even though their side lengths are different, these are going to be similar triangles. They're essentially scaled up versions of each other. All the corresponding angles. 60 corresponds to this 60."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "Even though their side lengths are different, these are going to be similar triangles. They're essentially scaled up versions of each other. All the corresponding angles. 60 corresponds to this 60. 30 corresponds to this 30. And 90 corresponds to this one. So these two triangles are similar."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "60 corresponds to this 60. 30 corresponds to this 30. And 90 corresponds to this one. So these two triangles are similar. And what's neat about similar triangles, if you can establish that two triangles are similar, then the ratio between corresponding sides is going to be the same. So if these two are similar, then the ratio of this side to this side is going to be the same as the ratio of this side to this side. And so you could see why that will be useful in proving that the slope is constant here."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "So these two triangles are similar. And what's neat about similar triangles, if you can establish that two triangles are similar, then the ratio between corresponding sides is going to be the same. So if these two are similar, then the ratio of this side to this side is going to be the same as the ratio of this side to this side. And so you could see why that will be useful in proving that the slope is constant here. Because all we have to do is, look, if these two triangles are similar, then the ratio between corresponding sides is always going to be the same. We've picked two arbitrary sets of points. Then this would be true really for any two arbitrary sets of points across the line."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "And so you could see why that will be useful in proving that the slope is constant here. Because all we have to do is, look, if these two triangles are similar, then the ratio between corresponding sides is always going to be the same. We've picked two arbitrary sets of points. Then this would be true really for any two arbitrary sets of points across the line. It would be true for the entire line. So let's try to prove similarity. So the first thing we know is that both of these are right triangles."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "Then this would be true really for any two arbitrary sets of points across the line. It would be true for the entire line. So let's try to prove similarity. So the first thing we know is that both of these are right triangles. These green lines are perfectly horizontal. These purple lines are perfectly vertical. Because the green lines literally go along the x-axis or go in the horizontal direction."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "So the first thing we know is that both of these are right triangles. These green lines are perfectly horizontal. These purple lines are perfectly vertical. Because the green lines literally go along the x-axis or go in the horizontal direction. The purple lines go in the vertical direction. So let me make sure that we mark that. So we know that these are both right angles."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "Because the green lines literally go along the x-axis or go in the horizontal direction. The purple lines go in the vertical direction. So let me make sure that we mark that. So we know that these are both right angles. So we have one corresponding angle that is congruent. Now we have to show that the other ones are. And we can show that the other ones are using our knowledge of parallel lines and transversals."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "So we know that these are both right angles. So we have one corresponding angle that is congruent. Now we have to show that the other ones are. And we can show that the other ones are using our knowledge of parallel lines and transversals. Let's look at these two green lines. So now I'll continue them. These are line segments."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "And we can show that the other ones are using our knowledge of parallel lines and transversals. Let's look at these two green lines. So now I'll continue them. These are line segments. But if we view them as lines and we just continue them on and on and on. So let me do that just like here. So this line is clearly parallel to that one."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "These are line segments. But if we view them as lines and we just continue them on and on and on. So let me do that just like here. So this line is clearly parallel to that one. They essentially are perfectly horizontal. And now you can view our orange line as a transversal. And if you view it as a transversal, then you know that this angle corresponds to this angle."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "So this line is clearly parallel to that one. They essentially are perfectly horizontal. And now you can view our orange line as a transversal. And if you view it as a transversal, then you know that this angle corresponds to this angle. And we know from transversals of parallel lines that corresponding angles are congruent. So this angle is going to be congruent to that angle right over there. Now we make a very similar argument for this angle."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "And if you view it as a transversal, then you know that this angle corresponds to this angle. And we know from transversals of parallel lines that corresponding angles are congruent. So this angle is going to be congruent to that angle right over there. Now we make a very similar argument for this angle. But now we use the two vertical lines. We know that this segment, we could continue it as a line. So we could continue it if we wanted as a line."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "Now we make a very similar argument for this angle. But now we use the two vertical lines. We know that this segment, we could continue it as a line. So we could continue it if we wanted as a line. So just like that, a vertical line. And we could continue this one as a vertical line. We know that these are both vertical."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "So we could continue it if we wanted as a line. So just like that, a vertical line. And we could continue this one as a vertical line. We know that these are both vertical. They're just measuring. They're exactly in the y direction, the vertical direction. So this line is parallel to this line right over here."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "We know that these are both vertical. They're just measuring. They're exactly in the y direction, the vertical direction. So this line is parallel to this line right over here. Once again, our orange line is a transversal of it. And this angle corresponds to this angle right over here. And there we have it."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "So this line is parallel to this line right over here. Once again, our orange line is a transversal of it. And this angle corresponds to this angle right over here. And there we have it. They're congruent. Corresponding angles of transversal of two parallel lines are congruent. We learned that in geometry class."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "And there we have it. They're congruent. Corresponding angles of transversal of two parallel lines are congruent. We learned that in geometry class. And there you have it. All of the corresponding, this angle is congruent to this angle. This angle is congruent to that angle."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "We learned that in geometry class. And there you have it. All of the corresponding, this angle is congruent to this angle. This angle is congruent to that angle. And then both of these are 90 degrees. So both of these are similar triangles. So both of these are similar."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "This angle is congruent to that angle. And then both of these are 90 degrees. So both of these are similar triangles. So both of these are similar. So let me write that down. So we know that these are both similar triangles. And now we can use the common ratio of both sides."}, {"video_title": "Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3", "Sentence": "So both of these are similar. So let me write that down. So we know that these are both similar triangles. And now we can use the common ratio of both sides. So for example, if we called this side length a, and we said that this side has length b, and we said this side has length c, and this side has length d, we know for a fact that the ratio, because these are similar triangles, the ratio between corresponding sides, the ratio of a to b, is going to be equal to the ratio of c to d. And that ratio is literally the definition of slope, your change in y over your change in x. And this is constant, because any triangle that you generate, this right triangle that you generate between these two points, we've just shown that they are going to be similar. And if they are similar, then the ratio of this vertical line, the length of this vertical line segment to this horizontal line segment is constant."}, {"video_title": "Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3", "Sentence": "We have three lines and we have to figure out which of the three are parallel. So line A, and it can't be parallel on its own, it has to be parallel to another of the three lines. So the equation for line A is y is equal to 3 4ths x minus 4. Line B is 4y minus 20 is equal to negative 3x. And then line C is negative 3x plus 4y is equal to 40. So to figure out if any of these lines are parallel to any of the other lines, we just have to compare their slopes. If any two of these lines have the same slope and they're different lines, they have different y-intercepts, then they're going to be parallel."}, {"video_title": "Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3", "Sentence": "Line B is 4y minus 20 is equal to negative 3x. And then line C is negative 3x plus 4y is equal to 40. So to figure out if any of these lines are parallel to any of the other lines, we just have to compare their slopes. If any two of these lines have the same slope and they're different lines, they have different y-intercepts, then they're going to be parallel. Now line A, it's very easy to figure out its slope. It's already in slope-intercept form. This is mx plus b."}, {"video_title": "Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3", "Sentence": "If any two of these lines have the same slope and they're different lines, they have different y-intercepts, then they're going to be parallel. Now line A, it's very easy to figure out its slope. It's already in slope-intercept form. This is mx plus b. The slope is 3 4ths and the y-intercept, which isn't as relevant when you're figuring out parallel lines, is negative 4. So let's see what the other character's slopes are. This isn't in any kind of standard form."}, {"video_title": "Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3", "Sentence": "This is mx plus b. The slope is 3 4ths and the y-intercept, which isn't as relevant when you're figuring out parallel lines, is negative 4. So let's see what the other character's slopes are. This isn't in any kind of standard form. It's not in standard form, slope-intercept, or point-slope form. But let's see what the slope of this line is. So to get it into slope-intercept form, which is really the easiest one to pick out the slope from, let's add 20 to both sides of this equation."}, {"video_title": "Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3", "Sentence": "This isn't in any kind of standard form. It's not in standard form, slope-intercept, or point-slope form. But let's see what the slope of this line is. So to get it into slope-intercept form, which is really the easiest one to pick out the slope from, let's add 20 to both sides of this equation. So let's add 20 to both sides. The left-hand side, those cancel out, that was the whole point. You get 4y is equal to negative 3x plus 20."}, {"video_title": "Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3", "Sentence": "So to get it into slope-intercept form, which is really the easiest one to pick out the slope from, let's add 20 to both sides of this equation. So let's add 20 to both sides. The left-hand side, those cancel out, that was the whole point. You get 4y is equal to negative 3x plus 20. And now we can divide everything by 4. Just dividing both sides of this equation by 4. We are left with y is equal to negative 3 4ths x plus 5."}, {"video_title": "Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3", "Sentence": "You get 4y is equal to negative 3x plus 20. And now we can divide everything by 4. Just dividing both sides of this equation by 4. We are left with y is equal to negative 3 4ths x plus 5. So in this case, y-intercept is 5, but most importantly, the slope is negative 3 4ths. So it's different than this guy. This is negative 3 4ths, this is positive 3 4ths."}, {"video_title": "Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3", "Sentence": "We are left with y is equal to negative 3 4ths x plus 5. So in this case, y-intercept is 5, but most importantly, the slope is negative 3 4ths. So it's different than this guy. This is negative 3 4ths, this is positive 3 4ths. So these two guys definitely aren't parallel. Let's move on to this guy. This guy written in standard form."}, {"video_title": "Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3", "Sentence": "This is negative 3 4ths, this is positive 3 4ths. So these two guys definitely aren't parallel. Let's move on to this guy. This guy written in standard form. Let's get the x term on the other side. So let's add 3x to both sides of this equation. Left-hand side, these cancel out."}, {"video_title": "Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3", "Sentence": "This guy written in standard form. Let's get the x term on the other side. So let's add 3x to both sides of this equation. Left-hand side, these cancel out. We're just left with 4y is equal to 3x plus 40, or 40 plus 3x, either way. Now we can divide both sides by 4. We have to divide every term by 4."}, {"video_title": "Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3", "Sentence": "Left-hand side, these cancel out. We're just left with 4y is equal to 3x plus 40, or 40 plus 3x, either way. Now we can divide both sides by 4. We have to divide every term by 4. The left-hand side, you're left with y. The right-hand side, you have 3 4ths x plus 10. So here, our slope is 3 4ths, and our y-intercept, if we care about it, is 10."}, {"video_title": "Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "What I want to do in this video is get some practice visualizing what happens if we were to try to rotate two-dimensional shapes in three dimensions. What do I mean by that? Let's say I started with a right triangle. So let's say my right triangle looks like this. So let's say it looks like that right over there. And so this is a right angle. And let's say that this width right over here is three units."}, {"video_title": "Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "So let's say my right triangle looks like this. So let's say it looks like that right over there. And so this is a right angle. And let's say that this width right over here is three units. And let's say that this length is five units. And I want to do something interesting. I'm going to take this two-dimensional right triangle and I'm going to try to rotate it in three dimensions around this line, around the line that I'm doing as a dotted magenta line."}, {"video_title": "Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "And let's say that this width right over here is three units. And let's say that this length is five units. And I want to do something interesting. I'm going to take this two-dimensional right triangle and I'm going to try to rotate it in three dimensions around this line, around the line that I'm doing as a dotted magenta line. So I'm going to rotate it around this line right over there. So if I were to rotate it around this line, what type of a shape am I going to get? And I encourage you, once again, it's going to be a three-dimensional shape."}, {"video_title": "Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "I'm going to take this two-dimensional right triangle and I'm going to try to rotate it in three dimensions around this line, around the line that I'm doing as a dotted magenta line. So I'm going to rotate it around this line right over there. So if I were to rotate it around this line, what type of a shape am I going to get? And I encourage you, once again, it's going to be a three-dimensional shape. I encourage you to think about it. Maybe take out a piece of paper, draw it, or just try to imagine it in your head. Well, to think about it in three dimensions, what I'm going to do is try to look at this thing in three dimensions."}, {"video_title": "Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "And I encourage you, once again, it's going to be a three-dimensional shape. I encourage you to think about it. Maybe take out a piece of paper, draw it, or just try to imagine it in your head. Well, to think about it in three dimensions, what I'm going to do is try to look at this thing in three dimensions. So let me draw this same line, but I'm going to draw it in an angle so we can visualize the whole thing in three dimensions. So imagine if this was sitting on the ground. So that's our magenta line."}, {"video_title": "Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "Well, to think about it in three dimensions, what I'm going to do is try to look at this thing in three dimensions. So let me draw this same line, but I'm going to draw it in an angle so we can visualize the whole thing in three dimensions. So imagine if this was sitting on the ground. So that's our magenta line. And then I can draw my triangle. So my triangle would look something like this. So it would look like this."}, {"video_title": "Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "So that's our magenta line. And then I can draw my triangle. So my triangle would look something like this. So it would look like this. So once again, this is five units, this is three units, this is a right triangle. I'm going to rotate it around the line. So what's it going to look like?"}, {"video_title": "Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "So it would look like this. So once again, this is five units, this is three units, this is a right triangle. I'm going to rotate it around the line. So what's it going to look like? Well, this end right over here is going to rotate around, and it's going to form a circle with a radius of three. Right? So it's going to form, so when it intersects, if that was on the ground, it's going to be three again."}, {"video_title": "Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "So what's it going to look like? Well, this end right over here is going to rotate around, and it's going to form a circle with a radius of three. Right? So it's going to form, so when it intersects, if that was on the ground, it's going to be three again. And let me draw it down. So it's going to keep going down. Whoops, don't want to press the wrong button."}, {"video_title": "Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "So it's going to form, so when it intersects, if that was on the ground, it's going to be three again. And let me draw it down. So it's going to keep going down. Whoops, don't want to press the wrong button. So it's going to look something like this. That's what the base is going to look like. But then this end right over here is just going to stay at a point, because this is right on that magenta line."}, {"video_title": "Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "Whoops, don't want to press the wrong button. So it's going to look something like this. That's what the base is going to look like. But then this end right over here is just going to stay at a point, because this is right on that magenta line. So it's just going to stay at a point. And so if you were to look at the intersect, so it would look something like this. So it would look like this, and then you'd have another thing that goes like this."}, {"video_title": "Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "But then this end right over here is just going to stay at a point, because this is right on that magenta line. So it's just going to stay at a point. And so if you were to look at the intersect, so it would look something like this. So it would look like this, and then you'd have another thing that goes like this. And so if you were to take a section like this, you would have a little smaller circle here based on what this distance is. So what is the shape? What is the shape that I am drawing?"}, {"video_title": "Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "So it would look like this, and then you'd have another thing that goes like this. And so if you were to take a section like this, you would have a little smaller circle here based on what this distance is. So what is the shape? What is the shape that I am drawing? Well, what you see, what it is, it's a cone. It's a cone, and if I shade it in, you might see the cone a little bit better. So let me shade it in."}, {"video_title": "Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "What is the shape that I am drawing? Well, what you see, what it is, it's a cone. It's a cone, and if I shade it in, you might see the cone a little bit better. So let me shade it in. So you see the cone. So what you end up getting is a cone where it's base, so I'm shading it in. So that hopefully helps a little bit."}, {"video_title": "Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "So let me shade it in. So you see the cone. So what you end up getting is a cone where it's base, so I'm shading it in. So that hopefully helps a little bit. So what you end up getting is a cone where the base has a radius of three units. So let me draw this. So this right over here is the radius of the base, and it is three units."}, {"video_title": "Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "So that hopefully helps a little bit. So what you end up getting is a cone where the base has a radius of three units. So let me draw this. So this right over here is the radius of the base, and it is three units. I could also draw it like this. So the cone is going to look like this, and this is the tip of the cone, and it's going to look just like this. And once again, let me shade it a little bit so that you can appreciate that this is a three-dimensional shape."}, {"video_title": "Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "So this right over here is the radius of the base, and it is three units. I could also draw it like this. So the cone is going to look like this, and this is the tip of the cone, and it's going to look just like this. And once again, let me shade it a little bit so that you can appreciate that this is a three-dimensional shape. So draw the cone so you can shade it, and we can even construct the original so that, well, or we can construct this original shape so you see how it constructs, so it makes this, the line, that magenta line, is going to do this type of thing. It's going to go through the center of the base. It's going to go through the center of the base just like that, and our original shape, our original right triangle, if you just took a cross-section of it that included that line, you would have your original shape."}, {"video_title": "Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "And once again, let me shade it a little bit so that you can appreciate that this is a three-dimensional shape. So draw the cone so you can shade it, and we can even construct the original so that, well, or we can construct this original shape so you see how it constructs, so it makes this, the line, that magenta line, is going to do this type of thing. It's going to go through the center of the base. It's going to go through the center of the base just like that, and our original shape, our original right triangle, if you just took a cross-section of it that included that line, you would have your original shape. Let me just say in orange. So the original shape is right over there. So what do you get?"}, {"video_title": "Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3", "Sentence": "The equation of a circle C is x plus 3 squared plus y minus 4 squared is equal to 49. What are its center hk and its radius r? So let's just remind ourselves what a circle is. You have some point. Let's call that hk. The circle is a set of all points that are equidistant from that point. So let's take the set of all points that are, say, r away from hk."}, {"video_title": "Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3", "Sentence": "You have some point. Let's call that hk. The circle is a set of all points that are equidistant from that point. So let's take the set of all points that are, say, r away from hk. So let's say that this distance right over here, this distance right over here is r. And so we want all the set of points that are exactly r away. So all the points x comma y that are exactly r away. And so you could imagine you could rotate around and all of these points are going to be exactly r away."}, {"video_title": "Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3", "Sentence": "So let's take the set of all points that are, say, r away from hk. So let's say that this distance right over here, this distance right over here is r. And so we want all the set of points that are exactly r away. So all the points x comma y that are exactly r away. And so you could imagine you could rotate around and all of these points are going to be exactly r away. And I'm going to try my best to draw at least a somewhat perfect looking circle. I won't be able to do a perfect job of it. But you get a sense."}, {"video_title": "Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3", "Sentence": "And so you could imagine you could rotate around and all of these points are going to be exactly r away. And I'm going to try my best to draw at least a somewhat perfect looking circle. I won't be able to do a perfect job of it. But you get a sense. All of these are exactly r away, at least if I were to draw it properly. They are r away. So how do we find an equation in terms of r and hk and x and y that describes all these points?"}, {"video_title": "Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3", "Sentence": "But you get a sense. All of these are exactly r away, at least if I were to draw it properly. They are r away. So how do we find an equation in terms of r and hk and x and y that describes all these points? Well, we know how to find the distance between two points on a coordinate plane. And in fact, it comes straight out of the Pythagorean theorem. If we were to draw a vertical line right over here that essentially is the change in the vertical axis between these two points, up here we're at y, here we're k. So this distance is going to be y minus k. We can do the exact same thing on the horizontal axis."}, {"video_title": "Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3", "Sentence": "So how do we find an equation in terms of r and hk and x and y that describes all these points? Well, we know how to find the distance between two points on a coordinate plane. And in fact, it comes straight out of the Pythagorean theorem. If we were to draw a vertical line right over here that essentially is the change in the vertical axis between these two points, up here we're at y, here we're k. So this distance is going to be y minus k. We can do the exact same thing on the horizontal axis. This x-coordinate is x, while this x-coordinate is h. So this is going to be x minus h is this distance. And this is a right triangle because by definition, we're saying, hey, we're measuring vertical distance here. We're measuring horizontal distance here."}, {"video_title": "Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3", "Sentence": "If we were to draw a vertical line right over here that essentially is the change in the vertical axis between these two points, up here we're at y, here we're k. So this distance is going to be y minus k. We can do the exact same thing on the horizontal axis. This x-coordinate is x, while this x-coordinate is h. So this is going to be x minus h is this distance. And this is a right triangle because by definition, we're saying, hey, we're measuring vertical distance here. We're measuring horizontal distance here. So these two things are perpendicular. And so from the Pythagorean theorem, we know that this squared plus this squared must be equal to our distance squared. And this is where the distance formula comes from."}, {"video_title": "Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3", "Sentence": "We're measuring horizontal distance here. So these two things are perpendicular. And so from the Pythagorean theorem, we know that this squared plus this squared must be equal to our distance squared. And this is where the distance formula comes from. So we know that x minus h squared plus y minus k squared must be equal to r squared. This is the equation for the set. This describes any x and y that satisfies this equation will sit on this circle."}, {"video_title": "Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3", "Sentence": "And this is where the distance formula comes from. So we know that x minus h squared plus y minus k squared must be equal to r squared. This is the equation for the set. This describes any x and y that satisfies this equation will sit on this circle. Now with that out of the way, let's go answer their question. The equation of the circle is this thing. And this looks awfully close to what we just wrote."}, {"video_title": "Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3", "Sentence": "This describes any x and y that satisfies this equation will sit on this circle. Now with that out of the way, let's go answer their question. The equation of the circle is this thing. And this looks awfully close to what we just wrote. We just have to make sure that we don't get confused with the negatives. Remember, it has to be in the form x minus h, y minus k. So let's write it a little bit differently. Instead of x plus 3 squared, we can write that as x minus negative 3 squared."}, {"video_title": "Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3", "Sentence": "And this looks awfully close to what we just wrote. We just have to make sure that we don't get confused with the negatives. Remember, it has to be in the form x minus h, y minus k. So let's write it a little bit differently. Instead of x plus 3 squared, we can write that as x minus negative 3 squared. And then plus, well, this is already in the form, plus y minus 4 squared is equal to, instead of 49, we can just call that 7 squared. And so now it becomes pretty clear that our h is negative 3. I wanted to do that in the red color."}, {"video_title": "Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3", "Sentence": "Instead of x plus 3 squared, we can write that as x minus negative 3 squared. And then plus, well, this is already in the form, plus y minus 4 squared is equal to, instead of 49, we can just call that 7 squared. And so now it becomes pretty clear that our h is negative 3. I wanted to do that in the red color. That our h is negative 3, and that our k is positive 4, and that our r is 7. So we could say h comma k is equal to negative 3 comma positive 4. You might say, hey, there's a negative 4 here."}, {"video_title": "Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3", "Sentence": "I wanted to do that in the red color. That our h is negative 3, and that our k is positive 4, and that our r is 7. So we could say h comma k is equal to negative 3 comma positive 4. You might say, hey, there's a negative 4 here. No, but look, it's minus k, minus 4. So k is 4. Likewise, it's minus h. You might say, hey, maybe h is a positive 3."}, {"video_title": "Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3", "Sentence": "What we're going to do in this video is think about a way to measure angles. And there's several ways to do this. You might have seen this leveraging things like degrees in other videos, but now we're going to introduce a new concept, or maybe you know this concept, but another way of looking at this concept. So we have this angle ABC, and we wanna think about what is a way to figure out a measure of angle ABC. Now, one way to think about it would be, well, this angle subtends some arc. In this case, it subtends arc AC. And we could see if this angle were smaller, if its measure were smaller, it would subtend a smaller arc."}, {"video_title": "Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3", "Sentence": "So we have this angle ABC, and we wanna think about what is a way to figure out a measure of angle ABC. Now, one way to think about it would be, well, this angle subtends some arc. In this case, it subtends arc AC. And we could see if this angle were smaller, if its measure were smaller, it would subtend a smaller arc. The length of that arc would be smaller. And if the angle were wider or had a larger measure, if it looks something like that, then the arc length would be larger. So should we define the measure of an angle like this as being equal to the length of the arc that it subtends?"}, {"video_title": "Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3", "Sentence": "And we could see if this angle were smaller, if its measure were smaller, it would subtend a smaller arc. The length of that arc would be smaller. And if the angle were wider or had a larger measure, if it looks something like that, then the arc length would be larger. So should we define the measure of an angle like this as being equal to the length of the arc that it subtends? Is this a good measure? Well, some of you might immediately see a problem with that because this length, the length of the arc that is subtended is not just dependent on the angle, the measure of the angle, it also depends on how big of a circle you're dealing with. If the radius is larger, then you're gonna have a larger arc length."}, {"video_title": "Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3", "Sentence": "So should we define the measure of an angle like this as being equal to the length of the arc that it subtends? Is this a good measure? Well, some of you might immediately see a problem with that because this length, the length of the arc that is subtended is not just dependent on the angle, the measure of the angle, it also depends on how big of a circle you're dealing with. If the radius is larger, then you're gonna have a larger arc length. For example, let me introduce another circle here. So we have the same angle measure, the central angle right over here. You could say angle ABC is still the same, but now it subtends a different arc in these two different circles."}, {"video_title": "Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3", "Sentence": "If the radius is larger, then you're gonna have a larger arc length. For example, let me introduce another circle here. So we have the same angle measure, the central angle right over here. You could say angle ABC is still the same, but now it subtends a different arc in these two different circles. You have this arc right over here. Let's call this arc DE. And the length of arc DE is not equal to the length of AC."}, {"video_title": "Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3", "Sentence": "You could say angle ABC is still the same, but now it subtends a different arc in these two different circles. You have this arc right over here. Let's call this arc DE. And the length of arc DE is not equal to the length of AC. And so we can't measure an angle just by the length of the arc that it subtends if that angle is a central angle in a circle. So we can get rid of that equality here. But what could we do?"}, {"video_title": "Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3", "Sentence": "And the length of arc DE is not equal to the length of AC. And so we can't measure an angle just by the length of the arc that it subtends if that angle is a central angle in a circle. So we can get rid of that equality here. But what could we do? Well, you might realize that these two pi's that I just created, you could kind of say pi ABC and pi DBE, these are similar pi's. Now we're not used to talking in terms of similar pi's, but what does it mean to be similar? Well, you have similarity if you can map one thing onto another, one shape onto another through not just rigid transformations, but also through dilations."}, {"video_title": "Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3", "Sentence": "But what could we do? Well, you might realize that these two pi's that I just created, you could kind of say pi ABC and pi DBE, these are similar pi's. Now we're not used to talking in terms of similar pi's, but what does it mean to be similar? Well, you have similarity if you can map one thing onto another, one shape onto another through not just rigid transformations, but also through dilations. And in this situation, if you were to take pi ABC and just dilate it by a scale factor larger than one, there's some scale factor where you would dilate it out to pi DBE. And what's interesting about that is if two things are similar, that means the ratio between corresponding parts are going to be the same. So for example, the ratio of the length of arc AC to the length of segment BC is going to be equal to the ratio of the length of arc DE, DE, to the length of segment BE."}, {"video_title": "Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3", "Sentence": "Well, you have similarity if you can map one thing onto another, one shape onto another through not just rigid transformations, but also through dilations. And in this situation, if you were to take pi ABC and just dilate it by a scale factor larger than one, there's some scale factor where you would dilate it out to pi DBE. And what's interesting about that is if two things are similar, that means the ratio between corresponding parts are going to be the same. So for example, the ratio of the length of arc AC to the length of segment BC is going to be equal to the ratio of the length of arc DE, DE, to the length of segment BE. So maybe this is a good measure, maybe this is a good measure for an angle. And it is indeed a measure that we use in geometry and trigonometry and throughout mathematics. And we call it the radian measure of an angle, of an angle."}, {"video_title": "Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3", "Sentence": "So for example, the ratio of the length of arc AC to the length of segment BC is going to be equal to the ratio of the length of arc DE, DE, to the length of segment BE. So maybe this is a good measure, maybe this is a good measure for an angle. And it is indeed a measure that we use in geometry and trigonometry and throughout mathematics. And we call it the radian measure of an angle, of an angle. And it equals, equals the ratio of the arc length subtended by that angle subtended by that angle to the radius. We just saw that in both of these situations. So let's see if we can make this a little bit more tangible."}, {"video_title": "Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3", "Sentence": "And we call it the radian measure of an angle, of an angle. And it equals, equals the ratio of the arc length subtended by that angle subtended by that angle to the radius. We just saw that in both of these situations. So let's see if we can make this a little bit more tangible. Let's say we had a circle here and it has a central point, let's just call that point F. And then let me create an angle here. And actually I can make a right angle. So let's call it F, let's call this point G and let's call this H. And let's say that the radius over here is two meters."}, {"video_title": "Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3", "Sentence": "So let's see if we can make this a little bit more tangible. Let's say we had a circle here and it has a central point, let's just call that point F. And then let me create an angle here. And actually I can make a right angle. So let's call it F, let's call this point G and let's call this H. And let's say that the radius over here is two meters. And now what would be the length of the arc subtended by angle GFH? Well, it's going to be 1 4th the circumference of this entire circle, the way that I've drawn it. So the entire circumference, I could write it here, the circumference is going to be equal to two pi times the radius, which is going to be two pi times two meters, is going to be four pi meters."}, {"video_title": "Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3", "Sentence": "So let's call it F, let's call this point G and let's call this H. And let's say that the radius over here is two meters. And now what would be the length of the arc subtended by angle GFH? Well, it's going to be 1 4th the circumference of this entire circle, the way that I've drawn it. So the entire circumference, I could write it here, the circumference is going to be equal to two pi times the radius, which is going to be two pi times two meters, is going to be four pi meters. And so if this arc length is 1 4th of that, this is going to be pi meters. And so based on this arc length and this radius, what is going to be the measure of angle GFH in radians? Well, we could say the measure of angle GFH in radians is going to be the ratio between the length of the arc subtended and the radius."}, {"video_title": "Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3", "Sentence": "So the entire circumference, I could write it here, the circumference is going to be equal to two pi times the radius, which is going to be two pi times two meters, is going to be four pi meters. And so if this arc length is 1 4th of that, this is going to be pi meters. And so based on this arc length and this radius, what is going to be the measure of angle GFH in radians? Well, we could say the measure of angle GFH in radians is going to be the ratio between the length of the arc subtended and the radius. And so it's going to be pi meters over two meters. The meters, you could view those as canceling out, which equals pi over two. And pi over two, what?"}, {"video_title": "Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3", "Sentence": "Well, we could say the measure of angle GFH in radians is going to be the ratio between the length of the arc subtended and the radius. And so it's going to be pi meters over two meters. The meters, you could view those as canceling out, which equals pi over two. And pi over two, what? Well, we would say this is equal to pi over two radians. Now, one thing to think about is why do we call it radians? It seems close to the word radius."}, {"video_title": "Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3", "Sentence": "And pi over two, what? Well, we would say this is equal to pi over two radians. Now, one thing to think about is why do we call it radians? It seems close to the word radius. And one way to think about it is when you divide this length by the length of the radius, you figure out how many of the radii is equivalent to the arc length in question. So for example, in this situation, one radii would look something like this. If you took the same length and you just went around like this, so you can see it's going to be one point something radii."}, {"video_title": "Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3", "Sentence": "It seems close to the word radius. And one way to think about it is when you divide this length by the length of the radius, you figure out how many of the radii is equivalent to the arc length in question. So for example, in this situation, one radii would look something like this. If you took the same length and you just went around like this, so you can see it's going to be one point something radii. And that's why you could also say it's one point something radians. If you took pi divided by two, you're going to get a little bit over one. You're going to get 1.07 something."}, {"video_title": "Focus and directrix introduction Conic sections Algebra II Khan Academy.mp3", "Sentence": "So what are these things? Well, a parabola can be defined as the set of all points. Let me draw an arbitrary axis right over here. So that's my y-axis. This is my x-axis. This is my x-axis. And so a parabola can be defined as the set of all points that are equidistant to a point and a line."}, {"video_title": "Focus and directrix introduction Conic sections Algebra II Khan Academy.mp3", "Sentence": "So that's my y-axis. This is my x-axis. This is my x-axis. And so a parabola can be defined as the set of all points that are equidistant to a point and a line. And that point is the focus of that parabola, and that line is the directrix of the parabola. So what am I talking about? So let's give ourselves a point."}, {"video_title": "Focus and directrix introduction Conic sections Algebra II Khan Academy.mp3", "Sentence": "And so a parabola can be defined as the set of all points that are equidistant to a point and a line. And that point is the focus of that parabola, and that line is the directrix of the parabola. So what am I talking about? So let's give ourselves a point. So let's say this point right over here. And we could even say that that is the point, let's say that's the point, let's say the x-coordinate is a, and the y-coordinate is b right over here. So that is the point a, comma, b."}, {"video_title": "Focus and directrix introduction Conic sections Algebra II Khan Academy.mp3", "Sentence": "So let's give ourselves a point. So let's say this point right over here. And we could even say that that is the point, let's say that's the point, let's say the x-coordinate is a, and the y-coordinate is b right over here. So that is the point a, comma, b. And then let's give ourselves a line for the directrix. And actually, let me do this in a different color instead of just white, because I already did the coordinates in white. So I will do it in this magenta color."}, {"video_title": "Focus and directrix introduction Conic sections Algebra II Khan Academy.mp3", "Sentence": "So that is the point a, comma, b. And then let's give ourselves a line for the directrix. And actually, let me do this in a different color instead of just white, because I already did the coordinates in white. So I will do it in this magenta color. So that's a, comma, b is the focus. And let's say y equals c is the directrix. So this right over here is the line."}, {"video_title": "Focus and directrix introduction Conic sections Algebra II Khan Academy.mp3", "Sentence": "So I will do it in this magenta color. So that's a, comma, b is the focus. And let's say y equals c is the directrix. So this right over here is the line. This right over here is the line y is equal to c. So this on the y-axis right over there, that is c. This is the line y is equal to c. So a parabola, what does it mean to be the set of all points that are equidistant between a point and this line? Let's think about what those points might be. Well, this point right over here would be halfway between this point, between the focus and the directrix."}, {"video_title": "Focus and directrix introduction Conic sections Algebra II Khan Academy.mp3", "Sentence": "So this right over here is the line. This right over here is the line y is equal to c. So this on the y-axis right over there, that is c. This is the line y is equal to c. So a parabola, what does it mean to be the set of all points that are equidistant between a point and this line? Let's think about what those points might be. Well, this point right over here would be halfway between this point, between the focus and the directrix. And then as we move away from x equals a, you're going to get points anywhere along this curve, which is a parabola. And you might be saying, wait, I don't get this. I don't get why points along this curve are going to be equidistant."}, {"video_title": "Focus and directrix introduction Conic sections Algebra II Khan Academy.mp3", "Sentence": "Well, this point right over here would be halfway between this point, between the focus and the directrix. And then as we move away from x equals a, you're going to get points anywhere along this curve, which is a parabola. And you might be saying, wait, I don't get this. I don't get why points along this curve are going to be equidistant. Well, let's just eyeball the distances. So this one, this distance, and obviously I'm drawing it by hand, so it's not going to be completely precise. That distance needs to be equal to that distance."}, {"video_title": "Focus and directrix introduction Conic sections Algebra II Khan Academy.mp3", "Sentence": "I don't get why points along this curve are going to be equidistant. Well, let's just eyeball the distances. So this one, this distance, and obviously I'm drawing it by hand, so it's not going to be completely precise. That distance needs to be equal to that distance. Well, that seems believable. And now if we take this point right over here on the parabola, this distance needs to be the same as that distance. Well, that seems believable."}, {"video_title": "Focus and directrix introduction Conic sections Algebra II Khan Academy.mp3", "Sentence": "That distance needs to be equal to that distance. Well, that seems believable. And now if we take this point right over here on the parabola, this distance needs to be the same as that distance. Well, that seems believable. If you take this point on the parabola, this distance needs to be the same as this distance. So hopefully you get what I'm talking about when I say that the parabola is a set of all points that are equidistant to this focus and this directrix. So any point along this parabola, this point right over here, the distance to the focus should be the same as the distance to the directrix."}, {"video_title": "Focus and directrix introduction Conic sections Algebra II Khan Academy.mp3", "Sentence": "Well, that seems believable. If you take this point on the parabola, this distance needs to be the same as this distance. So hopefully you get what I'm talking about when I say that the parabola is a set of all points that are equidistant to this focus and this directrix. So any point along this parabola, this point right over here, the distance to the focus should be the same as the distance to the directrix. Now what you might realize is, when you're taking the distance between a point and a point, the distance, it could be at an angle. This one's straight up and down. This one is going from the top left to the bottom right."}, {"video_title": "Focus and directrix introduction Conic sections Algebra II Khan Academy.mp3", "Sentence": "So any point along this parabola, this point right over here, the distance to the focus should be the same as the distance to the directrix. Now what you might realize is, when you're taking the distance between a point and a point, the distance, it could be at an angle. This one's straight up and down. This one is going from the top left to the bottom right. But when you take the distance from a point to a line, you essentially drop a perpendicular. You essentially go straight down, or if the parabola was down here, you would go straight up to find that distance. These are all right angles right over here."}, {"video_title": "Focus and directrix introduction Conic sections Algebra II Khan Academy.mp3", "Sentence": "This one is going from the top left to the bottom right. But when you take the distance from a point to a line, you essentially drop a perpendicular. You essentially go straight down, or if the parabola was down here, you would go straight up to find that distance. These are all right angles right over here. So that's all a focus and a directrix is. And every parabola is going to have a focus and directrix because every parabola is the set of all points that are equidistant to some focus and some directrix. So that's what they are."}, {"video_title": "Proof Right triangles inscribed in circles High School Math Khan Academy.mp3", "Sentence": "Let me draw my best diameter. That's pretty good. This right here is the diameter of the circle, or it's a diameter of the circle. That's a diameter. And let's say I have a triangle where the diameter is one side of the triangle and the angle opposite that side, its vertex, sits someplace on the circumference. So let's say the angle opposite of this diameter sits on that circumference, so the triangle looks like this. What I'm going to show you in this video is that this triangle is going to be a right triangle."}, {"video_title": "Proof Right triangles inscribed in circles High School Math Khan Academy.mp3", "Sentence": "That's a diameter. And let's say I have a triangle where the diameter is one side of the triangle and the angle opposite that side, its vertex, sits someplace on the circumference. So let's say the angle opposite of this diameter sits on that circumference, so the triangle looks like this. What I'm going to show you in this video is that this triangle is going to be a right triangle. And the 90 degree side is going to be the side that is opposite this diameter. I don't want to label it just yet because that would ruin the fun of the proof. Now let's see what we can do to show this."}, {"video_title": "Proof Right triangles inscribed in circles High School Math Khan Academy.mp3", "Sentence": "What I'm going to show you in this video is that this triangle is going to be a right triangle. And the 90 degree side is going to be the side that is opposite this diameter. I don't want to label it just yet because that would ruin the fun of the proof. Now let's see what we can do to show this. We have in our toolkit the notion of an inscribed angle. It's a relation to a central angle that subtends the same arc. So let's look at that."}, {"video_title": "Proof Right triangles inscribed in circles High School Math Khan Academy.mp3", "Sentence": "Now let's see what we can do to show this. We have in our toolkit the notion of an inscribed angle. It's a relation to a central angle that subtends the same arc. So let's look at that. Let's say that this is an inscribed angle right here. Let's call this theta. Let's say that's the center of my circle right there."}, {"video_title": "Proof Right triangles inscribed in circles High School Math Khan Academy.mp3", "Sentence": "So let's look at that. Let's say that this is an inscribed angle right here. Let's call this theta. Let's say that's the center of my circle right there. Then this angle right here would be a central angle. Let me draw another triangle right here, another line right there. This is a central angle right here."}, {"video_title": "Proof Right triangles inscribed in circles High School Math Khan Academy.mp3", "Sentence": "Let's say that's the center of my circle right there. Then this angle right here would be a central angle. Let me draw another triangle right here, another line right there. This is a central angle right here. This is a radius. This is the same radius. Actually, this distance is the same."}, {"video_title": "Proof Right triangles inscribed in circles High School Math Khan Academy.mp3", "Sentence": "This is a central angle right here. This is a radius. This is the same radius. Actually, this distance is the same. But we've learned several videos ago that this angle, this inscribed angle, it subtends this arc up here. The central angle that subtends that same arc is going to be twice this angle. We proved that several videos ago."}, {"video_title": "Proof Right triangles inscribed in circles High School Math Khan Academy.mp3", "Sentence": "Actually, this distance is the same. But we've learned several videos ago that this angle, this inscribed angle, it subtends this arc up here. The central angle that subtends that same arc is going to be twice this angle. We proved that several videos ago. So this is going to be 2 theta. It's the central angle subtending the same arc. Now this triangle right here, this one right here, this is an isosceles triangle."}, {"video_title": "Proof Right triangles inscribed in circles High School Math Khan Academy.mp3", "Sentence": "We proved that several videos ago. So this is going to be 2 theta. It's the central angle subtending the same arc. Now this triangle right here, this one right here, this is an isosceles triangle. I could rotate it and draw it like this. If I rotated it, I could draw it like this. If I flipped it over, it would look like that, that, and then the green side would be down like that."}, {"video_title": "Proof Right triangles inscribed in circles High School Math Khan Academy.mp3", "Sentence": "Now this triangle right here, this one right here, this is an isosceles triangle. I could rotate it and draw it like this. If I rotated it, I could draw it like this. If I flipped it over, it would look like that, that, and then the green side would be down like that. Both of these sides are of length r. This top angle is 2 theta. All I did is I took it and I rotated it around to draw it for you this way. This side is that side right there."}, {"video_title": "Proof Right triangles inscribed in circles High School Math Khan Academy.mp3", "Sentence": "If I flipped it over, it would look like that, that, and then the green side would be down like that. Both of these sides are of length r. This top angle is 2 theta. All I did is I took it and I rotated it around to draw it for you this way. This side is that side right there. Since its two sides are equal, this is isosceles, so these two base angles must be the same. These two base angles must be the same. That and that must be the same, or if I were to draw it up here, that and that must be the exact same base angle."}, {"video_title": "Proof Right triangles inscribed in circles High School Math Khan Academy.mp3", "Sentence": "This side is that side right there. Since its two sides are equal, this is isosceles, so these two base angles must be the same. These two base angles must be the same. That and that must be the same, or if I were to draw it up here, that and that must be the exact same base angle. Now let me see. I already used theta. Maybe I'll use x for these angles."}, {"video_title": "Proof Right triangles inscribed in circles High School Math Khan Academy.mp3", "Sentence": "That and that must be the same, or if I were to draw it up here, that and that must be the exact same base angle. Now let me see. I already used theta. Maybe I'll use x for these angles. This has to be x and that has to be x. What is x going to be equal to? x plus x plus 2 theta have to equal 180 degrees."}, {"video_title": "Proof Right triangles inscribed in circles High School Math Khan Academy.mp3", "Sentence": "Maybe I'll use x for these angles. This has to be x and that has to be x. What is x going to be equal to? x plus x plus 2 theta have to equal 180 degrees. They're all in the same triangle. Let me write that down. We get x plus x plus 2 theta all have to be equal to 180 degrees."}, {"video_title": "Proof Right triangles inscribed in circles High School Math Khan Academy.mp3", "Sentence": "x plus x plus 2 theta have to equal 180 degrees. They're all in the same triangle. Let me write that down. We get x plus x plus 2 theta all have to be equal to 180 degrees. We get 2x plus 2 theta is equal to 180 degrees. We get 2x is equal to 180 minus 2 theta. Divide both sides by 2, you get x is equal to 90 minus theta."}, {"video_title": "Proof Right triangles inscribed in circles High School Math Khan Academy.mp3", "Sentence": "We get x plus x plus 2 theta all have to be equal to 180 degrees. We get 2x plus 2 theta is equal to 180 degrees. We get 2x is equal to 180 minus 2 theta. Divide both sides by 2, you get x is equal to 90 minus theta. x is equal to 90 minus theta. Let's see what else we could do with this. We could look at this triangle right here."}, {"video_title": "Proof Right triangles inscribed in circles High School Math Khan Academy.mp3", "Sentence": "Divide both sides by 2, you get x is equal to 90 minus theta. x is equal to 90 minus theta. Let's see what else we could do with this. We could look at this triangle right here. This triangle, this side over here, also has this distance right here is also a radius of the circle. This distance over here, we've already labeled it, is a radius of a circle. Once again, this is also an isosceles triangle."}, {"video_title": "Proof Right triangles inscribed in circles High School Math Khan Academy.mp3", "Sentence": "We could look at this triangle right here. This triangle, this side over here, also has this distance right here is also a radius of the circle. This distance over here, we've already labeled it, is a radius of a circle. Once again, this is also an isosceles triangle. These two sides are equal. These two base angles have to be equal. This is theta."}, {"video_title": "Proof Right triangles inscribed in circles High School Math Khan Academy.mp3", "Sentence": "Once again, this is also an isosceles triangle. These two sides are equal. These two base angles have to be equal. This is theta. This is also going to be equal to theta. Actually, we use that information. We use that to actually show that first result about inscribed angles and the relation between them and central angles subtending the same arc."}, {"video_title": "Proof Right triangles inscribed in circles High School Math Khan Academy.mp3", "Sentence": "This is theta. This is also going to be equal to theta. Actually, we use that information. We use that to actually show that first result about inscribed angles and the relation between them and central angles subtending the same arc. If this is theta, that's theta because this is an isosceles triangle. What is this whole angle over here? What is that whole angle over here?"}, {"video_title": "Proof Right triangles inscribed in circles High School Math Khan Academy.mp3", "Sentence": "We use that to actually show that first result about inscribed angles and the relation between them and central angles subtending the same arc. If this is theta, that's theta because this is an isosceles triangle. What is this whole angle over here? What is that whole angle over here? It's going to be theta plus 90 minus theta. That angle right there is going to be theta plus 90 minus theta. The thetas cancel out."}, {"video_title": "Proof Right triangles inscribed in circles High School Math Khan Academy.mp3", "Sentence": "What is that whole angle over here? It's going to be theta plus 90 minus theta. That angle right there is going to be theta plus 90 minus theta. The thetas cancel out. No matter what, as long as one side of my triangle is a diameter and then the angle or the vertex of the angle opposite sits on the circumference, then this angle right here is going to be a right angle and this is going to be a right triangle. If I were to draw something random like this, if I were to just take a point right there like that and draw it just like that, this is a right angle. If I were to draw something like that and go out like that, this is a right angle."}, {"video_title": "Proof Opposite angles of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "What I want to do in this video is prove that the opposite angles of a parallelogram are congruent. So for example, we want to prove that CAB is congruent to BDC. So that that angle is equal to that angle, and that ABD, which is this angle, is congruent to DCA, which is this angle over here. And to do that, we just have to realize that we have some parallel lines and we have some transversals, and the parallel lines and the transversals actually switch roles. So let's just continue these so it looks a little bit more like transversals intersecting parallel lines. And really you could just pause it for yourself and try to prove it, because it really just comes out of alternate interior angles and corresponding angles of transversals intersecting parallel lines. So let's say that this angle right over here, let me do it in a new color since I've already used that yellow."}, {"video_title": "Proof Opposite angles of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "And to do that, we just have to realize that we have some parallel lines and we have some transversals, and the parallel lines and the transversals actually switch roles. So let's just continue these so it looks a little bit more like transversals intersecting parallel lines. And really you could just pause it for yourself and try to prove it, because it really just comes out of alternate interior angles and corresponding angles of transversals intersecting parallel lines. So let's say that this angle right over here, let me do it in a new color since I've already used that yellow. So let's start right here with angle BDC. So angle BDC, and I'm just going to mark this up here, angle BDC right over here, it is an alternate interior angle with this angle right over here. And actually we could extend this point over here."}, {"video_title": "Proof Opposite angles of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So let's say that this angle right over here, let me do it in a new color since I've already used that yellow. So let's start right here with angle BDC. So angle BDC, and I'm just going to mark this up here, angle BDC right over here, it is an alternate interior angle with this angle right over here. And actually we could extend this point over here. I could call that point E if I want. So I could say angle CDB is congruent to angle EBD by alternate interior angles. This is a transversal, these two lines are parallel."}, {"video_title": "Proof Opposite angles of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "And actually we could extend this point over here. I could call that point E if I want. So I could say angle CDB is congruent to angle EBD by alternate interior angles. This is a transversal, these two lines are parallel. AB or AE is parallel to CD. Fair enough. Now if we kind of change our thinking a little bit and instead we now view BD and AC as the parallel lines and now view AB as the transversal, then we see that angle EBD is going to be congruent to angle BAC because they are corresponding angles."}, {"video_title": "Proof Opposite angles of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "This is a transversal, these two lines are parallel. AB or AE is parallel to CD. Fair enough. Now if we kind of change our thinking a little bit and instead we now view BD and AC as the parallel lines and now view AB as the transversal, then we see that angle EBD is going to be congruent to angle BAC because they are corresponding angles. So angle EBD is going to be congruent to angle BAC, or I could say CAB, they are corresponding angles. And so if this angle is congruent to that angle and that angle is congruent to that angle, then they are congruent to each other. So angle CDB is congruent to angle CAB."}, {"video_title": "Proof Opposite angles of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "Now if we kind of change our thinking a little bit and instead we now view BD and AC as the parallel lines and now view AB as the transversal, then we see that angle EBD is going to be congruent to angle BAC because they are corresponding angles. So angle EBD is going to be congruent to angle BAC, or I could say CAB, they are corresponding angles. And so if this angle is congruent to that angle and that angle is congruent to that angle, then they are congruent to each other. So angle CDB is congruent to angle CAB. So we've proven this first part right over here. And then to prove that these two are congruent, we use the exact same logic. So first of all, we view this as a transversal, we view AC as a transversal of AB and CD."}, {"video_title": "Proof Opposite angles of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So angle CDB is congruent to angle CAB. So we've proven this first part right over here. And then to prove that these two are congruent, we use the exact same logic. So first of all, we view this as a transversal, we view AC as a transversal of AB and CD. Let me go here and let me create another point here. Let me call this point F right over here. So we know that angle ACD is going to be congruent to angle FAC because they are alternate interior angles."}, {"video_title": "Proof Opposite angles of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So first of all, we view this as a transversal, we view AC as a transversal of AB and CD. Let me go here and let me create another point here. Let me call this point F right over here. So we know that angle ACD is going to be congruent to angle FAC because they are alternate interior angles. And then we change our thinking a little bit and we view AC and BD as parallel lines and AB as a transversal. And then angle FAC is going to be congruent to angle ABD because they are corresponding angles. Angle F to angle ABD and they are corresponding angles."}, {"video_title": "Proof Opposite angles of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So we know that angle ACD is going to be congruent to angle FAC because they are alternate interior angles. And then we change our thinking a little bit and we view AC and BD as parallel lines and AB as a transversal. And then angle FAC is going to be congruent to angle ABD because they are corresponding angles. Angle F to angle ABD and they are corresponding angles. So the first time we view this as a transversal, AC is a transversal of AB and CD which are parallel lines. Now AB is a transversal and BD and AC are the parallel lines. And obviously if this is congruent to that and that is congruent to that, then these two have to be congruent to each other."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "And if we're talking about a right triangle, like the one that I've drawn here, one of them is going to be a right angle. And so we have two other angles to deal with. And what I want to explore in this video is the relationship between the sine of one of these angles and the cosine of the other, the cosine of one of these angles and the sine of the other. So to do that, let's just say that this angle, I guess we could call it angle A, let's say it's equal to theta. If this is equal to theta, if its measure is equal to theta degrees, say, what is the measure of angle B going to be? Well, the thing that will jump out at you, and we've looked at this in other problems, is the sum of the angles of a triangle are going to be 180 degrees. And this one right over here, it's a right triangle."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "So to do that, let's just say that this angle, I guess we could call it angle A, let's say it's equal to theta. If this is equal to theta, if its measure is equal to theta degrees, say, what is the measure of angle B going to be? Well, the thing that will jump out at you, and we've looked at this in other problems, is the sum of the angles of a triangle are going to be 180 degrees. And this one right over here, it's a right triangle. So this right angle takes up 90 of those 180 degrees. So you have 90 degrees left. So these two are going to have to add up to 90 degrees."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "And this one right over here, it's a right triangle. So this right angle takes up 90 of those 180 degrees. So you have 90 degrees left. So these two are going to have to add up to 90 degrees. This one and this one, angle A and angle B, are going to be complements of each other. They're going to be complementary. Or another way of thinking about it is B could be written as 90 minus theta."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "So these two are going to have to add up to 90 degrees. This one and this one, angle A and angle B, are going to be complements of each other. They're going to be complementary. Or another way of thinking about it is B could be written as 90 minus theta. If you add theta to 90 minus theta, you're going to get, you're going to get, or theta to 90 degrees minus theta, you're going to get 90 degrees. Now why is this interesting? Well, let's think about, let's think about what the sine, let's think about what the sine of theta is equal to."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "Or another way of thinking about it is B could be written as 90 minus theta. If you add theta to 90 minus theta, you're going to get, you're going to get, or theta to 90 degrees minus theta, you're going to get 90 degrees. Now why is this interesting? Well, let's think about, let's think about what the sine, let's think about what the sine of theta is equal to. Sine is opposite over hypotenuse. The opposite side is BC. So this is going to be the length of BC over the hypotenuse."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "Well, let's think about, let's think about what the sine, let's think about what the sine of theta is equal to. Sine is opposite over hypotenuse. The opposite side is BC. So this is going to be the length of BC over the hypotenuse. The hypotenuse is side AB. So the length of BC over the length of AB. Over the length of AB."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "So this is going to be the length of BC over the hypotenuse. The hypotenuse is side AB. So the length of BC over the length of AB. Over the length of AB. Now, what is that ratio if we were to look at this angle right over here? Well, for angle B, BC is the adjacent side, and AB is the hypotenuse. From angle B's perspective, this is the adjacent over the hypotenuse."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "Over the length of AB. Now, what is that ratio if we were to look at this angle right over here? Well, for angle B, BC is the adjacent side, and AB is the hypotenuse. From angle B's perspective, this is the adjacent over the hypotenuse. Now what trig ratio is adjacent over hypotenuse? Well, that's cosine. So ka toa, let me write that down."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "From angle B's perspective, this is the adjacent over the hypotenuse. Now what trig ratio is adjacent over hypotenuse? Well, that's cosine. So ka toa, let me write that down. Sine, doesn't hurt. Sine is opposite over hypotenuse. We see that right over there."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "So ka toa, let me write that down. Sine, doesn't hurt. Sine is opposite over hypotenuse. We see that right over there. Cosine is adjacent over hypotenuse, ka, and toa. Tangent is opposite over adjacent. So from this angle's perspective, taking the length of BC, which is BC is its adjacent side, and the hypotenuse is still AB."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "We see that right over there. Cosine is adjacent over hypotenuse, ka, and toa. Tangent is opposite over adjacent. So from this angle's perspective, taking the length of BC, which is BC is its adjacent side, and the hypotenuse is still AB. So from this angle's perspective, this is adjacent over hypotenuse. Or another way of thinking about it, it's the cosine of this angle. So that's going to be equal to the cosine of 90 degrees minus theta."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "So from this angle's perspective, taking the length of BC, which is BC is its adjacent side, and the hypotenuse is still AB. So from this angle's perspective, this is adjacent over hypotenuse. Or another way of thinking about it, it's the cosine of this angle. So that's going to be equal to the cosine of 90 degrees minus theta. That's a pretty neat relationship. The sine of an angle is equal to the cosine of its complement. So one way to think about it, the sine of, we could just pick any arbitrary angle."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "So that's going to be equal to the cosine of 90 degrees minus theta. That's a pretty neat relationship. The sine of an angle is equal to the cosine of its complement. So one way to think about it, the sine of, we could just pick any arbitrary angle. Let's say the sine of 60 degrees is going to be equal to the cosine of what? And I encourage you to pause the video and think about it. Well, it's going to be the cosine of 90 minus 60."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "So one way to think about it, the sine of, we could just pick any arbitrary angle. Let's say the sine of 60 degrees is going to be equal to the cosine of what? And I encourage you to pause the video and think about it. Well, it's going to be the cosine of 90 minus 60. It's going to be the cosine of 30 degrees. 30 plus 60 is 90. And of course, you could go the other way around."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "Well, it's going to be the cosine of 90 minus 60. It's going to be the cosine of 30 degrees. 30 plus 60 is 90. And of course, you could go the other way around. You could think about the cosine of theta is going to be equal to the adjacent side to theta, to angle A, I should say. So the adjacent side is right over here. That's AC."}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "And of course, you could go the other way around. You could think about the cosine of theta is going to be equal to the adjacent side to theta, to angle A, I should say. So the adjacent side is right over here. That's AC. So it's going to be AC over the hypotenuse, adjacent over hypotenuse. The hypotenuse is AB. But what is this ratio from angle B's point of view?"}, {"video_title": "Showing relationship between cosine and sine of complements Trigonometry Khan Academy.mp3", "Sentence": "That's AC. So it's going to be AC over the hypotenuse, adjacent over hypotenuse. The hypotenuse is AB. But what is this ratio from angle B's point of view? Well, the sine of angle B is going to be its opposite side, AC over the hypotenuse, AB. So this right over here, from angle B's perspective, this is angle B's sine. So this is equal to the sine of 90 degrees minus theta."}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "Let's do some example problems using our newly acquired knowledge of isosceles and equilateral triangles. So over here I have a kind of a triangle within a triangle, and we need to figure out this orange angle right over here and this blue angle right over here. And we know that side AB is equal to, or segment AB is equal to segment BC, which is equal to segment CD, or we could also call that DC. So first of all we see that triangle ABC is isosceles, and because it's isosceles the two base angles are going to be congruent. This is one leg, this is the other leg right over there, so the two base angles are going to be congruent. So we know that this angle right over here is also 31 degrees. Well if we know two of the angles in a triangle we can always figure out the third angle."}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "So first of all we see that triangle ABC is isosceles, and because it's isosceles the two base angles are going to be congruent. This is one leg, this is the other leg right over there, so the two base angles are going to be congruent. So we know that this angle right over here is also 31 degrees. Well if we know two of the angles in a triangle we can always figure out the third angle. They have to add up to 180 degrees. So if we call it, we could say 31 degrees plus 31 degrees plus the measure of angle ABC ABC is equal to 180 degrees. You can subtract 62, this right here is 62 degrees."}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "Well if we know two of the angles in a triangle we can always figure out the third angle. They have to add up to 180 degrees. So if we call it, we could say 31 degrees plus 31 degrees plus the measure of angle ABC ABC is equal to 180 degrees. You can subtract 62, this right here is 62 degrees. You subtract 62 from both sides you get the measure of angle ABC is equal to, let's see, 180 minus 60 would be 120. You subtract another two you get 118 degrees. So this angle right over here is 118 degrees."}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "You can subtract 62, this right here is 62 degrees. You subtract 62 from both sides you get the measure of angle ABC is equal to, let's see, 180 minus 60 would be 120. You subtract another two you get 118 degrees. So this angle right over here is 118 degrees. Well this angle right over here is supplementary to that 118 degrees. So that angle plus 118 is going to be equal to 180. We already know that that's 62 degrees."}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "So this angle right over here is 118 degrees. Well this angle right over here is supplementary to that 118 degrees. So that angle plus 118 is going to be equal to 180. We already know that that's 62 degrees. 62 plus 118 is 180. So this right over here is 62 degrees. Now this angle is one of the base angles for triangle BCD."}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "We already know that that's 62 degrees. 62 plus 118 is 180. So this right over here is 62 degrees. Now this angle is one of the base angles for triangle BCD. I didn't draw it that way but this side and this side are congruent. BC has the same length as CD. Those are the two legs of an isosceles triangle."}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "Now this angle is one of the base angles for triangle BCD. I didn't draw it that way but this side and this side are congruent. BC has the same length as CD. Those are the two legs of an isosceles triangle. You can kind of imagine it was turned upside down. This is the vertex. This is one base angle."}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "Those are the two legs of an isosceles triangle. You can kind of imagine it was turned upside down. This is the vertex. This is one base angle. This is the other base angle. Well the base angles are going to be congruent. They're going to be 62 degrees as well."}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "This is one base angle. This is the other base angle. Well the base angles are going to be congruent. They're going to be 62 degrees as well. Then finally if you want to figure out this blue angle, the blue angle plus these two 62 degree angles are going to have to add up to 180 degrees. So you get 62 plus 62 plus the blue angle, which is the measure of angle BCD, is going to have to be equal to 180 degrees. These two characters, let's see, 62 plus 62 is 124."}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "They're going to be 62 degrees as well. Then finally if you want to figure out this blue angle, the blue angle plus these two 62 degree angles are going to have to add up to 180 degrees. So you get 62 plus 62 plus the blue angle, which is the measure of angle BCD, is going to have to be equal to 180 degrees. These two characters, let's see, 62 plus 62 is 124. You subtract 124 from both sides. You get the measure of angle BCD is equal to, let's see, if you subtract 120 you get 60 and then you have to subtract another 4. So you get 56 degrees."}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "These two characters, let's see, 62 plus 62 is 124. You subtract 124 from both sides. You get the measure of angle BCD is equal to, let's see, if you subtract 120 you get 60 and then you have to subtract another 4. So you get 56 degrees. So this is equal to 56 degrees. And we're done. Now we can do either of these."}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "So you get 56 degrees. So this is equal to 56 degrees. And we're done. Now we can do either of these. Let's do this one right over here. So what is the measure of angle ABE? So they haven't even drawn segment BE here."}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "Now we can do either of these. Let's do this one right over here. So what is the measure of angle ABE? So they haven't even drawn segment BE here. So let me draw that for us. So we have to figure out the measure of angle ABE. So we have a bunch of congruent segments here."}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "So they haven't even drawn segment BE here. So let me draw that for us. So we have to figure out the measure of angle ABE. So we have a bunch of congruent segments here. In particular we see that triangle ABD, all of its sides are equal. So it's an equilateral triangle, which means all of the angles are equal. And if all of the angles are equal in a triangle, they all have to be 60 degrees."}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "So we have a bunch of congruent segments here. In particular we see that triangle ABD, all of its sides are equal. So it's an equilateral triangle, which means all of the angles are equal. And if all of the angles are equal in a triangle, they all have to be 60 degrees. So all of these characters are going to be 60 degrees. Well, that's part of angle ABE, but we have to figure out this other part right over here. And to do that, we can see that we're actually dealing with an isosceles triangle kind of tipped over to the left."}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "And if all of the angles are equal in a triangle, they all have to be 60 degrees. So all of these characters are going to be 60 degrees. Well, that's part of angle ABE, but we have to figure out this other part right over here. And to do that, we can see that we're actually dealing with an isosceles triangle kind of tipped over to the left. This is the vertex angle. This is one base angle. This is the other base angle."}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "And to do that, we can see that we're actually dealing with an isosceles triangle kind of tipped over to the left. This is the vertex angle. This is one base angle. This is the other base angle. And the vertex angle right here is 90 degrees. And once again, we know it's isosceles because this side, segment BD is equal to segment DE. And once again, these two angles plus this angle right over here is going to have to add up to 180 degrees."}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "This is the other base angle. And the vertex angle right here is 90 degrees. And once again, we know it's isosceles because this side, segment BD is equal to segment DE. And once again, these two angles plus this angle right over here is going to have to add up to 180 degrees. So if you call that an X, you get X plus X plus 90 is going to be 180 degrees. So you get 2X plus X plus 90 is going to be equal to 180 degrees. X plus X is the same thing as 2X plus 90 is equal to 180."}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "And once again, these two angles plus this angle right over here is going to have to add up to 180 degrees. So if you call that an X, you get X plus X plus 90 is going to be 180 degrees. So you get 2X plus X plus 90 is going to be equal to 180 degrees. X plus X is the same thing as 2X plus 90 is equal to 180. And then we can subtract 90 from both sides. You get 2X is equal to 90. Or divide both sides by 2."}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "X plus X is the same thing as 2X plus 90 is equal to 180. And then we can subtract 90 from both sides. You get 2X is equal to 90. Or divide both sides by 2. You get X is equal to 45 degrees. And then we're done because angle ABE is going to be equal to the 60 degrees plus the 45 degrees. So it's going to be this whole angle, which is what we care about."}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "Or divide both sides by 2. You get X is equal to 45 degrees. And then we're done because angle ABE is going to be equal to the 60 degrees plus the 45 degrees. So it's going to be this whole angle, which is what we care about. Angle ABE is going to be 60 plus 45, which is 105 degrees. And now we have this last problem over here. This one looks a little bit simpler."}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "So it's going to be this whole angle, which is what we care about. Angle ABE is going to be 60 plus 45, which is 105 degrees. And now we have this last problem over here. This one looks a little bit simpler. I have an isosceles triangle. This leg is equal to that leg. This is the vertex angle."}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "This one looks a little bit simpler. I have an isosceles triangle. This leg is equal to that leg. This is the vertex angle. I have to figure out B. And the trick here is, wait, how do I figure out one side of a triangle if I only know one other side? Don't I need to know two other sides?"}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "This is the vertex angle. I have to figure out B. And the trick here is, wait, how do I figure out one side of a triangle if I only know one other side? Don't I need to know two other sides? We'll do it the exact same way we just did that second part of that problem. If this is an isosceles triangle, which we know it is, then this angle is going to be equal to that angle there. And so if we call this X, then this is X as well."}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "Don't I need to know two other sides? We'll do it the exact same way we just did that second part of that problem. If this is an isosceles triangle, which we know it is, then this angle is going to be equal to that angle there. And so if we call this X, then this is X as well. We get X plus X plus 36 degrees plus 36 is equal to 180. The two X's, when you add them up, you get 2X. And then I'll just skip steps here."}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "And so if we call this X, then this is X as well. We get X plus X plus 36 degrees plus 36 is equal to 180. The two X's, when you add them up, you get 2X. And then I'll just skip steps here. 2X plus 36 is equal to 180. Subtract 36 from both sides. We get 2X."}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "And then I'll just skip steps here. 2X plus 36 is equal to 180. Subtract 36 from both sides. We get 2X. That 2 looks a little bit funny. We get 2X is equal to 180 minus 30 is 150. And then you want to subtract another 6 from 150."}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "We get 2X. That 2 looks a little bit funny. We get 2X is equal to 180 minus 30 is 150. And then you want to subtract another 6 from 150. It gets us to 144. Did I do that right? 180 minus 30 is 150."}, {"video_title": "Equilateral and isosceles example problems Congruence Geometry Khan Academy.mp3", "Sentence": "And then you want to subtract another 6 from 150. It gets us to 144. Did I do that right? 180 minus 30 is 150. Yep, 144. Divide both sides by 2. You get X is equal to 72 degrees."}, {"video_title": "Line of reflection example.mp3", "Sentence": "We're asked to draw the line of reflection that reflects triangle ABC, so that's this blue triangle, onto triangle A prime, B prime, C prime, which is this red triangle right over here. And they give us a little line drawing tool in order to draw the line of reflection. So the way I'm gonna think about it is, well, when I just eyeball it, they look, it looks like I'm just flipped over some type of a horizontal line here. But let's see if we can actually construct a horizontal line where it does actually look like the line of reflection. So let's see, C and C prime, how far apart are they from each other? So if we go one, two, three, four, five, six down, so they are six apart. So let's see if we just put this three above C prime and three below C. Let's see if this horizontal line works as a line of reflection."}, {"video_title": "Line of reflection example.mp3", "Sentence": "But let's see if we can actually construct a horizontal line where it does actually look like the line of reflection. So let's see, C and C prime, how far apart are they from each other? So if we go one, two, three, four, five, six down, so they are six apart. So let's see if we just put this three above C prime and three below C. Let's see if this horizontal line works as a line of reflection. So C, or C prime, is definitely the reflection of C across this line. C is exactly three units above it, and C prime is exactly three units below it. Let's see if it works for A and A prime."}, {"video_title": "Line of reflection example.mp3", "Sentence": "So let's see if we just put this three above C prime and three below C. Let's see if this horizontal line works as a line of reflection. So C, or C prime, is definitely the reflection of C across this line. C is exactly three units above it, and C prime is exactly three units below it. Let's see if it works for A and A prime. A is one, two, three, four, five units above it. A prime is one, two, three, four, five units below it. So that's looking good."}, {"video_title": "Line of reflection example.mp3", "Sentence": "Let's see if it works for A and A prime. A is one, two, three, four, five units above it. A prime is one, two, three, four, five units below it. So that's looking good. Now let's just check out B. So B, we can see it's at, the y-coordinate here is seven. This line right over here is y is equal to one."}, {"video_title": "Line of reflection example.mp3", "Sentence": "So that's looking good. Now let's just check out B. So B, we can see it's at, the y-coordinate here is seven. This line right over here is y is equal to one. And so what we would have here is, let's see, this is, looks like it's six units above this line, and B prime is six units below the line. So this indeed works. We've just constructed the line of reflection that reflects the blue triangle, triangle ABC, onto triangle A prime, B prime, C prime."}, {"video_title": "Determining angle of rotation.mp3", "Sentence": "So I'm just gonna think about how did each of these points have to be rotated to go from A to A prime or B to B prime or from C to C prime? So let's just start with A, so this is where A starts, remember we're rotating about the origin, that's why I'm drawing this line from the origin to A, and where does it get rotated to? Well, it gets rotated to right over here. So the rotation is going in the counterclockwise direction, so it's going to have a positive angle, so we can rule out these two right over here, and the key question is, is this 30 degrees or 60 degrees? And there's a bunch of ways that you could think about it. One, 60 degrees would be 2 3rds of a right angle, while 30 degrees would be 1 3rd of a right angle. A right angle would look something like this, so this looks much more like 2 3rds of a right angle, so I'll go with 60 degrees."}, {"video_title": "Determining angle of rotation.mp3", "Sentence": "So the rotation is going in the counterclockwise direction, so it's going to have a positive angle, so we can rule out these two right over here, and the key question is, is this 30 degrees or 60 degrees? And there's a bunch of ways that you could think about it. One, 60 degrees would be 2 3rds of a right angle, while 30 degrees would be 1 3rd of a right angle. A right angle would look something like this, so this looks much more like 2 3rds of a right angle, so I'll go with 60 degrees. Another way to think about it is that 60 degrees is 1 3rd of 180 degrees, which this also looks like right over here. And if you do that with any of the points, you would see a similar thing. So just looking at A to A prime makes me feel good that this was a 60 degree rotation."}, {"video_title": "Determining angle of rotation.mp3", "Sentence": "A right angle would look something like this, so this looks much more like 2 3rds of a right angle, so I'll go with 60 degrees. Another way to think about it is that 60 degrees is 1 3rd of 180 degrees, which this also looks like right over here. And if you do that with any of the points, you would see a similar thing. So just looking at A to A prime makes me feel good that this was a 60 degree rotation. Let's do another example. So we are told quadrilateral A prime, B prime, C prime, D prime in red here is the image of quadrilateral A, B, C, D in blue here under rotation about point Q. Determine the angle of rotation."}, {"video_title": "Determining angle of rotation.mp3", "Sentence": "So just looking at A to A prime makes me feel good that this was a 60 degree rotation. Let's do another example. So we are told quadrilateral A prime, B prime, C prime, D prime in red here is the image of quadrilateral A, B, C, D in blue here under rotation about point Q. Determine the angle of rotation. So once again, pause this video and see if you can figure it out. Well, I'm gonna tackle this the same way. I don't have a coordinate plane here, but it's the same notion."}, {"video_title": "Determining angle of rotation.mp3", "Sentence": "Determine the angle of rotation. So once again, pause this video and see if you can figure it out. Well, I'm gonna tackle this the same way. I don't have a coordinate plane here, but it's the same notion. I can take some initial point and then look at its image and think about, well, how much did I have to rotate it? I could do B to B prime, although this might be a little bit too close. So I'm going from B to, let me do a new color here just because this color is too close to, I'll use black."}, {"video_title": "Determining angle of rotation.mp3", "Sentence": "I don't have a coordinate plane here, but it's the same notion. I can take some initial point and then look at its image and think about, well, how much did I have to rotate it? I could do B to B prime, although this might be a little bit too close. So I'm going from B to, let me do a new color here just because this color is too close to, I'll use black. So we're going from B to B prime right over here. We are going clockwise, so it's going to be a negative rotation, so we can rule that and that out. And it looks like a right angle."}, {"video_title": "Determining angle of rotation.mp3", "Sentence": "So I'm going from B to, let me do a new color here just because this color is too close to, I'll use black. So we're going from B to B prime right over here. We are going clockwise, so it's going to be a negative rotation, so we can rule that and that out. And it looks like a right angle. This looks like a right angle, so I feel good about picking negative 90 degrees. We could try another point and feel good that that also meets that negative 90 degrees. Let's say D to D prime."}, {"video_title": "Determining angle of rotation.mp3", "Sentence": "And it looks like a right angle. This looks like a right angle, so I feel good about picking negative 90 degrees. We could try another point and feel good that that also meets that negative 90 degrees. Let's say D to D prime. So this is where D is initially. This is where D is, and this is where D prime is. And once again, we are moving clockwise, so it's a negative rotation, and this looks like a right angle, definitely more like a right angle than a 60-degree angle, and so this would be negative 90 degrees."}, {"video_title": "Rational equations intro Algebra 2 Khan Academy.mp3", "Sentence": "We have x plus one over nine minus x is equal to 2 3rds. Pause this video and see if you can try this before we work through it together. All right, now let's work through this together. Now, the first thing that we might want to do, there's several ways that you could approach this, but the one thing I like to do is get rid of this x here in the denominator, and the easiest way I can think of doing that is by multiplying both sides of this equation by nine minus x. Now, when you do that, it's important that you then put the qualifier that the x cannot be equal to the value that would have made this denominator zero, because clearly if somehow you do all this algebraic manipulation and you got x is equal to nine, that still wouldn't be a valid solution, because if you were to substitute nine back into the original equation, you would be dividing by zero in the denominator. So let's just put that right over here. X cannot be equal to nine."}, {"video_title": "Rational equations intro Algebra 2 Khan Academy.mp3", "Sentence": "Now, the first thing that we might want to do, there's several ways that you could approach this, but the one thing I like to do is get rid of this x here in the denominator, and the easiest way I can think of doing that is by multiplying both sides of this equation by nine minus x. Now, when you do that, it's important that you then put the qualifier that the x cannot be equal to the value that would have made this denominator zero, because clearly if somehow you do all this algebraic manipulation and you got x is equal to nine, that still wouldn't be a valid solution, because if you were to substitute nine back into the original equation, you would be dividing by zero in the denominator. So let's just put that right over here. X cannot be equal to nine. And so then we can safely move ahead with our algebraic manipulations. So on the left-hand side, as long as x does not equal nine, if we multiply and divide by nine minus x, they cancel out, and we'll just be left with an x plus one. And on the right-hand side, if you multiply 2 3rds times nine minus x, we get 2 3rds times nine is six, and then 2 3rds times negative x is negative 2 3rds x."}, {"video_title": "Rational equations intro Algebra 2 Khan Academy.mp3", "Sentence": "X cannot be equal to nine. And so then we can safely move ahead with our algebraic manipulations. So on the left-hand side, as long as x does not equal nine, if we multiply and divide by nine minus x, they cancel out, and we'll just be left with an x plus one. And on the right-hand side, if you multiply 2 3rds times nine minus x, we get 2 3rds times nine is six, and then 2 3rds times negative x is negative 2 3rds x. And once again, let's remind ourselves that x cannot be equal to nine. And then we can get all of our x's on the same side. So let's put that on the left."}, {"video_title": "Rational equations intro Algebra 2 Khan Academy.mp3", "Sentence": "And on the right-hand side, if you multiply 2 3rds times nine minus x, we get 2 3rds times nine is six, and then 2 3rds times negative x is negative 2 3rds x. And once again, let's remind ourselves that x cannot be equal to nine. And then we can get all of our x's on the same side. So let's put that on the left. So let's add 2 3rds x to both sides. So plus 2 3rds, 2 3rds x, plus 2 3rds x. And then what do we have?"}, {"video_title": "Rational equations intro Algebra 2 Khan Academy.mp3", "Sentence": "So let's put that on the left. So let's add 2 3rds x to both sides. So plus 2 3rds, 2 3rds x, plus 2 3rds x. And then what do we have? Well, on the left-hand side, we have one x, which is the same thing as 3 3rds x plus 2 3rds x is going to give us 5 3rds x plus one is equal to six, and then these characters cancel out. And then we can just subtract one from both sides, and we get 5 3rds x, 5 3rds x is equal to five. And then last but not least, we can multiply both sides of this equation times the reciprocal of 5 3rds, which is, of course, 3 5ths, and I'm doing that, so I just have an x isolated on the left-hand side."}, {"video_title": "Rational equations intro Algebra 2 Khan Academy.mp3", "Sentence": "And then what do we have? Well, on the left-hand side, we have one x, which is the same thing as 3 3rds x plus 2 3rds x is going to give us 5 3rds x plus one is equal to six, and then these characters cancel out. And then we can just subtract one from both sides, and we get 5 3rds x, 5 3rds x is equal to five. And then last but not least, we can multiply both sides of this equation times the reciprocal of 5 3rds, which is, of course, 3 5ths, and I'm doing that, so I just have an x isolated on the left-hand side. So times 3 5ths, and we are left with 3 5ths times 5 3rds is, of course, equal to one, so we're left with x is equal to five times 3 5ths is three. And so we're feeling pretty good about x equals three, but we have to make sure that that's consistent with our original expression. Well, if we look up here, if you substitute back x equals three, you don't get a zero in the denominator."}, {"video_title": "Features of a circle from its graph Mathematics II High School Math Khan Academy.mp3", "Sentence": "So we have a circle right over here, and the first question we'll ask ourselves is what are the coordinates of the center of that circle? Well, we can eyeball that. We can see, look, it looks like the center, it looks like the circle is centered on that point right over there, and the coordinates of that point, the x-coordinate is negative four, and the y-coordinate is negative seven. So the center of that circle would be the point negative four comma negative seven. Now let's say on top of that, someone were to tell us, someone were to tell us that this point, negative five comma negative nine, is also on the circle. So negative five comma negative nine is on the circle. So based on this information, the coordinate of the center and a point that sits on the circle, can we figure out the radius?"}, {"video_title": "Features of a circle from its graph Mathematics II High School Math Khan Academy.mp3", "Sentence": "So the center of that circle would be the point negative four comma negative seven. Now let's say on top of that, someone were to tell us, someone were to tell us that this point, negative five comma negative nine, is also on the circle. So negative five comma negative nine is on the circle. So based on this information, the coordinate of the center and a point that sits on the circle, can we figure out the radius? Well, the radius is just the distance between the center of the circle and any point on the circle. In fact, one of the most typical definitions of a circle is all of the points that are the same distance or that are the radius away from another point, from the, and that other point would be the center of the circle. So how do we find out the distance between these two points?"}, {"video_title": "Features of a circle from its graph Mathematics II High School Math Khan Academy.mp3", "Sentence": "So based on this information, the coordinate of the center and a point that sits on the circle, can we figure out the radius? Well, the radius is just the distance between the center of the circle and any point on the circle. In fact, one of the most typical definitions of a circle is all of the points that are the same distance or that are the radius away from another point, from the, and that other point would be the center of the circle. So how do we find out the distance between these two points? Between these two points, so the length of that orange line? Well, we can use the distance formula, which is essentially the Pythagorean theorem. The distance squared, so if this is, if the length of that is the distance, so we could say the distance squared is going to be equal to, is going to be equal to our change in x squared, so that right there is our change in x, I don't have to write really small, but that's our change in x, our change in x squared plus our change in y squared, our change in y squared, change in y squared."}, {"video_title": "Features of a circle from its graph Mathematics II High School Math Khan Academy.mp3", "Sentence": "So how do we find out the distance between these two points? Between these two points, so the length of that orange line? Well, we can use the distance formula, which is essentially the Pythagorean theorem. The distance squared, so if this is, if the length of that is the distance, so we could say the distance squared is going to be equal to, is going to be equal to our change in x squared, so that right there is our change in x, I don't have to write really small, but that's our change in x, our change in x squared plus our change in y squared, our change in y squared, change in y squared. Now what is our change in x? Our change in x, and you can even eyeball it here, it looks like it's one, but let's verify it. We could view this point as the, it doesn't matter which one you use, the start or the end, as long as you're consistent."}, {"video_title": "Features of a circle from its graph Mathematics II High School Math Khan Academy.mp3", "Sentence": "The distance squared, so if this is, if the length of that is the distance, so we could say the distance squared is going to be equal to, is going to be equal to our change in x squared, so that right there is our change in x, I don't have to write really small, but that's our change in x, our change in x squared plus our change in y squared, our change in y squared, change in y squared. Now what is our change in x? Our change in x, and you can even eyeball it here, it looks like it's one, but let's verify it. We could view this point as the, it doesn't matter which one you use, the start or the end, as long as you're consistent. So let's see, if we view this as the end, we'd say negative five, it'd be negative five, minus negative four, minus negative four, and so this would be equal to negative one. So when you go from the center to this outer point, negative five comma nine, negative five comma negative nine, you go one back in the x direction. Now the actual, this distance would just be the absolute value of that, but it doesn't matter that this is a negative because we're about to square it and so that negative sign will go away."}, {"video_title": "Features of a circle from its graph Mathematics II High School Math Khan Academy.mp3", "Sentence": "We could view this point as the, it doesn't matter which one you use, the start or the end, as long as you're consistent. So let's see, if we view this as the end, we'd say negative five, it'd be negative five, minus negative four, minus negative four, and so this would be equal to negative one. So when you go from the center to this outer point, negative five comma nine, negative five comma negative nine, you go one back in the x direction. Now the actual, this distance would just be the absolute value of that, but it doesn't matter that this is a negative because we're about to square it and so that negative sign will go away. Now what is our change in y? Our change in y? Well if this is the finishing y, negative nine, minus negative seven, minus our initial y, is equal to negative two."}, {"video_title": "Features of a circle from its graph Mathematics II High School Math Khan Academy.mp3", "Sentence": "Now the actual, this distance would just be the absolute value of that, but it doesn't matter that this is a negative because we're about to square it and so that negative sign will go away. Now what is our change in y? Our change in y? Well if this is the finishing y, negative nine, minus negative seven, minus our initial y, is equal to negative two. And notice, just to go from that point, that y to that y, we go to negative two. So actually we could call the length of that side as the absolute value of our change in y. And we could view this as the absolute value of our change in x."}, {"video_title": "Features of a circle from its graph Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well if this is the finishing y, negative nine, minus negative seven, minus our initial y, is equal to negative two. And notice, just to go from that point, that y to that y, we go to negative two. So actually we could call the length of that side as the absolute value of our change in y. And we could view this as the absolute value of our change in x. And it doesn't really matter because once we square them, the negatives go away. So our distance squared, our distance squared, I really could call this the radius squared, is going to be equal to our change in x squared. Well it's negative one squared, which is just going to be one, plus our change in y squared."}, {"video_title": "Features of a circle from its graph Mathematics II High School Math Khan Academy.mp3", "Sentence": "And we could view this as the absolute value of our change in x. And it doesn't really matter because once we square them, the negatives go away. So our distance squared, our distance squared, I really could call this the radius squared, is going to be equal to our change in x squared. Well it's negative one squared, which is just going to be one, plus our change in y squared. Negative two squared is just positive four. One plus four. And so you have your distance squared is equal to five, or that the distance is equal to the square root of five."}, {"video_title": "Features of a circle from its graph Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well it's negative one squared, which is just going to be one, plus our change in y squared. Negative two squared is just positive four. One plus four. And so you have your distance squared is equal to five, or that the distance is equal to the square root of five. And I could have just called this variable the radius. So we could say the radius is equal to the square root of five. And we're done."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And what I'm curious about is all of the points on my screen right over here that are exactly 2 centimeters away from A. So 2 centimeters on my screen is about that far. So clearly, if I started A and I go 2 centimeters in that direction, this point right over there is 2 centimeters from A. If I call that point point B, then I could say line segment AB is 2 centimeters. The length is 2 centimeters. Remember, this would refer to the actual line segment. I could say this looks nice."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "If I call that point point B, then I could say line segment AB is 2 centimeters. The length is 2 centimeters. Remember, this would refer to the actual line segment. I could say this looks nice. But if I talk about its length, I would get rid of that line on top. And I would just say AB is equal to 2. If I wanted to put units, I could say 2 centimeters."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "I could say this looks nice. But if I talk about its length, I would get rid of that line on top. And I would just say AB is equal to 2. If I wanted to put units, I could say 2 centimeters. But I'm not curious just about B. I want to think about all the points, the set of all of the points that are exactly 2 centimeters away from A. So I could go 2 centimeters in the other direction, maybe get to point C right over here. So AC is also going to be equal to 2 centimeters."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "If I wanted to put units, I could say 2 centimeters. But I'm not curious just about B. I want to think about all the points, the set of all of the points that are exactly 2 centimeters away from A. So I could go 2 centimeters in the other direction, maybe get to point C right over here. So AC is also going to be equal to 2 centimeters. But I could go 2 centimeters in any direction. And so if I find the set of all of the points that are exactly 2 centimeters away from A, I will get a very familiar-looking shape like this. And I'm trying to draw it freehand."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So AC is also going to be equal to 2 centimeters. But I could go 2 centimeters in any direction. And so if I find the set of all of the points that are exactly 2 centimeters away from A, I will get a very familiar-looking shape like this. And I'm trying to draw it freehand. So I would get a shape that looks like this. And actually, let me draw it in. I don't want to make you think it's only the points where that there's white."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And I'm trying to draw it freehand. So I would get a shape that looks like this. And actually, let me draw it in. I don't want to make you think it's only the points where that there's white. It's all of these points right over here. I want to draw a dashed line over there. Maybe I should just clear out all of these."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "I don't want to make you think it's only the points where that there's white. It's all of these points right over here. I want to draw a dashed line over there. Maybe I should just clear out all of these. And I'll just draw a solid line. So this could look something like that, my best attempt. And this set of all of the points that are exactly 2 centimeters away from A, this is a circle, as I'm sure you are already familiar with."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Maybe I should just clear out all of these. And I'll just draw a solid line. So this could look something like that, my best attempt. And this set of all of the points that are exactly 2 centimeters away from A, this is a circle, as I'm sure you are already familiar with. But that is the formal definition, the set of all points that have a fixed distance from A, that are a given distance from A. If I said the set of all points that are 3 centimeters from A, it might look something like this. It might look something like that."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And this set of all of the points that are exactly 2 centimeters away from A, this is a circle, as I'm sure you are already familiar with. But that is the formal definition, the set of all points that have a fixed distance from A, that are a given distance from A. If I said the set of all points that are 3 centimeters from A, it might look something like this. It might look something like that. That would give us another circle. It would give us another circle. I think you get the general idea."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "It might look something like that. That would give us another circle. It would give us another circle. I think you get the general idea. Now, what I want to do in this video is introduce ourselves into some of the concepts and words that we use when dealing with circles. So let me get rid of that 3 centimeter circle. So first of all, let's think about this distance."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "I think you get the general idea. Now, what I want to do in this video is introduce ourselves into some of the concepts and words that we use when dealing with circles. So let me get rid of that 3 centimeter circle. So first of all, let's think about this distance. This distance or one of these line segments that join A, which we would call the center of the circle. So we'll call A the center of the circle. And it makes sense just from the way we use the word center in everyday life."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So first of all, let's think about this distance. This distance or one of these line segments that join A, which we would call the center of the circle. So we'll call A the center of the circle. And it makes sense just from the way we use the word center in everyday life. What I want to do is think about what line segment AB is. So AB. AB connects the center, and it connects a point on the circle itself."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And it makes sense just from the way we use the word center in everyday life. What I want to do is think about what line segment AB is. So AB. AB connects the center, and it connects a point on the circle itself. Remember, the circle itself is all the points that are equal distance from the center. So AB, any point, any line segment, I should say, that connects the center to a point on the circle, we would call a radius. Is a radius."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "AB connects the center, and it connects a point on the circle itself. Remember, the circle itself is all the points that are equal distance from the center. So AB, any point, any line segment, I should say, that connects the center to a point on the circle, we would call a radius. Is a radius. And so the length of the radius, AB, over here is equal to 2 centimeters. And you're probably already familiar with the word radius, but I'm just being a little bit more formal. And what's interesting about geometry, at least when you start learning it at the high school level, is it's probably the first class where you're introduced into a slightly more formal mathematics, where we're a little bit more careful."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Is a radius. And so the length of the radius, AB, over here is equal to 2 centimeters. And you're probably already familiar with the word radius, but I'm just being a little bit more formal. And what's interesting about geometry, at least when you start learning it at the high school level, is it's probably the first class where you're introduced into a slightly more formal mathematics, where we're a little bit more careful. About giving our definitions, and then building on those definitions to come up with interesting results, and proving to ourselves that we definitely know what we think we know. And so that's why we're being a little bit more careful with our language over here. So AB is a radius, line segment AB."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And what's interesting about geometry, at least when you start learning it at the high school level, is it's probably the first class where you're introduced into a slightly more formal mathematics, where we're a little bit more careful. About giving our definitions, and then building on those definitions to come up with interesting results, and proving to ourselves that we definitely know what we think we know. And so that's why we're being a little bit more careful with our language over here. So AB is a radius, line segment AB. And so is line segment, let me draw another. So let me put another point on here. Let's say this is x."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So AB is a radius, line segment AB. And so is line segment, let me draw another. So let me put another point on here. Let's say this is x. So line segment AX is also a radius. Now you can also have other forms of lines and line segments that interact in interesting ways with the circle. So you could have a line that just intersects that circle at exactly one point."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Let's say this is x. So line segment AX is also a radius. Now you can also have other forms of lines and line segments that interact in interesting ways with the circle. So you could have a line that just intersects that circle at exactly one point. So let's call that point right over there. And let's call that D. And let's say you have a line. And the only point on the circle that the only point in the set of all the points that are equal distance from A, the only point on that circle that is also on that line is point D. And we could call that line L. So sometimes you'll see lines specified by some of the points on them."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So you could have a line that just intersects that circle at exactly one point. So let's call that point right over there. And let's call that D. And let's say you have a line. And the only point on the circle that the only point in the set of all the points that are equal distance from A, the only point on that circle that is also on that line is point D. And we could call that line L. So sometimes you'll see lines specified by some of the points on them. So for example, I've got another point right over here called E. We could call this line DE. Or we could just put a little script letter here with an L and say this is line L. But this line that only has one point in common with our circle, we call this a tangent line. So line L is a tangent."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And the only point on the circle that the only point in the set of all the points that are equal distance from A, the only point on that circle that is also on that line is point D. And we could call that line L. So sometimes you'll see lines specified by some of the points on them. So for example, I've got another point right over here called E. We could call this line DE. Or we could just put a little script letter here with an L and say this is line L. But this line that only has one point in common with our circle, we call this a tangent line. So line L is a tangent. It is tangent to the circle. So let me write it this way. Line L is tangent."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So line L is a tangent. It is tangent to the circle. So let me write it this way. Line L is tangent. And you normally wouldn't write it in caps like this. I'm just doing that for emphasis. Is tangent to, instead of writing the circle centered at A, you'll sometimes see this notation, to the circle centered at A."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Line L is tangent. And you normally wouldn't write it in caps like this. I'm just doing that for emphasis. Is tangent to, instead of writing the circle centered at A, you'll sometimes see this notation, to the circle centered at A. So this tells us that this is the circle we're talking about. Because who knows? Maybe we had another circle over here that is centered at M. Another circle."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Is tangent to, instead of writing the circle centered at A, you'll sometimes see this notation, to the circle centered at A. So this tells us that this is the circle we're talking about. Because who knows? Maybe we had another circle over here that is centered at M. Another circle. So we have to specify it's not tangent to that one. It's tangent to this one. So this circle with a dot in the middle tells we're talking about a circle."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Maybe we had another circle over here that is centered at M. Another circle. So we have to specify it's not tangent to that one. It's tangent to this one. So this circle with a dot in the middle tells we're talking about a circle. And this is a circle centered at point A. I want to be very clear. Point A is not on the circle. Point A is the center of the circle."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So this circle with a dot in the middle tells we're talking about a circle. And this is a circle centered at point A. I want to be very clear. Point A is not on the circle. Point A is the center of the circle. The points on the circle are the points equal distant from point A. Now, L is tangent because it only intersects the circle on one point. You could just as easily imagine a line that intersects the circle at two points."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Point A is the center of the circle. The points on the circle are the points equal distant from point A. Now, L is tangent because it only intersects the circle on one point. You could just as easily imagine a line that intersects the circle at two points. So we could call, maybe this is F and this is G. You could call that line FG. So we could write it like this. Line FG."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "You could just as easily imagine a line that intersects the circle at two points. So we could call, maybe this is F and this is G. You could call that line FG. So we could write it like this. Line FG. And this line that intersects at two points, we call this a secant of circle A. Is a secant. It is a secant line to this circle right over here."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Line FG. And this line that intersects at two points, we call this a secant of circle A. Is a secant. It is a secant line to this circle right over here. Because it intersects it in two points. Now, if FG was just a segment, if it didn't keep on going forever like lines like to do, if we only spoke about this line segment, let me do this in a new color. If we only spoke about this line segment between F and G and not thinking about going on forever, then all of a sudden we have a line segment, which we would specify there."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "It is a secant line to this circle right over here. Because it intersects it in two points. Now, if FG was just a segment, if it didn't keep on going forever like lines like to do, if we only spoke about this line segment, let me do this in a new color. If we only spoke about this line segment between F and G and not thinking about going on forever, then all of a sudden we have a line segment, which we would specify there. And we would call this a chord of the circle. Is a chord of circle A. It starts at a point on the circle, a point that is exactly, in this case, 2 centimeters away."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "If we only spoke about this line segment between F and G and not thinking about going on forever, then all of a sudden we have a line segment, which we would specify there. And we would call this a chord of the circle. Is a chord of circle A. It starts at a point on the circle, a point that is exactly, in this case, 2 centimeters away. And then it finishes at a point on the circle. So it connects two points on the circle. Now, you can have chords like this."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "It starts at a point on the circle, a point that is exactly, in this case, 2 centimeters away. And then it finishes at a point on the circle. So it connects two points on the circle. Now, you can have chords like this. And you can also have a chord, as you can imagine, a chord that actually goes through the center of the circle. So let's call this point H. And you have a straight line connecting F to H through A. So that's about as straight as I could draw it."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Now, you can have chords like this. And you can also have a chord, as you can imagine, a chord that actually goes through the center of the circle. So let's call this point H. And you have a straight line connecting F to H through A. So that's about as straight as I could draw it. So if you have a chord like that, that contains the actual center of the circle, a chord that goes from one point on the circle to another point on the circle, and it goes through the center of the circle, we call that a diameter of A. Is a diameter of circle A. And you've probably seen this in tons of problems before when we were not talking about geometry as formally."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So that's about as straight as I could draw it. So if you have a chord like that, that contains the actual center of the circle, a chord that goes from one point on the circle to another point on the circle, and it goes through the center of the circle, we call that a diameter of A. Is a diameter of circle A. And you've probably seen this in tons of problems before when we were not talking about geometry as formally. But a diameter is made up of exactly two radiuses. We already know that a radius connects a point to the center. So you have one radius right over here that connects F and A."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And you've probably seen this in tons of problems before when we were not talking about geometry as formally. But a diameter is made up of exactly two radiuses. We already know that a radius connects a point to the center. So you have one radius right over here that connects F and A. That's one radius. And then you have another radius connecting A and H, the center to a point on the circle. So a diameter is made up of these two radiuses, or radii I should call."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So you have one radius right over here that connects F and A. That's one radius. And then you have another radius connecting A and H, the center to a point on the circle. So a diameter is made up of these two radiuses, or radii I should call. I think that's the plural for radius. And so the length of a diameter is going to be twice the length of a radius. So we could say the length of the diameter, so the length of FH."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So a diameter is made up of these two radiuses, or radii I should call. I think that's the plural for radius. And so the length of a diameter is going to be twice the length of a radius. So we could say the length of the diameter, so the length of FH. And once again, I don't put the line on top of it when I'm just talking about the length, is going to be equal to FA, the length of segment FA, plus the length of segment AH. Now there's one last thing I want to talk about when we're dealing with circles. And that's the idea of an arc."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So we could say the length of the diameter, so the length of FH. And once again, I don't put the line on top of it when I'm just talking about the length, is going to be equal to FA, the length of segment FA, plus the length of segment AH. Now there's one last thing I want to talk about when we're dealing with circles. And that's the idea of an arc. So we also have the parts of the circle itself. So let me draw another circle over here. Let's center this circle right over here at B."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And that's the idea of an arc. So we also have the parts of the circle itself. So let me draw another circle over here. Let's center this circle right over here at B. And I'm going to find all the points that are a given distance from B. So it has some radius. I'm not going to specify it right over here."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Let's center this circle right over here at B. And I'm going to find all the points that are a given distance from B. So it has some radius. I'm not going to specify it right over here. And let me pick some random points on this circle. So let's call this J, let's call that K, let's call that S, let's call this T, let's call this U right over here. And I know it doesn't look that centered."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "I'm not going to specify it right over here. And let me pick some random points on this circle. So let's call this J, let's call that K, let's call that S, let's call this T, let's call this U right over here. And I know it doesn't look that centered. Let me try to center B a little bit better. Let me erase that and put B a little bit closer to the center of the circle. So that's my best shot."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And I know it doesn't look that centered. Let me try to center B a little bit better. Let me erase that and put B a little bit closer to the center of the circle. So that's my best shot. So let's put B right over there. Now, one interesting thing is, what do you call the length of the circle that goes between two points? So what would you call this?"}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So that's my best shot. So let's put B right over there. Now, one interesting thing is, what do you call the length of the circle that goes between two points? So what would you call this? Well, you could imagine in everyday language we would call something that looks like that an arc. And we would also call that an arc in geometry. And to specify this arc, we would call this, we would say JK, the two endpoints of the arc, the two points on the circle that are the endpoints of the arc."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So what would you call this? Well, you could imagine in everyday language we would call something that looks like that an arc. And we would also call that an arc in geometry. And to specify this arc, we would call this, we would say JK, the two endpoints of the arc, the two points on the circle that are the endpoints of the arc. And you use a little notation like that. So you put a little curve on top instead of a straight line. Now, you can also have another arc that connects J and K. This is the minor arc."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And to specify this arc, we would call this, we would say JK, the two endpoints of the arc, the two points on the circle that are the endpoints of the arc. And you use a little notation like that. So you put a little curve on top instead of a straight line. Now, you can also have another arc that connects J and K. This is the minor arc. It is the shortest way along the circle to connect J and K. But you could also go the other way around. You could also have this thing that goes all the way around the circle. And we would call that the major arc."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Now, you can also have another arc that connects J and K. This is the minor arc. It is the shortest way along the circle to connect J and K. But you could also go the other way around. You could also have this thing that goes all the way around the circle. And we would call that the major arc. And normally when you specify a major arc, just to show that you're going kind of the long way around, the way that you, it's not the shortest way to get between J and K, you'll often specify another point that you're going through. So for example, this major arc we could specify. We started J."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And we would call that the major arc. And normally when you specify a major arc, just to show that you're going kind of the long way around, the way that you, it's not the shortest way to get between J and K, you'll often specify another point that you're going through. So for example, this major arc we could specify. We started J. We went through. We could have said U, T, or S. But I'll put T right over there. We went through T. And then we went all the way to K. And so this specifies the major arc."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "We started J. We went through. We could have said U, T, or S. But I'll put T right over there. We went through T. And then we went all the way to K. And so this specifies the major arc. And this thing could have been the same thing as if I wrote JUK. These are specifying the same thing. Or JSK."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "We went through T. And then we went all the way to K. And so this specifies the major arc. And this thing could have been the same thing as if I wrote JUK. These are specifying the same thing. Or JSK. So there's multiple ways to specify this major arc right over here. But the one thing I want to make clear is that the minor arc is the shortest distance. So this is the minor arc."}, {"video_title": "Language and notation of the circle Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Or JSK. So there's multiple ways to specify this major arc right over here. But the one thing I want to make clear is that the minor arc is the shortest distance. So this is the minor arc. And the longer distance around is the major arc. And I'll leave you there. And maybe in the next few videos, we'll start playing with some of this notation a little bit more."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "It's isosceles, which means it has two equal sides. And we also know from isosceles triangles that the base angles must be equal. So these two base angles are going to be equal, and this side right over here is going to be equal in length to this side over here. We can say AC is going to be equal to CE. So we get all of that from this first statement right over there. Then they give us some more clues or some more information. They say CG is equal to 24."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "We can say AC is going to be equal to CE. So we get all of that from this first statement right over there. Then they give us some more clues or some more information. They say CG is equal to 24. So this is CG right over here. It has length 24. They tell us BH is equal to DF."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "They say CG is equal to 24. So this is CG right over here. It has length 24. They tell us BH is equal to DF. BH is equal to DF, so those two things are going to be congruent. They're going to be the same length. Then they tell us that GF is equal to 12."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "They tell us BH is equal to DF. BH is equal to DF, so those two things are going to be congruent. They're going to be the same length. Then they tell us that GF is equal to 12. So this is GF right over here. So GF is equal to 12. That's that distance right over there."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "Then they tell us that GF is equal to 12. So this is GF right over here. So GF is equal to 12. That's that distance right over there. Then finally they tell us that FE is equal to 6. So this is FE. Then finally they ask us what is the area of CBHFD."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "That's that distance right over there. Then finally they tell us that FE is equal to 6. So this is FE. Then finally they ask us what is the area of CBHFD. So CBHFD. So they're asking us for the area of this part right over here. That part and that part right over there."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "Then finally they ask us what is the area of CBHFD. So CBHFD. So they're asking us for the area of this part right over here. That part and that part right over there. That is CBHFD. So let's think about how we can do this. We can figure out the area of the larger triangle."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "That part and that part right over there. That is CBHFD. So let's think about how we can do this. We can figure out the area of the larger triangle. Then from that we can subtract the areas of these little pieces at the end. Then we'll be able to figure out this middle area, this area that I've shaded. We don't have all the information yet to solve that."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "We can figure out the area of the larger triangle. Then from that we can subtract the areas of these little pieces at the end. Then we'll be able to figure out this middle area, this area that I've shaded. We don't have all the information yet to solve that. We know what the height or the altitude of this triangle is, but we don't know its base. If we knew its base, we'd say 1 half base times height, we'd get the area of this triangle. Then we'd have to subtract out these areas."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "We don't have all the information yet to solve that. We know what the height or the altitude of this triangle is, but we don't know its base. If we knew its base, we'd say 1 half base times height, we'd get the area of this triangle. Then we'd have to subtract out these areas. We don't have full information there either. We don't know this height. Once we know that height, then we can figure out this height, but we also don't quite yet know what this length right over here is."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "Then we'd have to subtract out these areas. We don't have full information there either. We don't know this height. Once we know that height, then we can figure out this height, but we also don't quite yet know what this length right over here is. Let's just take it piece by piece. The first thing we might want to do, and you might guess because we've been talking a lot about similarity, is making some type of argument about similarity here because there's a bunch of similar triangles. For example, triangle CGE shares this angle with triangle DFE."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "Once we know that height, then we can figure out this height, but we also don't quite yet know what this length right over here is. Let's just take it piece by piece. The first thing we might want to do, and you might guess because we've been talking a lot about similarity, is making some type of argument about similarity here because there's a bunch of similar triangles. For example, triangle CGE shares this angle with triangle DFE. They both share this orange angle right here. They both have this right angle right over here. They have two angles in common."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "For example, triangle CGE shares this angle with triangle DFE. They both share this orange angle right here. They both have this right angle right over here. They have two angles in common. They are going to be similar by angle-angle. You can actually show that there's going to be a third angle in common because these two are parallel lines. We can write that triangle CGE is similar to triangle DFE."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "They have two angles in common. They are going to be similar by angle-angle. You can actually show that there's going to be a third angle in common because these two are parallel lines. We can write that triangle CGE is similar to triangle DFE. We know that by angle-angle. We have one set of corresponding angles congruent, and then this angle is in both triangles, so it is a set of corresponding congruent angles right over there. Then once we know that they are similar, we can set up the ratio between sides because we have some information about some of the sides."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "We can write that triangle CGE is similar to triangle DFE. We know that by angle-angle. We have one set of corresponding angles congruent, and then this angle is in both triangles, so it is a set of corresponding congruent angles right over there. Then once we know that they are similar, we can set up the ratio between sides because we have some information about some of the sides. We know that the ratio between DF and this side right over here, which is a corresponding side, the ratio between DF and CG, which is 24, is going to be the same thing as the ratio between FE, which is 6, and GE, which is not 12. It's 12 plus 6. It is 18."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "Then once we know that they are similar, we can set up the ratio between sides because we have some information about some of the sides. We know that the ratio between DF and this side right over here, which is a corresponding side, the ratio between DF and CG, which is 24, is going to be the same thing as the ratio between FE, which is 6, and GE, which is not 12. It's 12 plus 6. It is 18. Then let's see. 6 over 18, this is just 1 over 3. You get 3DF."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "It is 18. Then let's see. 6 over 18, this is just 1 over 3. You get 3DF. You get 3DF is equal to 24. I just cross-multiplied, or you could multiply both sides by 24, multiply both sides by 3. You would get this."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "You get 3DF. You get 3DF is equal to 24. I just cross-multiplied, or you could multiply both sides by 24, multiply both sides by 3. You would get this. Actually, you could just multiply both sides times 24, and you'll get 24 times 1 third, but we'll just do it this way. Divide both sides by 3. You get DF is equal to 8."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "You would get this. Actually, you could just multiply both sides times 24, and you'll get 24 times 1 third, but we'll just do it this way. Divide both sides by 3. You get DF is equal to 8. We found out that DF is equal to 8, that length right over there. That's useful for us because we know that this length right over here is also equal to 8. Now what can we do?"}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "You get DF is equal to 8. We found out that DF is equal to 8, that length right over there. That's useful for us because we know that this length right over here is also equal to 8. Now what can we do? It seems like we can make another similarity argument because we have this angle right over here. It is congruent to that angle right over there. We also have this angle, which is going to be 90 degrees."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "Now what can we do? It seems like we can make another similarity argument because we have this angle right over here. It is congruent to that angle right over there. We also have this angle, which is going to be 90 degrees. We have a 90-degree angle there. That by itself is actually enough to say that we have two similar triangles. We don't even have to show that they have a congruent side here."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "We also have this angle, which is going to be 90 degrees. We have a 90-degree angle there. That by itself is actually enough to say that we have two similar triangles. We don't even have to show that they have a congruent side here. Actually, we're going to show that these are actually congruent triangles that we're dealing with right over here. We have two angles. Actually, we could just go straight to that because when we talk about congruency, if you have an angle that's congruent to another angle, another angle that's congruent to another angle, and then a side that's congruent to another side, you're dealing with two congruent triangles."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "We don't even have to show that they have a congruent side here. Actually, we're going to show that these are actually congruent triangles that we're dealing with right over here. We have two angles. Actually, we could just go straight to that because when we talk about congruency, if you have an angle that's congruent to another angle, another angle that's congruent to another angle, and then a side that's congruent to another side, you're dealing with two congruent triangles. We can write triangle A, H, B is congruent to triangle E, F, D. We know that because we have angle, angle, side. Angle, angle, side postulate for congruency. If the two triangles are congruent, that makes things convenient."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "Actually, we could just go straight to that because when we talk about congruency, if you have an angle that's congruent to another angle, another angle that's congruent to another angle, and then a side that's congruent to another side, you're dealing with two congruent triangles. We can write triangle A, H, B is congruent to triangle E, F, D. We know that because we have angle, angle, side. Angle, angle, side postulate for congruency. If the two triangles are congruent, that makes things convenient. That means if this side is 8, that side is 8. We already knew that. That's how we established our congruency."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "If the two triangles are congruent, that makes things convenient. That means if this side is 8, that side is 8. We already knew that. That's how we established our congruency. That means if this side has length 6, then the corresponding side on this triangle is also going to have length 6. We can write this length right over here is also going to be 6. I can imagine you can imagine where all of this is going to go, but we want to prove to ourselves."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "That's how we established our congruency. That means if this side has length 6, then the corresponding side on this triangle is also going to have length 6. We can write this length right over here is also going to be 6. I can imagine you can imagine where all of this is going to go, but we want to prove to ourselves. We want to know for sure what the area is. We don't want to say, hey, maybe this is the same thing as that. Let's just actually prove it to ourselves."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "I can imagine you can imagine where all of this is going to go, but we want to prove to ourselves. We want to know for sure what the area is. We don't want to say, hey, maybe this is the same thing as that. Let's just actually prove it to ourselves. How do we figure out? We've almost figured out the entire base of this triangle, but we still haven't figured out the length of HG. Now we can use a similarity argument again because we can see that triangle ABH is actually similar to triangle ACG."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "Let's just actually prove it to ourselves. How do we figure out? We've almost figured out the entire base of this triangle, but we still haven't figured out the length of HG. Now we can use a similarity argument again because we can see that triangle ABH is actually similar to triangle ACG. They both have this angle here, and then they both have a right angle. They have one. ABH has a right angle there."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "Now we can use a similarity argument again because we can see that triangle ABH is actually similar to triangle ACG. They both have this angle here, and then they both have a right angle. They have one. ABH has a right angle there. ACG has a right angle right over there. So you have two angles, two corresponding angles are equal to each other. You're now dealing with similar triangles."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "ABH has a right angle there. ACG has a right angle right over there. So you have two angles, two corresponding angles are equal to each other. You're now dealing with similar triangles. We know that triangle ABH, I'll just write it as AHB since I already wrote it this way, AHB is similar to triangle AGC. You want to make sure you get the vertices in the right order. A is the orange angle, G is the right angle, and C is the unlabeled angle."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "You're now dealing with similar triangles. We know that triangle ABH, I'll just write it as AHB since I already wrote it this way, AHB is similar to triangle AGC. You want to make sure you get the vertices in the right order. A is the orange angle, G is the right angle, and C is the unlabeled angle. This is similar to triangle AGC. What that does for us is now we can use the ratios to figure out what HG is equal to. What can we say over here?"}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "A is the orange angle, G is the right angle, and C is the unlabeled angle. This is similar to triangle AGC. What that does for us is now we can use the ratios to figure out what HG is equal to. What can we say over here? We can say that 8 over 24, BH over its corresponding side of the larger triangle, so we say 8 over 24 is equal to 6 over not HG but over AG. I think you can see where this is going. You have 1 third is equal to 6 over AG, or we can cross multiply here and we can get AG is equal to 18."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "What can we say over here? We can say that 8 over 24, BH over its corresponding side of the larger triangle, so we say 8 over 24 is equal to 6 over not HG but over AG. I think you can see where this is going. You have 1 third is equal to 6 over AG, or we can cross multiply here and we can get AG is equal to 18. This entire length right over here is 18. If AG is 18 and AH is 6, then HG is 12. This is what you might have guessed if you were just trying to guess the answer right over here."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "You have 1 third is equal to 6 over AG, or we can cross multiply here and we can get AG is equal to 18. This entire length right over here is 18. If AG is 18 and AH is 6, then HG is 12. This is what you might have guessed if you were just trying to guess the answer right over here. Now we have proven to ourselves that this base has length of 18 here and then we have another 18 here. It has a length of 36. The entire base here is 36."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "This is what you might have guessed if you were just trying to guess the answer right over here. Now we have proven to ourselves that this base has length of 18 here and then we have another 18 here. It has a length of 36. The entire base here is 36. Now we can figure out the area of this larger, of the entire isosceles triangle. The area of ACE is going to be equal to 1 half times the base, which is 36, times 24. This is going to be the same thing as 1 half times 36 is 18 times 24."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "The entire base here is 36. Now we can figure out the area of this larger, of the entire isosceles triangle. The area of ACE is going to be equal to 1 half times the base, which is 36, times 24. This is going to be the same thing as 1 half times 36 is 18 times 24. I'll just do that over here on the top. 18 times 24, 8 times 4 is 32, 1 times 4 is 4 plus 3 is 7. We put a zero here because we're not dealing with 2 but 20."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "This is going to be the same thing as 1 half times 36 is 18 times 24. I'll just do that over here on the top. 18 times 24, 8 times 4 is 32, 1 times 4 is 4 plus 3 is 7. We put a zero here because we're not dealing with 2 but 20. You have 2 times 8 is 16, 2 times 1 is 2 plus 1, so it's 360. Then you have the 2, 7 plus 6 is 13, 1 plus 3 is 4. The area of ACE is equal to 432, but we're not done yet."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "We put a zero here because we're not dealing with 2 but 20. You have 2 times 8 is 16, 2 times 1 is 2 plus 1, so it's 360. Then you have the 2, 7 plus 6 is 13, 1 plus 3 is 4. The area of ACE is equal to 432, but we're not done yet. This area that we care about is the area of the entire triangle minus this area and minus this area right over here. What is the area of each of these little wedges right over here? It's going to be 1 half times 8 times 6."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "The area of ACE is equal to 432, but we're not done yet. This area that we care about is the area of the entire triangle minus this area and minus this area right over here. What is the area of each of these little wedges right over here? It's going to be 1 half times 8 times 6. 1 half times 8 is 4 times 6, so this is going to be 24 right over there. This is going to be another 24 right over there. This is going to be equal to 432 minus 24 minus 24 or minus 48, which is equal to, and we could try this to do this in our head, if we subtract 32, we're going to get to 400."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "It's going to be 1 half times 8 times 6. 1 half times 8 is 4 times 6, so this is going to be 24 right over there. This is going to be another 24 right over there. This is going to be equal to 432 minus 24 minus 24 or minus 48, which is equal to, and we could try this to do this in our head, if we subtract 32, we're going to get to 400. Then we're going to have to subtract another 16. If you subtract 10 from 400, you get 390, so you get to 384, whatever the units were. If these were in meters, then this would be meters squared."}, {"video_title": "Finding area using similarity and congruence Similarity Geometry Khan Academy.mp3", "Sentence": "This is going to be equal to 432 minus 24 minus 24 or minus 48, which is equal to, and we could try this to do this in our head, if we subtract 32, we're going to get to 400. Then we're going to have to subtract another 16. If you subtract 10 from 400, you get 390, so you get to 384, whatever the units were. If these were in meters, then this would be meters squared. If this was centimeters, this would be centimeters squared. Did I do that right? Let me go the other way."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "We're told the triangle PIN is rotated negative 270 degrees about the origin. So this is the triangle PIN. We're gonna rotate it negative 270 degrees about the origin. Draw the image of this rotation using the interactive graph. The direction of rotation by a positive angle is counterclockwise. So positive is counterclockwise, which is the standard convention. And this is negative, so negative degree would be clockwise."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "Draw the image of this rotation using the interactive graph. The direction of rotation by a positive angle is counterclockwise. So positive is counterclockwise, which is the standard convention. And this is negative, so negative degree would be clockwise. And we wanna use this tool here. And this tool, I can put points in, or I could delete points, I can draw a point by clicking on it. So what we wanna do is think about, well look, if we rotate the points of this triangle around the origin by negative 270 degrees, where is it gonna put these points?"}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "And this is negative, so negative degree would be clockwise. And we wanna use this tool here. And this tool, I can put points in, or I could delete points, I can draw a point by clicking on it. So what we wanna do is think about, well look, if we rotate the points of this triangle around the origin by negative 270 degrees, where is it gonna put these points? And to help us think about that, I've copied and pasted this on our scratch pad. So actually, let me go over here so I can actually draw on it. So let's just first think about what a negative 270 degree rotation actually is."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "So what we wanna do is think about, well look, if we rotate the points of this triangle around the origin by negative 270 degrees, where is it gonna put these points? And to help us think about that, I've copied and pasted this on our scratch pad. So actually, let me go over here so I can actually draw on it. So let's just first think about what a negative 270 degree rotation actually is. So if I were to start, if I were to, let me draw some coordinate axes here. So that's a x-axis and that's a y-axis. If you were to start right over here, and you were to rotate around the origin by negative, so this is the origin here, by negative 270 degrees, what would that be?"}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "So let's just first think about what a negative 270 degree rotation actually is. So if I were to start, if I were to, let me draw some coordinate axes here. So that's a x-axis and that's a y-axis. If you were to start right over here, and you were to rotate around the origin by negative, so this is the origin here, by negative 270 degrees, what would that be? Well let's see, this would be rotating negative 90. This would be rotating negative, another negative 90, which together would be negative 180. And then this would be another negative 90, which would give you in total negative 270 degrees."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "If you were to start right over here, and you were to rotate around the origin by negative, so this is the origin here, by negative 270 degrees, what would that be? Well let's see, this would be rotating negative 90. This would be rotating negative, another negative 90, which together would be negative 180. And then this would be another negative 90, which would give you in total negative 270 degrees. That's negative 270 degrees. Now notice, that would get that point here, which we could have also gotten there by just rotating it by positive 90 degrees. We could have just said that this is equivalent to a positive 90 degree rotation."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "And then this would be another negative 90, which would give you in total negative 270 degrees. That's negative 270 degrees. Now notice, that would get that point here, which we could have also gotten there by just rotating it by positive 90 degrees. We could have just said that this is equivalent to a positive 90 degree rotation. So if they want us to rotate the points here around the origin by negative 270 degrees, that's equivalent to just rotating all of the points, and I'll just focus on the vertices, because those are the easiest ones to think about, to visualize. I can just rotate each of those around the origin by positive 90 degrees. But how do we do that?"}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "We could have just said that this is equivalent to a positive 90 degree rotation. So if they want us to rotate the points here around the origin by negative 270 degrees, that's equivalent to just rotating all of the points, and I'll just focus on the vertices, because those are the easiest ones to think about, to visualize. I can just rotate each of those around the origin by positive 90 degrees. But how do we do that? And to do that, what I am going to do, to do that, what I'm gonna do is I'm gonna draw a series of right triangles. So let's first focus on, actually let's first focus on point I right over here. And if I were to, let me draw a right triangle."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "But how do we do that? And to do that, what I am going to do, to do that, what I'm gonna do is I'm gonna draw a series of right triangles. So let's first focus on, actually let's first focus on point I right over here. And if I were to, let me draw a right triangle. And I could draw it several ways, but let me draw it like this. So it's a right triangle, where the line between the origin and I is its hypotenuse. So let me see if I can, it's, I could probably draw, I could use a line tool for that."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "And if I were to, let me draw a right triangle. And I could draw it several ways, but let me draw it like this. So it's a right triangle, where the line between the origin and I is its hypotenuse. So let me see if I can, it's, I could probably draw, I could use a line tool for that. So let me, so that's the hypotenuse of the line. Now if I'm gonna rotate I 90 degrees about the origin, that's equivalent to rotating this right triangle 90 degrees. So what's going to happen there?"}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "So let me see if I can, it's, I could probably draw, I could use a line tool for that. So let me, so that's the hypotenuse of the line. Now if I'm gonna rotate I 90 degrees about the origin, that's equivalent to rotating this right triangle 90 degrees. So what's going to happen there? Well, this side, right over here, if I rotate this 90 degrees, where is that gonna go? Well, instead of going seven along the x-axis, it's gonna go seven along the y-axis. So it's going to, if you rotate it positive 90 degrees, that side is going to look like this."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "So what's going to happen there? Well, this side, right over here, if I rotate this 90 degrees, where is that gonna go? Well, instead of going seven along the x-axis, it's gonna go seven along the y-axis. So it's going to, if you rotate it positive 90 degrees, that side is going to look like this. So that's rotating it 90 degrees, just like that. Now what about this side over here? What about this side?"}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "So it's going to, if you rotate it positive 90 degrees, that side is going to look like this. So that's rotating it 90 degrees, just like that. Now what about this side over here? What about this side? Let me do this in a different color. What about, what about this side right over here? Well, this side over here, notice we've gone down from the origin, we've gone down seven."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "What about this side? Let me do this in a different color. What about, what about this side right over here? Well, this side over here, notice we've gone down from the origin, we've gone down seven. But if you were to rotate it up, notice this forms a right angle between this magenta side and this blue side. So you're gonna form a right angle again. And so from this point, you're gonna go straight as a right angle, and instead of going down seven, you're gonna go to the right seven."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "Well, this side over here, notice we've gone down from the origin, we've gone down seven. But if you were to rotate it up, notice this forms a right angle between this magenta side and this blue side. So you're gonna form a right angle again. And so from this point, you're gonna go straight as a right angle, and instead of going down seven, you're gonna go to the right seven. So you're gonna go to the right seven, just like this. And so the point I, or the corresponding point in the image after the rotation, is going to be right over here. So that green line, let me draw the hypotenuse now, it's gonna look, it's gonna look like, whoops, I wanted to do that in a different color, I wanted to do that in the green, I have trouble changing colors."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "And so from this point, you're gonna go straight as a right angle, and instead of going down seven, you're gonna go to the right seven. So you're gonna go to the right seven, just like this. And so the point I, or the corresponding point in the image after the rotation, is going to be right over here. So that green line, let me draw the hypotenuse now, it's gonna look, it's gonna look like, whoops, I wanted to do that in a different color, I wanted to do that in the green, I have trouble changing colors. All right, so there you go. It's gonna look, it's gonna look like that. So your new point I, if you rotate this triangle, this right triangle 90 degrees, your new point I, maybe I shouldn't say, I'll call it I prime, which is, what is the image of this point after I've done the 90 degree rotation, is going to be right over here."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "So that green line, let me draw the hypotenuse now, it's gonna look, it's gonna look like, whoops, I wanted to do that in a different color, I wanted to do that in the green, I have trouble changing colors. All right, so there you go. It's gonna look, it's gonna look like that. So your new point I, if you rotate this triangle, this right triangle 90 degrees, your new point I, maybe I shouldn't say, I'll call it I prime, which is, what is the image of this point after I've done the 90 degree rotation, is going to be right over here. And now we can do that for each of the points. We can do that for N here. Let's draw a right triangle."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "So your new point I, if you rotate this triangle, this right triangle 90 degrees, your new point I, maybe I shouldn't say, I'll call it I prime, which is, what is the image of this point after I've done the 90 degree rotation, is going to be right over here. And now we can do that for each of the points. We can do that for N here. Let's draw a right triangle. Let's draw a right triangle. And I could do it a bunch of different ways. I could draw it, I could draw a right triangle like this."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "Let's draw a right triangle. Let's draw a right triangle. And I could do it a bunch of different ways. I could draw it, I could draw a right triangle like this. Let me draw it like this. So that's one side of my right triangle. Actually, let me draw the hypotenuse first."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "I could draw it, I could draw a right triangle like this. Let me draw it like this. So that's one side of my right triangle. Actually, let me draw the hypotenuse first. So I have the hypotenuse, connects the origin to my point, just like that. And then I could, I could either draw it up here, or I could draw it down here, like this. I could draw it like this."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "Actually, let me draw the hypotenuse first. So I have the hypotenuse, connects the origin to my point, just like that. And then I could, I could either draw it up here, or I could draw it down here, like this. I could draw it like this. So if you rotate this 90 degrees, if you rotate this 90 degrees, this side, which is seven units long, and we're going seven below the origin, it's now going to be seven to the right, right? If I rotate it by 90 degrees, it's gonna be right over here. It's gonna be here."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "I could draw it like this. So if you rotate this 90 degrees, if you rotate this 90 degrees, this side, which is seven units long, and we're going seven below the origin, it's now going to be seven to the right, right? If I rotate it by 90 degrees, it's gonna be right over here. It's gonna be here. And this side, which has length, this, let me switch colors. This side, which has length two, it forms a right angle. So we're gonna form a right angle and have length two right over here."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "It's gonna be here. And this side, which has length, this, let me switch colors. This side, which has length two, it forms a right angle. So we're gonna form a right angle and have length two right over here. And so your, the image of point N is going to be, let me get the right color, it is going to be just like that. It is going to be, the image of point N is going to be right here. I'll call that N prime."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "So we're gonna form a right angle and have length two right over here. And so your, the image of point N is going to be, let me get the right color, it is going to be just like that. It is going to be, the image of point N is going to be right here. I'll call that N prime. So we know where the new, where the image of I is, the image of N is, so now we just have to think about where does the image of P sit? And once again, we can do our little right triangle idea. So let's draw a right triangle."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "I'll call that N prime. So we know where the new, where the image of I is, the image of N is, so now we just have to think about where does the image of P sit? And once again, we can do our little right triangle idea. So let's draw a right triangle. So just like that, I can draw that side, and I can do this side right over here. So if I were to rotate, if I were to focus on this, let me do this in a color I haven't. So if I were to focus on, I've already used that color."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "So let's draw a right triangle. So just like that, I can draw that side, and I can do this side right over here. So if I were to rotate, if I were to focus on this, let me do this in a color I haven't. So if I were to focus on, I've already used that color. If I were to focus on this right over here, and if I were to rotate it by 90 degrees, instead of going two to the right, it's gonna go two straight up. If I rotate this by 90 degrees, it's gonna be just like this. Now, this side on, let me pick another color."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "So if I were to focus on, I've already used that color. If I were to focus on this right over here, and if I were to rotate it by 90 degrees, instead of going two to the right, it's gonna go two straight up. If I rotate this by 90 degrees, it's gonna be just like this. Now, this side on, let me pick another color. This side right over here forms a right angle, and it has a length of three. So we're gonna form a right angle here and have a length of three. And just like that, we know where the image of P is going to be."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "Now, this side on, let me pick another color. This side right over here forms a right angle, and it has a length of three. So we're gonna form a right angle here and have a length of three. And just like that, we know where the image of P is going to be. It is going to be, it is going to be right over here. So this is P prime. And I know it's kind of confusing, but now we just have to, we now have to connect the P prime, the I prime, and the N prime to figure out what the image of my triangle is after rotation."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "And just like that, we know where the image of P is going to be. It is going to be, it is going to be right over here. So this is P prime. And I know it's kind of confusing, but now we just have to, we now have to connect the P prime, the I prime, and the N prime to figure out what the image of my triangle is after rotation. So let me do that. So if I connect these two, I get that. If I connect these two, I get that."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "And I know it's kind of confusing, but now we just have to, we now have to connect the P prime, the I prime, and the N prime to figure out what the image of my triangle is after rotation. So let me do that. So if I connect these two, I get that. If I connect these two, I get that. And if I connect these two, I connect those two, I have that. And there you have it. I have the image."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "If I connect these two, I get that. And if I connect these two, I connect those two, I have that. And there you have it. I have the image. And now I just have to input it on the actual problem. So let's see. Point negative three comma two."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "I have the image. And now I just have to input it on the actual problem. So let's see. Point negative three comma two. Let me get it out. So negative three comma two is there. I have seven comma two."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "Point negative three comma two. Let me get it out. So negative three comma two is there. I have seven comma two. So let me put that in there. Seven comma two. And then I have seven comma seven is there."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "I have seven comma two. So let me put that in there. Seven comma two. And then I have seven comma seven is there. So let me get this out. So seven comma seven. And then I'll draw it here again to connect the lines."}, {"video_title": "Points after rotation Transformations Geometry Khan Academy.mp3", "Sentence": "And then I have seven comma seven is there. So let me get this out. So seven comma seven. And then I'll draw it here again to connect the lines. And I'm done. I've rotated it through an angle of 90 degrees. Or negative 270 degrees, which is what they originally, they originally asked me for."}, {"video_title": "Points inside outside on a circle Mathematics I High School Math Khan Academy.mp3", "Sentence": "Where does the point P, which has a coordinates negative six comma negative six lie? And we have three options. Inside the circle, on the circle, or outside the circle. And the key realization here is just what a circle is all about. If we have the point C, which is the center of a circle, a circle of radius six, so let me draw that radius. So let's say that is its radius. It is six units."}, {"video_title": "Points inside outside on a circle Mathematics I High School Math Khan Academy.mp3", "Sentence": "And the key realization here is just what a circle is all about. If we have the point C, which is the center of a circle, a circle of radius six, so let me draw that radius. So let's say that is its radius. It is six units. The circle will look something like this. Remember, the circle is a set of all points that are exactly six units away from that center. So that's the definition of a circle."}, {"video_title": "Points inside outside on a circle Mathematics I High School Math Khan Academy.mp3", "Sentence": "It is six units. The circle will look something like this. Remember, the circle is a set of all points that are exactly six units away from that center. So that's the definition of a circle. It's a set of all points that are exactly six units away from the center. So if, for example, P is less than six units away, it's going to be inside the circle. If it's exactly six units away, it's going to be on the circle."}, {"video_title": "Points inside outside on a circle Mathematics I High School Math Khan Academy.mp3", "Sentence": "So that's the definition of a circle. It's a set of all points that are exactly six units away from the center. So if, for example, P is less than six units away, it's going to be inside the circle. If it's exactly six units away, it's going to be on the circle. And if it's more than six units away, it's going to be outside of the circle. So the key is is let's find the distance between these two points. If the distance is less than six, inside, distance equals six, we're on the circle, distance more than six, we are outside of the circle."}, {"video_title": "Points inside outside on a circle Mathematics I High School Math Khan Academy.mp3", "Sentence": "If it's exactly six units away, it's going to be on the circle. And if it's more than six units away, it's going to be outside of the circle. So the key is is let's find the distance between these two points. If the distance is less than six, inside, distance equals six, we're on the circle, distance more than six, we are outside of the circle. So let's do that. So if we wanted to find, and there's different notations for the distance, well, I'll just write D, or I could write the distance between C and P is going to be equal to, and the distance formula comes straight out of the Pythagorean theorem, but it's going to be the square root of our change in X squared plus our change in Y squared. So what is our change in X?"}, {"video_title": "Points inside outside on a circle Mathematics I High School Math Khan Academy.mp3", "Sentence": "If the distance is less than six, inside, distance equals six, we're on the circle, distance more than six, we are outside of the circle. So let's do that. So if we wanted to find, and there's different notations for the distance, well, I'll just write D, or I could write the distance between C and P is going to be equal to, and the distance formula comes straight out of the Pythagorean theorem, but it's going to be the square root of our change in X squared plus our change in Y squared. So what is our change in X? So our change in X, if we view C as our starting point and P as our end point, but we could do it either way, our change in X, our change in X is negative six minus negative one. So negative six minus negative one, and we're going to square it. So what we have inside here, that is change in X."}, {"video_title": "Points inside outside on a circle Mathematics I High School Math Khan Academy.mp3", "Sentence": "So what is our change in X? So our change in X, if we view C as our starting point and P as our end point, but we could do it either way, our change in X, our change in X is negative six minus negative one. So negative six minus negative one, and we're going to square it. So what we have inside here, that is change in X. So we're taking our change in X squared and then plus our change in Y squared. So we are going, we're going from negative three to negative six. So our change in Y is negative six minus negative three."}, {"video_title": "Points inside outside on a circle Mathematics I High School Math Khan Academy.mp3", "Sentence": "So what we have inside here, that is change in X. So we're taking our change in X squared and then plus our change in Y squared. So we are going, we're going from negative three to negative six. So our change in Y is negative six minus negative three. Negative six minus negative three, and we're going to square everything. So that is our change in Y inside the parentheses, and we're going to square it. So this is equal to, this is equal to negative six, negative six plus positive one is one way to think about it."}, {"video_title": "Points inside outside on a circle Mathematics I High School Math Khan Academy.mp3", "Sentence": "So our change in Y is negative six minus negative three. Negative six minus negative three, and we're going to square everything. So that is our change in Y inside the parentheses, and we're going to square it. So this is equal to, this is equal to negative six, negative six plus positive one is one way to think about it. So this is negative five squared, and then this is negative six plus three. So plus negative three squared. And once again, you can see our change in X is negative five."}, {"video_title": "Points inside outside on a circle Mathematics I High School Math Khan Academy.mp3", "Sentence": "So this is equal to, this is equal to negative six, negative six plus positive one is one way to think about it. So this is negative five squared, and then this is negative six plus three. So plus negative three squared. And once again, you can see our change in X is negative five. We go five lower in X, and we're going three lower in Y. So our change in Y is negative three. So this is equal to the square root of 25, square root of 25 plus nine."}, {"video_title": "Points inside outside on a circle Mathematics I High School Math Khan Academy.mp3", "Sentence": "And once again, you can see our change in X is negative five. We go five lower in X, and we're going three lower in Y. So our change in Y is negative three. So this is equal to the square root of 25, square root of 25 plus nine. Square root of 25 plus nine which is equal to the square root of 34. Now, the key is, is the square root of 34 less than six, greater than six, or equal to six? Well, we know that six is equal to the square root of 36."}, {"video_title": "Points inside outside on a circle Mathematics I High School Math Khan Academy.mp3", "Sentence": "So this is equal to the square root of 25, square root of 25 plus nine. Square root of 25 plus nine which is equal to the square root of 34. Now, the key is, is the square root of 34 less than six, greater than six, or equal to six? Well, we know that six is equal to the square root of 36. So the square root of 34 is less than the square root of 36. So I could write the square root of 34 is less than the square root of 36, and so the square root of 34 is less than six. Square root of 36 is six."}, {"video_title": "Points inside outside on a circle Mathematics I High School Math Khan Academy.mp3", "Sentence": "Well, we know that six is equal to the square root of 36. So the square root of 34 is less than the square root of 36. So I could write the square root of 34 is less than the square root of 36, and so the square root of 34 is less than six. Square root of 36 is six. And so since the distance between C and P is less than six, we are going to be on the inside of the circle. If I somehow got square root of 36 here, then we'd be on the circle. And if I somehow got square root of 37 here, or something larger, we would have been outside the circle."}, {"video_title": "Non-congruent shapes & transformations.mp3", "Sentence": "Here's the center of it. And this is circle M. This circle right over here. And so it looks like at first she translates it. So the center goes from this point to this point here. And so after the translation, we have the circle right over here. And then she dilates it. And the center of dilation looks like it is point N. And so she dilates it with some type of a scale factor in order to map it exactly onto N. Alright, so that all seems right."}, {"video_title": "Non-congruent shapes & transformations.mp3", "Sentence": "So the center goes from this point to this point here. And so after the translation, we have the circle right over here. And then she dilates it. And the center of dilation looks like it is point N. And so she dilates it with some type of a scale factor in order to map it exactly onto N. Alright, so that all seems right. Now, Brenda concluded, I was able to map circle M onto circle N using a sequence of rigid transformations. So the figures are congruent. So is she correct?"}, {"video_title": "Non-congruent shapes & transformations.mp3", "Sentence": "And the center of dilation looks like it is point N. And so she dilates it with some type of a scale factor in order to map it exactly onto N. Alright, so that all seems right. Now, Brenda concluded, I was able to map circle M onto circle N using a sequence of rigid transformations. So the figures are congruent. So is she correct? Pause that video, or pause this video, and think about that. Alright, now let's work on this together. So she was able to map circle M onto circle N using a sequence of transformations."}, {"video_title": "Non-congruent shapes & transformations.mp3", "Sentence": "So is she correct? Pause that video, or pause this video, and think about that. Alright, now let's work on this together. So she was able to map circle M onto circle N using a sequence of transformations. She did a translation and then a dilation. Those are all transformations. But they are not all rigid transformations."}, {"video_title": "Non-congruent shapes & transformations.mp3", "Sentence": "So she was able to map circle M onto circle N using a sequence of transformations. She did a translation and then a dilation. Those are all transformations. But they are not all rigid transformations. I'll put a question mark right over there. A translation is a rigid transformation. Remember, rigid transformations are ones that preserve distances, preserve angle measures, preserve lengths."}, {"video_title": "Non-congruent shapes & transformations.mp3", "Sentence": "But they are not all rigid transformations. I'll put a question mark right over there. A translation is a rigid transformation. Remember, rigid transformations are ones that preserve distances, preserve angle measures, preserve lengths. While a dilation is not a rigid transformation. As you can see very clearly, it is not preserving lengths. It is not, for example, preserving the radius of the circle."}, {"video_title": "Non-congruent shapes & transformations.mp3", "Sentence": "Remember, rigid transformations are ones that preserve distances, preserve angle measures, preserve lengths. While a dilation is not a rigid transformation. As you can see very clearly, it is not preserving lengths. It is not, for example, preserving the radius of the circle. In order for two figures to be congruent, the mapping has to be only with rigid transformations. So because she used a dilation, in fact, you have to use a dilation if you wanna be able to map M onto N because they have different radii, well, then she's not correct. These are not congruent figures."}, {"video_title": "Angle congruence equivalent to having same measure Congruence Geometry Khan Academy.mp3", "Sentence": "And for our definition of congruence, we will use the rigid transformation definition, which tells us two figures are congruent if and only if there exists a series of rigid transformations which will map one figure onto the other. And then what are rigid transformations? Those are transformations that preserve distance between points and angle measures. So let's get to it. So let's start with two angles that are congruent, and I'm going to show that they have the same measure. I'm gonna demonstrate that. So they start congruent."}, {"video_title": "Angle congruence equivalent to having same measure Congruence Geometry Khan Academy.mp3", "Sentence": "So let's get to it. So let's start with two angles that are congruent, and I'm going to show that they have the same measure. I'm gonna demonstrate that. So they start congruent. So these two angles are congruent to each other. Now, this means by the rigid transformation definition of congruence, there is a series of rigid transformations that map angle ABC onto angle, I'll do it here, onto angle DEF. By definition, by definition of rigid transformations, they preserve angle measure, preserve angle measure."}, {"video_title": "Angle congruence equivalent to having same measure Congruence Geometry Khan Academy.mp3", "Sentence": "So they start congruent. So these two angles are congruent to each other. Now, this means by the rigid transformation definition of congruence, there is a series of rigid transformations that map angle ABC onto angle, I'll do it here, onto angle DEF. By definition, by definition of rigid transformations, they preserve angle measure, preserve angle measure. So if you're able to map the left angle onto the right angle, and in doing so, you did it with transformations that preserved angle measure, they must now have the same angle measure. We now know that the measure of angle ABC is equal to the measure of angle DEF. So we've demonstrated this green statement the first way, that if things are congruent, they will have the same measure."}, {"video_title": "Angle congruence equivalent to having same measure Congruence Geometry Khan Academy.mp3", "Sentence": "By definition, by definition of rigid transformations, they preserve angle measure, preserve angle measure. So if you're able to map the left angle onto the right angle, and in doing so, you did it with transformations that preserved angle measure, they must now have the same angle measure. We now know that the measure of angle ABC is equal to the measure of angle DEF. So we've demonstrated this green statement the first way, that if things are congruent, they will have the same measure. Now let's prove it the other way around. So now let's start with the idea that measure of angle ABC is equal to the measure of angle DEF. And to demonstrate that these are going to be congruent, we just have to show that there's always a series of rigid transformations that will map angle ABC onto angle DEF."}, {"video_title": "Angle congruence equivalent to having same measure Congruence Geometry Khan Academy.mp3", "Sentence": "So we've demonstrated this green statement the first way, that if things are congruent, they will have the same measure. Now let's prove it the other way around. So now let's start with the idea that measure of angle ABC is equal to the measure of angle DEF. And to demonstrate that these are going to be congruent, we just have to show that there's always a series of rigid transformations that will map angle ABC onto angle DEF. And to help us there, let's just visualize these angles. So draw this really fast, angle ABC. An angle is defined by two rays that start at a point."}, {"video_title": "Angle congruence equivalent to having same measure Congruence Geometry Khan Academy.mp3", "Sentence": "And to demonstrate that these are going to be congruent, we just have to show that there's always a series of rigid transformations that will map angle ABC onto angle DEF. And to help us there, let's just visualize these angles. So draw this really fast, angle ABC. An angle is defined by two rays that start at a point. That point is the vertex. So it's ABC, and then let me draw angle DEF. So it might look something like this."}, {"video_title": "Angle congruence equivalent to having same measure Congruence Geometry Khan Academy.mp3", "Sentence": "An angle is defined by two rays that start at a point. That point is the vertex. So it's ABC, and then let me draw angle DEF. So it might look something like this. D-E-F. And what we will now do is, let's do our first rigid transformation. Let's translate, translate angle ABC so that B mapped to point E. And if we did that, so we're gonna translate it like that, then ABC is going to look something like, ABC is gonna look something like this. It's gonna look something like this."}, {"video_title": "Angle congruence equivalent to having same measure Congruence Geometry Khan Academy.mp3", "Sentence": "So it might look something like this. D-E-F. And what we will now do is, let's do our first rigid transformation. Let's translate, translate angle ABC so that B mapped to point E. And if we did that, so we're gonna translate it like that, then ABC is going to look something like, ABC is gonna look something like this. It's gonna look something like this. B is now mapped onto E. This would be where A would get mapped to. This would be where C would get mapped to. Sometimes you might see a notation A prime, C prime."}, {"video_title": "Angle congruence equivalent to having same measure Congruence Geometry Khan Academy.mp3", "Sentence": "It's gonna look something like this. B is now mapped onto E. This would be where A would get mapped to. This would be where C would get mapped to. Sometimes you might see a notation A prime, C prime. And this is where B would get mapped to. And then the next thing I would do is I would rotate angle ABC about its vertex, about B, so that ray BC, ray BC coincides, coincides with ray EF. Now, so you're just gonna rotate the whole angle that way so that now ray BC coincides with ray EF."}, {"video_title": "Angle congruence equivalent to having same measure Congruence Geometry Khan Academy.mp3", "Sentence": "Sometimes you might see a notation A prime, C prime. And this is where B would get mapped to. And then the next thing I would do is I would rotate angle ABC about its vertex, about B, so that ray BC, ray BC coincides, coincides with ray EF. Now, so you're just gonna rotate the whole angle that way so that now ray BC coincides with ray EF. Well, you might be saying, hey, C doesn't necessarily have to sit on F because they might be different distances from their vertices, but that's all right. The ray can be defined by any point that sits on that ray. So now if you do this rotation and ray BC coincides with ray EF, now those two rays would be equivalent because measure of angle ABC is equal to the measure of angle DEF."}, {"video_title": "Angle congruence equivalent to having same measure Congruence Geometry Khan Academy.mp3", "Sentence": "Now, so you're just gonna rotate the whole angle that way so that now ray BC coincides with ray EF. Well, you might be saying, hey, C doesn't necessarily have to sit on F because they might be different distances from their vertices, but that's all right. The ray can be defined by any point that sits on that ray. So now if you do this rotation and ray BC coincides with ray EF, now those two rays would be equivalent because measure of angle ABC is equal to the measure of angle DEF. That will also tell us that ray BA, ray BA now coincides, coincides with ray ED. And just like that, I've given you a series of rigid transformations that will always work. If you translate so that the vertices are mapped onto each other, and then you rotate it so that the bottom ray of one angle coincides with the bottom ray of the other angle, then you could say the top ray of the two angles will now coincide because the angles have the same measure."}, {"video_title": "Angle congruence equivalent to having same measure Congruence Geometry Khan Academy.mp3", "Sentence": "So now if you do this rotation and ray BC coincides with ray EF, now those two rays would be equivalent because measure of angle ABC is equal to the measure of angle DEF. That will also tell us that ray BA, ray BA now coincides, coincides with ray ED. And just like that, I've given you a series of rigid transformations that will always work. If you translate so that the vertices are mapped onto each other, and then you rotate it so that the bottom ray of one angle coincides with the bottom ray of the other angle, then you could say the top ray of the two angles will now coincide because the angles have the same measure. And because of that, the angles now completely coincide. And so we know that angle ABC is congruent to angle DEF. And we're now done."}, {"video_title": "Angle congruence equivalent to having same measure Congruence Geometry Khan Academy.mp3", "Sentence": "If you translate so that the vertices are mapped onto each other, and then you rotate it so that the bottom ray of one angle coincides with the bottom ray of the other angle, then you could say the top ray of the two angles will now coincide because the angles have the same measure. And because of that, the angles now completely coincide. And so we know that angle ABC is congruent to angle DEF. And we're now done. We've proven both sides of the statement. If they're congruent, they have the same measure. If they have the same measure, then they are congruent."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy (2).mp3", "Sentence": "So that's one line, and then let me draw another line that is parallel to that. I'm claiming that these are parallel lines. And now I'm gonna draw some transversals here. So first let me draw a horizontal transversal. So just like that. And then let me do a vertical transversal. So just like that."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy (2).mp3", "Sentence": "So first let me draw a horizontal transversal. So just like that. And then let me do a vertical transversal. So just like that. And I'm assuming that the green one is horizontal and the blue one is vertical. So we assume that they are perpendicular to each other, that these intersect at right angles. And from this, I'm going to figure out, I'm gonna use some parallel line angle properties to establish that this triangle and this triangle are similar, and then use that to establish that both of these lines, both of these yellow lines have the same slope."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy (2).mp3", "Sentence": "So just like that. And I'm assuming that the green one is horizontal and the blue one is vertical. So we assume that they are perpendicular to each other, that these intersect at right angles. And from this, I'm going to figure out, I'm gonna use some parallel line angle properties to establish that this triangle and this triangle are similar, and then use that to establish that both of these lines, both of these yellow lines have the same slope. So actually let me label some points here. So let's call that point A, point B, point C, point D, and point E. So let's see. First of all, we know that angle CED is going to be congruent to angle AEB because they're both right angles."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy (2).mp3", "Sentence": "And from this, I'm going to figure out, I'm gonna use some parallel line angle properties to establish that this triangle and this triangle are similar, and then use that to establish that both of these lines, both of these yellow lines have the same slope. So actually let me label some points here. So let's call that point A, point B, point C, point D, and point E. So let's see. First of all, we know that angle CED is going to be congruent to angle AEB because they're both right angles. So that's a right angle, and then that is a right angle right over there. We also know some things about corresponding angles where a transversal intersects parallel lines. This angle corresponds to this angle if we look at the blue transversal as it intersects those two lines."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy (2).mp3", "Sentence": "First of all, we know that angle CED is going to be congruent to angle AEB because they're both right angles. So that's a right angle, and then that is a right angle right over there. We also know some things about corresponding angles where a transversal intersects parallel lines. This angle corresponds to this angle if we look at the blue transversal as it intersects those two lines. And so they're going to have the same measure. They're going to be congruent. Now this angle on one side of this point B is going to also be congruent to that because they are vertical angles, and we've seen that multiple times before."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy (2).mp3", "Sentence": "This angle corresponds to this angle if we look at the blue transversal as it intersects those two lines. And so they're going to have the same measure. They're going to be congruent. Now this angle on one side of this point B is going to also be congruent to that because they are vertical angles, and we've seen that multiple times before. And so we know that this angle, angle ABE, is congruent to angle ECD. Sometimes this is called alternate interior angles of a transversal and parallel lines. Well, if you look at triangle CED and triangle ABE, we see they already have two angles in common."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy (2).mp3", "Sentence": "Now this angle on one side of this point B is going to also be congruent to that because they are vertical angles, and we've seen that multiple times before. And so we know that this angle, angle ABE, is congruent to angle ECD. Sometimes this is called alternate interior angles of a transversal and parallel lines. Well, if you look at triangle CED and triangle ABE, we see they already have two angles in common. So if they have two angles in common, well, their third angle has to be in common because this third angle's just going to be 180 minus these other two. And so this third angle's just going to be 180 minus the other two. And so just like that, we notice we have all three angles are the same in both of these triangles, or they're not all the same, but all of the corresponding angles, I should say, are the same."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy (2).mp3", "Sentence": "Well, if you look at triangle CED and triangle ABE, we see they already have two angles in common. So if they have two angles in common, well, their third angle has to be in common because this third angle's just going to be 180 minus these other two. And so this third angle's just going to be 180 minus the other two. And so just like that, we notice we have all three angles are the same in both of these triangles, or they're not all the same, but all of the corresponding angles, I should say, are the same. Angle, this blue angle has the same measure as this blue angle. This magenta angle has the same measure as this magenta angle. And then the other angles are right angles."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy (2).mp3", "Sentence": "And so just like that, we notice we have all three angles are the same in both of these triangles, or they're not all the same, but all of the corresponding angles, I should say, are the same. Angle, this blue angle has the same measure as this blue angle. This magenta angle has the same measure as this magenta angle. And then the other angles are right angles. These are right triangles here. So we could say triangle AEB, triangle AEB, is similar, similar, is similar to triangle DEC, triangle DEC, by, and we could say by angle, angle, angle. All the corresponding angles are congruent."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy (2).mp3", "Sentence": "And then the other angles are right angles. These are right triangles here. So we could say triangle AEB, triangle AEB, is similar, similar, is similar to triangle DEC, triangle DEC, by, and we could say by angle, angle, angle. All the corresponding angles are congruent. So we are dealing with similar triangles. And so we know similar triangles, the ratio of corresponding sides are going to be the same. So we could say that the ratio of, let's say the ratio of BE, the ratio of BE, let me write this down."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy (2).mp3", "Sentence": "All the corresponding angles are congruent. So we are dealing with similar triangles. And so we know similar triangles, the ratio of corresponding sides are going to be the same. So we could say that the ratio of, let's say the ratio of BE, the ratio of BE, let me write this down. This is this side right over here. The ratio of BE to AE, to AE, to AE, is going to be equal to, so that side over that side, well what is the corresponding side? The corresponding side to BE is side CE."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy (2).mp3", "Sentence": "So we could say that the ratio of, let's say the ratio of BE, the ratio of BE, let me write this down. This is this side right over here. The ratio of BE to AE, to AE, to AE, is going to be equal to, so that side over that side, well what is the corresponding side? The corresponding side to BE is side CE. So that's going to be the same as the ratio between CE and DE, and DE. And this just comes out of similar, the similarity of the triangles, CE to DE. So once again, once we establish these triangles are similar we can say the ratio of corresponding sides are going to be the same."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy (2).mp3", "Sentence": "The corresponding side to BE is side CE. So that's going to be the same as the ratio between CE and DE, and DE. And this just comes out of similar, the similarity of the triangles, CE to DE. So once again, once we establish these triangles are similar we can say the ratio of corresponding sides are going to be the same. Now what is the ratio between BE and AE? The ratio between BE and AE. Well that is the slope of this top line right over here."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy (2).mp3", "Sentence": "So once again, once we establish these triangles are similar we can say the ratio of corresponding sides are going to be the same. Now what is the ratio between BE and AE? The ratio between BE and AE. Well that is the slope of this top line right over here. We could say that's the slope of line AB. Slope, slope of line connecting, connecting A to B. Or actually let me just use, I could write it like this."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy (2).mp3", "Sentence": "Well that is the slope of this top line right over here. We could say that's the slope of line AB. Slope, slope of line connecting, connecting A to B. Or actually let me just use, I could write it like this. That is slope of, slope of A, slope of line AB. Remember slope is, when you're going from A to B, it's change in Y over change in X. So when you're going from A to B, your change in X is AE, and your change in Y is BE, or EB, however you want to refer to it."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy (2).mp3", "Sentence": "Or actually let me just use, I could write it like this. That is slope of, slope of A, slope of line AB. Remember slope is, when you're going from A to B, it's change in Y over change in X. So when you're going from A to B, your change in X is AE, and your change in Y is BE, or EB, however you want to refer to it. So this right over here is change in Y, and this over here is change in X. Well now let's look at this second expression right over here, CE over DE. CE over DE."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy (2).mp3", "Sentence": "So when you're going from A to B, your change in X is AE, and your change in Y is BE, or EB, however you want to refer to it. So this right over here is change in Y, and this over here is change in X. Well now let's look at this second expression right over here, CE over DE. CE over DE. Well now this is going to be change in Y over change in X between point C and D. So this is, this right over here, this is the slope of line, of line CD. And so just like that, by establishing similarity, we were able to see the ratio of corresponding sides are congruent, which shows us that the slopes of these two lines are going to be the same. And we are done."}, {"video_title": "Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3", "Sentence": "And what I want to prove is that the sum of the measures of the interior angles of a triangle, that x plus y plus z, is equal to 180 degrees. And the way that I'm going to do it is using our knowledge of parallel lines or transversals of parallel lines and corresponding angles. And to do that, I'm going to extend each of these sides of the triangle, which right now are line segments, but extend them into lines. So this side down here, if I keep going on and on forever in the same directions, then now all of a sudden I have an orange line. And what I want to do is construct another line that is parallel to the orange line that goes through this vertex of the triangle right over here. And I can always do that. I can just start from this point and go in the same direction as this line, and I will never intersect."}, {"video_title": "Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3", "Sentence": "So this side down here, if I keep going on and on forever in the same directions, then now all of a sudden I have an orange line. And what I want to do is construct another line that is parallel to the orange line that goes through this vertex of the triangle right over here. And I can always do that. I can just start from this point and go in the same direction as this line, and I will never intersect. I'm not getting any closer or further away from that line. So I'm never going to intersect that line. So these two lines right over here are parallel."}, {"video_title": "Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3", "Sentence": "I can just start from this point and go in the same direction as this line, and I will never intersect. I'm not getting any closer or further away from that line. So I'm never going to intersect that line. So these two lines right over here are parallel. This is parallel to that. Now I'm going to go to the other two sides of my original triangle and extend them into lines. So I'm going to extend this one into a line."}, {"video_title": "Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3", "Sentence": "So these two lines right over here are parallel. This is parallel to that. Now I'm going to go to the other two sides of my original triangle and extend them into lines. So I'm going to extend this one into a line. So I'll do that as neatly as I can. So I'm going to extend that into a line. And you see that this is clearly a transversal of these two parallel lines."}, {"video_title": "Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3", "Sentence": "So I'm going to extend this one into a line. So I'll do that as neatly as I can. So I'm going to extend that into a line. And you see that this is clearly a transversal of these two parallel lines. Now, if we have a transversal here of two parallel lines, then we must have some corresponding angles. And we see that this angle is formed when the transversal intersects the bottom orange line. Well, what's the corresponding angle when the transversal intersects this top blue line?"}, {"video_title": "Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3", "Sentence": "And you see that this is clearly a transversal of these two parallel lines. Now, if we have a transversal here of two parallel lines, then we must have some corresponding angles. And we see that this angle is formed when the transversal intersects the bottom orange line. Well, what's the corresponding angle when the transversal intersects this top blue line? What's the angle on the top right of the intersection? Angle on the top right of the intersection must also be x. The other thing that pops out at you is there's another vertical angle with x, another angle that must be equivalent."}, {"video_title": "Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3", "Sentence": "Well, what's the corresponding angle when the transversal intersects this top blue line? What's the angle on the top right of the intersection? Angle on the top right of the intersection must also be x. The other thing that pops out at you is there's another vertical angle with x, another angle that must be equivalent. On the opposite side of this intersection, you have this angle right over here. These two angles are vertical. So if this has measure x, then this one must have measure x as well."}, {"video_title": "Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3", "Sentence": "The other thing that pops out at you is there's another vertical angle with x, another angle that must be equivalent. On the opposite side of this intersection, you have this angle right over here. These two angles are vertical. So if this has measure x, then this one must have measure x as well. Let's do the same thing with the last side of the triangle that we have not extended into a line yet. So let's do that. So if we take this one, so we just keep going."}, {"video_title": "Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3", "Sentence": "So if this has measure x, then this one must have measure x as well. Let's do the same thing with the last side of the triangle that we have not extended into a line yet. So let's do that. So if we take this one, so we just keep going. So it becomes a line. So now it becomes a transversal of the two parallel lines, just like the magenta line did. And we say, hey, look, this angle y right over here, this angle is formed from the intersection of the transversal and the bottom parallel line."}, {"video_title": "Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3", "Sentence": "So if we take this one, so we just keep going. So it becomes a line. So now it becomes a transversal of the two parallel lines, just like the magenta line did. And we say, hey, look, this angle y right over here, this angle is formed from the intersection of the transversal and the bottom parallel line. What angle does it correspond to up here? Well, this is kind of on the left side of the intersection. It corresponds to this angle right over here, where the green line, the green transversal intersects the blue parallel line."}, {"video_title": "Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3", "Sentence": "And we say, hey, look, this angle y right over here, this angle is formed from the intersection of the transversal and the bottom parallel line. What angle does it correspond to up here? Well, this is kind of on the left side of the intersection. It corresponds to this angle right over here, where the green line, the green transversal intersects the blue parallel line. Well, what angle is vertical to it? Well, this angle. So this is going to have measure y as well."}, {"video_title": "Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3", "Sentence": "It corresponds to this angle right over here, where the green line, the green transversal intersects the blue parallel line. Well, what angle is vertical to it? Well, this angle. So this is going to have measure y as well. So now we're really at the home stretch of our proof, because we will see that the measure, we have this angle and this angle. This has measure angle x. This has measure z."}, {"video_title": "Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3", "Sentence": "So this is going to have measure y as well. So now we're really at the home stretch of our proof, because we will see that the measure, we have this angle and this angle. This has measure angle x. This has measure z. They're both adjacent angles. If we take the two outer rays that form the angle, and we think about this angle right over here, what's this measure of this wide angle right over there? Well, it's going to be x plus z."}, {"video_title": "Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3", "Sentence": "This has measure z. They're both adjacent angles. If we take the two outer rays that form the angle, and we think about this angle right over here, what's this measure of this wide angle right over there? Well, it's going to be x plus z. And that angle is supplementary to this angle right over here that has measure y. So the measure of this wide angle, which is x plus z, plus the measure of this magenta angle, which is y, must be equal to 180 degrees, because these two angles are supplementary. So the measure of the wide angle, x plus z plus the measure of the magenta angle, which is supplementary to the wide angle, it must be equal to 180 degrees, because they are supplementary."}, {"video_title": "Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3", "Sentence": "Well, it's going to be x plus z. And that angle is supplementary to this angle right over here that has measure y. So the measure of this wide angle, which is x plus z, plus the measure of this magenta angle, which is y, must be equal to 180 degrees, because these two angles are supplementary. So the measure of the wide angle, x plus z plus the measure of the magenta angle, which is supplementary to the wide angle, it must be equal to 180 degrees, because they are supplementary. Well, we could just reorder this if we want to put it in alphabetical order, but we've just completed our proof. The measure of the interior angles of the triangle, x plus z plus y. We could write this as x plus y plus z, if the lack of alphabetical order is making you uncomfortable."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "I will now do a proof for which we credit the 12th century Indian mathematician Bhaskara. So what we're gonna do is we're gonna start with a square. So let me see if I can draw a square. I'm gonna draw it tilted at a bit of an angle, just because I think it'll make it a little bit easier on me. So let me, my best attempt at drawing something that reasonably looks like a square. You have to bear with me if it's not exactly a tilted square. So that looks pretty good."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "I'm gonna draw it tilted at a bit of an angle, just because I think it'll make it a little bit easier on me. So let me, my best attempt at drawing something that reasonably looks like a square. You have to bear with me if it's not exactly a tilted square. So that looks pretty good. And I'm assuming it's a square. So this is a right angle, this is a right angle, that's a right angle, that's a right angle. I'm assuming the lengths of all of these sides are the same."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "So that looks pretty good. And I'm assuming it's a square. So this is a right angle, this is a right angle, that's a right angle, that's a right angle. I'm assuming the lengths of all of these sides are the same. So let's just assume that they're all of length C. I'll write that in yellow. So all of the sides of the square are of length C. And now I'm gonna construct four triangles inside of this square. And the way I'm gonna do it is I'm going to be dropping, so here I'm gonna go straight down."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "I'm assuming the lengths of all of these sides are the same. So let's just assume that they're all of length C. I'll write that in yellow. So all of the sides of the square are of length C. And now I'm gonna construct four triangles inside of this square. And the way I'm gonna do it is I'm going to be dropping, so here I'm gonna go straight down. I'm gonna drop a line straight down and draw a triangle that looks like this. So I'm gonna go straight down here. Here, I'm gonna go straight across."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "And the way I'm gonna do it is I'm going to be dropping, so here I'm gonna go straight down. I'm gonna drop a line straight down and draw a triangle that looks like this. So I'm gonna go straight down here. Here, I'm gonna go straight across. And so since this is straight down this is straight across, we know that this is a right angle. Then from this vertex on our square, I'm going to go straight up. And since this is straight up and this is straight across, we know that this is a right angle."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "Here, I'm gonna go straight across. And so since this is straight down this is straight across, we know that this is a right angle. Then from this vertex on our square, I'm going to go straight up. And since this is straight up and this is straight across, we know that this is a right angle. And then from this vertex right over here, I'm going to go straight horizontally. I'm assuming that's what I'm doing. And so we know that this is going to be a right angle, and then we know this is going to be a right angle."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "And since this is straight up and this is straight across, we know that this is a right angle. And then from this vertex right over here, I'm going to go straight horizontally. I'm assuming that's what I'm doing. And so we know that this is going to be a right angle, and then we know this is going to be a right angle. So we see that we've constructed, from our square, we've constructed four right triangles. and in between we have something that at minimum looks like a rectangle or possibly a square. We haven't quite proven to ourselves yet that this is a square."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "And so we know that this is going to be a right angle, and then we know this is going to be a right angle. So we see that we've constructed, from our square, we've constructed four right triangles. and in between we have something that at minimum looks like a rectangle or possibly a square. We haven't quite proven to ourselves yet that this is a square. Now the next thing I want to think about is whether these triangles are congruent. So they definitely all have the same length of their hypotenuse. All of the hypotenuses, I don't know what the plural of hypotenuse is, hypotenai, hypotenuses, they have all length c. The side opposite the right angle is always length c. So if we can show that all the corresponding angles are the same, then we know it's congruent."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "We haven't quite proven to ourselves yet that this is a square. Now the next thing I want to think about is whether these triangles are congruent. So they definitely all have the same length of their hypotenuse. All of the hypotenuses, I don't know what the plural of hypotenuse is, hypotenai, hypotenuses, they have all length c. The side opposite the right angle is always length c. So if we can show that all the corresponding angles are the same, then we know it's congruent. If you have something where all the angles are the same and you have a side that is also the corresponding side that's also congruent, then the whole triangles are congruent. And we can show that if we assume that this angle is theta, then this angle right over here has to be 90 minus theta because together they are complementary. We know that because they combine to form this angle of the square, this right angle."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "All of the hypotenuses, I don't know what the plural of hypotenuse is, hypotenai, hypotenuses, they have all length c. The side opposite the right angle is always length c. So if we can show that all the corresponding angles are the same, then we know it's congruent. If you have something where all the angles are the same and you have a side that is also the corresponding side that's also congruent, then the whole triangles are congruent. And we can show that if we assume that this angle is theta, then this angle right over here has to be 90 minus theta because together they are complementary. We know that because they combine to form this angle of the square, this right angle. If this is 90 minus theta, we know this angle and this angle have to add up to 90 because we only have 90 left when we subtract the right angle from 180. So we know this has to be theta. And if that's theta, then that's 90 minus theta."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "We know that because they combine to form this angle of the square, this right angle. If this is 90 minus theta, we know this angle and this angle have to add up to 90 because we only have 90 left when we subtract the right angle from 180. So we know this has to be theta. And if that's theta, then that's 90 minus theta. I think you see where this is going. If that's 90 minus theta, this has to be theta. And if that's theta, then this is 90 minus theta."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "And if that's theta, then that's 90 minus theta. I think you see where this is going. If that's 90 minus theta, this has to be theta. And if that's theta, then this is 90 minus theta. If this is 90 minus theta, then this is theta. And then this would have to be 90 minus theta. So we see in all four of these triangles, the three angles are theta, 90 minus theta, and 90 degrees."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "And if that's theta, then this is 90 minus theta. If this is 90 minus theta, then this is theta. And then this would have to be 90 minus theta. So we see in all four of these triangles, the three angles are theta, 90 minus theta, and 90 degrees. So they all have the same exact angle. So at minimum, they are similar. And their hypotenuses are the same."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "So we see in all four of these triangles, the three angles are theta, 90 minus theta, and 90 degrees. So they all have the same exact angle. So at minimum, they are similar. And their hypotenuses are the same. So we know that all four of these triangles are completely congruent triangles. So with that assumption, let's just assume that the longer side of these triangles, that these are of length b. So the longer side of these triangles, I'm just going to assume."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "And their hypotenuses are the same. So we know that all four of these triangles are completely congruent triangles. So with that assumption, let's just assume that the longer side of these triangles, that these are of length b. So the longer side of these triangles, I'm just going to assume. So this length right over here, I'll call that lowercase b. And let's assume that the shorter side, so this distance right over here, this distance right over here, this distance right over here, that these are all this distance right over here, that these are of length a. So if I were to say this height right over here, this height is of length a."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "So the longer side of these triangles, I'm just going to assume. So this length right over here, I'll call that lowercase b. And let's assume that the shorter side, so this distance right over here, this distance right over here, this distance right over here, that these are all this distance right over here, that these are of length a. So if I were to say this height right over here, this height is of length a. Now we will do something interesting. Well, first, let's think about the area of the entire square. What's the area of the entire square in terms of c?"}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "So if I were to say this height right over here, this height is of length a. Now we will do something interesting. Well, first, let's think about the area of the entire square. What's the area of the entire square in terms of c? Well, that's pretty straightforward. It's a c by c square. So the area here is equal to c squared."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "What's the area of the entire square in terms of c? Well, that's pretty straightforward. It's a c by c square. So the area here is equal to c squared. Now what I'm going to do is rearrange two of these triangles and then come up with the area of that other figure in terms of a's and b's. And hopefully it gets us to the Pythagorean theorem. And to do that, just so we don't lose our starting point, because our starting point is interesting, let me just copy and paste this entire thing."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "So the area here is equal to c squared. Now what I'm going to do is rearrange two of these triangles and then come up with the area of that other figure in terms of a's and b's. And hopefully it gets us to the Pythagorean theorem. And to do that, just so we don't lose our starting point, because our starting point is interesting, let me just copy and paste this entire thing. So I don't want it to clip off. So let me just copy and paste this. Copy and paste."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "And to do that, just so we don't lose our starting point, because our starting point is interesting, let me just copy and paste this entire thing. So I don't want it to clip off. So let me just copy and paste this. Copy and paste. So this is our original diagram. And what I will now do, and actually let me clear that out, clear. I'm now going to shift."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "Copy and paste. So this is our original diagram. And what I will now do, and actually let me clear that out, clear. I'm now going to shift. This is the fun part. I'm going to shift this triangle here in the top left. I'm going to shift it below this triangle on the bottom right."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "I'm now going to shift. This is the fun part. I'm going to shift this triangle here in the top left. I'm going to shift it below this triangle on the bottom right. And I'm going to attempt to do that by copying and pasting. So let's see how much. Well, the way I drew it, it's not that easy."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "I'm going to shift it below this triangle on the bottom right. And I'm going to attempt to do that by copying and pasting. So let's see how much. Well, the way I drew it, it's not that easy. Well, that might do the trick. I want to retain a little bit of the. So let me copy, or let me actually cut it, and then let me paste it."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "Well, the way I drew it, it's not that easy. Well, that might do the trick. I want to retain a little bit of the. So let me copy, or let me actually cut it, and then let me paste it. So that triangle, I'm going to stick right over there. Stick that right over there. And let me draw in the lines that I just erased."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "So let me copy, or let me actually cut it, and then let me paste it. So that triangle, I'm going to stick right over there. Stick that right over there. And let me draw in the lines that I just erased. So just to be clear, we had a line over there. And we also had this right over here. And this was straight up and down."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "And let me draw in the lines that I just erased. So just to be clear, we had a line over there. And we also had this right over here. And this was straight up and down. And these were straight side to side. Now, so I moved this part over down here. So I moved that over down there."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "And this was straight up and down. And these were straight side to side. Now, so I moved this part over down here. So I moved that over down there. And now I'm going to move this top right triangle down to the bottom left. So I'm just rearranging the exact same area. So let me just capture the whole thing as best as I can."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "So I moved that over down there. And now I'm going to move this top right triangle down to the bottom left. So I'm just rearranging the exact same area. So let me just capture the whole thing as best as I can. So let me cut, and then let me paste. And I'm going to move it right over here. And while I went through that process, I kind of lost its floor."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "So let me just capture the whole thing as best as I can. So let me cut, and then let me paste. And I'm going to move it right over here. And while I went through that process, I kind of lost its floor. So let me redraw the floor. So I just moved it right over here. So this thing, this triangle, let me color it in, is now right over there."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "And while I went through that process, I kind of lost its floor. So let me redraw the floor. So I just moved it right over here. So this thing, this triangle, let me color it in, is now right over there. And this triangle is now right over here. That center square, and it is a square, is now right over here. So hopefully, you can appreciate how we rearranged it."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "So this thing, this triangle, let me color it in, is now right over there. And this triangle is now right over here. That center square, and it is a square, is now right over here. So hopefully, you can appreciate how we rearranged it. Now, my question for you is, how can we express the area of this new figure, which has the exact same area as the old figure? I just shifted parts of it around. How can we express this in terms of a's and b's?"}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "So hopefully, you can appreciate how we rearranged it. Now, my question for you is, how can we express the area of this new figure, which has the exact same area as the old figure? I just shifted parts of it around. How can we express this in terms of a's and b's? Well, the key insight here is to recognize the length of this bottom side. What's the length of this bottom side right over here? The length of this bottom side, well, this length right over here is b."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "How can we express this in terms of a's and b's? Well, the key insight here is to recognize the length of this bottom side. What's the length of this bottom side right over here? The length of this bottom side, well, this length right over here is b. This length right over here is a. So the length of this entire bottom is a plus b. Well, that by itself is kind of interesting."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "The length of this bottom side, well, this length right over here is b. This length right over here is a. So the length of this entire bottom is a plus b. Well, that by itself is kind of interesting. But what we can realize is that this length right over here, this length right over here, which is the exact same thing as this length over here, was also a. So we can construct an a by a square. So we can construct an a by a square."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "Well, that by itself is kind of interesting. But what we can realize is that this length right over here, this length right over here, which is the exact same thing as this length over here, was also a. So we can construct an a by a square. So we can construct an a by a square. So this square right over here is a by a. And so it has area a squared. Let me do that in a color that you can actually see."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "So we can construct an a by a square. So this square right over here is a by a. And so it has area a squared. Let me do that in a color that you can actually see. So this has area of a squared. And then what's the area of what's left over? Well, if this is length a, then this is length a as well."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "Let me do that in a color that you can actually see. So this has area of a squared. And then what's the area of what's left over? Well, if this is length a, then this is length a as well. If this entire bottom is a plus b, then we know that what's left over after subtracting the a out has to be b. If this whole thing is a plus b, this is a, then this right over here is b. And so the rest of this newly oriented figure, this new figure, everything that I'm shading in over here, this is just a b by b square."}, {"video_title": "Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3", "Sentence": "Well, if this is length a, then this is length a as well. If this entire bottom is a plus b, then we know that what's left over after subtracting the a out has to be b. If this whole thing is a plus b, this is a, then this right over here is b. And so the rest of this newly oriented figure, this new figure, everything that I'm shading in over here, this is just a b by b square. So the area here is b squared. So the entire area of this figure is a squared plus b squared, which, lucky for us, is equal to the area of this expressed in terms of c, because they're the exact same figure, just rearranged. So it's going to be equal to c squared."}, {"video_title": "Reflecting segments over line Transformations Geometry Khan Academy.mp3", "Sentence": "Line segments IN, this is segment IN over here, and TO, this is TO here, are reflected over the line Y is equal to negative X minus two. So this is the line that they're reflected about, this dashed purple line, and it is indeed Y equals negative X minus two. This right over here is in slope intercept form. The slope should be negative one, and we see that the slope of this purple line is indeed negative one. If X changes by a certain amount, Y changes by the negative of that. If X changes by one, Y changes by negative one to get back to that line. If X changes by positive two, Y changes by negative two to get back to another point on that line."}, {"video_title": "Reflecting segments over line Transformations Geometry Khan Academy.mp3", "Sentence": "The slope should be negative one, and we see that the slope of this purple line is indeed negative one. If X changes by a certain amount, Y changes by the negative of that. If X changes by one, Y changes by negative one to get back to that line. If X changes by positive two, Y changes by negative two to get back to another point on that line. And the Y intercept, we see when X is equal to zero, Y should be negative two. When X is equal to zero, Y is indeed negative two. So we validated that."}, {"video_title": "Reflecting segments over line Transformations Geometry Khan Academy.mp3", "Sentence": "If X changes by positive two, Y changes by negative two to get back to another point on that line. And the Y intercept, we see when X is equal to zero, Y should be negative two. When X is equal to zero, Y is indeed negative two. So we validated that. Now they say draw the image of this reflection using the interactive graph. All right, so we can move these lines around, and we wanna reflect these, and I could try to eyeball it, you know, maybe it's something like this. I don't know, this doesn't seem exactly right."}, {"video_title": "Reflecting segments over line Transformations Geometry Khan Academy.mp3", "Sentence": "So we validated that. Now they say draw the image of this reflection using the interactive graph. All right, so we can move these lines around, and we wanna reflect these, and I could try to eyeball it, you know, maybe it's something like this. I don't know, this doesn't seem exactly right. That looks close to the reflection of IN, and for TO, I'd wanna move this down here. TO looks like it would be, I don't know, I'm eyeballing it. This is close, but I can't be close."}, {"video_title": "Reflecting segments over line Transformations Geometry Khan Academy.mp3", "Sentence": "I don't know, this doesn't seem exactly right. That looks close to the reflection of IN, and for TO, I'd wanna move this down here. TO looks like it would be, I don't know, I'm eyeballing it. This is close, but I can't be close. I wanna get exact. So let's, I've copied and pasted the original problem on my scratch pad, so we can find the exact points, and so I don't just have to estimate this. So let's go to the scratch pad."}, {"video_title": "Reflecting segments over line Transformations Geometry Khan Academy.mp3", "Sentence": "This is close, but I can't be close. I wanna get exact. So let's, I've copied and pasted the original problem on my scratch pad, so we can find the exact points, and so I don't just have to estimate this. So let's go to the scratch pad. So exactly what we just saw. And the main realization is, is if we wanna reflect a given point, if we wanna reflect a given point, say point I right over here, what we wanna do is we wanna drop a perpendicular. We wanna find a line that's perpendicular, a line that has the point I on it, and it's perpendicular to this line right over here."}, {"video_title": "Reflecting segments over line Transformations Geometry Khan Academy.mp3", "Sentence": "So let's go to the scratch pad. So exactly what we just saw. And the main realization is, is if we wanna reflect a given point, if we wanna reflect a given point, say point I right over here, what we wanna do is we wanna drop a perpendicular. We wanna find a line that's perpendicular, a line that has the point I on it, and it's perpendicular to this line right over here. Remember, this is the line, let me do this in the purple color. This is the line Y is equal to negative X minus two. Its slope is equal to negative one."}, {"video_title": "Reflecting segments over line Transformations Geometry Khan Academy.mp3", "Sentence": "We wanna find a line that's perpendicular, a line that has the point I on it, and it's perpendicular to this line right over here. Remember, this is the line, let me do this in the purple color. This is the line Y is equal to negative X minus two. Its slope is equal to negative one. So I wanna align that goes through I, point I, that is perpendicular to this line, and I wanna drop it to, I wanna drop it to the line that I'm gonna reflect on, and then I wanna go the same distance onto the other side to find the corresponding point in the image. So how do I do that? Well, if this line, if this purple line has a slope of negative one, a line that is perpendicular to it, a line that is perpendicular to it, so this thing that I'm drawing in purple right over here, its slope is going to be the negative reciprocal of this."}, {"video_title": "Reflecting segments over line Transformations Geometry Khan Academy.mp3", "Sentence": "Its slope is equal to negative one. So I wanna align that goes through I, point I, that is perpendicular to this line, and I wanna drop it to, I wanna drop it to the line that I'm gonna reflect on, and then I wanna go the same distance onto the other side to find the corresponding point in the image. So how do I do that? Well, if this line, if this purple line has a slope of negative one, a line that is perpendicular to it, a line that is perpendicular to it, so this thing that I'm drawing in purple right over here, its slope is going to be the negative reciprocal of this. So the reciprocal of negative one is still just negative one. One over negative one is still negative one, but we want the negative of that. So the slope here needs to be one, and luckily, that's how I drew it."}, {"video_title": "Reflecting segments over line Transformations Geometry Khan Academy.mp3", "Sentence": "Well, if this line, if this purple line has a slope of negative one, a line that is perpendicular to it, a line that is perpendicular to it, so this thing that I'm drawing in purple right over here, its slope is going to be the negative reciprocal of this. So the reciprocal of negative one is still just negative one. One over negative one is still negative one, but we want the negative of that. So the slope here needs to be one, and luckily, that's how I drew it. The slope here needs to be equal to one, which is however much I change in the X direction, I change in the Y direction. We see that. To go from this point to this point right over here, we decrease Y by four, and we decrease X by four."}, {"video_title": "Reflecting segments over line Transformations Geometry Khan Academy.mp3", "Sentence": "So the slope here needs to be one, and luckily, that's how I drew it. The slope here needs to be equal to one, which is however much I change in the X direction, I change in the Y direction. We see that. To go from this point to this point right over here, we decrease Y by four, and we decrease X by four. Now, if we wanna stay on this line to find the reflection, we just do the same thing. We could decrease X by four, so we'll go from negative two to negative six, and decrease Y by four, and we end up at this point right over here. So we end up at the point, this is X equals negative six, Y is equal to negative four."}, {"video_title": "Reflecting segments over line Transformations Geometry Khan Academy.mp3", "Sentence": "To go from this point to this point right over here, we decrease Y by four, and we decrease X by four. Now, if we wanna stay on this line to find the reflection, we just do the same thing. We could decrease X by four, so we'll go from negative two to negative six, and decrease Y by four, and we end up at this point right over here. So we end up at the point, this is X equals negative six, Y is equal to negative four. So this point corresponds to this point right over there. Now, let's do the same thing for point N. For point N, we already know if we drop a perpendicular, if this is perpendicular, it's going to have a slope of one, because this purple line has a slope of negative one, and the negative reciprocal of negative one is positive one. And let's see, to go from this point to this point of intersection, we have to go down one and a half."}, {"video_title": "Reflecting segments over line Transformations Geometry Khan Academy.mp3", "Sentence": "So we end up at the point, this is X equals negative six, Y is equal to negative four. So this point corresponds to this point right over there. Now, let's do the same thing for point N. For point N, we already know if we drop a perpendicular, if this is perpendicular, it's going to have a slope of one, because this purple line has a slope of negative one, and the negative reciprocal of negative one is positive one. And let's see, to go from this point to this point of intersection, we have to go down one and a half. We're going down one and a half, and we're going to the left one and a half. So we wanna do that on the other side. We wanna stay on this perpendicular line, so we wanna go left one and a half, and down one and a half."}, {"video_title": "Reflecting segments over line Transformations Geometry Khan Academy.mp3", "Sentence": "And let's see, to go from this point to this point of intersection, we have to go down one and a half. We're going down one and a half, and we're going to the left one and a half. So we wanna do that on the other side. We wanna stay on this perpendicular line, so we wanna go left one and a half, and down one and a half. And we get to this point right over here, which is the point X equals three, Y is equal to negative eight. And so we are now equidistant. We're on this perpendicular line still, but we're equidistant on the other side."}, {"video_title": "Reflecting segments over line Transformations Geometry Khan Academy.mp3", "Sentence": "We wanna stay on this perpendicular line, so we wanna go left one and a half, and down one and a half. And we get to this point right over here, which is the point X equals three, Y is equal to negative eight. And so we are now equidistant. We're on this perpendicular line still, but we're equidistant on the other side. So the image of IN is gonna go through negative six comma negative four and three comma negative eight. So let me draw that. So let me see if I can remember."}, {"video_title": "Reflecting segments over line Transformations Geometry Khan Academy.mp3", "Sentence": "We're on this perpendicular line still, but we're equidistant on the other side. So the image of IN is gonna go through negative six comma negative four and three comma negative eight. So let me draw that. So let me see if I can remember. Negative six, negative four, three comma negative eight. So I have a bad memory. So negative six, negative four, and three comma negative eight."}, {"video_title": "Reflecting segments over line Transformations Geometry Khan Academy.mp3", "Sentence": "So let me see if I can remember. Negative six, negative four, three comma negative eight. So I have a bad memory. So negative six, negative four, and three comma negative eight. And I was close when I estimated, but I wasn't exactly right. So that's looking pretty good. And then actually we can do the exact same thing with points T and point O."}, {"video_title": "Reflecting segments over line Transformations Geometry Khan Academy.mp3", "Sentence": "So negative six, negative four, and three comma negative eight. And I was close when I estimated, but I wasn't exactly right. So that's looking pretty good. And then actually we can do the exact same thing with points T and point O. Now let me do that. So point T, to get from point T to the line in the shortest distance, once again we drop a perpendicular. This line is gonna have a slope of one because it's perpendicular to the line that has a slope of negative one."}, {"video_title": "Reflecting segments over line Transformations Geometry Khan Academy.mp3", "Sentence": "And then actually we can do the exact same thing with points T and point O. Now let me do that. So point T, to get from point T to the line in the shortest distance, once again we drop a perpendicular. This line is gonna have a slope of one because it's perpendicular to the line that has a slope of negative one. And so to get there, we have to decrease our X by, we have to decrease our X. We're going from X equals five to X equals, looks like half. So X went down by four and a half in the X direction, and Y also needs to go down by four and a half."}, {"video_title": "Reflecting segments over line Transformations Geometry Khan Academy.mp3", "Sentence": "This line is gonna have a slope of one because it's perpendicular to the line that has a slope of negative one. And so to get there, we have to decrease our X by, we have to decrease our X. We're going from X equals five to X equals, looks like half. So X went down by four and a half in the X direction, and Y also needs to go down by four and a half. So if we wanna stay on that line, let's decrease our X by four and a half. So that's half, one, two, three, four. And Y needs to go down by four and a half."}, {"video_title": "Reflecting segments over line Transformations Geometry Khan Academy.mp3", "Sentence": "So X went down by four and a half in the X direction, and Y also needs to go down by four and a half. So if we wanna stay on that line, let's decrease our X by four and a half. So that's half, one, two, three, four. And Y needs to go down by four and a half. So that's half, one, two, three, four. And we get to this point right over here, which is the point X equals negative four, Y is equal to negative seven. Negative four comma negative seven."}, {"video_title": "Reflecting segments over line Transformations Geometry Khan Academy.mp3", "Sentence": "And Y needs to go down by four and a half. So that's half, one, two, three, four. And we get to this point right over here, which is the point X equals negative four, Y is equal to negative seven. Negative four comma negative seven. So this should be at X equals negative four, Y equals negative seven. And there's a couple of things you could do here. You could just say, hey, this is too long, or two units long, not like too long in length somehow."}, {"video_title": "Reflecting segments over line Transformations Geometry Khan Academy.mp3", "Sentence": "Negative four comma negative seven. So this should be at X equals negative four, Y equals negative seven. And there's a couple of things you could do here. You could just say, hey, this is too long, or two units long, not like too long in length somehow. This is two units long, so maybe this is two units long. So this is feeling pretty good. But let's just go through the exercise for point O as well."}, {"video_title": "Reflecting segments over line Transformations Geometry Khan Academy.mp3", "Sentence": "You could just say, hey, this is too long, or two units long, not like too long in length somehow. This is two units long, so maybe this is two units long. So this is feeling pretty good. But let's just go through the exercise for point O as well. So point O, once again, this is going to have, if we drop a perpendicular, it's gonna have a slope of one. So whatever our change in X between this point and this point, we're gonna have the same change in Y. And our change in X to go from seven to one and a half, our change in X is five and a half."}, {"video_title": "Reflecting segments over line Transformations Geometry Khan Academy.mp3", "Sentence": "But let's just go through the exercise for point O as well. So point O, once again, this is going to have, if we drop a perpendicular, it's gonna have a slope of one. So whatever our change in X between this point and this point, we're gonna have the same change in Y. And our change in X to go from seven to one and a half, our change in X is five and a half. So let me do it this way. So the change in X here, so the change in X is equal to negative five and a half. 5.5, if you subtract 5.5 from seven, you get to 1.5."}, {"video_title": "Reflecting segments over line Transformations Geometry Khan Academy.mp3", "Sentence": "And our change in X to go from seven to one and a half, our change in X is five and a half. So let me do it this way. So the change in X here, so the change in X is equal to negative five and a half. 5.5, if you subtract 5.5 from seven, you get to 1.5. And our change in Y, our change in Y, is also negative 5.5. Change in Y is negative, you're not gonna see that, so I'm gonna do it a different way, negative 5.5. And so we need to stay on this line."}, {"video_title": "Reflecting segments over line Transformations Geometry Khan Academy.mp3", "Sentence": "5.5, if you subtract 5.5 from seven, you get to 1.5. And our change in Y, our change in Y, is also negative 5.5. Change in Y is negative, you're not gonna see that, so I'm gonna do it a different way, negative 5.5. And so we need to stay on this line. So we wanna change by those same amounts onto the other side of that line. So if we decrease our X by five and a half, so half, one, two, three, four, five, we get there. And Y by five and a half, half, one, two, three, four, five, three, four, five, we get to the point X equals negative four, Y is equal to negative nine."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "What I want to do in this video is I want to prove that segment AC is perpendicular to segment DB based on the information that we have in this diagram over here. This side has the same length as that side. This side has the same length as that side. I'll give you a hint. We're going to use one or more of our congruence postulates. I'll just stick with calling them postulates from now on. The ones that we know, let me draw a little line here."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "I'll give you a hint. We're going to use one or more of our congruence postulates. I'll just stick with calling them postulates from now on. The ones that we know, let me draw a little line here. This is kind of our tool kit. We have the side-side-side postulate. If the three sides are congruent, then the two triangles are congruent."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "The ones that we know, let me draw a little line here. This is kind of our tool kit. We have the side-side-side postulate. If the three sides are congruent, then the two triangles are congruent. We have side-angle-side. If the two sides and the angle in between are congruent, then the two triangles are congruent. We have ASA, two angles with a side in between, and then we have AAS, two angles and then a side."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "If the three sides are congruent, then the two triangles are congruent. We have side-angle-side. If the two sides and the angle in between are congruent, then the two triangles are congruent. We have ASA, two angles with a side in between, and then we have AAS, two angles and then a side. Any of these things we've established, these are our postulates. We're going to assume that they imply congruency. I'm also going to do this as what we call a two-column proof."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "We have ASA, two angles with a side in between, and then we have AAS, two angles and then a side. Any of these things we've established, these are our postulates. We're going to assume that they imply congruency. I'm also going to do this as what we call a two-column proof. You don't have to do something as a two-column proof, but this is what you normally see in an introductory geometry class. I thought I would expose you to it. It's a pretty basic idea."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "I'm also going to do this as what we call a two-column proof. You don't have to do something as a two-column proof, but this is what you normally see in an introductory geometry class. I thought I would expose you to it. It's a pretty basic idea. You make a statement, and you just have to give the reason for your statement, which is what we've been doing with any proof, but we haven't always put it in a very structured way. I'm just going to do it like this. I'll have two columns, write this like that, and I'll have a statement, and then I will give the reason for the statement."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "It's a pretty basic idea. You make a statement, and you just have to give the reason for your statement, which is what we've been doing with any proof, but we haven't always put it in a very structured way. I'm just going to do it like this. I'll have two columns, write this like that, and I'll have a statement, and then I will give the reason for the statement. The strategy that I'm going to try to do is it looks like, right off the bat, it seems like I can prove that triangle CDA is congruent to triangle CBA based on side, side, side. That's a pretty good starting point, because once I can base congruency, then I can start to have angles be the same. The reason why I can do that is because this side is the same as that side, this side is the same as that side, and they both share that side."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "I'll have two columns, write this like that, and I'll have a statement, and then I will give the reason for the statement. The strategy that I'm going to try to do is it looks like, right off the bat, it seems like I can prove that triangle CDA is congruent to triangle CBA based on side, side, side. That's a pretty good starting point, because once I can base congruency, then I can start to have angles be the same. The reason why I can do that is because this side is the same as that side, this side is the same as that side, and they both share that side. I don't want to just do it verbally this time. I want to write it out properly in this two-column proof. We have CD, we have the length of segment CD is equal to the length of CB."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "The reason why I can do that is because this side is the same as that side, this side is the same as that side, and they both share that side. I don't want to just do it verbally this time. I want to write it out properly in this two-column proof. We have CD, we have the length of segment CD is equal to the length of CB. CD is equal to CB, and that is given. These two characters have the same length. We also know that DA, the length of segment DA, is the same as the length of segment BA."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "We have CD, we have the length of segment CD is equal to the length of CB. CD is equal to CB, and that is given. These two characters have the same length. We also know that DA, the length of segment DA, is the same as the length of segment BA. DA is equal to BA, that's also given in the diagram. Then we also know that CA is equal to a CA, I guess we could say. CA is equal to itself, and it's obviously in both triangles."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "We also know that DA, the length of segment DA, is the same as the length of segment BA. DA is equal to BA, that's also given in the diagram. Then we also know that CA is equal to a CA, I guess we could say. CA is equal to itself, and it's obviously in both triangles. This is also given, or it's obvious from the diagram. It's a bit obvious, both triangles share that side. We have two triangles, their corresponding sides have the same length, and so we know that they are congruent."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "CA is equal to itself, and it's obviously in both triangles. This is also given, or it's obvious from the diagram. It's a bit obvious, both triangles share that side. We have two triangles, their corresponding sides have the same length, and so we know that they are congruent. We know that triangle CDA is congruent to triangle CBA. We know that by the side-side-side postulate, and the statements given up here. Let me number our statements so we can refer back to this."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "We have two triangles, their corresponding sides have the same length, and so we know that they are congruent. We know that triangle CDA is congruent to triangle CBA. We know that by the side-side-side postulate, and the statements given up here. Let me number our statements so we can refer back to this. 1, 2, 3, and 4. Side-side-side postulate, and 1, 2, and 3. Statements 1, 2, and 3."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "Let me number our statements so we can refer back to this. 1, 2, 3, and 4. Side-side-side postulate, and 1, 2, and 3. Statements 1, 2, and 3. Statements 1, 2, and 3 in the side-side postulate let us know that these two triangles are congruent. If these are congruent, then we know, for example, that all of their corresponding angles are equivalent. For example, this angle is going to be equal to that angle."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "Statements 1, 2, and 3. Statements 1, 2, and 3 in the side-side postulate let us know that these two triangles are congruent. If these are congruent, then we know, for example, that all of their corresponding angles are equivalent. For example, this angle is going to be equal to that angle. Let's make that statement right over there. This angle DCE, this is going to be statement 5. We know that angle DCE, that's this angle right over here, is going to have the same measure."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "For example, this angle is going to be equal to that angle. Let's make that statement right over there. This angle DCE, this is going to be statement 5. We know that angle DCE, that's this angle right over here, is going to have the same measure. We can even say they're congruent. I'll say the measure of angle DCE is going to be equal to the measure of angle BCE. This comes straight out of statement 4."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "We know that angle DCE, that's this angle right over here, is going to have the same measure. We can even say they're congruent. I'll say the measure of angle DCE is going to be equal to the measure of angle BCE. This comes straight out of statement 4. I can put in parentheses, congruency of those triangles. This implies straight, because they're both part of this larger triangle, they are the corresponding angles, so they are going to have the exact same measure. Now it seems like we can do something pretty interesting with these two smaller triangles, at the top left and the top right of this kite-like figure."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "This comes straight out of statement 4. I can put in parentheses, congruency of those triangles. This implies straight, because they're both part of this larger triangle, they are the corresponding angles, so they are going to have the exact same measure. Now it seems like we can do something pretty interesting with these two smaller triangles, at the top left and the top right of this kite-like figure. Because we have a side, two corresponding sides are congruent, two corresponding angles are congruent, and they have a side in common. They have this side in common right over here. Let's first establish that they have this side in common right over here."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "Now it seems like we can do something pretty interesting with these two smaller triangles, at the top left and the top right of this kite-like figure. Because we have a side, two corresponding sides are congruent, two corresponding angles are congruent, and they have a side in common. They have this side in common right over here. Let's first establish that they have this side in common right over here. I'll just write statement 6. We have CE, the measure or the length of that line is equal to itself. Once again, this is just obvious."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "Let's first establish that they have this side in common right over here. I'll just write statement 6. We have CE, the measure or the length of that line is equal to itself. Once again, this is just obvious. It's the same obvious from diagram. It's the same line, obvious from diagram. But now we can use that information."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "Once again, this is just obvious. It's the same obvious from diagram. It's the same line, obvious from diagram. But now we can use that information. We don't have three sides. We haven't proven to ourselves that this side is the same as this side, that DE has the same length as EB, but we do have a side, an angle between the sides, and then another side. This looks pretty interesting for our side-angle-side postulate."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "But now we can use that information. We don't have three sides. We haven't proven to ourselves that this side is the same as this side, that DE has the same length as EB, but we do have a side, an angle between the sides, and then another side. This looks pretty interesting for our side-angle-side postulate. We can say by the side-angle-side postulate that triangle DCE is congruent to triangle BCE. When I write the labels for the triangles, I'm making sure that I'm putting the corresponding points. I started at D, then went to C, then to E. The corresponding angle or the corresponding point or vertex, I could say, for this triangle right over here is B."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "This looks pretty interesting for our side-angle-side postulate. We can say by the side-angle-side postulate that triangle DCE is congruent to triangle BCE. When I write the labels for the triangles, I'm making sure that I'm putting the corresponding points. I started at D, then went to C, then to E. The corresponding angle or the corresponding point or vertex, I could say, for this triangle right over here is B. If I start with D, I start with B. C in the middle is the corresponding vertex for either of these triangles, so I put it in the middle, and then they both go to E. That's just to make sure that we are specifying what's corresponding to what. We know this is true by side-angle-side, and the information we got from this side is established that these two sides are congruent was from statement 1."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "I started at D, then went to C, then to E. The corresponding angle or the corresponding point or vertex, I could say, for this triangle right over here is B. If I start with D, I start with B. C in the middle is the corresponding vertex for either of these triangles, so I put it in the middle, and then they both go to E. That's just to make sure that we are specifying what's corresponding to what. We know this is true by side-angle-side, and the information we got from this side is established that these two sides are congruent was from statement 1. Then that these angles are congruent is from statement 5, right over here. Then statement 6 gave us the other side, statement 6, just like that. If we know that these triangles are congruent, that means that all of their corresponding angles are congruent."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "We know this is true by side-angle-side, and the information we got from this side is established that these two sides are congruent was from statement 1. Then that these angles are congruent is from statement 5, right over here. Then statement 6 gave us the other side, statement 6, just like that. If we know that these triangles are congruent, that means that all of their corresponding angles are congruent. We know, for example, that this angle right over here is going to be congruent to that angle over there. Let's write that down. We know statement 8, the measure of angle, let's call that DEC, is equal to the measure of angle BEC."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "If we know that these triangles are congruent, that means that all of their corresponding angles are congruent. We know, for example, that this angle right over here is going to be congruent to that angle over there. Let's write that down. We know statement 8, the measure of angle, let's call that DEC, is equal to the measure of angle BEC. This comes straight from statement 7. Once again, they're congruent. Then we also know, we'll make statement 9, that the measure of angle DEC, or maybe we should just write it this way, angle DEC and angle BEC are supplementary."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "We know statement 8, the measure of angle, let's call that DEC, is equal to the measure of angle BEC. This comes straight from statement 7. Once again, they're congruent. Then we also know, we'll make statement 9, that the measure of angle DEC, or maybe we should just write it this way, angle DEC and angle BEC are supplementary. They are supplementary, and you can just look at that from inspection, but I'll write it decently. Supplementary, which means their measures add up to 180 degrees. We know that because they are adjacent, and outer sides form straight angle."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "Then we also know, we'll make statement 9, that the measure of angle DEC, or maybe we should just write it this way, angle DEC and angle BEC are supplementary. They are supplementary, and you can just look at that from inspection, but I'll write it decently. Supplementary, which means their measures add up to 180 degrees. We know that because they are adjacent, and outer sides form straight angle. Then we can essentially, the next step, if we know that these two angles are equal to each other, and if we know that they are complementary, our next step means that we can actually deduce that they must be 90 degrees. 10, measure of angle DEC equals measure of angle BEC, which equals 90 degrees. Then for the reason, it might be a little bit more involved, we could put these two statements together."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "We know that because they are adjacent, and outer sides form straight angle. Then we can essentially, the next step, if we know that these two angles are equal to each other, and if we know that they are complementary, our next step means that we can actually deduce that they must be 90 degrees. 10, measure of angle DEC equals measure of angle BEC, which equals 90 degrees. Then for the reason, it might be a little bit more involved, we could put these two statements together. It would be statements 8 and 9. Then statements 8 and 9 mean that DEC, so I could write this, measure of angle DEC plus measure of angle, of angle, actually let me just, since I don't want to do too many steps all at once, let me just take it little bit by little bit. Let me just do it all like this."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "Then for the reason, it might be a little bit more involved, we could put these two statements together. It would be statements 8 and 9. Then statements 8 and 9 mean that DEC, so I could write this, measure of angle DEC plus measure of angle, of angle, actually let me just, since I don't want to do too many steps all at once, let me just take it little bit by little bit. Let me just do it all like this. Let me say measure of angle DEC plus measure of angle BEC is equal to 180, and this comes straight from.9, that they are supplementary. Then we could say statement, I'm taking up a lot of space now, statement 11, we could say measure of angle DEC plus measure of angle DEC is equal to 180 degrees. We know that from statement 9, we know that from statement 9 and statement 8."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "Let me just do it all like this. Let me say measure of angle DEC plus measure of angle BEC is equal to 180, and this comes straight from.9, that they are supplementary. Then we could say statement, I'm taking up a lot of space now, statement 11, we could say measure of angle DEC plus measure of angle DEC is equal to 180 degrees. We know that from statement 9, we know that from statement 9 and statement 8. We essentially just took statement 9 and substituted that BEC, the measure of BEC is the same as the measure of DEC. Then if we want statement 12, we could say measure of angle DEC is equal to 90, which is equal to the measure of angle BEC. Then this comes once again straight out of point number 11 and 8. You can see I'm taking a little bit more time, going a little bit more granular through the steps."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "We know that from statement 9, we know that from statement 9 and statement 8. We essentially just took statement 9 and substituted that BEC, the measure of BEC is the same as the measure of DEC. Then if we want statement 12, we could say measure of angle DEC is equal to 90, which is equal to the measure of angle BEC. Then this comes once again straight out of point number 11 and 8. You can see I'm taking a little bit more time, going a little bit more granular through the steps. Some of the other proofs I would have just said, obviously this implies this or that. Then we're done, because if these are 90 degrees, let me write the last statement, statement 13, which is what we wanted to prove. We wanted to prove that AC is perpendicular to DB."}, {"video_title": "Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3", "Sentence": "You can see I'm taking a little bit more time, going a little bit more granular through the steps. Some of the other proofs I would have just said, obviously this implies this or that. Then we're done, because if these are 90 degrees, let me write the last statement, statement 13, which is what we wanted to prove. We wanted to prove that AC is perpendicular to DB. AC is perpendicular to segment DB, and it comes straight out of point 12. We're done. We've done a two-column proof, and we have proven that this line segment right over here is perpendicular to that line segment right over there."}, {"video_title": "Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "I have a pyramid here on the left and I have a cone here on the right. And we know a few things about these two figures. First of all, they have the exact same height. So this length right over here is H and this length right over here, going from the peak to the center of the base here, is H as well. We also know that the area of the bases is the same. So for example, in this left pyramid, the area of the base would be X times, and let's just assume that it is a square. So X times X."}, {"video_title": "Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So this length right over here is H and this length right over here, going from the peak to the center of the base here, is H as well. We also know that the area of the bases is the same. So for example, in this left pyramid, the area of the base would be X times, and let's just assume that it is a square. So X times X. So the area here is going to be equal to X squared. And the area of the base, so that's area of this base is equal to X squared. And the area of this base right over here would be equal to, area is equal to pi times R squared."}, {"video_title": "Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So X times X. So the area here is going to be equal to X squared. And the area of the base, so that's area of this base is equal to X squared. And the area of this base right over here would be equal to, area is equal to pi times R squared. And I'm saying that these two things are the same. So we also know that X squared is equal to pi R squared. Now, my question to you is, do these two figures have the same volume or is it different?"}, {"video_title": "Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "And the area of this base right over here would be equal to, area is equal to pi times R squared. And I'm saying that these two things are the same. So we also know that X squared is equal to pi R squared. Now, my question to you is, do these two figures have the same volume or is it different? And if they are different, which one has a larger volume? Pause this video and try to think about that. All right, now let's do this together."}, {"video_title": "Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "Now, my question to you is, do these two figures have the same volume or is it different? And if they are different, which one has a larger volume? Pause this video and try to think about that. All right, now let's do this together. Now, given that we're talking about two figures that have the same height and at least the area of the base is the same, you might be thinking that Cavalieri's principle might be useful. Just a reminder of what that is, Cavalieri's principle tells us that if you have two figures and we're thinking in three-dimensional version of Cavalieri's principle, if you have two figures that have the same height and at any point along that height, the cross-sectional area is the same, then the figures have the same volume. So what we need to do is figure out, is it true that at any point in this height, do these figures have the same cross-sectional area?"}, {"video_title": "Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "All right, now let's do this together. Now, given that we're talking about two figures that have the same height and at least the area of the base is the same, you might be thinking that Cavalieri's principle might be useful. Just a reminder of what that is, Cavalieri's principle tells us that if you have two figures and we're thinking in three-dimensional version of Cavalieri's principle, if you have two figures that have the same height and at any point along that height, the cross-sectional area is the same, then the figures have the same volume. So what we need to do is figure out, is it true that at any point in this height, do these figures have the same cross-sectional area? Well, to think about that, let's pick an arbitrary point along this height. And just for simplicity, let's pick halfway along the height, although we could do this analysis at any point along the height. So halfway along the height there, halfway along the height there."}, {"video_title": "Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So what we need to do is figure out, is it true that at any point in this height, do these figures have the same cross-sectional area? Well, to think about that, let's pick an arbitrary point along this height. And just for simplicity, let's pick halfway along the height, although we could do this analysis at any point along the height. So halfway along the height there, halfway along the height there. So this distance right over here, that would be h over two. This distance right over here would be h over two. This whole thing is h. And what we can do is construct what look like similar triangles, and we can even prove it to ourselves that these are similar triangles."}, {"video_title": "Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So halfway along the height there, halfway along the height there. So this distance right over here, that would be h over two. This distance right over here would be h over two. This whole thing is h. And what we can do is construct what look like similar triangles, and we can even prove it to ourselves that these are similar triangles. So let me construct them right over here. And the reason why we know they're similar is that this line is going to be parallel to this line, and that this line is parallel to that line, to that radius. And how do we know that?"}, {"video_title": "Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "This whole thing is h. And what we can do is construct what look like similar triangles, and we can even prove it to ourselves that these are similar triangles. So let me construct them right over here. And the reason why we know they're similar is that this line is going to be parallel to this line, and that this line is parallel to that line, to that radius. And how do we know that? Well, we're taking cross-sectional areas that are parallel to the base, that are parallel to the surface on which it sits in this situation. So in either case, these cross-sections are going to be parallel. So these lines which sit in these cross-sections, or sit on the base and sit in the cross-section, have to be parallel as well."}, {"video_title": "Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "And how do we know that? Well, we're taking cross-sectional areas that are parallel to the base, that are parallel to the surface on which it sits in this situation. So in either case, these cross-sections are going to be parallel. So these lines which sit in these cross-sections, or sit on the base and sit in the cross-section, have to be parallel as well. Well, because these are parallel lines, this angle is congruent to that angle. This angle is congruent to this angle, because these are transversals across parallel lines, and these are just corresponding angles. And of course, they share this angle in common."}, {"video_title": "Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So these lines which sit in these cross-sections, or sit on the base and sit in the cross-section, have to be parallel as well. Well, because these are parallel lines, this angle is congruent to that angle. This angle is congruent to this angle, because these are transversals across parallel lines, and these are just corresponding angles. And of course, they share this angle in common. And here you see very clearly, right angle, right angle. This angle is congruent to that angle, and then both triangles share that. And so the smaller triangle in either case is similar to the larger triangle."}, {"video_title": "Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "And of course, they share this angle in common. And here you see very clearly, right angle, right angle. This angle is congruent to that angle, and then both triangles share that. And so the smaller triangle in either case is similar to the larger triangle. And what that helps us realize is that the ratio between corresponding sides is going to be the same. So if this side is h over two, and the entire height is h, so this is half of the entire height, that tells us that this side is going to be half of r. So this right over here is going to be r over two. And this side over here by the same argument is going to be x over two."}, {"video_title": "Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "And so the smaller triangle in either case is similar to the larger triangle. And what that helps us realize is that the ratio between corresponding sides is going to be the same. So if this side is h over two, and the entire height is h, so this is half of the entire height, that tells us that this side is going to be half of r. So this right over here is going to be r over two. And this side over here by the same argument is going to be x over two. And so what's the cross-sectional area here? Well, it's going to be x over two squared. So it's going to be x over two squared, which is equal to x squared over four, which is 1 1\u20444 of the base's area, which is equal to 1 1\u20444 of the base's area."}, {"video_title": "Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "And this side over here by the same argument is going to be x over two. And so what's the cross-sectional area here? Well, it's going to be x over two squared. So it's going to be x over two squared, which is equal to x squared over four, which is 1 1\u20444 of the base's area, which is equal to 1 1\u20444 of the base's area. And what about over here? Well, this cross-sectional area is going to be pi times r over two squared, which is the same thing as pi r squared over four, or we could say that as 1 1\u20444 pi r squared, which is the same thing as 1 1\u20444 of the area of the base. The area of the base is pi r squared."}, {"video_title": "Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So it's going to be x over two squared, which is equal to x squared over four, which is 1 1\u20444 of the base's area, which is equal to 1 1\u20444 of the base's area. And what about over here? Well, this cross-sectional area is going to be pi times r over two squared, which is the same thing as pi r squared over four, or we could say that as 1 1\u20444 pi r squared, which is the same thing as 1 1\u20444 of the area of the base. The area of the base is pi r squared. Now we're saying 1 1\u20444 pi r squared. So this is going to be equal to 1 1\u20444 the area. And we already said that these areas are the same."}, {"video_title": "Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "The area of the base is pi r squared. Now we're saying 1 1\u20444 pi r squared. So this is going to be equal to 1 1\u20444 the area. And we already said that these areas are the same. And so we've just seen that the cross-sectional area at that point of the height of both of these figures is the same. And you could do that 1\u20444 along the height, 3\u20444 along the height, you're going to get the same exact analysis. You're gonna have two similar triangles, and you're going to see that you have the same areas, same cross-sectional areas at that point of the height."}, {"video_title": "Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "And we already said that these areas are the same. And so we've just seen that the cross-sectional area at that point of the height of both of these figures is the same. And you could do that 1\u20444 along the height, 3\u20444 along the height, you're going to get the same exact analysis. You're gonna have two similar triangles, and you're going to see that you have the same areas, same cross-sectional areas at that point of the height. And so therefore, we see by Cavalieri's principle in three dimensions, that these two figures have the same volume. And what's interesting about that is it allows us to take the formula, which we've proven and gotten the intuition for in other videos for the volume of a pyramid. We've learned that the volume of a pyramid is equal to 1\u20443 times base times height, and say, well, this one must have the exact same volume."}, {"video_title": "Dilating in 3D Solid geometry High school geometry Khan Academy.mp3", "Sentence": "Let's say I have some type of a surface. Let's say that this right over here is the top of your desk. And I were to draw a triangle on that surface. So maybe the triangle looks like this. Something like this. It doesn't have to be a right triangle. And so I'm not implying that this is necessarily a right triangle, although it looks a little bit like one."}, {"video_title": "Dilating in 3D Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So maybe the triangle looks like this. Something like this. It doesn't have to be a right triangle. And so I'm not implying that this is necessarily a right triangle, although it looks a little bit like one. And let's call it triangle A, B, and then C. Now what I'm going to do is something interesting. I'm gonna take a fourth point P that's not on the surface of this desk, and it's going to be right above point B. So let me just take that point, go straight up, and I'm going to get to point P right over here."}, {"video_title": "Dilating in 3D Solid geometry High school geometry Khan Academy.mp3", "Sentence": "And so I'm not implying that this is necessarily a right triangle, although it looks a little bit like one. And let's call it triangle A, B, and then C. Now what I'm going to do is something interesting. I'm gonna take a fourth point P that's not on the surface of this desk, and it's going to be right above point B. So let me just take that point, go straight up, and I'm going to get to point P right over here. Now what I can do is construct a pyramid using point P as the peak of that pyramid. Now what we're going to start thinking about is what happens if I take cross sections of this pyramid? So in this case, the length of segment PB is the height of this pyramid."}, {"video_title": "Dilating in 3D Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So let me just take that point, go straight up, and I'm going to get to point P right over here. Now what I can do is construct a pyramid using point P as the peak of that pyramid. Now what we're going to start thinking about is what happens if I take cross sections of this pyramid? So in this case, the length of segment PB is the height of this pyramid. Now if we were to go halfway along that height, and if we were to take a cross section of this pyramid that is parallel to the surface of our original desk, what would that look like? Well, it would look something, it would look something like this. Now you might be noticing something really interesting."}, {"video_title": "Dilating in 3D Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So in this case, the length of segment PB is the height of this pyramid. Now if we were to go halfway along that height, and if we were to take a cross section of this pyramid that is parallel to the surface of our original desk, what would that look like? Well, it would look something, it would look something like this. Now you might be noticing something really interesting. If you were to translate that blue triangle straight down onto the surface of the table, it would look like this. And when you see it that way, it looks like it is a dilation of our original triangle centered at point B. And in fact, it is a dilation centered at point B with a scale factor of 0.5."}, {"video_title": "Dilating in 3D Solid geometry High school geometry Khan Academy.mp3", "Sentence": "Now you might be noticing something really interesting. If you were to translate that blue triangle straight down onto the surface of the table, it would look like this. And when you see it that way, it looks like it is a dilation of our original triangle centered at point B. And in fact, it is a dilation centered at point B with a scale factor of 0.5. And you can see it right over here. This length right over here, what BC was dilated down to is half the length of the original BC. This is half the length of the original AB."}, {"video_title": "Dilating in 3D Solid geometry High school geometry Khan Academy.mp3", "Sentence": "And in fact, it is a dilation centered at point B with a scale factor of 0.5. And you can see it right over here. This length right over here, what BC was dilated down to is half the length of the original BC. This is half the length of the original AB. And then this is half the length of the original AC. But you could do it at other heights along this pyramid. What if we were to go 0.75 of the way between P and B?"}, {"video_title": "Dilating in 3D Solid geometry High school geometry Khan Academy.mp3", "Sentence": "This is half the length of the original AB. And then this is half the length of the original AC. But you could do it at other heights along this pyramid. What if we were to go 0.75 of the way between P and B? So if we were to go right over here. So it's closer to our original triangle, closer to our surface. So then the cross section would look like, would look like this."}, {"video_title": "Dilating in 3D Solid geometry High school geometry Khan Academy.mp3", "Sentence": "What if we were to go 0.75 of the way between P and B? So if we were to go right over here. So it's closer to our original triangle, closer to our surface. So then the cross section would look like, would look like this. Now if we were to translate that down onto our original surface, what would that look like? Well, it would look like this. It would look like a dilation of our original triangle centered at point B, but this time with a scale factor of 0.75."}, {"video_title": "Dilating in 3D Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So then the cross section would look like, would look like this. Now if we were to translate that down onto our original surface, what would that look like? Well, it would look like this. It would look like a dilation of our original triangle centered at point B, but this time with a scale factor of 0.75. And then what if you were to go only a quarter of the way between point P and point B? Well then, you would see something like this. A quarter of the way."}, {"video_title": "Dilating in 3D Solid geometry High school geometry Khan Academy.mp3", "Sentence": "It would look like a dilation of our original triangle centered at point B, but this time with a scale factor of 0.75. And then what if you were to go only a quarter of the way between point P and point B? Well then, you would see something like this. A quarter of the way. If you were to take the cross section parallel to our original surface, it would look like this. If you were to translate that straight down onto our table, it would look something like this. And it looks like a dilation centered at point B with a scale factor of 0.25."}, {"video_title": "Dilating in 3D Solid geometry High school geometry Khan Academy.mp3", "Sentence": "A quarter of the way. If you were to take the cross section parallel to our original surface, it would look like this. If you were to translate that straight down onto our table, it would look something like this. And it looks like a dilation centered at point B with a scale factor of 0.25. And the reason why all of these dilations look like dilations centered at point B is because point P is directly above point B. But this is a way to conceptualize dilations or see the relationship between cross sections of a three-dimensional shape, in this case like a pyramid, and how those cross sections relate to the base of the pyramid. Now let me ask you an interesting question."}, {"video_title": "Dilating in 3D Solid geometry High school geometry Khan Academy.mp3", "Sentence": "And it looks like a dilation centered at point B with a scale factor of 0.25. And the reason why all of these dilations look like dilations centered at point B is because point P is directly above point B. But this is a way to conceptualize dilations or see the relationship between cross sections of a three-dimensional shape, in this case like a pyramid, and how those cross sections relate to the base of the pyramid. Now let me ask you an interesting question. What if I were to try to take a cross section right at point P? Well then, I would just get a point. I would not get an actual triangle, but you could view that as a dilation with a scale factor of 0."}, {"video_title": "Dilating in 3D Solid geometry High school geometry Khan Academy.mp3", "Sentence": "Now let me ask you an interesting question. What if I were to try to take a cross section right at point P? Well then, I would just get a point. I would not get an actual triangle, but you could view that as a dilation with a scale factor of 0. And what if I were to take a cross section at the base? Well then, that would be my original triangle, triangle ABC, and then you could view that as a dilation with a scale factor of one because you've gone all the way down to the base. So hopefully this connects some dots for you between cross sections of a three-dimensional shape that is parallel to the base and notions of dilation."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "Actually, you probably recognize a few of them already, and I'll write them out. They're the circle, the ellipse, the parabola, and the hyperbola. Hyperbola. And you know what these are already. When I first learned conic sections, I was like, oh, I know what a circle is, I know what a parabola is, and I even know a little bit about ellipses and hyperbolas. Why on earth are they called conic sections? To put things simply, because they're the intersection of a plane and a cone."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "And you know what these are already. When I first learned conic sections, I was like, oh, I know what a circle is, I know what a parabola is, and I even know a little bit about ellipses and hyperbolas. Why on earth are they called conic sections? To put things simply, because they're the intersection of a plane and a cone. I'll draw you down in a second, but just before I do that, it probably makes sense to just draw them by themselves, and I'll switch colors. Circle, we all know what that is. Let me pick a thicker line for my circles."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "To put things simply, because they're the intersection of a plane and a cone. I'll draw you down in a second, but just before I do that, it probably makes sense to just draw them by themselves, and I'll switch colors. Circle, we all know what that is. Let me pick a thicker line for my circles. So a circle looks something like that. It's all the points that are equidistant from some center, and that distance that they all are, that's the radius. So this is r, and this is the center."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "Let me pick a thicker line for my circles. So a circle looks something like that. It's all the points that are equidistant from some center, and that distance that they all are, that's the radius. So this is r, and this is the center. The circle is all of the points that are exactly r away from this center. We learned that early in our education, what a circle is. It makes the world go round, literally."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "So this is r, and this is the center. The circle is all of the points that are exactly r away from this center. We learned that early in our education, what a circle is. It makes the world go round, literally. Ellipse, in layman's terms, is kind of a squished circle. It could look something like this. It could look like, let me do an ellipse in another color."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "It makes the world go round, literally. Ellipse, in layman's terms, is kind of a squished circle. It could look something like this. It could look like, let me do an ellipse in another color. So an ellipse could be like that, could be like that. It's harder to draw using the tool I'm drawing, but it could also be tilted and rotated around, but this is the general sense. Actually, circles are a special case of an ellipse."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "It could look like, let me do an ellipse in another color. So an ellipse could be like that, could be like that. It's harder to draw using the tool I'm drawing, but it could also be tilted and rotated around, but this is the general sense. Actually, circles are a special case of an ellipse. It's an ellipse where it's not stretched in one dimension more than the other. It's kind of perfectly symmetric in every way. Parabola, you've learned that if you've taken Algebra 2, and you probably have if you care about conic sections."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "Actually, circles are a special case of an ellipse. It's an ellipse where it's not stretched in one dimension more than the other. It's kind of perfectly symmetric in every way. Parabola, you've learned that if you've taken Algebra 2, and you probably have if you care about conic sections. But a parabola, let me draw lines here so you know where to separate things. A parabola looks something like this, kind of a U shape. I won't go into the equations right now, but the classic, well I will because you're probably familiar with it."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "Parabola, you've learned that if you've taken Algebra 2, and you probably have if you care about conic sections. But a parabola, let me draw lines here so you know where to separate things. A parabola looks something like this, kind of a U shape. I won't go into the equations right now, but the classic, well I will because you're probably familiar with it. It's y is equal to x squared. And then it could be, you could shift it around, and you could even have a parabola that goes like this. That would be x is equal to y squared."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "I won't go into the equations right now, but the classic, well I will because you're probably familiar with it. It's y is equal to x squared. And then it could be, you could shift it around, and you could even have a parabola that goes like this. That would be x is equal to y squared. You could also rotate these things around. But I think you know the general shape of a parabola. We'll talk more about how do you graph it, or how do you know what the interesting points on a parabola actually are."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "That would be x is equal to y squared. You could also rotate these things around. But I think you know the general shape of a parabola. We'll talk more about how do you graph it, or how do you know what the interesting points on a parabola actually are. And then the last one, you might have seen this before, is a hyperbola. It almost looks like two parabolas, but not quite, because the curves, they look a little less U-ish, and a little more open, but I'll explain what I mean by that. So a hyperbola usually looks something like this."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "We'll talk more about how do you graph it, or how do you know what the interesting points on a parabola actually are. And then the last one, you might have seen this before, is a hyperbola. It almost looks like two parabolas, but not quite, because the curves, they look a little less U-ish, and a little more open, but I'll explain what I mean by that. So a hyperbola usually looks something like this. So if these are the axes, and if I were to draw, let me draw some asymptotes. I want to go right through. That's pretty good."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "So a hyperbola usually looks something like this. So if these are the axes, and if I were to draw, let me draw some asymptotes. I want to go right through. That's pretty good. These are asymptotes. Those aren't the actual hyperbola. But a hyperbola would look something like this."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "That's pretty good. These are asymptotes. Those aren't the actual hyperbola. But a hyperbola would look something like this. It could either be on, it would be like, they could be like right here, and they get really close. The asymptote, they get closer and closer to those blue lines, like that. And it would happen on this side, too."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "But a hyperbola would look something like this. It could either be on, it would be like, they could be like right here, and they get really close. The asymptote, they get closer and closer to those blue lines, like that. And it would happen on this side, too. So it kind of, the graphs show up here, and then they pop over, and they show up there. So maybe this magenta could be one hyperbola. I haven't done true justice to it."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "And it would happen on this side, too. So it kind of, the graphs show up here, and then they pop over, and they show up there. So maybe this magenta could be one hyperbola. I haven't done true justice to it. Or another hyperbola could be on kind of a, you could kind of call it a vertical hyperbola, and that's not the exact word, but it would look something like that, where it's below the asymptote here, and it's above the asymptote there. So this blue one would be one hyperbola, and then the magenta one would be a different hyperbola. So those are the different graphs."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "I haven't done true justice to it. Or another hyperbola could be on kind of a, you could kind of call it a vertical hyperbola, and that's not the exact word, but it would look something like that, where it's below the asymptote here, and it's above the asymptote there. So this blue one would be one hyperbola, and then the magenta one would be a different hyperbola. So those are the different graphs. So then, you know, the one thing that I'm sure you're asking is, why are they called conic sections? Why are they not called bolas, or variations of circles, or whatever? And in fact, what's even the relationship?"}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "So those are the different graphs. So then, you know, the one thing that I'm sure you're asking is, why are they called conic sections? Why are they not called bolas, or variations of circles, or whatever? And in fact, what's even the relationship? It's pretty clear that circles and ellipses are somehow related, that an ellipse is just a squished circle. And maybe it even seems that parabolas and hyperbolas are somewhat related. This is a P, once again."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "And in fact, what's even the relationship? It's pretty clear that circles and ellipses are somehow related, that an ellipse is just a squished circle. And maybe it even seems that parabolas and hyperbolas are somewhat related. This is a P, once again. They both have bola in their name, and they both kind of look like opened U's, although a hyperbola has two of these, kind of opening in different directions, but they look related. But what is the connection behind all these? And that's, frankly, where the word conic comes from."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "This is a P, once again. They both have bola in their name, and they both kind of look like opened U's, although a hyperbola has two of these, kind of opening in different directions, but they look related. But what is the connection behind all these? And that's, frankly, where the word conic comes from. So let me see if I can draw a three-dimensional cone. So if this is a cone, let me see. So that's the top."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "And that's, frankly, where the word conic comes from. So let me see if I can draw a three-dimensional cone. So if this is a cone, let me see. So that's the top. Let me draw... I could have used an ellipse for the top. Looks like that."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "So that's the top. Let me draw... I could have used an ellipse for the top. Looks like that. And it has no top. It would actually keep going on forever in that direction. I'm just kind of slicing it so you see that it's a cone."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "Looks like that. And it has no top. It would actually keep going on forever in that direction. I'm just kind of slicing it so you see that it's a cone. This could be the bottom part of it. So let's take different intersections of a plane with this cone and see if we can at least generate the different shapes that we talked about just now. So if we have a plane that goes directly..."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "I'm just kind of slicing it so you see that it's a cone. This could be the bottom part of it. So let's take different intersections of a plane with this cone and see if we can at least generate the different shapes that we talked about just now. So if we have a plane that goes directly... I guess if you call this the axis of this three-dimensional cone. So this is the axis. So if we have a plane that's exactly perpendicular to that axis... Let's see if I can draw it in three dimensions."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "So if we have a plane that goes directly... I guess if you call this the axis of this three-dimensional cone. So this is the axis. So if we have a plane that's exactly perpendicular to that axis... Let's see if I can draw it in three dimensions. So this would look something like this. So it would have a line. This is the front line that's closer to you."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "So if we have a plane that's exactly perpendicular to that axis... Let's see if I can draw it in three dimensions. So this would look something like this. So it would have a line. This is the front line that's closer to you. And then they'll have another line back here. That's close enough. And then, of course, you know these are infinite planes."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "This is the front line that's closer to you. And then they'll have another line back here. That's close enough. And then, of course, you know these are infinite planes. It goes off in every direction. So if this plane is directly perpendicular to the axis of these... So this is where the plane goes behind it."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "And then, of course, you know these are infinite planes. It goes off in every direction. So if this plane is directly perpendicular to the axis of these... So this is where the plane goes behind it. The intersection of this plane and this cone is going to look like this. And we're looking at it from an angle, but if you were to look at it straight down, if you were sitting here and you were to look at this plane, if you were to look at it right above, if I were to just flip this over like this, so we're looking straight down at this plane, that intersection would be a circle. Now, if we take the plane and we tilt it down a little bit, so if instead of that, we have a situation like this, let me see if I can do it justice."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "So this is where the plane goes behind it. The intersection of this plane and this cone is going to look like this. And we're looking at it from an angle, but if you were to look at it straight down, if you were sitting here and you were to look at this plane, if you were to look at it right above, if I were to just flip this over like this, so we're looking straight down at this plane, that intersection would be a circle. Now, if we take the plane and we tilt it down a little bit, so if instead of that, we have a situation like this, let me see if I can do it justice. We have a situation where it's... Let me undo that. Edit, undo. Where it's like this, and it has another side like this, and then I connect them."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "Now, if we take the plane and we tilt it down a little bit, so if instead of that, we have a situation like this, let me see if I can do it justice. We have a situation where it's... Let me undo that. Edit, undo. Where it's like this, and it has another side like this, and then I connect them. So that's the plane. Now the intersection of this plane, which is now not orthogonal, or it's not perpendicular to the axis of this 3-dimensional cone, if you take the intersection of that plane and that cone, and in future videos, and you don't do this in your Algebra 2 class, but eventually we'll kind of do the 3-dimensional intersection and prove that this is definitely the case, and you definitely do get the equations, which I'll show you in the not-too-far-off future, this intersection would look something like this. I think you can visualize it right now."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "Where it's like this, and it has another side like this, and then I connect them. So that's the plane. Now the intersection of this plane, which is now not orthogonal, or it's not perpendicular to the axis of this 3-dimensional cone, if you take the intersection of that plane and that cone, and in future videos, and you don't do this in your Algebra 2 class, but eventually we'll kind of do the 3-dimensional intersection and prove that this is definitely the case, and you definitely do get the equations, which I'll show you in the not-too-far-off future, this intersection would look something like this. I think you can visualize it right now. It would look something like this. And if you were to look straight down on this plane, if you were to look right above the plane, this would look something... This figure I just drew in purple would look something like this."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "I think you can visualize it right now. It would look something like this. And if you were to look straight down on this plane, if you were to look right above the plane, this would look something... This figure I just drew in purple would look something like this. Oh, I didn't draw it that well. It would be an ellipse. You know what an ellipse looks like."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "This figure I just drew in purple would look something like this. Oh, I didn't draw it that well. It would be an ellipse. You know what an ellipse looks like. And if I tilted it the other way, the ellipse would be... would kind of... would squeeze the other way. But that just gives you a general sense of why both of these are conic sections. Now something very interesting, if we keep tilting this plane..."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "You know what an ellipse looks like. And if I tilted it the other way, the ellipse would be... would kind of... would squeeze the other way. But that just gives you a general sense of why both of these are conic sections. Now something very interesting, if we keep tilting this plane... So if we tilt the plane so it's... so let's say we're pivoting around that point. So now my plane... Let me see if I can do this. This is a good exercise in 3-dimensional drawing."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "Now something very interesting, if we keep tilting this plane... So if we tilt the plane so it's... so let's say we're pivoting around that point. So now my plane... Let me see if I can do this. This is a good exercise in 3-dimensional drawing. Let's say it looks something like this. I want to go through that point. Oh."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "This is a good exercise in 3-dimensional drawing. Let's say it looks something like this. I want to go through that point. Oh. So this is my 3-dimensional plane. And I'm drawing it in such a way that it only intersects this bottom cone. And the surface of the plane is parallel to the side of this top cone."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "Oh. So this is my 3-dimensional plane. And I'm drawing it in such a way that it only intersects this bottom cone. And the surface of the plane is parallel to the side of this top cone. In this case, the intersection of the plane and the cone is going to intersect right at that point. I'm kind of... You can almost view that I'm pivoting around this point, the intersection of this point and the plane and the cone. Well, this now would look... the intersection would look something like this."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "And the surface of the plane is parallel to the side of this top cone. In this case, the intersection of the plane and the cone is going to intersect right at that point. I'm kind of... You can almost view that I'm pivoting around this point, the intersection of this point and the plane and the cone. Well, this now would look... the intersection would look something like this. It would look like that. And it would keep going down. So if I were to draw it, it would look like this."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "Well, this now would look... the intersection would look something like this. It would look like that. And it would keep going down. So if I were to draw it, it would look like this. If I was right above the plane, if I were to just draw the plane. And there you get your parabola. So that's interesting."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "So if I were to draw it, it would look like this. If I was right above the plane, if I were to just draw the plane. And there you get your parabola. So that's interesting. If you keep kind of tilting... If you start with a circle, tilt a little bit, you get an ellipse, you get an ellipse, you get kind of a more and more skewed ellipse, more and more skewed ellipse. And at some point, the ellipse kind of... You know, the ellipse keeps getting more and more skewed like that."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "So that's interesting. If you keep kind of tilting... If you start with a circle, tilt a little bit, you get an ellipse, you get an ellipse, you get kind of a more and more skewed ellipse, more and more skewed ellipse. And at some point, the ellipse kind of... You know, the ellipse keeps getting more and more skewed like that. At some point, it kind of pops. Right when you... Right when you become exactly parallel to the side of this top cone."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "And at some point, the ellipse kind of... You know, the ellipse keeps getting more and more skewed like that. At some point, it kind of pops. Right when you... Right when you become exactly parallel to the side of this top cone. And I'm doing it all very inexact right now, but I think I want to give you the intuition. It pops, and it turns into a parabola. So you can kind of view a parabola."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "Right when you become exactly parallel to the side of this top cone. And I'm doing it all very inexact right now, but I think I want to give you the intuition. It pops, and it turns into a parabola. So you can kind of view a parabola. There is this relationship. It's kind of... The parabola is what happens when one side of an ellipse pops open, and you get this parabola."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "So you can kind of view a parabola. There is this relationship. It's kind of... The parabola is what happens when one side of an ellipse pops open, and you get this parabola. And then if you keep tilting this plane, if you keep tilting the plane... I'll do it in another color. So it intersects both sides of the cone."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "The parabola is what happens when one side of an ellipse pops open, and you get this parabola. And then if you keep tilting this plane, if you keep tilting the plane... I'll do it in another color. So it intersects both sides of the cone. So let me see if I can draw that. So if this is my new plane... Whoops. That's good enough."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "So it intersects both sides of the cone. So let me see if I can draw that. So if this is my new plane... Whoops. That's good enough. So if my plane looks like this... I know it's very hard to read now. And you wanted the intersection of this plane, this green plane in the cone."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "That's good enough. So if my plane looks like this... I know it's very hard to read now. And you wanted the intersection of this plane, this green plane in the cone. I should probably redraw it all, but hopefully you're not getting overwhelmingly confused. The intersection would look like this. It would intersect the bottom cone there, and it would intersect the top cone over there."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "And you wanted the intersection of this plane, this green plane in the cone. I should probably redraw it all, but hopefully you're not getting overwhelmingly confused. The intersection would look like this. It would intersect the bottom cone there, and it would intersect the top cone over there. And then you would have something like this. You would have... This would be the intersection of the plane in the bottom cone, and then up here would be the intersection of the plane in the top cone."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "It would intersect the bottom cone there, and it would intersect the top cone over there. And then you would have something like this. You would have... This would be the intersection of the plane in the bottom cone, and then up here would be the intersection of the plane in the top cone. Remember, this plane goes off in every direction infinitely. So that's just a general sense of what the conic sections are and why, frankly, they're called conic sections. And let me know if this got confusing, because maybe I'll do another video where I'll redraw it a little bit cleaner, or maybe I can find some kind of neat 3-D application that can do it better than I can do it."}, {"video_title": "Introduction to conic sections Conic sections Algebra II Khan Academy.mp3", "Sentence": "This would be the intersection of the plane in the bottom cone, and then up here would be the intersection of the plane in the top cone. Remember, this plane goes off in every direction infinitely. So that's just a general sense of what the conic sections are and why, frankly, they're called conic sections. And let me know if this got confusing, because maybe I'll do another video where I'll redraw it a little bit cleaner, or maybe I can find some kind of neat 3-D application that can do it better than I can do it. But this is kind of just the reason why they all are conic sections, and why they really are related to each other. And we'll do that in a little more depth mathematically in a few videos. But in the next video, now that you know what they are and why they're all called conic sections, I'll actually talk about the formulas about these and how do you recognize the formulas, and given a formula, how do you actually plot the graphs of these conic sections?"}, {"video_title": "Proof Parallel lines divide triangle sides proportionally Similarity Geometry Khan Academy.mp3", "Sentence": "So pause this video and see if you can do that, and you might wanna leverage this diagram. All right, so let's work through this together. So we can start with this diagram, and what we know is that segment ED is parallel to segment CB. So we can write that down. Segment ED is parallel to segment CB. And so segment ED is what they're talking about. That is a line or a line segment that is parallel to one side of the triangle."}, {"video_title": "Proof Parallel lines divide triangle sides proportionally Similarity Geometry Khan Academy.mp3", "Sentence": "So we can write that down. Segment ED is parallel to segment CB. And so segment ED is what they're talking about. That is a line or a line segment that is parallel to one side of the triangle. So really, given what we know and what's already been written over here on this triangle, we need to prove another way of writing it, another way of saying it divides the other two sides proportionately is that the ratio between the part of the original triangle side that is on one side of the dividing line to the length on the other side is going to be the same on both sides that it is intersecting. So another way to say that it divides the other two sides proportionately, if we look at this triangle over here, it would mean that the length of segment AE over the length of segment EC is going to be equal to the length of segment AD over the length of segment DB. This statement right over here and what I underlined up here are equivalent given this triangle."}, {"video_title": "Proof Parallel lines divide triangle sides proportionally Similarity Geometry Khan Academy.mp3", "Sentence": "That is a line or a line segment that is parallel to one side of the triangle. So really, given what we know and what's already been written over here on this triangle, we need to prove another way of writing it, another way of saying it divides the other two sides proportionately is that the ratio between the part of the original triangle side that is on one side of the dividing line to the length on the other side is going to be the same on both sides that it is intersecting. So another way to say that it divides the other two sides proportionately, if we look at this triangle over here, it would mean that the length of segment AE over the length of segment EC is going to be equal to the length of segment AD over the length of segment DB. This statement right over here and what I underlined up here are equivalent given this triangle. So the way that we can try to do it is to establish a similarity between triangle AED and triangle ACB. So how do we do that? Well, because these two lines are parallel, we can view segment AC as a transversal intersecting two parallel lines."}, {"video_title": "Proof Parallel lines divide triangle sides proportionally Similarity Geometry Khan Academy.mp3", "Sentence": "This statement right over here and what I underlined up here are equivalent given this triangle. So the way that we can try to do it is to establish a similarity between triangle AED and triangle ACB. So how do we do that? Well, because these two lines are parallel, we can view segment AC as a transversal intersecting two parallel lines. So that tells us that these two corresponding angles are going to be congruent. So we could say that angle one is congruent to angle three. And the reason why is because they are corresponding corresponding angles."}, {"video_title": "Proof Parallel lines divide triangle sides proportionally Similarity Geometry Khan Academy.mp3", "Sentence": "Well, because these two lines are parallel, we can view segment AC as a transversal intersecting two parallel lines. So that tells us that these two corresponding angles are going to be congruent. So we could say that angle one is congruent to angle three. And the reason why is because they are corresponding corresponding angles. I'm just trying to write a little bit of shorthand. This is short for corresponding angles. That's the rationale."}, {"video_title": "Proof Parallel lines divide triangle sides proportionally Similarity Geometry Khan Academy.mp3", "Sentence": "And the reason why is because they are corresponding corresponding angles. I'm just trying to write a little bit of shorthand. This is short for corresponding angles. That's the rationale. And we also know that angle two is congruent to angle four for the same reason. So angle two is congruent to angle four, once again, because they are corresponding angles. This time we have a different transversal, corresponding angles where a transversal intersects two parallel lines."}, {"video_title": "Proof Parallel lines divide triangle sides proportionally Similarity Geometry Khan Academy.mp3", "Sentence": "That's the rationale. And we also know that angle two is congruent to angle four for the same reason. So angle two is congruent to angle four, once again, because they are corresponding angles. This time we have a different transversal, corresponding angles where a transversal intersects two parallel lines. And so now if you look at triangle AED and triangle ACB, you see that they have two sets of corresponding angles that are congruent. And if you have two sets of corresponding angles, that means that all of the angles are congruent. And you actually see that over here if you care about it."}, {"video_title": "Proof Parallel lines divide triangle sides proportionally Similarity Geometry Khan Academy.mp3", "Sentence": "This time we have a different transversal, corresponding angles where a transversal intersects two parallel lines. And so now if you look at triangle AED and triangle ACB, you see that they have two sets of corresponding angles that are congruent. And if you have two sets of corresponding angles, that means that all of the angles are congruent. And you actually see that over here if you care about it. But two is enough, but you actually have a third because angle, I guess you call it BAC, is common to both triangles. And so we can say that triangle AED is similar to triangle ACB, ACB, by angle similarity, similarity. And then given that these two are similar, then we can set up a proportion."}, {"video_title": "Proof Parallel lines divide triangle sides proportionally Similarity Geometry Khan Academy.mp3", "Sentence": "And you actually see that over here if you care about it. But two is enough, but you actually have a third because angle, I guess you call it BAC, is common to both triangles. And so we can say that triangle AED is similar to triangle ACB, ACB, by angle similarity, similarity. And then given that these two are similar, then we can set up a proportion. That tells us that the ratio of the length of segment AE to this entire side to AC, AC, is equal to the ratio of AD, the length of that segment, to the length of the entire thing, to AB. Now, this implies, and I'm just gonna start writing it to the right here to save space. This is the same thing as the ratio of AE over, AC is AE plus EC's length."}, {"video_title": "Proof Parallel lines divide triangle sides proportionally Similarity Geometry Khan Academy.mp3", "Sentence": "And then given that these two are similar, then we can set up a proportion. That tells us that the ratio of the length of segment AE to this entire side to AC, AC, is equal to the ratio of AD, the length of that segment, to the length of the entire thing, to AB. Now, this implies, and I'm just gonna start writing it to the right here to save space. This is the same thing as the ratio of AE over, AC is AE plus EC's length. So AE's length plus EC's length. And then this is going to be equal to the length of segment AD over segment AB. Its length is the length of segment AD, AD plus segment DB, plus DB."}, {"video_title": "Proof Parallel lines divide triangle sides proportionally Similarity Geometry Khan Academy.mp3", "Sentence": "This is the same thing as the ratio of AE over, AC is AE plus EC's length. So AE's length plus EC's length. And then this is going to be equal to the length of segment AD over segment AB. Its length is the length of segment AD, AD plus segment DB, plus DB. Now, really what I need to do is figure out how do I algebraically manipulate it so I get what I have up here. Let me scroll down a little bit. So one way I could try to simplify this is to essentially cross multiply."}, {"video_title": "Proof Parallel lines divide triangle sides proportionally Similarity Geometry Khan Academy.mp3", "Sentence": "Its length is the length of segment AD, AD plus segment DB, plus DB. Now, really what I need to do is figure out how do I algebraically manipulate it so I get what I have up here. Let me scroll down a little bit. So one way I could try to simplify this is to essentially cross multiply. That's equivalent to multiplying both sides by both of these denominators, and we've covered that in other videos. And so this is going to be equal to the length of segment AE times AD plus DB, those segment lengths. That's gotta be equal to length of AD times AE plus the length of segment EC."}, {"video_title": "Proof Parallel lines divide triangle sides proportionally Similarity Geometry Khan Academy.mp3", "Sentence": "So one way I could try to simplify this is to essentially cross multiply. That's equivalent to multiplying both sides by both of these denominators, and we've covered that in other videos. And so this is going to be equal to the length of segment AE times AD plus DB, those segment lengths. That's gotta be equal to length of AD times AE plus the length of segment EC. And I can distribute this over here. I have length of segment AE times length of segment AD plus length of segment AE times length of segment DB is equal to length of segment AD times length of segment AE plus length of segment AD times length of segment EC. And let's see, is there anything that I can simplify here?"}, {"video_title": "Proof Parallel lines divide triangle sides proportionally Similarity Geometry Khan Academy.mp3", "Sentence": "That's gotta be equal to length of AD times AE plus the length of segment EC. And I can distribute this over here. I have length of segment AE times length of segment AD plus length of segment AE times length of segment DB is equal to length of segment AD times length of segment AE plus length of segment AD times length of segment EC. And let's see, is there anything that I can simplify here? Well, I have an AE times AD on both sides, so let me just subtract AE times AD from both sides. And so then I'm just left with that this is equal to that. So scroll down a little bit more, and let me actually just rewrite this cleanly."}, {"video_title": "Proof Parallel lines divide triangle sides proportionally Similarity Geometry Khan Academy.mp3", "Sentence": "And let's see, is there anything that I can simplify here? Well, I have an AE times AD on both sides, so let me just subtract AE times AD from both sides. And so then I'm just left with that this is equal to that. So scroll down a little bit more, and let me actually just rewrite this cleanly. So I have AE times DB is equal to AD times EC. These are all the segment lengths right over here. Now, if you divide both sides by EC, you're going to get an EC down here, and then this would cancel out."}, {"video_title": "Transformations - dilation.mp3", "Sentence": "In previous videos, we started talking about the idea of transformations. In particular, we talked about rigid transformations. So, for example, you can shift something. This would be a translation. So the thing that I'm moving around is a translation of our original triangle. You could have a rotation. So that thing that I translated, I am now rotating it, as you see right over there."}, {"video_title": "Transformations - dilation.mp3", "Sentence": "This would be a translation. So the thing that I'm moving around is a translation of our original triangle. You could have a rotation. So that thing that I translated, I am now rotating it, as you see right over there. And you could also have a reflection, and the tool that I'm using doesn't make reflection too easy but that's essentially flipping it over a line. But what we're going to talk about in this video is a non-rigid transformation. And what makes something a rigid transformation is that lengths between points are preserved."}, {"video_title": "Transformations - dilation.mp3", "Sentence": "So that thing that I translated, I am now rotating it, as you see right over there. And you could also have a reflection, and the tool that I'm using doesn't make reflection too easy but that's essentially flipping it over a line. But what we're going to talk about in this video is a non-rigid transformation. And what makes something a rigid transformation is that lengths between points are preserved. But in a non-rigid transformation, those lengths do not need to be preserved. So, for example, this rotated and translated triangle that I'm moving around right here, in fact, I'm continuing to translate it as I talk, I can dilate it. And one way to think about dilation is that we're just scaling it down or scaling it up."}, {"video_title": "Transformations - dilation.mp3", "Sentence": "And what makes something a rigid transformation is that lengths between points are preserved. But in a non-rigid transformation, those lengths do not need to be preserved. So, for example, this rotated and translated triangle that I'm moving around right here, in fact, I'm continuing to translate it as I talk, I can dilate it. And one way to think about dilation is that we're just scaling it down or scaling it up. So, for example, here, I am scaling it down. That is a dilation. Or I could scale it up."}, {"video_title": "Transformations - dilation.mp3", "Sentence": "And one way to think about dilation is that we're just scaling it down or scaling it up. So, for example, here, I am scaling it down. That is a dilation. Or I could scale it up. This is also a dilation. We're even going off of the graph paper. So the whole point here is just to appreciate that we don't just have the rigid transformations, we can have other types of transformations."}, {"video_title": "Using right triangle ratios to approximate angle measure High school geometry Khan Academy.mp3", "Sentence": "So what they're saying here is if you were to take the adjacent leg length over the hypotenuse leg length for 25 degree angle, it would be a ratio of approximately 0.91. For a 35 degree angle, it would be a ratio of 0.82, and then they do this for 45 degrees, and they do the different ratios right over here. So we're gonna use the table to approximate the measure of angle D in the triangle below. So pause this video and see if you can figure that out. All right, now let's work through this together. Now what information do they give us about angle D in this triangle? Well, we are given the opposite length right over here."}, {"video_title": "Using right triangle ratios to approximate angle measure High school geometry Khan Academy.mp3", "Sentence": "So pause this video and see if you can figure that out. All right, now let's work through this together. Now what information do they give us about angle D in this triangle? Well, we are given the opposite length right over here. Let me label that. That is the opposite leg length, which is 3.4. We're also given, what is this right over here?"}, {"video_title": "Using right triangle ratios to approximate angle measure High school geometry Khan Academy.mp3", "Sentence": "Well, we are given the opposite length right over here. Let me label that. That is the opposite leg length, which is 3.4. We're also given, what is this right over here? Is this adjacent or is this a hypotenuse? You might be tempted to say, well, this is right next to the angle, or this is one of the lines, or it's on the ray that helps form the angle, so maybe it's adjacent. But remember, adjacent is the adjacent side that is not the hypotenuse."}, {"video_title": "Using right triangle ratios to approximate angle measure High school geometry Khan Academy.mp3", "Sentence": "We're also given, what is this right over here? Is this adjacent or is this a hypotenuse? You might be tempted to say, well, this is right next to the angle, or this is one of the lines, or it's on the ray that helps form the angle, so maybe it's adjacent. But remember, adjacent is the adjacent side that is not the hypotenuse. And this is clearly the hypotenuse. It is the longest side. It is the side opposite the 90 degree angle."}, {"video_title": "Using right triangle ratios to approximate angle measure High school geometry Khan Academy.mp3", "Sentence": "But remember, adjacent is the adjacent side that is not the hypotenuse. And this is clearly the hypotenuse. It is the longest side. It is the side opposite the 90 degree angle. So this right over here is the hypotenuse. Hypotenuse. So we're given the opposite leg length and the hypotenuse length."}, {"video_title": "Using right triangle ratios to approximate angle measure High school geometry Khan Academy.mp3", "Sentence": "It is the side opposite the 90 degree angle. So this right over here is the hypotenuse. Hypotenuse. So we're given the opposite leg length and the hypotenuse length. And so let's see, which of these ratios deal with the opposite and the hypotenuse? And if we, let's see, this first one is adjacent and hypotenuse. The second one here is hypotenuse, sorry, opposite and hypotenuse."}, {"video_title": "Using right triangle ratios to approximate angle measure High school geometry Khan Academy.mp3", "Sentence": "So we're given the opposite leg length and the hypotenuse length. And so let's see, which of these ratios deal with the opposite and the hypotenuse? And if we, let's see, this first one is adjacent and hypotenuse. The second one here is hypotenuse, sorry, opposite and hypotenuse. So that's exactly what we're talking about. We're talking about the opposite leg length over the hypotenuse, over the hypotenuse length. So in this case, what is going to be our opposite leg length over our hypotenuse leg length?"}, {"video_title": "Using right triangle ratios to approximate angle measure High school geometry Khan Academy.mp3", "Sentence": "The second one here is hypotenuse, sorry, opposite and hypotenuse. So that's exactly what we're talking about. We're talking about the opposite leg length over the hypotenuse, over the hypotenuse length. So in this case, what is going to be our opposite leg length over our hypotenuse leg length? It's going to be 3.4 over eight. 3.4 over eight, which is approximately going to be equal to, let me do this down here, this is eight goes into 3.4. Eight goes in, doesn't go into three."}, {"video_title": "Using right triangle ratios to approximate angle measure High school geometry Khan Academy.mp3", "Sentence": "So in this case, what is going to be our opposite leg length over our hypotenuse leg length? It's going to be 3.4 over eight. 3.4 over eight, which is approximately going to be equal to, let me do this down here, this is eight goes into 3.4. Eight goes in, doesn't go into three. Eight goes into 34 four times. Four times eight is 32. If I subtract, and I could scroll down a little bit, I get a two."}, {"video_title": "Using right triangle ratios to approximate angle measure High school geometry Khan Academy.mp3", "Sentence": "Eight goes in, doesn't go into three. Eight goes into 34 four times. Four times eight is 32. If I subtract, and I could scroll down a little bit, I get a two. I can bring down a zero. Eight goes into 20 two times. And that's about as much precision as any of these have."}, {"video_title": "Using right triangle ratios to approximate angle measure High school geometry Khan Academy.mp3", "Sentence": "If I subtract, and I could scroll down a little bit, I get a two. I can bring down a zero. Eight goes into 20 two times. And that's about as much precision as any of these have. And so it looks like for this particular triangle and this angle of the triangle, if I were to take a ratio of the opposite length and the hypotenuse length, opposite over hypotenuse, I get 0.42. So that looks like this situation right over here. So that would imply that this is a 25 degree, 25 degree angle approximately."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "You might recognize the first part of geometry right over here. You have the root word geo, the same word that you see in things like geography and geology. And this comes, this refers to the Earth. This refers, my E looks like a C right over there. This refers to the Earth. And then you see this metri part. And you see metri in things like trigonometry as well."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "This refers, my E looks like a C right over there. This refers to the Earth. And then you see this metri part. And you see metri in things like trigonometry as well. And metri, or the metric system, and this comes from measurement. This comes from measurement or measure. So when someone's talking about geometry, the word itself comes from Earth measurement."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And you see metri in things like trigonometry as well. And metri, or the metric system, and this comes from measurement. This comes from measurement or measure. So when someone's talking about geometry, the word itself comes from Earth measurement. And that's kind of not so bad of a name because it is such a general subject. Geometry really is the study and trying to understand how shapes and space and things that we see relate to each other. So when you start learning about geometry, you learn about lines and triangles and circles."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So when someone's talking about geometry, the word itself comes from Earth measurement. And that's kind of not so bad of a name because it is such a general subject. Geometry really is the study and trying to understand how shapes and space and things that we see relate to each other. So when you start learning about geometry, you learn about lines and triangles and circles. And you learn about angles. And we'll define all of these things more and more precisely as we go further and further on. But also encapsulate things like patterns and three-dimensional shapes."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So when you start learning about geometry, you learn about lines and triangles and circles. And you learn about angles. And we'll define all of these things more and more precisely as we go further and further on. But also encapsulate things like patterns and three-dimensional shapes. So it's almost everything that we see, all of the visually mathematical things that we understand can in some way be categorized in geometry. Now with that out of the way, let's just start from the basics, a basic starting point from geometry. And then we can just grow from there."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "But also encapsulate things like patterns and three-dimensional shapes. So it's almost everything that we see, all of the visually mathematical things that we understand can in some way be categorized in geometry. Now with that out of the way, let's just start from the basics, a basic starting point from geometry. And then we can just grow from there. So if we just start at a dot, that dot right over there, it's just a point. It's just that little point on that screen right over there. We'd literally call that a point."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And then we can just grow from there. So if we just start at a dot, that dot right over there, it's just a point. It's just that little point on that screen right over there. We'd literally call that a point. And I'll call that a definition. And the fun thing about mathematics is that you can make definitions. We could have called this an armadillo."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "We'd literally call that a point. And I'll call that a definition. And the fun thing about mathematics is that you can make definitions. We could have called this an armadillo. But we decide to call this a point, which I think makes sense because it's what we would call it in just everyday language as well. That is a point. Now what's interesting about a point is that it is just a position, that you can't move on a point."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "We could have called this an armadillo. But we decide to call this a point, which I think makes sense because it's what we would call it in just everyday language as well. That is a point. Now what's interesting about a point is that it is just a position, that you can't move on a point. If you moved, if you were at this point, and if you moved in any direction at all, you would no longer be at that point. So you cannot move on a point. Now there are differences between points."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Now what's interesting about a point is that it is just a position, that you can't move on a point. If you moved, if you were at this point, and if you moved in any direction at all, you would no longer be at that point. So you cannot move on a point. Now there are differences between points. For example, that's one point there. Maybe I have another point over here. And then I have another point over here."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Now there are differences between points. For example, that's one point there. Maybe I have another point over here. And then I have another point over here. And then another point over there. And you want to be able to refer to the different points. And not everyone has the luxury of a nice colored pen like I do."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And then I have another point over here. And then another point over there. And you want to be able to refer to the different points. And not everyone has the luxury of a nice colored pen like I do. Otherwise, they could refer to the green point, or the blue point, or the pink point. And so in geometry, to refer to points, we tend to give them labels. And the labels tend to have letters."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And not everyone has the luxury of a nice colored pen like I do. Otherwise, they could refer to the green point, or the blue point, or the pink point. And so in geometry, to refer to points, we tend to give them labels. And the labels tend to have letters. So for example, this could be point A. This could be point B. This would be point C. And this right over here could be point D. So if someone says, hey, circle point C, you know which one to circle."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And the labels tend to have letters. So for example, this could be point A. This could be point B. This would be point C. And this right over here could be point D. So if someone says, hey, circle point C, you know which one to circle. You know that you would have to circle that point right over there. Well, that so far, it's kind of interesting. You have these things called points."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "This would be point C. And this right over here could be point D. So if someone says, hey, circle point C, you know which one to circle. You know that you would have to circle that point right over there. Well, that so far, it's kind of interesting. You have these things called points. You really can't move around on a point. All they do is specify a position. What if we want to move around a little bit more?"}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "You have these things called points. You really can't move around on a point. All they do is specify a position. What if we want to move around a little bit more? What if we want to get from one point to another? So what if we started at one point, and we wanted all of the points, including that point, that connect that point and another point? So all of these points right over here."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "What if we want to move around a little bit more? What if we want to get from one point to another? So what if we started at one point, and we wanted all of the points, including that point, that connect that point and another point? So all of these points right over here. So what would we call this thing? All of the points that connect A and B along a straight, and I'll use everyday language here, along kind of a straight line like this. Well, we'll call this a line segment."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So all of these points right over here. So what would we call this thing? All of the points that connect A and B along a straight, and I'll use everyday language here, along kind of a straight line like this. Well, we'll call this a line segment. In everyday language, you might call it a line, but we'll call it a line segment. Because we'll see when we talk in mathematical terms, a line means something slightly different. So this is a line segment."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Well, we'll call this a line segment. In everyday language, you might call it a line, but we'll call it a line segment. Because we'll see when we talk in mathematical terms, a line means something slightly different. So this is a line segment. And if we were to connect D and C, this would also be another line segment. And once again, because we always don't have the luxury of colors, this one is clearly the orange line segment. This is clearly the yellow line segment."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So this is a line segment. And if we were to connect D and C, this would also be another line segment. And once again, because we always don't have the luxury of colors, this one is clearly the orange line segment. This is clearly the yellow line segment. We want to have labels for these line segments. And the best way to label the line segments are with its end points. And that's another word here."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "This is clearly the yellow line segment. We want to have labels for these line segments. And the best way to label the line segments are with its end points. And that's another word here. So a point is just literally A or B. But A and B are also the end points of these line segments, because it starts and ends at A and B. So let me write this."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And that's another word here. So a point is just literally A or B. But A and B are also the end points of these line segments, because it starts and ends at A and B. So let me write this. A and B are end points. Another definition right over here. Once again, we could have called them aardvarks or end armadillos."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So let me write this. A and B are end points. Another definition right over here. Once again, we could have called them aardvarks or end armadillos. But we as mathematicians decide to call them end points, because that seems to be a good name for it. And once again, we need a way to label these line segments that have the end points. And what's a better way to label a line segment than with its actual end points?"}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Once again, we could have called them aardvarks or end armadillos. But we as mathematicians decide to call them end points, because that seems to be a good name for it. And once again, we need a way to label these line segments that have the end points. And what's a better way to label a line segment than with its actual end points? So we would refer to this line segment over here. We would put its end points there. And to show that it's a line segment, we would draw a line over it, just like that."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And what's a better way to label a line segment than with its actual end points? So we would refer to this line segment over here. We would put its end points there. And to show that it's a line segment, we would draw a line over it, just like that. This line segment down here, we would write it like this. And we could have just as easily written it like this. CD with a line over it would have referred to that same line segment."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And to show that it's a line segment, we would draw a line over it, just like that. This line segment down here, we would write it like this. And we could have just as easily written it like this. CD with a line over it would have referred to that same line segment. BA with a line over it would refer to that same line segment. Now you might be saying, well, I'm not satisfied just traveling in between A and B. And this is actually another interesting idea."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "CD with a line over it would have referred to that same line segment. BA with a line over it would refer to that same line segment. Now you might be saying, well, I'm not satisfied just traveling in between A and B. And this is actually another interesting idea. When you were just on A, when you were just on a point, and you couldn't travel at all, you couldn't travel in any direction while staying on that point, that means you have zero options to travel in. You can't go up or down, left or right, in or out of the page, and still be on that point. And so that's why we say a point has zero dimensions."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And this is actually another interesting idea. When you were just on A, when you were just on a point, and you couldn't travel at all, you couldn't travel in any direction while staying on that point, that means you have zero options to travel in. You can't go up or down, left or right, in or out of the page, and still be on that point. And so that's why we say a point has zero dimensions. Zero dimensions. Now all of a sudden, we have this thing, this line segment here. And this line segment, we can at least go to the left and the right along this line segment."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And so that's why we say a point has zero dimensions. Zero dimensions. Now all of a sudden, we have this thing, this line segment here. And this line segment, we can at least go to the left and the right along this line segment. We can go towards A or towards B. So we can go back or forward in one dimension. So the line segment is a one dimensional idea, almost, or one dimensional object, although these are more kind of abstract ideas."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And this line segment, we can at least go to the left and the right along this line segment. We can go towards A or towards B. So we can go back or forward in one dimension. So the line segment is a one dimensional idea, almost, or one dimensional object, although these are more kind of abstract ideas. There is no such thing as a perfect line segment. Because everything, a line segment, you can't move up or down on this line segment while being on it. While in reality, anything that we think is a line segment, even a stick of some type, a very straight stick, or a string that is taut, that still will have some width."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So the line segment is a one dimensional idea, almost, or one dimensional object, although these are more kind of abstract ideas. There is no such thing as a perfect line segment. Because everything, a line segment, you can't move up or down on this line segment while being on it. While in reality, anything that we think is a line segment, even a stick of some type, a very straight stick, or a string that is taut, that still will have some width. But the geometrical pure line segment has no width. It only has a length here. So you can only move along the line."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "While in reality, anything that we think is a line segment, even a stick of some type, a very straight stick, or a string that is taut, that still will have some width. But the geometrical pure line segment has no width. It only has a length here. So you can only move along the line. And that's why we say it's one dimensional. A point, you can't move at all. A line segment, you can only move in that back and forth along that same direction."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So you can only move along the line. And that's why we say it's one dimensional. A point, you can't move at all. A line segment, you can only move in that back and forth along that same direction. Now I just hinted that it can actually have a length. How do you refer to that? Well, you refer to that by not writing that line on it."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "A line segment, you can only move in that back and forth along that same direction. Now I just hinted that it can actually have a length. How do you refer to that? Well, you refer to that by not writing that line on it. So if I write AB with a line on top of it like that, that means I'm referring to the actual line segment. If I say that, let me do this in a new color, if I say that AB is equal to 5 units, it might be centimeters or meters or whatever, just the abstract units 5, that means that the distance between A and B is 5, that the length of line segment AB is actually 5. Now let's keep on extending it."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Well, you refer to that by not writing that line on it. So if I write AB with a line on top of it like that, that means I'm referring to the actual line segment. If I say that, let me do this in a new color, if I say that AB is equal to 5 units, it might be centimeters or meters or whatever, just the abstract units 5, that means that the distance between A and B is 5, that the length of line segment AB is actually 5. Now let's keep on extending it. Let's say we want to just keep going in one direction. So let's say that I started A, let me do this in a new color, let's say I started A and I want to go to D, but I want the option of keep on going. So I can't go further in A's direction than A, but I can go further in D's direction."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Now let's keep on extending it. Let's say we want to just keep going in one direction. So let's say that I started A, let me do this in a new color, let's say I started A and I want to go to D, but I want the option of keep on going. So I can't go further in A's direction than A, but I can go further in D's direction. So this idea that I just showed, essentially it's like a line segment, but I can keep on going past this end point, we call this a ray. And the starting point for a ray is called the vertex, not a term that you'll see too often. You'll see vertex later on in other contexts, but it's good to know."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So I can't go further in A's direction than A, but I can go further in D's direction. So this idea that I just showed, essentially it's like a line segment, but I can keep on going past this end point, we call this a ray. And the starting point for a ray is called the vertex, not a term that you'll see too often. You'll see vertex later on in other contexts, but it's good to know. This is the vertex of the ray. It's not the vertex of this line segment, so maybe I shouldn't label it just like that. And what's interesting about a ray, it's once again a one-dimensional figure, but you could keep on going in one of the direct, you can keep on going past one of the end points."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "You'll see vertex later on in other contexts, but it's good to know. This is the vertex of the ray. It's not the vertex of this line segment, so maybe I shouldn't label it just like that. And what's interesting about a ray, it's once again a one-dimensional figure, but you could keep on going in one of the direct, you can keep on going past one of the end points. And the way that we would specify a ray is we would say, we would call it AD, and we would put this little arrow over on top of it to show that it is a ray. And in this case, it matters the order that we put the letters in. If I put DA as a ray, this would mean a different ray."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And what's interesting about a ray, it's once again a one-dimensional figure, but you could keep on going in one of the direct, you can keep on going past one of the end points. And the way that we would specify a ray is we would say, we would call it AD, and we would put this little arrow over on top of it to show that it is a ray. And in this case, it matters the order that we put the letters in. If I put DA as a ray, this would mean a different ray. That would mean that we're starting at D and then we're going past A. So this is not ray DA, this is ray AD. Now the last idea that I'm sure you're thinking about is well, what if I could keep on going in both directions?"}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "If I put DA as a ray, this would mean a different ray. That would mean that we're starting at D and then we're going past A. So this is not ray DA, this is ray AD. Now the last idea that I'm sure you're thinking about is well, what if I could keep on going in both directions? So let's say I can keep going in, let me, my diagram is getting messy, so let me introduce some more points. So let's say I have point E, and then I have point F right over here. And let's say that I have this object that goes through both E and F, but just keeps on going in both directions."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Now the last idea that I'm sure you're thinking about is well, what if I could keep on going in both directions? So let's say I can keep going in, let me, my diagram is getting messy, so let me introduce some more points. So let's say I have point E, and then I have point F right over here. And let's say that I have this object that goes through both E and F, but just keeps on going in both directions. This is, when we talk in geometry terms, this is what we call a line. Now notice, a line never ends. You can keep going in either direction."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And let's say that I have this object that goes through both E and F, but just keeps on going in both directions. This is, when we talk in geometry terms, this is what we call a line. Now notice, a line never ends. You can keep going in either direction. A line segment does end, it has end points. A line does not. And actually a line segment can sometimes be called just a segment."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "You can keep going in either direction. A line segment does end, it has end points. A line does not. And actually a line segment can sometimes be called just a segment. And so you would specify line EF, you would specify line EF with these arrows just like that. Now the thing that you're gonna see most typically when we're studying geometry are these right over here. Because we're gonna be concerned with sides of shapes, distances between points, and when you're talking about any of those things, things that have finite length, things that have an actual length, things that don't go off forever in one or two directions, then you're talking about a segment or a line segment."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And actually a line segment can sometimes be called just a segment. And so you would specify line EF, you would specify line EF with these arrows just like that. Now the thing that you're gonna see most typically when we're studying geometry are these right over here. Because we're gonna be concerned with sides of shapes, distances between points, and when you're talking about any of those things, things that have finite length, things that have an actual length, things that don't go off forever in one or two directions, then you're talking about a segment or a line segment. Now, if we go back to a line segment, just to kind of keep talking about new words that you might confront in geometry, if we go back talking about a line, and I was drawing a ray. So let's say I have point X and point Y, and so this is line segment XY, so I could specify, denote it just like that. If I have another point, let's say I have another point right over here, let's call that point Z."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Because we're gonna be concerned with sides of shapes, distances between points, and when you're talking about any of those things, things that have finite length, things that have an actual length, things that don't go off forever in one or two directions, then you're talking about a segment or a line segment. Now, if we go back to a line segment, just to kind of keep talking about new words that you might confront in geometry, if we go back talking about a line, and I was drawing a ray. So let's say I have point X and point Y, and so this is line segment XY, so I could specify, denote it just like that. If I have another point, let's say I have another point right over here, let's call that point Z. And I'll introduce another word. X, Y, and Z are on the same, they all lie on the same line, if you would imagine that a line could keep going on and on forever and ever. So we can say that X, Y, and Z are collinear."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "If I have another point, let's say I have another point right over here, let's call that point Z. And I'll introduce another word. X, Y, and Z are on the same, they all lie on the same line, if you would imagine that a line could keep going on and on forever and ever. So we can say that X, Y, and Z are collinear. So those three points are collinear. They all sit on the same line, and they also all sit on line segment XY. Now let's say we know, we're told, that XZ is equal to ZY, and they are all collinear."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So we can say that X, Y, and Z are collinear. So those three points are collinear. They all sit on the same line, and they also all sit on line segment XY. Now let's say we know, we're told, that XZ is equal to ZY, and they are all collinear. So that means this is telling us that the distance between X and Z is the same as the distance between Z and Y. So sometimes we can mark it like that. This distance is the same as that distance over there."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Now let's say we know, we're told, that XZ is equal to ZY, and they are all collinear. So that means this is telling us that the distance between X and Z is the same as the distance between Z and Y. So sometimes we can mark it like that. This distance is the same as that distance over there. So that tells us that Z is exactly halfway between X and Y. So in this situation, we would call Z the midpoint, the midpoint of line segment XY, because it's exactly halfway between. Now to finish up, we've talked about things that have zero dimensions, points."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "This distance is the same as that distance over there. So that tells us that Z is exactly halfway between X and Y. So in this situation, we would call Z the midpoint, the midpoint of line segment XY, because it's exactly halfway between. Now to finish up, we've talked about things that have zero dimensions, points. We've talked about things that have one dimension, a line, a line segment, or a ray. You might say, well, what has two dimensions? Well, in order to have two dimensions, that means I can go backwards and forwards in two different directions."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Now to finish up, we've talked about things that have zero dimensions, points. We've talked about things that have one dimension, a line, a line segment, or a ray. You might say, well, what has two dimensions? Well, in order to have two dimensions, that means I can go backwards and forwards in two different directions. So this page right here, or this video, or the screen that you're looking at is a two-dimensional object. I can go right, left, that is one dimension, or I can go up, down. And so this surface of the monitor you're looking at is actually two dimensions."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Well, in order to have two dimensions, that means I can go backwards and forwards in two different directions. So this page right here, or this video, or the screen that you're looking at is a two-dimensional object. I can go right, left, that is one dimension, or I can go up, down. And so this surface of the monitor you're looking at is actually two dimensions. Two dimensions. You can go backwards or forwards in two directions. And things that are two dimensions, we call them planar, or we call them planes."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And so this surface of the monitor you're looking at is actually two dimensions. Two dimensions. You can go backwards or forwards in two directions. And things that are two dimensions, we call them planar, or we call them planes. So if you took a piece of paper that extended forever, it just extended in every direction forever, that in the geometrical sense was a plane. The piece of paper itself, the thing that's finite, and you'll never see this talked about in a typical geometry class, but I guess if we were to draw the analogy, you could call a piece of paper maybe a plane segment, because it's a segment of an entire plane. If you had a third dimension, then you're talking about kind of our three-dimensional space and three-dimensional space, not only could you move left or right along the screen or up and down, you could also move in and out of the screen."}, {"video_title": "Basic geometry language and labels Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And things that are two dimensions, we call them planar, or we call them planes. So if you took a piece of paper that extended forever, it just extended in every direction forever, that in the geometrical sense was a plane. The piece of paper itself, the thing that's finite, and you'll never see this talked about in a typical geometry class, but I guess if we were to draw the analogy, you could call a piece of paper maybe a plane segment, because it's a segment of an entire plane. If you had a third dimension, then you're talking about kind of our three-dimensional space and three-dimensional space, not only could you move left or right along the screen or up and down, you could also move in and out of the screen. You could also have this dimension that I'll try to draw. You could go into the screen or you could go out of the screen like that. And as we go into higher and higher mathematics, although it becomes very hard to visualize, you'll see that we can even start to study things that have more than three dimensions."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy.mp3", "Sentence": "So that's one line, and then let me draw another line that is parallel to that. I'm claiming that these are parallel lines. And now I'm gonna draw some transversals here. So first let me draw a horizontal transversal. So just like that. And then let me do a vertical transversal. So just like that."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy.mp3", "Sentence": "So first let me draw a horizontal transversal. So just like that. And then let me do a vertical transversal. So just like that. And I'm assuming that the green one is horizontal and the blue one is vertical. So we assume that they are perpendicular to each other, that these intersect at right angles. And from this, I'm going to figure out, I'm gonna use some parallel line angle properties to establish that this triangle and this triangle are similar, and then use that to establish that both of these lines, both of these yellow lines have the same slope."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy.mp3", "Sentence": "So just like that. And I'm assuming that the green one is horizontal and the blue one is vertical. So we assume that they are perpendicular to each other, that these intersect at right angles. And from this, I'm going to figure out, I'm gonna use some parallel line angle properties to establish that this triangle and this triangle are similar, and then use that to establish that both of these lines, both of these yellow lines have the same slope. So actually let me label some points here. So let's call that point A, point B, point C, point D, and point E. So let's see. First of all, we know that angle CED is going to be congruent to angle AEB because they're both right angles."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy.mp3", "Sentence": "And from this, I'm going to figure out, I'm gonna use some parallel line angle properties to establish that this triangle and this triangle are similar, and then use that to establish that both of these lines, both of these yellow lines have the same slope. So actually let me label some points here. So let's call that point A, point B, point C, point D, and point E. So let's see. First of all, we know that angle CED is going to be congruent to angle AEB because they're both right angles. So that's a right angle, and then that is a right angle right over there. We also know some things about corresponding angles where a transversal intersects parallel lines. This angle corresponds to this angle if we look at the blue transversal as it intersects those two lines."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy.mp3", "Sentence": "First of all, we know that angle CED is going to be congruent to angle AEB because they're both right angles. So that's a right angle, and then that is a right angle right over there. We also know some things about corresponding angles where a transversal intersects parallel lines. This angle corresponds to this angle if we look at the blue transversal as it intersects those two lines. And so they're going to have the same measure. They're going to be congruent. Now this angle on one side of this point B is going to also be congruent to that because they are vertical angles, and we've seen that multiple times before."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy.mp3", "Sentence": "This angle corresponds to this angle if we look at the blue transversal as it intersects those two lines. And so they're going to have the same measure. They're going to be congruent. Now this angle on one side of this point B is going to also be congruent to that because they are vertical angles, and we've seen that multiple times before. And so we know that this angle, angle ABE, is congruent to angle ECD. Sometimes this is called alternate interior angles of a transversal and parallel lines. Well, if you look at triangle CED and triangle ABE, we see they already have two angles in common."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy.mp3", "Sentence": "Now this angle on one side of this point B is going to also be congruent to that because they are vertical angles, and we've seen that multiple times before. And so we know that this angle, angle ABE, is congruent to angle ECD. Sometimes this is called alternate interior angles of a transversal and parallel lines. Well, if you look at triangle CED and triangle ABE, we see they already have two angles in common. So if they have two angles in common, well, their third angle has to be in common because this third angle's just going to be 180 minus these other two. And so this third angle's just going to be 180 minus the other two. And so just like that, we notice we have all three angles are the same in both of these triangles, or they're not all the same, but all of the corresponding angles, I should say, are the same."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy.mp3", "Sentence": "Well, if you look at triangle CED and triangle ABE, we see they already have two angles in common. So if they have two angles in common, well, their third angle has to be in common because this third angle's just going to be 180 minus these other two. And so this third angle's just going to be 180 minus the other two. And so just like that, we notice we have all three angles are the same in both of these triangles, or they're not all the same, but all of the corresponding angles, I should say, are the same. Angle, this blue angle has the same measure as this blue angle. This magenta angle has the same measure as this magenta angle. And then the other angles are right angles."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy.mp3", "Sentence": "And so just like that, we notice we have all three angles are the same in both of these triangles, or they're not all the same, but all of the corresponding angles, I should say, are the same. Angle, this blue angle has the same measure as this blue angle. This magenta angle has the same measure as this magenta angle. And then the other angles are right angles. These are right triangles here. So we could say triangle AEB, triangle AEB, is similar, similar, is similar to triangle DEC, triangle DEC, by, and we could say by angle, angle, angle. All the corresponding angles are congruent."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy.mp3", "Sentence": "And then the other angles are right angles. These are right triangles here. So we could say triangle AEB, triangle AEB, is similar, similar, is similar to triangle DEC, triangle DEC, by, and we could say by angle, angle, angle. All the corresponding angles are congruent. So we are dealing with similar triangles. And so we know similar triangles, the ratio of corresponding sides are going to be the same. So we could say that the ratio of, let's say the ratio of BE, the ratio of BE, let me write this down."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy.mp3", "Sentence": "All the corresponding angles are congruent. So we are dealing with similar triangles. And so we know similar triangles, the ratio of corresponding sides are going to be the same. So we could say that the ratio of, let's say the ratio of BE, the ratio of BE, let me write this down. This is this side right over here. The ratio of BE to AE, to AE, to AE, is going to be equal to, so that side over that side, well what is the corresponding side? The corresponding side to BE is side CE."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy.mp3", "Sentence": "So we could say that the ratio of, let's say the ratio of BE, the ratio of BE, let me write this down. This is this side right over here. The ratio of BE to AE, to AE, to AE, is going to be equal to, so that side over that side, well what is the corresponding side? The corresponding side to BE is side CE. So that's going to be the same as the ratio between CE and DE, and DE. And this just comes out of similar, the similarity of the triangles, CE to DE. So once again, once we establish these triangles are similar we can say the ratio of corresponding sides are going to be the same."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy.mp3", "Sentence": "The corresponding side to BE is side CE. So that's going to be the same as the ratio between CE and DE, and DE. And this just comes out of similar, the similarity of the triangles, CE to DE. So once again, once we establish these triangles are similar we can say the ratio of corresponding sides are going to be the same. Now what is the ratio between BE and AE? The ratio between BE and AE. Well that is the slope of this top line right over here."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy.mp3", "Sentence": "So once again, once we establish these triangles are similar we can say the ratio of corresponding sides are going to be the same. Now what is the ratio between BE and AE? The ratio between BE and AE. Well that is the slope of this top line right over here. We could say that's the slope of line AB. Slope, slope of line connecting, connecting A to B. Or actually let me just use, I could write it like this."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy.mp3", "Sentence": "Well that is the slope of this top line right over here. We could say that's the slope of line AB. Slope, slope of line connecting, connecting A to B. Or actually let me just use, I could write it like this. That is slope of, slope of A, slope of line AB. Remember slope is, when you're going from A to B, it's change in Y over change in X. So when you're going from A to B, your change in X is AE, and your change in Y is BE, or EB, however you want to refer to it."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy.mp3", "Sentence": "Or actually let me just use, I could write it like this. That is slope of, slope of A, slope of line AB. Remember slope is, when you're going from A to B, it's change in Y over change in X. So when you're going from A to B, your change in X is AE, and your change in Y is BE, or EB, however you want to refer to it. So this right over here is change in Y, and this over here is change in X. Well now let's look at this second expression right over here, CE over DE. CE over DE."}, {"video_title": "Proof parallel lines have the same slope High School Math Khan Academy.mp3", "Sentence": "So when you're going from A to B, your change in X is AE, and your change in Y is BE, or EB, however you want to refer to it. So this right over here is change in Y, and this over here is change in X. Well now let's look at this second expression right over here, CE over DE. CE over DE. Well now this is going to be change in Y over change in X between point C and D. So this is, this right over here, this is the slope of line, of line CD. And so just like that, by establishing similarity, we were able to see the ratio of corresponding sides are congruent, which shows us that the slopes of these two lines are going to be the same. And we are done."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "So side, side, side works. What about angle, angle, angle? So let me do that over here. What about angle, angle, angle? So what I'm saying is, is if, let's say I have a triangle like this, like I have a triangle like that, and I have a triangle like this, and if we know, if we know that this angle is congruent to that angle, if this angle is congruent to that angle, which means that their measures are equal, or and, I should say and, and that angle is congruent to that angle, can we say that this is, that these are two congruent triangles, and in the first case, it looks like maybe it is, at least the way I drew it here. But when you think about it, you can have the exact same corresponding angles being, having the same measure or being congruent, but you can actually scale one of these triangles up and down and still have that property. For example, if I had this triangle right over here, this triangle right over here, it looks similar, and I'm using that in just the everyday language size, it has the same shape as these triangles right over here, and it has the same, it has the same angles, that angle is congruent to that angle, this angle down here is congruent to this angle over here, and this angle over here is congruent to this angle over here."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "What about angle, angle, angle? So what I'm saying is, is if, let's say I have a triangle like this, like I have a triangle like that, and I have a triangle like this, and if we know, if we know that this angle is congruent to that angle, if this angle is congruent to that angle, which means that their measures are equal, or and, I should say and, and that angle is congruent to that angle, can we say that this is, that these are two congruent triangles, and in the first case, it looks like maybe it is, at least the way I drew it here. But when you think about it, you can have the exact same corresponding angles being, having the same measure or being congruent, but you can actually scale one of these triangles up and down and still have that property. For example, if I had this triangle right over here, this triangle right over here, it looks similar, and I'm using that in just the everyday language size, it has the same shape as these triangles right over here, and it has the same, it has the same angles, that angle is congruent to that angle, this angle down here is congruent to this angle over here, and this angle over here is congruent to this angle over here. So all of the angles in all three of these triangles are the same, the corresponding angles have the same measure but clearly, clearly this triangle right over here is not the same, it is not congruent to the other two, the sides have a very different length. This side is much shorter than this side right over here, this side is much shorter than that side over there, and this side is much shorter over here. So with just angle, angle, angle, you cannot say that a triangle has the same size and shape."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "For example, if I had this triangle right over here, this triangle right over here, it looks similar, and I'm using that in just the everyday language size, it has the same shape as these triangles right over here, and it has the same, it has the same angles, that angle is congruent to that angle, this angle down here is congruent to this angle over here, and this angle over here is congruent to this angle over here. So all of the angles in all three of these triangles are the same, the corresponding angles have the same measure but clearly, clearly this triangle right over here is not the same, it is not congruent to the other two, the sides have a very different length. This side is much shorter than this side right over here, this side is much shorter than that side over there, and this side is much shorter over here. So with just angle, angle, angle, you cannot say that a triangle has the same size and shape. It does have the same shape but not the same size. So this does not imply, this does not imply congruency. So angle, angle, angle does not imply congruency."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "So with just angle, angle, angle, you cannot say that a triangle has the same size and shape. It does have the same shape but not the same size. So this does not imply, this does not imply congruency. So angle, angle, angle does not imply congruency. What it does imply, and we haven't talked about this yet, is that these are similar triangles. So angle, angle, angle implies similar. So let me write it over here, it implies similar triangles."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "So angle, angle, angle does not imply congruency. What it does imply, and we haven't talked about this yet, is that these are similar triangles. So angle, angle, angle implies similar. So let me write it over here, it implies similar triangles. And similar, you probably are used to the word in just everyday language, but similar has a very specific meaning in geometry, and similar things have the same shape but not necessarily the same size. So anything that is congruent, because it has the same size and shape, is also similar, but not everything that is similar is also congruent. So for example, this triangle is similar, all of these triangles are similar to each other, but they aren't all congruent."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "So let me write it over here, it implies similar triangles. And similar, you probably are used to the word in just everyday language, but similar has a very specific meaning in geometry, and similar things have the same shape but not necessarily the same size. So anything that is congruent, because it has the same size and shape, is also similar, but not everything that is similar is also congruent. So for example, this triangle is similar, all of these triangles are similar to each other, but they aren't all congruent. These two are congruent if their sides are the same, I didn't make that assumption, but if we know that their sides are the same, then we can say that they're congruent, but neither of these are congruent to this one right over here because this is clearly much larger, has the same shape but a different size. So we can't have an AAA postulate or an AAA axiom to get to congruency. What about side-angle side?"}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "So for example, this triangle is similar, all of these triangles are similar to each other, but they aren't all congruent. These two are congruent if their sides are the same, I didn't make that assumption, but if we know that their sides are the same, then we can say that they're congruent, but neither of these are congruent to this one right over here because this is clearly much larger, has the same shape but a different size. So we can't have an AAA postulate or an AAA axiom to get to congruency. What about side-angle side? So let's try this out, side-angle side. So let's start off with one triangle right over here. So let's start off with a triangle that looks like this, I have my blue side, I have my pink side, and I have my magenta side."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "What about side-angle side? So let's try this out, side-angle side. So let's start off with one triangle right over here. So let's start off with a triangle that looks like this, I have my blue side, I have my pink side, and I have my magenta side. And let's say that I have another triangle that has this blue side, it has the same length as that blue side, so let me draw it like that, it has the same length as that blue side, so that length and that length are going to be the same, it has a congruent angle right after that, so this angle and the next angle for this triangle are going to have the same measure, they're going to be congruent, and then the next side is going to have the same length of this one over here, so that's going to be the same length as this over here. So it's going to be the same length. And we don't know, and because we only know that two of the corresponding sides have the same length, and the angle between them, and this is important, the angle between the two corresponding sides also have the same measure, we can do anything we want with this last side on this one."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "So let's start off with a triangle that looks like this, I have my blue side, I have my pink side, and I have my magenta side. And let's say that I have another triangle that has this blue side, it has the same length as that blue side, so let me draw it like that, it has the same length as that blue side, so that length and that length are going to be the same, it has a congruent angle right after that, so this angle and the next angle for this triangle are going to have the same measure, they're going to be congruent, and then the next side is going to have the same length of this one over here, so that's going to be the same length as this over here. So it's going to be the same length. And we don't know, and because we only know that two of the corresponding sides have the same length, and the angle between them, and this is important, the angle between the two corresponding sides also have the same measure, we can do anything we want with this last side on this one. We can essentially, it's going to have to start right over here. You could start from this point. And we can pivot it to form any triangle we want."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "And we don't know, and because we only know that two of the corresponding sides have the same length, and the angle between them, and this is important, the angle between the two corresponding sides also have the same measure, we can do anything we want with this last side on this one. We can essentially, it's going to have to start right over here. You could start from this point. And we can pivot it to form any triangle we want. But we can see the only way we can form a triangle is if we bring this side all the way over here and close this right over there. And so we can see just logically that if we have, if for two triangles, they have one side that has the length the same, the next side has the length the same, and the angle in between them. So this angle, let me do that in the same color."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "And we can pivot it to form any triangle we want. But we can see the only way we can form a triangle is if we bring this side all the way over here and close this right over there. And so we can see just logically that if we have, if for two triangles, they have one side that has the length the same, the next side has the length the same, and the angle in between them. So this angle, let me do that in the same color. This angle in between them, this is the angle. This a is this angle and that angle. It's the angle in between them."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "So this angle, let me do that in the same color. This angle in between them, this is the angle. This a is this angle and that angle. It's the angle in between them. This first side is in blue, and the second side right over here is in pink. And well, it's already written in pink. So we can see that if two sides have the same length, two corresponding sides have the same length, and the corresponding angle between them, they have to be congruent."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "It's the angle in between them. This first side is in blue, and the second side right over here is in pink. And well, it's already written in pink. So we can see that if two sides have the same length, two corresponding sides have the same length, and the corresponding angle between them, they have to be congruent. There's no other place to put this third side. So SAS, and sometimes it's once again called a postulate, an axiom, or if it's kind of proven, sometimes it's called a theorem. This does imply that the two triangles are congruent."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "So we can see that if two sides have the same length, two corresponding sides have the same length, and the corresponding angle between them, they have to be congruent. There's no other place to put this third side. So SAS, and sometimes it's once again called a postulate, an axiom, or if it's kind of proven, sometimes it's called a theorem. This does imply that the two triangles are congruent. So we will give ourselves this tool in our toolkit. We had the SSS postulate. Now we have the SAS postulate."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "This does imply that the two triangles are congruent. So we will give ourselves this tool in our toolkit. We had the SSS postulate. Now we have the SAS postulate. Two sides are equal and the angle in between them for two triangles, corresponding sides and angles, then we can say that these are congruent triangles. Now what about, and I'm just going to try to go through all of the different combinations here. What if I have something, what if I have angle, side, angle?"}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "Now we have the SAS postulate. Two sides are equal and the angle in between them for two triangles, corresponding sides and angles, then we can say that these are congruent triangles. Now what about, and I'm just going to try to go through all of the different combinations here. What if I have something, what if I have angle, side, angle? So let me try that. So what happens if I have angle, side, angle? So let's go back to this one right over here."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "What if I have something, what if I have angle, side, angle? So let me try that. So what happens if I have angle, side, angle? So let's go back to this one right over here. So actually, let me just redraw a new one for each of these cases. So angle, side, angle. So I'll draw a triangle here."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "So let's go back to this one right over here. So actually, let me just redraw a new one for each of these cases. So angle, side, angle. So I'll draw a triangle here. So I have this triangle. So this would be maybe the side. That would be the side."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "So I'll draw a triangle here. So I have this triangle. So this would be maybe the side. That would be the side. Let me draw the whole triangle actually first. So I have this triangle. Let me draw one side over here."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "That would be the side. Let me draw the whole triangle actually first. So I have this triangle. Let me draw one side over here. And then let me draw one side over there. And then this angle right over here, I'll do it in orange. And this angle over here, I will do it in yellow."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "Let me draw one side over here. And then let me draw one side over there. And then this angle right over here, I'll do it in orange. And this angle over here, I will do it in yellow. So if I have another triangle that has one side having equal measure, so I'll use it as this blue side right over here. So it has one side that has equal measure. And the two angles on either side of that side or either end of that side are the same, will this triangle necessarily be congruent?"}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "And this angle over here, I will do it in yellow. So if I have another triangle that has one side having equal measure, so I'll use it as this blue side right over here. So it has one side that has equal measure. And the two angles on either side of that side or either end of that side are the same, will this triangle necessarily be congruent? And we're just going to try to reason it out. These aren't formal proofs. We're really just trying to set up what are reasonable postulates or what are reasonable assumptions we can have in our toolkit as we try to prove other things."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "And the two angles on either side of that side or either end of that side are the same, will this triangle necessarily be congruent? And we're just going to try to reason it out. These aren't formal proofs. We're really just trying to set up what are reasonable postulates or what are reasonable assumptions we can have in our toolkit as we try to prove other things. So that angle, let's call it that angle right over there, is going to be the same measure in this triangle. And this angle right over here in yellow is going to have the same measure on this triangle right over here. So regardless, I'm not in any way constraining the sides over here."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "We're really just trying to set up what are reasonable postulates or what are reasonable assumptions we can have in our toolkit as we try to prove other things. So that angle, let's call it that angle right over there, is going to be the same measure in this triangle. And this angle right over here in yellow is going to have the same measure on this triangle right over here. So regardless, I'm not in any way constraining the sides over here. So this side right over here could have any length. But it has to form this angle with it. So it could have any length."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "So regardless, I'm not in any way constraining the sides over here. So this side right over here could have any length. But it has to form this angle with it. So it could have any length. And it can just go as far as it wants to go. No way have we constrained what the length of that is. Actually, let me mark this off too."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "So it could have any length. And it can just go as far as it wants to go. No way have we constrained what the length of that is. Actually, let me mark this off too. So this is the same as this. So that side can be anything. We haven't constrained it at all."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "Actually, let me mark this off too. So this is the same as this. So that side can be anything. We haven't constrained it at all. And once again, this side could be anything. We haven't constrained it at all, but we know it has to go at this angle. So it has to go at that angle."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "We haven't constrained it at all. And once again, this side could be anything. We haven't constrained it at all, but we know it has to go at this angle. So it has to go at that angle. Well, once again, there's only one triangle that can be formed this way. We can say all day that this length could be as long as we want or as short as we want. And this one could be as long as we want or as short as we want."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "So it has to go at that angle. Well, once again, there's only one triangle that can be formed this way. We can say all day that this length could be as long as we want or as short as we want. And this one could be as long as we want or as short as we want. But the only way that they can actually touch each other and form a triangle and have these two angles is if they are the exact same length as these two sides right over here. So this side will actually have to be the same as that side. And this would have to be the same as that side."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "And this one could be as long as we want or as short as we want. But the only way that they can actually touch each other and form a triangle and have these two angles is if they are the exact same length as these two sides right over here. So this side will actually have to be the same as that side. And this would have to be the same as that side. Once again, this isn't a proof. I'd call it more of a reasoning through it or an investigation, really just to establish what are reasonable baselines or axioms or assumptions or postulates that we can have. So for my purposes, I think ASA does show us that two triangles are congruent."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "And this would have to be the same as that side. Once again, this isn't a proof. I'd call it more of a reasoning through it or an investigation, really just to establish what are reasonable baselines or axioms or assumptions or postulates that we can have. So for my purposes, I think ASA does show us that two triangles are congruent. Now let's try another one. Let's try angle-angle-side. Let's try angle-angle-side."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "So for my purposes, I think ASA does show us that two triangles are congruent. Now let's try another one. Let's try angle-angle-side. Let's try angle-angle-side. And in some geometry classes, and maybe if you have to go through an exam quickly, you might memorize, OK, side-side-side implies congruency. And that's kind of logical. Side-angle-side implies congruency, and so on and so forth."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "Let's try angle-angle-side. And in some geometry classes, and maybe if you have to go through an exam quickly, you might memorize, OK, side-side-side implies congruency. And that's kind of logical. Side-angle-side implies congruency, and so on and so forth. I'm not a fan of memorizing it. It might be good for time pressure. It is good to sometimes even just go through this logic."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "Side-angle-side implies congruency, and so on and so forth. I'm not a fan of memorizing it. It might be good for time pressure. It is good to sometimes even just go through this logic. If you're like, wait, does angle-angle-angle work? Well, no. I can find this case that breaks down angle-angle-angle."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "It is good to sometimes even just go through this logic. If you're like, wait, does angle-angle-angle work? Well, no. I can find this case that breaks down angle-angle-angle. If these work, just try to verify for yourself that they make logical sense why they would imply congruency. Now let's try angle-angle-side. Let's try angle-angle-side."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "I can find this case that breaks down angle-angle-angle. If these work, just try to verify for yourself that they make logical sense why they would imply congruency. Now let's try angle-angle-side. Let's try angle-angle-side. So once again, let's have a triangle over here. It has some side. So this one's going to be a little bit more interesting."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "Let's try angle-angle-side. So once again, let's have a triangle over here. It has some side. So this one's going to be a little bit more interesting. So it has some side. That's the side right over there. And then it has two angles."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "So this one's going to be a little bit more interesting. So it has some side. That's the side right over there. And then it has two angles. So let me draw the other sides of this triangle. I'll draw one in magenta and then one in green. And there's two angles and then the side."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "And then it has two angles. So let me draw the other sides of this triangle. I'll draw one in magenta and then one in green. And there's two angles and then the side. So let's say you have this angle. You have that angle right over there. Actually, I didn't have to put a double because that's the first angle that I'm."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "And there's two angles and then the side. So let's say you have this angle. You have that angle right over there. Actually, I didn't have to put a double because that's the first angle that I'm. So I have that angle, which we refer to as that first a. Then we have this angle, which is that second a. So if I know that there's another triangle that has one side having the same length, so let me draw it like that."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "Actually, I didn't have to put a double because that's the first angle that I'm. So I have that angle, which we refer to as that first a. Then we have this angle, which is that second a. So if I know that there's another triangle that has one side having the same length, so let me draw it like that. It has one side having the same length. It has one angle on that side that has the same measure. So it has a measure like that."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "So if I know that there's another triangle that has one side having the same length, so let me draw it like that. It has one side having the same length. It has one angle on that side that has the same measure. So it has a measure like that. And so this side right over here could be of any length. We aren't constraining what the length of that side is. But whatever the angle is on the other side of that side is going to be the same as this green angle right over here."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "So it has a measure like that. And so this side right over here could be of any length. We aren't constraining what the length of that side is. But whatever the angle is on the other side of that side is going to be the same as this green angle right over here. So for example, it could be like that. And then you could have a green side go like that. It could be like that and have the green side go like that."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "But whatever the angle is on the other side of that side is going to be the same as this green angle right over here. So for example, it could be like that. And then you could have a green side go like that. It could be like that and have the green side go like that. And if we have, so the only thing we're assuming is that this is the same length as this, and that this angle is the same measure as that angle, and that this measure is the same measure as that angle. And this magenta line can be of any length, and this green line can be of any length. We in no way have constrained that."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "It could be like that and have the green side go like that. And if we have, so the only thing we're assuming is that this is the same length as this, and that this angle is the same measure as that angle, and that this measure is the same measure as that angle. And this magenta line can be of any length, and this green line can be of any length. We in no way have constrained that. But can we form any triangle that is not congruent to this? Because the bottom line is this green line is going to have to touch this one right over there. And the only way it's going to touch that one right over there is if it starts right over here."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "We in no way have constrained that. But can we form any triangle that is not congruent to this? Because the bottom line is this green line is going to have to touch this one right over there. And the only way it's going to touch that one right over there is if it starts right over here. Because we're constraining this angle right over here. We're constraining that angle. And so it looks like angle-angle side does indeed imply congruency."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "And the only way it's going to touch that one right over there is if it starts right over here. Because we're constraining this angle right over here. We're constraining that angle. And so it looks like angle-angle side does indeed imply congruency. So that does imply congruency. So let's just do one more just to kind of try out all of the different situations. What if we have, and I'm running out of a little bit of real estate right over here at the bottom."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "And so it looks like angle-angle side does indeed imply congruency. So that does imply congruency. So let's just do one more just to kind of try out all of the different situations. What if we have, and I'm running out of a little bit of real estate right over here at the bottom. What if we tried out side-side angle? So once again, draw a triangle. So it has one side there."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "What if we have, and I'm running out of a little bit of real estate right over here at the bottom. What if we tried out side-side angle? So once again, draw a triangle. So it has one side there. It has another side there. And then I don't have to do those hash marks just yet. So one side, then another side, and then another side."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "So it has one side there. It has another side there. And then I don't have to do those hash marks just yet. So one side, then another side, and then another side. And what happens if we know that there's another triangle that has two of the sides the same, and then the angle after it? So for example, it would have that side just like that. And then it has another side, but we're not constraining the angle."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "So one side, then another side, and then another side. And what happens if we know that there's another triangle that has two of the sides the same, and then the angle after it? So for example, it would have that side just like that. And then it has another side, but we're not constraining the angle. We aren't constraining this angle right over here, but we're constraining the length of that side. So let me color code it. So that blue side is that first side."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "And then it has another side, but we're not constraining the angle. We aren't constraining this angle right over here, but we're constraining the length of that side. So let me color code it. So that blue side is that first side. Then we have this magenta side right over there. So this is going to be the same length as this right over here. But let me make it at a different angle to see if I can disprove it."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "So that blue side is that first side. Then we have this magenta side right over there. So this is going to be the same length as this right over here. But let me make it at a different angle to see if I can disprove it. So let's say it looks like that. Or actually, let me make it even more interesting. Let me try to make it like that."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "But let me make it at a different angle to see if I can disprove it. So let's say it looks like that. Or actually, let me make it even more interesting. Let me try to make it like that. So it's a very different angle. But now it has to have the same angle out here. It has to have that same angle out here."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "Let me try to make it like that. So it's a very different angle. But now it has to have the same angle out here. It has to have that same angle out here. So it has to be roughly that angle. So it actually looks like we can draw a triangle that is not congruent, that has the same two sides being the same length, and then an angle is different. For example, this is pretty much that."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "It has to have that same angle out here. So it has to be roughly that angle. So it actually looks like we can draw a triangle that is not congruent, that has the same two sides being the same length, and then an angle is different. For example, this is pretty much that. I made this angle smaller than this angle. These two sides are the same. This angle is the same now, but what the byproduct of that is that this green side is going to be shorter on this triangle right over here."}, {"video_title": "Other triangle congruence postulates Congruence Geometry Khan Academy.mp3", "Sentence": "For example, this is pretty much that. I made this angle smaller than this angle. These two sides are the same. This angle is the same now, but what the byproduct of that is that this green side is going to be shorter on this triangle right over here. So you don't necessarily have congruent triangles with side-side angle. So this is not necessarily congruent, or similar. It gives us neither congruency nor similarity."}, {"video_title": "Corresponding parts of congruent triangles are congruent.mp3", "Sentence": "So in algebra, when something is equal to another thing, it means that their quantities are the same. But if we're now all of a sudden talking about shapes, and we say that those shapes are the same, the shapes are the same size and shape, then we say that they're congruent. And just to see a simple example here, I have this triangle right over there. And let's say I have this triangle right over here. And if you are able to shift this triangle and rotate this triangle and flip this triangle, you can make it look exactly like this triangle, as long as you're not changing the lengths of any of the sides or the angles here. But you can flip it. You can shift it."}, {"video_title": "Corresponding parts of congruent triangles are congruent.mp3", "Sentence": "And let's say I have this triangle right over here. And if you are able to shift this triangle and rotate this triangle and flip this triangle, you can make it look exactly like this triangle, as long as you're not changing the lengths of any of the sides or the angles here. But you can flip it. You can shift it. You can do any and rotate it. So you can shift. Let me write this."}, {"video_title": "Corresponding parts of congruent triangles are congruent.mp3", "Sentence": "You can shift it. You can do any and rotate it. So you can shift. Let me write this. You can shift it. You can flip it. You can flip it."}, {"video_title": "Corresponding parts of congruent triangles are congruent.mp3", "Sentence": "Let me write this. You can shift it. You can flip it. You can flip it. And you can rotate. If you can do those three procedures to make the exact same triangle, to make them look exactly the same, then they are congruent. And if you say that a triangle is congruent, and let me label these."}, {"video_title": "Corresponding parts of congruent triangles are congruent.mp3", "Sentence": "You can flip it. And you can rotate. If you can do those three procedures to make the exact same triangle, to make them look exactly the same, then they are congruent. And if you say that a triangle is congruent, and let me label these. So let's call this triangle A, B, and C. And let's call this D. Let me call it X, Y, and Z. So if we were to say, if we make the claim that both of these triangles are congruent, so if we say triangle A, B, C is congruent, and the way you specified it, it looks almost like an equal sign. But it's an equal sign with this little curly thing on top."}, {"video_title": "Corresponding parts of congruent triangles are congruent.mp3", "Sentence": "And if you say that a triangle is congruent, and let me label these. So let's call this triangle A, B, and C. And let's call this D. Let me call it X, Y, and Z. So if we were to say, if we make the claim that both of these triangles are congruent, so if we say triangle A, B, C is congruent, and the way you specified it, it looks almost like an equal sign. But it's an equal sign with this little curly thing on top. Let me write it a little bit neater. So we would write it like this. If we know that triangle A, B, C is congruent to triangle X, Y, Z, that means that their corresponding sides have the same length and their corresponding angles and their corresponding angles have the same measure."}, {"video_title": "Corresponding parts of congruent triangles are congruent.mp3", "Sentence": "But it's an equal sign with this little curly thing on top. Let me write it a little bit neater. So we would write it like this. If we know that triangle A, B, C is congruent to triangle X, Y, Z, that means that their corresponding sides have the same length and their corresponding angles and their corresponding angles have the same measure. So if we make this assumption, or if someone tells us that this is true, then we know, for example, that AB is going to be equal to XY. The length of segment AB is going to be equal to the length of segment XY. And we could denote it like this."}, {"video_title": "Corresponding parts of congruent triangles are congruent.mp3", "Sentence": "If we know that triangle A, B, C is congruent to triangle X, Y, Z, that means that their corresponding sides have the same length and their corresponding angles and their corresponding angles have the same measure. So if we make this assumption, or if someone tells us that this is true, then we know, for example, that AB is going to be equal to XY. The length of segment AB is going to be equal to the length of segment XY. And we could denote it like this. And I'm assuming that these are the corresponding sides. And you can see it, actually, by the way we've defined these triangles. A corresponds to X, B corresponds to Y, and then C corresponds to Z right over there."}, {"video_title": "Corresponding parts of congruent triangles are congruent.mp3", "Sentence": "And we could denote it like this. And I'm assuming that these are the corresponding sides. And you can see it, actually, by the way we've defined these triangles. A corresponds to X, B corresponds to Y, and then C corresponds to Z right over there. So AB, side AB, is going to have the same length as side XY. And you can sometimes, if you don't have the colors, you would denote it just like that. These two lengths are, or these two line segments have the same length."}, {"video_title": "Corresponding parts of congruent triangles are congruent.mp3", "Sentence": "A corresponds to X, B corresponds to Y, and then C corresponds to Z right over there. So AB, side AB, is going to have the same length as side XY. And you can sometimes, if you don't have the colors, you would denote it just like that. These two lengths are, or these two line segments have the same length. And you can actually say this, and you don't always see it written this way. You could also make the statement that line segment AB is congruent to line segment XY. But congruence of line segments really just means that their lengths are equivalent."}, {"video_title": "Corresponding parts of congruent triangles are congruent.mp3", "Sentence": "These two lengths are, or these two line segments have the same length. And you can actually say this, and you don't always see it written this way. You could also make the statement that line segment AB is congruent to line segment XY. But congruence of line segments really just means that their lengths are equivalent. So these two things mean the same thing. If one line segment is congruent to another line segment, that just means the measure of one line segment is equal to the measure of the other line segment. And so we can go through all the corresponding sides."}, {"video_title": "Corresponding parts of congruent triangles are congruent.mp3", "Sentence": "But congruence of line segments really just means that their lengths are equivalent. So these two things mean the same thing. If one line segment is congruent to another line segment, that just means the measure of one line segment is equal to the measure of the other line segment. And so we can go through all the corresponding sides. If these two characters are congruent, we also know that BC is going to be the length of YZ, assuming that those are the corresponding sides. And we could put these double hash marks right over here to show that these two lengths are the same. And then if we go to the third side, we also know that these are going to have the same length, or the line segments themselves are going to be congruent."}, {"video_title": "Corresponding parts of congruent triangles are congruent.mp3", "Sentence": "And so we can go through all the corresponding sides. If these two characters are congruent, we also know that BC is going to be the length of YZ, assuming that those are the corresponding sides. And we could put these double hash marks right over here to show that these two lengths are the same. And then if we go to the third side, we also know that these are going to have the same length, or the line segments themselves are going to be congruent. So we also know that the length of AC is going to be equal to the length of XZ. Not only do we know that all of the corresponding sides are going to have the same length, if someone tells us that a triangle is congruent, we also know that all the corresponding angles are going to have the same measure. So for example, we also know that this angle's measure is going to be the same as the corresponding angle's measure."}, {"video_title": "Corresponding parts of congruent triangles are congruent.mp3", "Sentence": "And then if we go to the third side, we also know that these are going to have the same length, or the line segments themselves are going to be congruent. So we also know that the length of AC is going to be equal to the length of XZ. Not only do we know that all of the corresponding sides are going to have the same length, if someone tells us that a triangle is congruent, we also know that all the corresponding angles are going to have the same measure. So for example, we also know that this angle's measure is going to be the same as the corresponding angle's measure. And the corresponding angle is right over here. It's between this orange side and this blue side, or this orange side and this purple side, I should say. And between the orange side and this purple side."}, {"video_title": "Corresponding parts of congruent triangles are congruent.mp3", "Sentence": "So for example, we also know that this angle's measure is going to be the same as the corresponding angle's measure. And the corresponding angle is right over here. It's between this orange side and this blue side, or this orange side and this purple side, I should say. And between the orange side and this purple side. And so it also tells us that the measure of angle is this BAC is equal to the measure of angle of angle YXZ. Let me write that angle symbol a little less. Like a measure of angle YXZ."}, {"video_title": "Corresponding parts of congruent triangles are congruent.mp3", "Sentence": "And between the orange side and this purple side. And so it also tells us that the measure of angle is this BAC is equal to the measure of angle of angle YXZ. Let me write that angle symbol a little less. Like a measure of angle YXZ. We could also write that as angle BAC is congruent to angle YXZ. And once again, like line segments, if one line segment is congruent to another line segment, it just means that their lengths are equal. And if one angle is congruent to another angle, it just means that their measures are equal."}, {"video_title": "Corresponding parts of congruent triangles are congruent.mp3", "Sentence": "Like a measure of angle YXZ. We could also write that as angle BAC is congruent to angle YXZ. And once again, like line segments, if one line segment is congruent to another line segment, it just means that their lengths are equal. And if one angle is congruent to another angle, it just means that their measures are equal. So we know that those two corresponding angles have the same measure. They're congruent. We also know that these two corresponding angles have the same measure."}, {"video_title": "Corresponding parts of congruent triangles are congruent.mp3", "Sentence": "And if one angle is congruent to another angle, it just means that their measures are equal. So we know that those two corresponding angles have the same measure. They're congruent. We also know that these two corresponding angles have the same measure. And I'll use a double arc to specify that this has the same measure as that. So we also know that the measure of angle ABC is equal to the measure of angle XYZ. And then finally, we know that this angle, if we know that these two characters are congruent, that this angle is going to have the same measure as this angle as its corresponding angle."}, {"video_title": "Corresponding parts of congruent triangles are congruent.mp3", "Sentence": "We also know that these two corresponding angles have the same measure. And I'll use a double arc to specify that this has the same measure as that. So we also know that the measure of angle ABC is equal to the measure of angle XYZ. And then finally, we know that this angle, if we know that these two characters are congruent, that this angle is going to have the same measure as this angle as its corresponding angle. So we know that the measure of angle ACB is going to be equal to the measure of angle XZY. Now what we're going to concern ourselves a lot with is how do we prove congruence? Because it's cool."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "So the first question I'll ask you, if you do one revolution, if you have an angle that went all the way around once, how many radians is that? Well, we know that that is 2 pi radians. Now, that exact same angle, if we were to measure it in degrees, how many degrees is that? Well, you've heard of people doing a 360, doing one full revolution. That is equal to 360 degrees. Now, can we simplify this? It's important to write this little superscript circle."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "Well, you've heard of people doing a 360, doing one full revolution. That is equal to 360 degrees. Now, can we simplify this? It's important to write this little superscript circle. That's literally the units under question. Sometimes it doesn't look like a unit, but it is a unit. You could literally write degrees instead of that little symbol."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "It's important to write this little superscript circle. That's literally the units under question. Sometimes it doesn't look like a unit, but it is a unit. You could literally write degrees instead of that little symbol. And the units right here, of course, are the word radians. Now, can we simplify this a little bit? Well, sure."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "You could literally write degrees instead of that little symbol. And the units right here, of course, are the word radians. Now, can we simplify this a little bit? Well, sure. Both 2 pi and 360 are divisible by 2, so let's divide things by 2. And if we do that, what do we get for what pi radians are equal to? Well, on the left-hand side here, we're just left with pi radians."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "Well, sure. Both 2 pi and 360 are divisible by 2, so let's divide things by 2. And if we do that, what do we get for what pi radians are equal to? Well, on the left-hand side here, we're just left with pi radians. And on the right-hand side here, 360 divided by 2 is 180, and we have still the units, which are degrees. So we get pi radians are equal to 180 degrees, which actually answered the first part of our question. We wanted to convert pi radians."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "Well, on the left-hand side here, we're just left with pi radians. And on the right-hand side here, 360 divided by 2 is 180, and we have still the units, which are degrees. So we get pi radians are equal to 180 degrees, which actually answered the first part of our question. We wanted to convert pi radians. Well, we just figured out pi radians are equal to 100, 180 degrees. Pi radians are equal to 180 degrees. And if you want to think about it, we know pi radians are halfway around a circle like that, and that's the same thing as 180 degrees."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "We wanted to convert pi radians. Well, we just figured out pi radians are equal to 100, 180 degrees. Pi radians are equal to 180 degrees. And if you want to think about it, we know pi radians are halfway around a circle like that, and that's the same thing as 180 degrees. So now let's think about the second part of it. We want to convert negative pi over 3 radians. Let me do this in a new color."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "And if you want to think about it, we know pi radians are halfway around a circle like that, and that's the same thing as 180 degrees. So now let's think about the second part of it. We want to convert negative pi over 3 radians. Let me do this in a new color. Negative pi over 3. So negative pi over 3 radians. How can we convert that to degrees?"}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "Let me do this in a new color. Negative pi over 3. So negative pi over 3 radians. How can we convert that to degrees? What do we get based on this information right over here? Well, to figure this out, we need to know how many degrees there are per radian. If we need to multiply this times degrees, and I'm going to write the word out because if I just wrote a little circle here, it would be hard to visualize that as a unit."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "How can we convert that to degrees? What do we get based on this information right over here? Well, to figure this out, we need to know how many degrees there are per radian. If we need to multiply this times degrees, and I'm going to write the word out because if I just wrote a little circle here, it would be hard to visualize that as a unit. Degrees per radian. So how many degrees are there per radian? Well, we know that for every 180 degrees, we have pi radians."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "If we need to multiply this times degrees, and I'm going to write the word out because if I just wrote a little circle here, it would be hard to visualize that as a unit. Degrees per radian. So how many degrees are there per radian? Well, we know that for every 180 degrees, we have pi radians. Or you could say that there are 180 over pi degrees per radian. And this is going to work out. We have however many radians we have times the number of degrees per radian."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "Well, we know that for every 180 degrees, we have pi radians. Or you could say that there are 180 over pi degrees per radian. And this is going to work out. We have however many radians we have times the number of degrees per radian. So of course, the units are going to work out. Radians cancel out. The pi also cancels out."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "We have however many radians we have times the number of degrees per radian. So of course, the units are going to work out. Radians cancel out. The pi also cancels out. So you're left with negative 180 divided by 3, leaving us with negative 60. And we don't want to forget the units. We could write them out."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "The pi also cancels out. So you're left with negative 180 divided by 3, leaving us with negative 60. And we don't want to forget the units. We could write them out. That's the only units that's left. Degrees, which we could write out. We could write out the word degrees or we could just put that little symbol there."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "And to graph a circle, you have to know where its center is, and you have to know what its radius is. So let me see if I can change that. And you have to know what its radius is. So what we need to do is put this in some form where we can pick out its center and its radius. Let me get my little scratch pad out and see if we can do that. So this is that same equation. And what I essentially want to do is I want to complete the square in terms of x and complete the square in terms of y to put it into a form that we can recognize."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "So what we need to do is put this in some form where we can pick out its center and its radius. Let me get my little scratch pad out and see if we can do that. So this is that same equation. And what I essentially want to do is I want to complete the square in terms of x and complete the square in terms of y to put it into a form that we can recognize. So first, let's take all of the x terms. So let's take all of the x terms. So you have x squared and 4x on the left-hand side."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "And what I essentially want to do is I want to complete the square in terms of x and complete the square in terms of y to put it into a form that we can recognize. So first, let's take all of the x terms. So let's take all of the x terms. So you have x squared and 4x on the left-hand side. So I could rewrite this as x squared plus 4x. And I'm going to put some parentheses around here because I want to complete the square. And then I have my y terms."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "So you have x squared and 4x on the left-hand side. So I could rewrite this as x squared plus 4x. And I'm going to put some parentheses around here because I want to complete the square. And then I have my y terms. I'll circle those in. Well, the red looks too much like the purple. I'll circle those in blue."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "And then I have my y terms. I'll circle those in. Well, the red looks too much like the purple. I'll circle those in blue. y squared and negative 4y. So we have plus y squared minus 4y. And then we have a minus 17."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "I'll circle those in blue. y squared and negative 4y. So we have plus y squared minus 4y. And then we have a minus 17. And I'll just do that in a neutral color. So minus 17 is equal to 0. Now, what I want to do is make each of these purple expressions perfect squares."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "And then we have a minus 17. And I'll just do that in a neutral color. So minus 17 is equal to 0. Now, what I want to do is make each of these purple expressions perfect squares. So how could I do that here? Well, this would be a perfect square if I took half of this 4 and I squared it. So if I made this plus 4, then this entire expression would be x plus 2 squared."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "Now, what I want to do is make each of these purple expressions perfect squares. So how could I do that here? Well, this would be a perfect square if I took half of this 4 and I squared it. So if I made this plus 4, then this entire expression would be x plus 2 squared. And you can verify that if you like. If you need review on completing the square, there's plenty of videos on Khan Academy on that. All we did is we took half of this coefficient and then squared it to get 4."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "So if I made this plus 4, then this entire expression would be x plus 2 squared. And you can verify that if you like. If you need review on completing the square, there's plenty of videos on Khan Academy on that. All we did is we took half of this coefficient and then squared it to get 4. Half of 4 is 2 squared to get 4. And that comes straight out of the idea. If you take x plus 2 and square it, it's going to be x squared plus twice the product of 2 and x plus 2 squared."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "All we did is we took half of this coefficient and then squared it to get 4. Half of 4 is 2 squared to get 4. And that comes straight out of the idea. If you take x plus 2 and square it, it's going to be x squared plus twice the product of 2 and x plus 2 squared. Now, we can't just willy-nilly add a 4 here. We had an equality before. Just adding a 4, it wouldn't be equal anymore."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "If you take x plus 2 and square it, it's going to be x squared plus twice the product of 2 and x plus 2 squared. Now, we can't just willy-nilly add a 4 here. We had an equality before. Just adding a 4, it wouldn't be equal anymore. So if we want to maintain the equality, we have to add 4 on the right-hand side as well. Now, let's do the same thing for the y's. Half of this coefficient right over here is a negative 2."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "Just adding a 4, it wouldn't be equal anymore. So if we want to maintain the equality, we have to add 4 on the right-hand side as well. Now, let's do the same thing for the y's. Half of this coefficient right over here is a negative 2. If we square a negative 2, it becomes a positive 4. We can't just do that on the left-hand side. We have to do that on the right-hand side as well."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "Half of this coefficient right over here is a negative 2. If we square a negative 2, it becomes a positive 4. We can't just do that on the left-hand side. We have to do that on the right-hand side as well. Now, what we have in blue becomes y minus 2 squared. And of course, we have the minus 17. But why don't we add 17 to both sides as well to get rid of this minus 17 here?"}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "We have to do that on the right-hand side as well. Now, what we have in blue becomes y minus 2 squared. And of course, we have the minus 17. But why don't we add 17 to both sides as well to get rid of this minus 17 here? So let's add 17 on the left and add 17 on the right. So on the left, we're just left with these two expressions. And on the right, we have 4 plus 4 plus 17."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "But why don't we add 17 to both sides as well to get rid of this minus 17 here? So let's add 17 on the left and add 17 on the right. So on the left, we're just left with these two expressions. And on the right, we have 4 plus 4 plus 17. Well, that's 8 plus 17, which is equal to 25. Now, this is a form that we recognize. If you have the form x minus a squared plus y minus b squared is equal to r squared, we know that the center is at the point a comma b, essentially the point that makes both of these equal to 0, and that the radius is going to be r. So if we look over here, what is our a?"}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "And on the right, we have 4 plus 4 plus 17. Well, that's 8 plus 17, which is equal to 25. Now, this is a form that we recognize. If you have the form x minus a squared plus y minus b squared is equal to r squared, we know that the center is at the point a comma b, essentially the point that makes both of these equal to 0, and that the radius is going to be r. So if we look over here, what is our a? We have to be careful here. Our a isn't 2. Our a is negative 2. x minus negative 2 is equal to 2."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "If you have the form x minus a squared plus y minus b squared is equal to r squared, we know that the center is at the point a comma b, essentially the point that makes both of these equal to 0, and that the radius is going to be r. So if we look over here, what is our a? We have to be careful here. Our a isn't 2. Our a is negative 2. x minus negative 2 is equal to 2. So the x-coordinate of our center is going to be negative 2. And the y-coordinate of our center is going to be 2. Remember, we care about the x value that makes this 0 and the y value that makes this 0."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "Our a is negative 2. x minus negative 2 is equal to 2. So the x-coordinate of our center is going to be negative 2. And the y-coordinate of our center is going to be 2. Remember, we care about the x value that makes this 0 and the y value that makes this 0. So the center is negative 2 comma 2. And this is the radius squared. So the radius is equal to 5."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "Remember, we care about the x value that makes this 0 and the y value that makes this 0. So the center is negative 2 comma 2. And this is the radius squared. So the radius is equal to 5. So let's go back to the exercise and actually plot this. So it's negative 2 comma 2. So our center is negative 2 comma 2."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "So the radius is equal to 5. So let's go back to the exercise and actually plot this. So it's negative 2 comma 2. So our center is negative 2 comma 2. So that's right over there. x is negative 2. y is positive 2. And the radius is 5."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "So our center is negative 2 comma 2. So that's right over there. x is negative 2. y is positive 2. And the radius is 5. So see, this would be 1, 2, 3, 4, 5. So we have to go a little bit wider than this. My pen is having trouble."}, {"video_title": "Completing the square to write equation in standard form of a circle Algebra II Khan Academy.mp3", "Sentence": "And the radius is 5. So see, this would be 1, 2, 3, 4, 5. So we have to go a little bit wider than this. My pen is having trouble. There you go. 1, 2, 3, 4, 5. Let's check our answer."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So parallel lines are lines that have the same slope, and they're different lines, so they never, ever intersect. So we need to look for different lines that have the exact same slope. And lucky for us, all of these lines are in y equals mx plus b, or slope intercept form. So you can really just look at these lines and figure out their slopes. The slope for line A, m is equal to 2. We see it right over there. For line B, our slope is equal to 3."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So you can really just look at these lines and figure out their slopes. The slope for line A, m is equal to 2. We see it right over there. For line B, our slope is equal to 3. So these two guys are not parallel. And I'll graph it in a second, and you'll see that. And then finally, for line C, I'll do it in purple, the slope is 2."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "For line B, our slope is equal to 3. So these two guys are not parallel. And I'll graph it in a second, and you'll see that. And then finally, for line C, I'll do it in purple, the slope is 2. So m is equal to 2. I don't know if that purple is too dark for you. So line C and line A have the same slope, but they're different lines."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "And then finally, for line C, I'll do it in purple, the slope is 2. So m is equal to 2. I don't know if that purple is too dark for you. So line C and line A have the same slope, but they're different lines. They have different y-intercepts. So they're going to be parallel. And to see that, let's actually graph all of these characters."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So line C and line A have the same slope, but they're different lines. They have different y-intercepts. So they're going to be parallel. And to see that, let's actually graph all of these characters. So line A, our y-intercept is negative 6, so the point 0, 1, 2, 3, 4, 5, 6. And our slope is 2. So if we move 1 in the positive x direction, we go up 2 in the positive y direction."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "And to see that, let's actually graph all of these characters. So line A, our y-intercept is negative 6, so the point 0, 1, 2, 3, 4, 5, 6. And our slope is 2. So if we move 1 in the positive x direction, we go up 2 in the positive y direction. 1 in x, up 2 in y. If we go 2 in x, we're going to go up 4 in y. We're going to go up 4 in y."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So if we move 1 in the positive x direction, we go up 2 in the positive y direction. 1 in x, up 2 in y. If we go 2 in x, we're going to go up 4 in y. We're going to go up 4 in y. And I could just do up 2, then we're going to go 2, 4. And you're going to see it's all on the same line. So line A is going to look something like, do my best to draw it as straight as possible."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "We're going to go up 4 in y. And I could just do up 2, then we're going to go 2, 4. And you're going to see it's all on the same line. So line A is going to look something like, do my best to draw it as straight as possible. Line A, I could do a better version than that. Line A is going to look like, well, that's about just as good as what I just drew. That is line A."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So line A is going to look something like, do my best to draw it as straight as possible. Line A, I could do a better version than that. Line A is going to look like, well, that's about just as good as what I just drew. That is line A. Now let's do line B. Line B, the y-intercept is negative 6. So it has the same y-intercept, but its slope is 3."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "That is line A. Now let's do line B. Line B, the y-intercept is negative 6. So it has the same y-intercept, but its slope is 3. So if x goes up by 1, y will go up by 3. So x goes up by 1, y goes up by 3. If x goes up by 2, y is going to go up by 6."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So it has the same y-intercept, but its slope is 3. So if x goes up by 1, y will go up by 3. So x goes up by 1, y goes up by 3. If x goes up by 2, y is going to go up by 6. 2, 4, 6. So 2, and then you go 2, 4, 6. So this line is going to look something like this."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "If x goes up by 2, y is going to go up by 6. 2, 4, 6. So 2, and then you go 2, 4, 6. So this line is going to look something like this. Trying my best to connect the dots. It has a steeper slope. And you see that when x increases, this blue line increases by more in the y direction."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So this line is going to look something like this. Trying my best to connect the dots. It has a steeper slope. And you see that when x increases, this blue line increases by more in the y direction. So that is line B. And notice they do intersect. There's definitely not two parallel lines."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "And you see that when x increases, this blue line increases by more in the y direction. So that is line B. And notice they do intersect. There's definitely not two parallel lines. And then finally, let's look at line C. The y-intercept is 5. So 0, 1, 2, 3, 4, 5. The point 0, 5, it's y-intercept."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "There's definitely not two parallel lines. And then finally, let's look at line C. The y-intercept is 5. So 0, 1, 2, 3, 4, 5. The point 0, 5, it's y-intercept. And its slope is 2. So you increase by 1 in the x direction. You're going to go up by 2 in the y direction."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "The point 0, 5, it's y-intercept. And its slope is 2. So you increase by 1 in the x direction. You're going to go up by 2 in the y direction. If you decrease by 1, you're going to go down 2 in the y direction. If you increase by, well, you're going to go to that point. You're going to have a bunch of these points."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "You're going to go up by 2 in the y direction. If you decrease by 1, you're going to go down 2 in the y direction. If you increase by, well, you're going to go to that point. You're going to have a bunch of these points. And then if I were to graph the line, let me do it one more time. If I were to decrease by 2, I'm going to have to go down by 4. Negative 4 over negative 2 is still a slope of 2."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "You're going to have a bunch of these points. And then if I were to graph the line, let me do it one more time. If I were to decrease by 2, I'm going to have to go down by 4. Negative 4 over negative 2 is still a slope of 2. So 1, 2, 3, 4. And I can do that one more time. Get right over there."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "Negative 4 over negative 2 is still a slope of 2. So 1, 2, 3, 4. And I can do that one more time. Get right over there. And then you'll see the line. The line will look like that. It will look just like that."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "Get right over there. And then you'll see the line. The line will look like that. It will look just like that. And notice that line C and line A never intersect. They have the exact same slope. Different y-intercepts, same slope."}, {"video_title": "Parallel lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "It will look just like that. And notice that line C and line A never intersect. They have the exact same slope. Different y-intercepts, same slope. So they're increasing at the exact same rate. But they're never going to intersect each other. So line A and line C are parallel."}, {"video_title": "Reflecting points across horizontal and vertical lines.mp3", "Sentence": "So if you're doing this on the Khan Academy exercise, you would actually just click on a point right over there, and it would show up, but this would be the reflection of point A across the line L. Let's do another example. So here we're asked, plot the image of point B under a reflection across the X axis. Alright, so this is point B, and we're going to reflect it across the X axis right over here. So to go from B to the X axis, it's exactly five units below the X axis, one, two, three, four, five. So if we were to reflect across the X axis, essentially create its mirror image, it's going to be five units above the X axis, one, two, three, four, five. So that's where the image would be. Maybe we could denote that with a B prime."}, {"video_title": "Reflecting points across horizontal and vertical lines.mp3", "Sentence": "So to go from B to the X axis, it's exactly five units below the X axis, one, two, three, four, five. So if we were to reflect across the X axis, essentially create its mirror image, it's going to be five units above the X axis, one, two, three, four, five. So that's where the image would be. Maybe we could denote that with a B prime. We are reflecting across the X axis. Let's do another example. So here, they tell us point C prime is the image of C, which is at the coordinates negative four comma negative two, under a reflection across the Y axis."}, {"video_title": "Reflecting points across horizontal and vertical lines.mp3", "Sentence": "Maybe we could denote that with a B prime. We are reflecting across the X axis. Let's do another example. So here, they tell us point C prime is the image of C, which is at the coordinates negative four comma negative two, under a reflection across the Y axis. What are the coordinates of C prime? So pause this video and see if you can figure it out on your own. So there's several ways to approach it."}, {"video_title": "Reflecting points across horizontal and vertical lines.mp3", "Sentence": "So here, they tell us point C prime is the image of C, which is at the coordinates negative four comma negative two, under a reflection across the Y axis. What are the coordinates of C prime? So pause this video and see if you can figure it out on your own. So there's several ways to approach it. It doesn't hurt to do a quick visual diagram. So that could be my X axis. This would be my Y axis."}, {"video_title": "Reflecting points across horizontal and vertical lines.mp3", "Sentence": "So there's several ways to approach it. It doesn't hurt to do a quick visual diagram. So that could be my X axis. This would be my Y axis. And it's the point negative four comma negative two. So that might look like this. Negative four, negative two."}, {"video_title": "Reflecting points across horizontal and vertical lines.mp3", "Sentence": "This would be my Y axis. And it's the point negative four comma negative two. So that might look like this. Negative four, negative two. So this is the point C right over here. And we want to reflect across the Y axis. So we want to reflect across the Y axis, which I am coloring in in red right over here."}, {"video_title": "Reflecting points across horizontal and vertical lines.mp3", "Sentence": "Negative four, negative two. So this is the point C right over here. And we want to reflect across the Y axis. So we want to reflect across the Y axis, which I am coloring in in red right over here. So let's see. The point C is four to the left of the Y axis. So its reflection is going to be four to the right of the Y axis."}, {"video_title": "Reflecting points across horizontal and vertical lines.mp3", "Sentence": "So we want to reflect across the Y axis, which I am coloring in in red right over here. So let's see. The point C is four to the left of the Y axis. So its reflection is going to be four to the right of the Y axis. So let me do it like this. So instead of being four to the left, we want to go four to the right. So plus four."}, {"video_title": "Reflecting points across horizontal and vertical lines.mp3", "Sentence": "So its reflection is going to be four to the right of the Y axis. So let me do it like this. So instead of being four to the left, we want to go four to the right. So plus four. So where would that put our C prime? So our C prime would be right over there. And what would its coordinates be?"}, {"video_title": "Reflecting points across horizontal and vertical lines.mp3", "Sentence": "So plus four. So where would that put our C prime? So our C prime would be right over there. And what would its coordinates be? Well, it would have the same Y coordinates. So C prime would have a Y coordinate of negative two. But what would its X coordinate be?"}, {"video_title": "Reflecting points across horizontal and vertical lines.mp3", "Sentence": "And what would its coordinates be? Well, it would have the same Y coordinates. So C prime would have a Y coordinate of negative two. But what would its X coordinate be? Well, instead of it being negative four, it gets flipped over the Y axis. So now it's going to have an X coordinate of positive four. So the coordinates here would be four comma negative two."}, {"video_title": "Reflecting points across horizontal and vertical lines.mp3", "Sentence": "But what would its X coordinate be? Well, instead of it being negative four, it gets flipped over the Y axis. So now it's going to have an X coordinate of positive four. So the coordinates here would be four comma negative two. Four comma negative two. You might have been able to do this in your head. Although for me, even if I tried to do it in my head, I would still have this visualization going on in my head."}, {"video_title": "Reflecting points across horizontal and vertical lines.mp3", "Sentence": "So the coordinates here would be four comma negative two. Four comma negative two. You might have been able to do this in your head. Although for me, even if I tried to do it in my head, I would still have this visualization going on in my head. I was okay, negative four comma negative two. I'm sitting there in the third quadrant. If I'm flipping over the Y axis, my Y coordinate wouldn't change, but my X coordinate would become the opposite, and I would end up in the fourth quadrant."}, {"video_title": "Proof Segments tangent to circle from outside point are congruent High School Math Khan Academy.mp3", "Sentence": "So I have a circle here with a center at point O, and let's pick an arbitrary point that sits outside of the circle. So let me just pick this point right over here, point A. And if I have an arbitrary point outside of the circle, I can actually draw two different tangent lines that contain A that are tangent to this circle. Let me draw them. So one of them would look like this. Actually, let me just start right over here. So I wanna make it tangent to the circle so it could look like that."}, {"video_title": "Proof Segments tangent to circle from outside point are congruent High School Math Khan Academy.mp3", "Sentence": "Let me draw them. So one of them would look like this. Actually, let me just start right over here. So I wanna make it tangent to the circle so it could look like that. And then the other one would look like this, would look like, would look like that. And let's say that the point that the tangent lines intersect the circle, let's say this point right over here is point B, and this point right over here is point C. This is point C right over here. What I wanna prove, what I wanna prove is that the segment AB is congruent to the segment AC, or another way of thinking about it."}, {"video_title": "Proof Segments tangent to circle from outside point are congruent High School Math Khan Academy.mp3", "Sentence": "So I wanna make it tangent to the circle so it could look like that. And then the other one would look like this, would look like, would look like that. And let's say that the point that the tangent lines intersect the circle, let's say this point right over here is point B, and this point right over here is point C. This is point C right over here. What I wanna prove, what I wanna prove is that the segment AB is congruent to the segment AC, or another way of thinking about it. I wanna prove, let me do this in a new color that I haven't used yet. I wanna prove that this segment right over here is congruent to this segment, is congruent to that segment. I'd encourage you to pause the video and try to work it out, and try to work it out on your own before I work through it."}, {"video_title": "Proof Segments tangent to circle from outside point are congruent High School Math Khan Academy.mp3", "Sentence": "What I wanna prove, what I wanna prove is that the segment AB is congruent to the segment AC, or another way of thinking about it. I wanna prove, let me do this in a new color that I haven't used yet. I wanna prove that this segment right over here is congruent to this segment, is congruent to that segment. I'd encourage you to pause the video and try to work it out, and try to work it out on your own before I work through it. All right, now let's try to work through this together. And to work through this together, I'm gonna actually set up two triangles, two triangles, and they're going to be right triangles as we'll see in a second. So let me draw some lines here to set up our triangles."}, {"video_title": "Proof Segments tangent to circle from outside point are congruent High School Math Khan Academy.mp3", "Sentence": "I'd encourage you to pause the video and try to work it out, and try to work it out on your own before I work through it. All right, now let's try to work through this together. And to work through this together, I'm gonna actually set up two triangles, two triangles, and they're going to be right triangles as we'll see in a second. So let me draw some lines here to set up our triangles. So I'm gonna draw one line just like that, and then let me draw that, and then let me draw that. Now what do we know, what do we know about these triangles? Well, as I mentioned, we're going to be dealing with right triangles."}, {"video_title": "Proof Segments tangent to circle from outside point are congruent High School Math Khan Academy.mp3", "Sentence": "So let me draw some lines here to set up our triangles. So I'm gonna draw one line just like that, and then let me draw that, and then let me draw that. Now what do we know, what do we know about these triangles? Well, as I mentioned, we're going to be dealing with right triangles. How do I know that? Well, in a previous video, we saw that if we have a radius intersecting a tangent line, that they intersect at right angles. We've proven this."}, {"video_title": "Proof Segments tangent to circle from outside point are congruent High School Math Khan Academy.mp3", "Sentence": "Well, as I mentioned, we're going to be dealing with right triangles. How do I know that? Well, in a previous video, we saw that if we have a radius intersecting a tangent line, that they intersect at right angles. We've proven this. So that's a radius, that's a tangent line. They're gonna intersect at a right angle. So radius, tangent line, they intersect at a right angle."}, {"video_title": "Proof Segments tangent to circle from outside point are congruent High School Math Khan Academy.mp3", "Sentence": "We've proven this. So that's a radius, that's a tangent line. They're gonna intersect at a right angle. So radius, tangent line, they intersect at a right angle. We also know since OB and OC are both radii, that they're both the length of the radius of the circle. So this side right over here, let me do some new colors here. So this side is going to be congruent to that side."}, {"video_title": "Proof Segments tangent to circle from outside point are congruent High School Math Khan Academy.mp3", "Sentence": "So radius, tangent line, they intersect at a right angle. We also know since OB and OC are both radii, that they're both the length of the radius of the circle. So this side right over here, let me do some new colors here. So this side is going to be congruent to that side. And you can see that the hypotenuse of both circles is the same side, side OA. So of course, it is equal to itself. This is equal to itself."}, {"video_title": "Proof Segments tangent to circle from outside point are congruent High School Math Khan Academy.mp3", "Sentence": "So this side is going to be congruent to that side. And you can see that the hypotenuse of both circles is the same side, side OA. So of course, it is equal to itself. This is equal to itself. And so we see triangle ABO and triangle ACO, they're both right triangles that have two sides in common. In particular, they both have a hypotenuse that are equal to each other, and they both have a base or a leg. And we know from hypotenuse leg congruence that if you have two right triangles where the hypotenuses are equal and you have a leg that are equal, then the two triangles are going to be congruent."}, {"video_title": "Proof Segments tangent to circle from outside point are congruent High School Math Khan Academy.mp3", "Sentence": "This is equal to itself. And so we see triangle ABO and triangle ACO, they're both right triangles that have two sides in common. In particular, they both have a hypotenuse that are equal to each other, and they both have a base or a leg. And we know from hypotenuse leg congruence that if you have two right triangles where the hypotenuses are equal and you have a leg that are equal, then the two triangles are going to be congruent. So triangle ABO is congruent to triangle ACO. And in that proof where we prove it is, well, the Pythagorean theorem tells us if you know two sides of a right triangle, that determines what the third side is. So the third side, so it's length of AB is going to be the same thing as the length of AC."}, {"video_title": "Proof Segments tangent to circle from outside point are congruent High School Math Khan Academy.mp3", "Sentence": "And we know from hypotenuse leg congruence that if you have two right triangles where the hypotenuses are equal and you have a leg that are equal, then the two triangles are going to be congruent. So triangle ABO is congruent to triangle ACO. And in that proof where we prove it is, well, the Pythagorean theorem tells us if you know two sides of a right triangle, that determines what the third side is. So the third side, so it's length of AB is going to be the same thing as the length of AC. And that's, once again, it just comes out of these are both right triangles. If two sides, if two corresponding sides of these two right triangles are congruent, then the third side has to. That comes straight from the Pythagorean theorem."}, {"video_title": "Proof Segments tangent to circle from outside point are congruent High School Math Khan Academy.mp3", "Sentence": "So the third side, so it's length of AB is going to be the same thing as the length of AC. And that's, once again, it just comes out of these are both right triangles. If two sides, if two corresponding sides of these two right triangles are congruent, then the third side has to. That comes straight from the Pythagorean theorem. And there you go. We've hopefully feel good about the fact that AB is going to be congruent to AC. Or another way to think about it, if I take a point outside of a circle and I construct segments that are tangent to the circle, that those two segments are going to be congruent to each other."}, {"video_title": "Using related volumes Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So that's a square base, that's a square, that's a square, and that's a square. All of the figures except for C are prisms. Yes, C is a pyramid right over here. All of the figures except for D are right. You can see D right over here is a little bit skewed or you could view it as oblique. All of the figures except for E have the same base area. The base of figure E is a dilation of the base of figure A by a scale factor of 1.5."}, {"video_title": "Using related volumes Solid geometry High school geometry Khan Academy.mp3", "Sentence": "All of the figures except for D are right. You can see D right over here is a little bit skewed or you could view it as oblique. All of the figures except for E have the same base area. The base of figure E is a dilation of the base of figure A by a scale factor of 1.5. All right. So they tell us if figure A has a volume of 28 cubic centimeters, what are the volumes of the other figures? So pause this video and see if you can have a go at that."}, {"video_title": "Using related volumes Solid geometry High school geometry Khan Academy.mp3", "Sentence": "The base of figure E is a dilation of the base of figure A by a scale factor of 1.5. All right. So they tell us if figure A has a volume of 28 cubic centimeters, what are the volumes of the other figures? So pause this video and see if you can have a go at that. All right, now let's work through this together. Now, they're telling us about the bases and the heights, that a lot of these have the same base area, figure E is going to be different, and they also tell us that they all have the same height. So one way to think about volume is it's going to deal with base and height."}, {"video_title": "Using related volumes Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So pause this video and see if you can have a go at that. All right, now let's work through this together. Now, they're telling us about the bases and the heights, that a lot of these have the same base area, figure E is going to be different, and they also tell us that they all have the same height. So one way to think about volume is it's going to deal with base and height. And so for figure A, it's pretty straightforward. If we call this area right over here, let's call that B for the area of the base, and then it has some height, H, right over here, we know that the base area times the height is going to be the volume. So we could say that based on figure A, base times height is going to be equal to 28 cubic centimeters."}, {"video_title": "Using related volumes Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So one way to think about volume is it's going to deal with base and height. And so for figure A, it's pretty straightforward. If we call this area right over here, let's call that B for the area of the base, and then it has some height, H, right over here, we know that the base area times the height is going to be the volume. So we could say that based on figure A, base times height is going to be equal to 28 cubic centimeters. Fair enough. Now, what's going on over here with figure B? Well, it's a cylinder."}, {"video_title": "Using related volumes Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So we could say that based on figure A, base times height is going to be equal to 28 cubic centimeters. Fair enough. Now, what's going on over here with figure B? Well, it's a cylinder. Now, for a cylinder, what is the volume of a cylinder? Well, it too is going to be base times height. So it's going to be the area of the base times the height."}, {"video_title": "Using related volumes Solid geometry High school geometry Khan Academy.mp3", "Sentence": "Well, it's a cylinder. Now, for a cylinder, what is the volume of a cylinder? Well, it too is going to be base times height. So it's going to be the area of the base times the height. And if you're wondering how is that possible that it's the same as the volume of a rectangular prism over here, it's actually Cavalieri's principle. If they have the same height, and if at any point along that height, they have the same cross-sectional area, then you're going to have the same volume. So this volume is also going to be base times height."}, {"video_title": "Using related volumes Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So it's going to be the area of the base times the height. And if you're wondering how is that possible that it's the same as the volume of a rectangular prism over here, it's actually Cavalieri's principle. If they have the same height, and if at any point along that height, they have the same cross-sectional area, then you're going to have the same volume. So this volume is also going to be base times height. So let me say this is figure A. Figure B, right over here, let me draw those dots a little better, these colons a little bit better. Figure B, the volume is also going to be base times height, which is equal to 28 cubic centimeters."}, {"video_title": "Using related volumes Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So this volume is also going to be base times height. So let me say this is figure A. Figure B, right over here, let me draw those dots a little better, these colons a little bit better. Figure B, the volume is also going to be base times height, which is equal to 28 cubic centimeters. Let me make that clear. That's the volume is equal to that. Volume is equal to that."}, {"video_title": "Using related volumes Solid geometry High school geometry Khan Academy.mp3", "Sentence": "Figure B, the volume is also going to be base times height, which is equal to 28 cubic centimeters. Let me make that clear. That's the volume is equal to that. Volume is equal to that. Now, what about for figure C? What is the volume going to be? What is the formula for the volume of a pyramid?"}, {"video_title": "Using related volumes Solid geometry High school geometry Khan Academy.mp3", "Sentence": "Volume is equal to that. Now, what about for figure C? What is the volume going to be? What is the formula for the volume of a pyramid? And we've gotten the intuition and proven this to ourselves in other videos. Well, we know that for a pyramid, the volume is going to be equal to 1 3rd times base times height. And we know that it has the same base area as these other characters here, it has the same height."}, {"video_title": "Using related volumes Solid geometry High school geometry Khan Academy.mp3", "Sentence": "What is the formula for the volume of a pyramid? And we've gotten the intuition and proven this to ourselves in other videos. Well, we know that for a pyramid, the volume is going to be equal to 1 3rd times base times height. And we know that it has the same base area as these other characters here, it has the same height. And so we know what base times height is, is 28 cubic centimeters. So this is going to be 1 3rd times 28 cubic centimeters. So this is going to be equal to 1 3rd times 28 cubic centimeters, which we could rewrite as 28 over three cubic centimeters."}, {"video_title": "Using related volumes Solid geometry High school geometry Khan Academy.mp3", "Sentence": "And we know that it has the same base area as these other characters here, it has the same height. And so we know what base times height is, is 28 cubic centimeters. So this is going to be 1 3rd times 28 cubic centimeters. So this is going to be equal to 1 3rd times 28 cubic centimeters, which we could rewrite as 28 over three cubic centimeters. You could also write that as 9 1 3rd cubic centimeters. So that's for figure C. Now let's think about figure D. I'll do that right over here. Well, for this oblique prism, I guess we could say, you're going to have the same idea, and it comes from Cavalier's principle again."}, {"video_title": "Using related volumes Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So this is going to be equal to 1 3rd times 28 cubic centimeters, which we could rewrite as 28 over three cubic centimeters. You could also write that as 9 1 3rd cubic centimeters. So that's for figure C. Now let's think about figure D. I'll do that right over here. Well, for this oblique prism, I guess we could say, you're going to have the same idea, and it comes from Cavalier's principle again. It's gonna have the same formula for volume as figure A. It's going to be the base times the height right over here. So I could write volume is going to be equal to base times height."}, {"video_title": "Using related volumes Solid geometry High school geometry Khan Academy.mp3", "Sentence": "Well, for this oblique prism, I guess we could say, you're going to have the same idea, and it comes from Cavalier's principle again. It's gonna have the same formula for volume as figure A. It's going to be the base times the height right over here. So I could write volume is going to be equal to base times height. And we already know what that is. They tell us the base times height is going to be the same as figure A. It's gonna be 28 cubic centimeters."}, {"video_title": "Using related volumes Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So I could write volume is going to be equal to base times height. And we already know what that is. They tell us the base times height is going to be the same as figure A. It's gonna be 28 cubic centimeters. Now let's go to figure E. This is an interesting one because it has a different base area. What is going to be the area right over here? Well, they tell us that the base of figure E is a dilation of the base of figure A by a scale factor of 1.5."}, {"video_title": "Using related volumes Solid geometry High school geometry Khan Academy.mp3", "Sentence": "It's gonna be 28 cubic centimeters. Now let's go to figure E. This is an interesting one because it has a different base area. What is going to be the area right over here? Well, they tell us that the base of figure E is a dilation of the base of figure A by a scale factor of 1.5. And these are both squares. So figure A will say X by X. This one over here is going to be 1.5X by 1.5X."}, {"video_title": "Using related volumes Solid geometry High school geometry Khan Academy.mp3", "Sentence": "Well, they tell us that the base of figure E is a dilation of the base of figure A by a scale factor of 1.5. And these are both squares. So figure A will say X by X. This one over here is going to be 1.5X by 1.5X. So let me write that down. 1.5X times 1.5X. Or another way to think about it, let me do it over here where I have some free space."}, {"video_title": "Using related volumes Solid geometry High school geometry Khan Academy.mp3", "Sentence": "This one over here is going to be 1.5X by 1.5X. So let me write that down. 1.5X times 1.5X. Or another way to think about it, let me do it over here where I have some free space. We know that B, which we know is the area of figure A, that would be equal to X times X. Now what's the area of the base of figure E? Well, it's going to be 1.5X times 1.5X or 1.5X squared, which is the same thing as 1.5 squared is 2.25X squared."}, {"video_title": "Using related volumes Solid geometry High school geometry Khan Academy.mp3", "Sentence": "Or another way to think about it, let me do it over here where I have some free space. We know that B, which we know is the area of figure A, that would be equal to X times X. Now what's the area of the base of figure E? Well, it's going to be 1.5X times 1.5X or 1.5X squared, which is the same thing as 1.5 squared is 2.25X squared. And we know X squared or X times X, that is equal to B. That is equal to our original base area and all of these other figures. So the area over here, this area right over here, is going to be 2.25 times B."}, {"video_title": "Using related volumes Solid geometry High school geometry Khan Academy.mp3", "Sentence": "Well, it's going to be 1.5X times 1.5X or 1.5X squared, which is the same thing as 1.5 squared is 2.25X squared. And we know X squared or X times X, that is equal to B. That is equal to our original base area and all of these other figures. So the area over here, this area right over here, is going to be 2.25 times B. 2.25 times B. I didn't, that wasn't so easy to read. Let me write that a little bit clearer. So 2.25B is the base area right over here."}, {"video_title": "Using related volumes Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So the area over here, this area right over here, is going to be 2.25 times B. 2.25 times B. I didn't, that wasn't so easy to read. Let me write that a little bit clearer. So 2.25B is the base area right over here. And so what's the volume of this figure? The volume is going to be the area of the base, which is 2.25 times the area of all these other figures' bases, times the height, which is the same, times H. Now we know what base time, we know what B times H is, where B is the area of figure, the base area of figure A. We know that B times H is 28 cubic centimeters."}, {"video_title": "Using related volumes Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So 2.25B is the base area right over here. And so what's the volume of this figure? The volume is going to be the area of the base, which is 2.25 times the area of all these other figures' bases, times the height, which is the same, times H. Now we know what base time, we know what B times H is, where B is the area of figure, the base area of figure A. We know that B times H is 28 cubic centimeters. So the volume for figure E is going to be 2.25 times 28 cubic centimeters, times 28 cubic centimeters. And I don't have a calculator here in front of me and I could do it by hand, but I think you get the general point. You just have to multiply 2.25 times 28 to get the cubic, to get the volume of figure E. And that's because its base has been scaled in each dimension by 1.5."}, {"video_title": "Similar triangle basics Similarity Geometry Khan Academy.mp3", "Sentence": "One, all of their corresponding angles are the same. So the angle right here, angle BAC, is congruent to angle YXZ. Angle BCA is congruent to angle YZX. And angle ABC is congruent to angle XYZ. So all of their angles, their corresponding angles, are the same. And we also see that the sides are just scaled up versions of each other. So to go from the length of XZ to AC, we can multiply by 3."}, {"video_title": "Similar triangle basics Similarity Geometry Khan Academy.mp3", "Sentence": "And angle ABC is congruent to angle XYZ. So all of their angles, their corresponding angles, are the same. And we also see that the sides are just scaled up versions of each other. So to go from the length of XZ to AC, we can multiply by 3. We multiplied by 3 there. To go from XY, the length of XY, to the length of AB, which is the corresponding side, we are multiplying by 3. We had to multiply by 3."}, {"video_title": "Similar triangle basics Similarity Geometry Khan Academy.mp3", "Sentence": "So to go from the length of XZ to AC, we can multiply by 3. We multiplied by 3 there. To go from XY, the length of XY, to the length of AB, which is the corresponding side, we are multiplying by 3. We had to multiply by 3. And then to go from the length of YZ to the length of BC, we also multiplied by 3. So essentially, triangle ABC is just a scaled up version of triangle XYZ. If they were the same scale, they would be the exact same triangles."}, {"video_title": "Similar triangle basics Similarity Geometry Khan Academy.mp3", "Sentence": "We had to multiply by 3. And then to go from the length of YZ to the length of BC, we also multiplied by 3. So essentially, triangle ABC is just a scaled up version of triangle XYZ. If they were the same scale, they would be the exact same triangles. But one is just a bigger, a blown up version of the other one. Or this is a miniaturized version of that one over there. If you just multiply all the sides by 3, you get to this triangle."}, {"video_title": "Similar triangle basics Similarity Geometry Khan Academy.mp3", "Sentence": "If they were the same scale, they would be the exact same triangles. But one is just a bigger, a blown up version of the other one. Or this is a miniaturized version of that one over there. If you just multiply all the sides by 3, you get to this triangle. And so we can't call them congruent. But this does seem to be a bit of a special relationship. So we call this special relationship similarity."}, {"video_title": "Similar triangle basics Similarity Geometry Khan Academy.mp3", "Sentence": "If you just multiply all the sides by 3, you get to this triangle. And so we can't call them congruent. But this does seem to be a bit of a special relationship. So we call this special relationship similarity. So we can write that triangle ABC is similar to triangle. And we want to make sure we get the corresponding sides right. ABC is going to be similar to XYZ."}, {"video_title": "Similar triangle basics Similarity Geometry Khan Academy.mp3", "Sentence": "So we call this special relationship similarity. So we can write that triangle ABC is similar to triangle. And we want to make sure we get the corresponding sides right. ABC is going to be similar to XYZ. And so based on what we just saw, there's actually kind of three ideas here. And they're all equivalent ways of thinking about similarity. One way to think about it is that one is a scaled up version of the other."}, {"video_title": "Similar triangle basics Similarity Geometry Khan Academy.mp3", "Sentence": "ABC is going to be similar to XYZ. And so based on what we just saw, there's actually kind of three ideas here. And they're all equivalent ways of thinking about similarity. One way to think about it is that one is a scaled up version of the other. So scaled up or down version of the other. When we talked about congruency, they had to be exactly the same. You could rotate it."}, {"video_title": "Similar triangle basics Similarity Geometry Khan Academy.mp3", "Sentence": "One way to think about it is that one is a scaled up version of the other. So scaled up or down version of the other. When we talked about congruency, they had to be exactly the same. You could rotate it. You could shift it. You could flip it. But when you do all those things, they would have to essentially be identical."}, {"video_title": "Similar triangle basics Similarity Geometry Khan Academy.mp3", "Sentence": "You could rotate it. You could shift it. You could flip it. But when you do all those things, they would have to essentially be identical. With similarity, you can rotate it. You can shift it. You can flip it."}, {"video_title": "Similar triangle basics Similarity Geometry Khan Academy.mp3", "Sentence": "But when you do all those things, they would have to essentially be identical. With similarity, you can rotate it. You can shift it. You can flip it. And you can also scale it up and down in order for something to be similar. So for example, if you say if something is congruent, if let me say triangle CDE, if we know that triangle CDE is congruent to triangle FGH, then we definitely know that they are similar. They're scaled up by a factor of 1."}, {"video_title": "Similar triangle basics Similarity Geometry Khan Academy.mp3", "Sentence": "You can flip it. And you can also scale it up and down in order for something to be similar. So for example, if you say if something is congruent, if let me say triangle CDE, if we know that triangle CDE is congruent to triangle FGH, then we definitely know that they are similar. They're scaled up by a factor of 1. Then we know for a fact that CDE is also similar to triangle FGH. But we can't say it the other way around. If triangle ABC is similar to XYZ, we can't say that it's necessarily congruent."}, {"video_title": "Similar triangle basics Similarity Geometry Khan Academy.mp3", "Sentence": "They're scaled up by a factor of 1. Then we know for a fact that CDE is also similar to triangle FGH. But we can't say it the other way around. If triangle ABC is similar to XYZ, we can't say that it's necessarily congruent. And we see for this particular example, they definitely are not congruent. So this is one way to think about similarity. The other way to think about similarity is that all of the corresponding angles will be equal."}, {"video_title": "Similar triangle basics Similarity Geometry Khan Academy.mp3", "Sentence": "If triangle ABC is similar to XYZ, we can't say that it's necessarily congruent. And we see for this particular example, they definitely are not congruent. So this is one way to think about similarity. The other way to think about similarity is that all of the corresponding angles will be equal. So if something is similar, then all of the corresponding angles are going to be congruent. I always have trouble spelling this. It is two Rs, one S. Corresponding angles are congruent."}, {"video_title": "Similar triangle basics Similarity Geometry Khan Academy.mp3", "Sentence": "The other way to think about similarity is that all of the corresponding angles will be equal. So if something is similar, then all of the corresponding angles are going to be congruent. I always have trouble spelling this. It is two Rs, one S. Corresponding angles are congruent. So if we say that triangle ABC is similar to triangle XYZ, that is equivalent to saying that angle ABC is congruent, or we can say that their measures are equal to angle XYZ. That angle BAC is going to be congruent to angle YXZ. YXZ to angle YXZ."}, {"video_title": "Similar triangle basics Similarity Geometry Khan Academy.mp3", "Sentence": "It is two Rs, one S. Corresponding angles are congruent. So if we say that triangle ABC is similar to triangle XYZ, that is equivalent to saying that angle ABC is congruent, or we can say that their measures are equal to angle XYZ. That angle BAC is going to be congruent to angle YXZ. YXZ to angle YXZ. And then finally, angle ACB is going to be congruent to angle XZY. So if you have two triangles, all of their angles are the same, then you can say that they're similar. Or if you find two triangles and you're told that they are similar triangles, then you know that all of their corresponding angles are the same."}, {"video_title": "Similar triangle basics Similarity Geometry Khan Academy.mp3", "Sentence": "YXZ to angle YXZ. And then finally, angle ACB is going to be congruent to angle XZY. So if you have two triangles, all of their angles are the same, then you can say that they're similar. Or if you find two triangles and you're told that they are similar triangles, then you know that all of their corresponding angles are the same. And the last way to think about it is that the sides are all just scaled up versions of each other. So the sides scaled by the same factor. In the example we did here, the scaling factor was 3."}, {"video_title": "Similar triangle basics Similarity Geometry Khan Academy.mp3", "Sentence": "Or if you find two triangles and you're told that they are similar triangles, then you know that all of their corresponding angles are the same. And the last way to think about it is that the sides are all just scaled up versions of each other. So the sides scaled by the same factor. In the example we did here, the scaling factor was 3. It doesn't have to be 3. It just has to be the same scaling factor for every side. If we scale this side up by 3 and we only scale this side up by 2, then we would not be dealing with a similar triangle."}, {"video_title": "Similar triangle basics Similarity Geometry Khan Academy.mp3", "Sentence": "In the example we did here, the scaling factor was 3. It doesn't have to be 3. It just has to be the same scaling factor for every side. If we scale this side up by 3 and we only scale this side up by 2, then we would not be dealing with a similar triangle. But if we scaled all of these sides up by 7, then that's still a similar, as long as you have all of them scaled up or scaled down by the exact same factor. So one way to think about it is, and I want to keep having, well, I want to still visualize those triangles. Let me redraw them right over here a little bit simpler, because I'm not talking now in general terms, not even for that specific case."}, {"video_title": "Similar triangle basics Similarity Geometry Khan Academy.mp3", "Sentence": "If we scale this side up by 3 and we only scale this side up by 2, then we would not be dealing with a similar triangle. But if we scaled all of these sides up by 7, then that's still a similar, as long as you have all of them scaled up or scaled down by the exact same factor. So one way to think about it is, and I want to keep having, well, I want to still visualize those triangles. Let me redraw them right over here a little bit simpler, because I'm not talking now in general terms, not even for that specific case. So if we say that this is A, B, and C, and this right over here is X, Y, and Z. I just redrew them so I can refer to them when we write down here. If we're saying that these two things right over here are similar, that means that corresponding sides are scaled up versions of each other. So we could say that the length of AB is equal to some scaling factor, and this thing could be less than 1, times the length of XY, the corresponding sides."}, {"video_title": "Similar triangle basics Similarity Geometry Khan Academy.mp3", "Sentence": "Let me redraw them right over here a little bit simpler, because I'm not talking now in general terms, not even for that specific case. So if we say that this is A, B, and C, and this right over here is X, Y, and Z. I just redrew them so I can refer to them when we write down here. If we're saying that these two things right over here are similar, that means that corresponding sides are scaled up versions of each other. So we could say that the length of AB is equal to some scaling factor, and this thing could be less than 1, times the length of XY, the corresponding sides. And I know that AB corresponds to XY because of the order in which I wrote this similarity statement. So some scaling factor times XY. We know that BC, the length of BC, needs to be that same scaling factor times the length of YZ."}, {"video_title": "Similar triangle basics Similarity Geometry Khan Academy.mp3", "Sentence": "So we could say that the length of AB is equal to some scaling factor, and this thing could be less than 1, times the length of XY, the corresponding sides. And I know that AB corresponds to XY because of the order in which I wrote this similarity statement. So some scaling factor times XY. We know that BC, the length of BC, needs to be that same scaling factor times the length of YZ. So that same scaling factor. And then we know the length of AC is going to be equal to that same scaling factor times XZ. So that's XZ, and this could be a scaling factor."}, {"video_title": "Similar triangle basics Similarity Geometry Khan Academy.mp3", "Sentence": "We know that BC, the length of BC, needs to be that same scaling factor times the length of YZ. So that same scaling factor. And then we know the length of AC is going to be equal to that same scaling factor times XZ. So that's XZ, and this could be a scaling factor. So if ABC is larger than XYZ, then these Ks will be larger than 1. If they're the exact same size, if they're essentially congruent triangles, then these Ks will be 1. And if XYZ is bigger than ABC, then these scaling factors will be less than 1."}, {"video_title": "Similar triangle basics Similarity Geometry Khan Academy.mp3", "Sentence": "So that's XZ, and this could be a scaling factor. So if ABC is larger than XYZ, then these Ks will be larger than 1. If they're the exact same size, if they're essentially congruent triangles, then these Ks will be 1. And if XYZ is bigger than ABC, then these scaling factors will be less than 1. But another way to write the same statement, notice all I'm saying is corresponding sides are scaled up versions of each other. This first statement right here, if you divide both sides by XY, you get AB over XY is equal to our scaling factor. And then the second statement right over here, if you divide both sides by YZ, you get B."}, {"video_title": "Similar triangle basics Similarity Geometry Khan Academy.mp3", "Sentence": "And if XYZ is bigger than ABC, then these scaling factors will be less than 1. But another way to write the same statement, notice all I'm saying is corresponding sides are scaled up versions of each other. This first statement right here, if you divide both sides by XY, you get AB over XY is equal to our scaling factor. And then the second statement right over here, if you divide both sides by YZ, you get B. Let me do that same color. You get BC divided by YZ is equal to that scaling factor. Remember, in the example we just showed, that scaling factor was 3."}, {"video_title": "Similar triangle basics Similarity Geometry Khan Academy.mp3", "Sentence": "And then the second statement right over here, if you divide both sides by YZ, you get B. Let me do that same color. You get BC divided by YZ is equal to that scaling factor. Remember, in the example we just showed, that scaling factor was 3. But now we're saying in the more general terms, similarity, as long as you have the same scaling factor. And then finally, if you divide both sides here by the length between X and XZ, or segment XZ's length, you get AC over XZ is equal to K as well. Or another way to think about it is the ratio between corresponding sides."}, {"video_title": "Similar triangle basics Similarity Geometry Khan Academy.mp3", "Sentence": "Remember, in the example we just showed, that scaling factor was 3. But now we're saying in the more general terms, similarity, as long as you have the same scaling factor. And then finally, if you divide both sides here by the length between X and XZ, or segment XZ's length, you get AC over XZ is equal to K as well. Or another way to think about it is the ratio between corresponding sides. Notice, this is the ratio between AB and XY. The ratio between BC and YZ. The ratio between AC and XZ."}, {"video_title": "Similar triangle basics Similarity Geometry Khan Academy.mp3", "Sentence": "Or another way to think about it is the ratio between corresponding sides. Notice, this is the ratio between AB and XY. The ratio between BC and YZ. The ratio between AC and XZ. That the ratio between corresponding sides all gives us the same constant. Or you could rewrite this as AB over XY is equal to BC over YZ is equal to AC over XZ, which would be equal to some scaling factor, which is equal to K. So if you have similar triangles, let me draw an arrow right over here. Similar triangles means that they're scaled up versions, and you can also flip and rotate and do all this stuff with congruency."}, {"video_title": "Tangents of circles problem (example 1) Mathematics II High School Math Khan Academy.mp3", "Sentence": "So we're told that angle A is circumscribed about circle O. So this is angle A right over here. We're talking about this angle right over there. And when they say it's circumscribed about circle O, that means that the two sides of the angle, they're segments that would be part of tangent lines. So if we were to continue, so for example, that right over there, that line is tangent to the circle. And let me see if I could, and this line is also tangent to the circle. So you see that the sides of angle A are parts of those tangents."}, {"video_title": "Tangents of circles problem (example 1) Mathematics II High School Math Khan Academy.mp3", "Sentence": "And when they say it's circumscribed about circle O, that means that the two sides of the angle, they're segments that would be part of tangent lines. So if we were to continue, so for example, that right over there, that line is tangent to the circle. And let me see if I could, and this line is also tangent to the circle. So you see that the sides of angle A are parts of those tangents. And point B and point C are where those tangents actually sit on the circle. So given all of that, they're asking us, what is the measure of angle A? So we're trying to figure this out right here."}, {"video_title": "Tangents of circles problem (example 1) Mathematics II High School Math Khan Academy.mp3", "Sentence": "So you see that the sides of angle A are parts of those tangents. And point B and point C are where those tangents actually sit on the circle. So given all of that, they're asking us, what is the measure of angle A? So we're trying to figure this out right here. And I encourage you to pause the video and figure it out on your own. So the key insight here, and there's multiple ways that you could approach this, is to realize that a radius is going to be perpendicular to a tangent, or a radius that intersects a tangent is going to be perpendicular to it. So let me label this."}, {"video_title": "Tangents of circles problem (example 1) Mathematics II High School Math Khan Academy.mp3", "Sentence": "So we're trying to figure this out right here. And I encourage you to pause the video and figure it out on your own. So the key insight here, and there's multiple ways that you could approach this, is to realize that a radius is going to be perpendicular to a tangent, or a radius that intersects a tangent is going to be perpendicular to it. So let me label this. So this is going to be a right angle, and this is going to be a right angle. And any quadrilateral, the sum of the angles are going to add up to 360 degrees. And if you wonder where that comes from, well you could divide a quadrilateral into two triangles, where the sum of all the interior angles of a triangle are 180, and since you have two of them, 360 degrees."}, {"video_title": "Tangents of circles problem (example 1) Mathematics II High School Math Khan Academy.mp3", "Sentence": "So let me label this. So this is going to be a right angle, and this is going to be a right angle. And any quadrilateral, the sum of the angles are going to add up to 360 degrees. And if you wonder where that comes from, well you could divide a quadrilateral into two triangles, where the sum of all the interior angles of a triangle are 180, and since you have two of them, 360 degrees. So this 92 plus 90 plus 90 plus question mark is going to be equal to 360 degrees. So let me write that down. 92 plus 90 plus 90, so plus 180, plus, let's just call this x, x degrees, plus x."}, {"video_title": "Tangents of circles problem (example 1) Mathematics II High School Math Khan Academy.mp3", "Sentence": "And if you wonder where that comes from, well you could divide a quadrilateral into two triangles, where the sum of all the interior angles of a triangle are 180, and since you have two of them, 360 degrees. So this 92 plus 90 plus 90 plus question mark is going to be equal to 360 degrees. So let me write that down. 92 plus 90 plus 90, so plus 180, plus, let's just call this x, x degrees, plus x. They all have to add up to be 360 degrees. So let's see, we could subtract 180 from both sides, and so if we do that we would have 92 plus x is equal to 180. And if we subtract 92 from both sides, we get x is equal to, let's see, 180 minus 90 would be 90, and then we're going to subtract two more, so it's going to be x is equal to 88."}, {"video_title": "Tangents of circles problem (example 1) Mathematics II High School Math Khan Academy.mp3", "Sentence": "92 plus 90 plus 90, so plus 180, plus, let's just call this x, x degrees, plus x. They all have to add up to be 360 degrees. So let's see, we could subtract 180 from both sides, and so if we do that we would have 92 plus x is equal to 180. And if we subtract 92 from both sides, we get x is equal to, let's see, 180 minus 90 would be 90, and then we're going to subtract two more, so it's going to be x is equal to 88. So the measure of angle A is 88 degrees. Now let's do one more of these. These are surprisingly fun."}, {"video_title": "Tangents of circles problem (example 1) Mathematics II High School Math Khan Academy.mp3", "Sentence": "And if we subtract 92 from both sides, we get x is equal to, let's see, 180 minus 90 would be 90, and then we're going to subtract two more, so it's going to be x is equal to 88. So the measure of angle A is 88 degrees. Now let's do one more of these. These are surprisingly fun. All right, so it says angle A is circumscribed about circle O. We've seen that before in the last question, and they said what is the measure of angle D? So we want to find, let me make sure I'm on the right layer, we want to find the measure of that angle, and let's call that x again."}, {"video_title": "Tangents of circles problem (example 1) Mathematics II High School Math Khan Academy.mp3", "Sentence": "These are surprisingly fun. All right, so it says angle A is circumscribed about circle O. We've seen that before in the last question, and they said what is the measure of angle D? So we want to find, let me make sure I'm on the right layer, we want to find the measure of that angle, and let's call that x again. So what can we figure out? Well, just like in the last question, we have a quadrilateral here. Quadrilateral A, B, O, C, and we know two of the angles."}, {"video_title": "Tangents of circles problem (example 1) Mathematics II High School Math Khan Academy.mp3", "Sentence": "So we want to find, let me make sure I'm on the right layer, we want to find the measure of that angle, and let's call that x again. So what can we figure out? Well, just like in the last question, we have a quadrilateral here. Quadrilateral A, B, O, C, and we know two of the angles. We know this is going to be a right angle. We have a radius intersecting with a tangent, or part of a tangent, I guess you could say, and then this would be a right angle. And so by the same logic as we saw in the last question, this angle plus this angle plus this angle plus the central angle are going to add up to 360 degrees."}, {"video_title": "Tangents of circles problem (example 1) Mathematics II High School Math Khan Academy.mp3", "Sentence": "Quadrilateral A, B, O, C, and we know two of the angles. We know this is going to be a right angle. We have a radius intersecting with a tangent, or part of a tangent, I guess you could say, and then this would be a right angle. And so by the same logic as we saw in the last question, this angle plus this angle plus this angle plus the central angle are going to add up to 360 degrees. So let's call the measure of the central angle, let's call that y over there. So we have y plus 80 degrees, or we'll just assume everything's in degrees, so y plus 80 plus 90 plus 90, so I could say plus another 180, is going to be equal to 360 degrees, sum of the interior angles of a quadrilateral. And so let's see, you have y plus 80 is equal to 180."}, {"video_title": "Tangents of circles problem (example 1) Mathematics II High School Math Khan Academy.mp3", "Sentence": "And so by the same logic as we saw in the last question, this angle plus this angle plus this angle plus the central angle are going to add up to 360 degrees. So let's call the measure of the central angle, let's call that y over there. So we have y plus 80 degrees, or we'll just assume everything's in degrees, so y plus 80 plus 90 plus 90, so I could say plus another 180, is going to be equal to 360 degrees, sum of the interior angles of a quadrilateral. And so let's see, you have y plus 80 is equal to 180. If I just subtract 180 from both sides, I'm going to subtract 80 from both sides, and we get y is equal to 100, or the measure of this is 100, the measure of this interior angle right over here is 100 degrees, which also tells us that the measure of this arc, because that interior angle intercepts arc CB right over here, that tells us that the measure of arc CB is also 100 degrees. And so if we're trying to find this angle, the measure of angle D, is the inscribed angle that intercepts the same arc. And we've seen in previous videos that that inscribed angle that intercepts that arc is going to have half the arc's measure."}, {"video_title": "Tangents of circles problem (example 1) Mathematics II High School Math Khan Academy.mp3", "Sentence": "And so let's see, you have y plus 80 is equal to 180. If I just subtract 180 from both sides, I'm going to subtract 80 from both sides, and we get y is equal to 100, or the measure of this is 100, the measure of this interior angle right over here is 100 degrees, which also tells us that the measure of this arc, because that interior angle intercepts arc CB right over here, that tells us that the measure of arc CB is also 100 degrees. And so if we're trying to find this angle, the measure of angle D, is the inscribed angle that intercepts the same arc. And we've seen in previous videos that that inscribed angle that intercepts that arc is going to have half the arc's measure. So this is 100 degree measured arc, then the measure of this angle right over here is going to be 50 degrees. So the measure of angle D is 50 degrees. And we are done."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We've done some examples on this playlist where if you had an angle like that, you might call that a 30-degree angle. If you have an angle like this, you could call that a 90-degree angle, and we'd often use this symbol just like that. If you were to go 180 degrees, you'd essentially form a straight line. Let me make these proper angles. If you go 360 degrees, you've essentially done one full rotation. If you watch figure skating on the Olympics and someone does a rotation, they'll say, oh, they did a 360, or especially in some skateboarding competitions and things like that. But the one thing to realize, and it might not be obvious right from the get-go, is this whole notion of degrees, this is a human-constructed system."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let me make these proper angles. If you go 360 degrees, you've essentially done one full rotation. If you watch figure skating on the Olympics and someone does a rotation, they'll say, oh, they did a 360, or especially in some skateboarding competitions and things like that. But the one thing to realize, and it might not be obvious right from the get-go, is this whole notion of degrees, this is a human-constructed system. This is not the only way that you can measure angles. If you think about it, you say, well, why do we call a full rotation 360 degrees? There are some possible theories, and I encourage you to think about them."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "But the one thing to realize, and it might not be obvious right from the get-go, is this whole notion of degrees, this is a human-constructed system. This is not the only way that you can measure angles. If you think about it, you say, well, why do we call a full rotation 360 degrees? There are some possible theories, and I encourage you to think about them. Why does 360 degrees show up in our culture as a full rotation? Well, there's a couple of theories there. One is ancient calendars, and even our calendar is close to this, but ancient calendars were based on 360 days in a year."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "There are some possible theories, and I encourage you to think about them. Why does 360 degrees show up in our culture as a full rotation? Well, there's a couple of theories there. One is ancient calendars, and even our calendar is close to this, but ancient calendars were based on 360 days in a year. Some ancient astronomers observed that things seemed to move 1 360th of the sky per day. Another theory is the ancient Babylonians liked equilateral triangles a lot, and they had a base 60 number system, so they had 60 symbols. We only have 10."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "One is ancient calendars, and even our calendar is close to this, but ancient calendars were based on 360 days in a year. Some ancient astronomers observed that things seemed to move 1 360th of the sky per day. Another theory is the ancient Babylonians liked equilateral triangles a lot, and they had a base 60 number system, so they had 60 symbols. We only have 10. We have a base 10. They had 60. In our system, we like to divide things into 10."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We only have 10. We have a base 10. They had 60. In our system, we like to divide things into 10. They probably like to divide things into 60. If you had a circle and you divided it into 6 equilateral triangles, and each of those equilateral triangles you divided into 60 sections because you have a base 60 number system, then you might end up with 360 degrees. What I want to think about in this video is an alternate way of measuring angles."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "In our system, we like to divide things into 10. They probably like to divide things into 60. If you had a circle and you divided it into 6 equilateral triangles, and each of those equilateral triangles you divided into 60 sections because you have a base 60 number system, then you might end up with 360 degrees. What I want to think about in this video is an alternate way of measuring angles. That alternate way, even though it might not seem as intuitive to you from the get-go, in some ways is much more mathematically pure than degrees. It's not based on these cultural artifacts of base 60 number systems or astronomical patterns. To some degree, an alien on another planet would not use degrees, especially if the degrees are motivated by these astronomical phenomena, but they might use what we're going to define as a radian."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "What I want to think about in this video is an alternate way of measuring angles. That alternate way, even though it might not seem as intuitive to you from the get-go, in some ways is much more mathematically pure than degrees. It's not based on these cultural artifacts of base 60 number systems or astronomical patterns. To some degree, an alien on another planet would not use degrees, especially if the degrees are motivated by these astronomical phenomena, but they might use what we're going to define as a radian. There's a certain degree of purity here. Radians. Let's just cut to the chase and define what a radian is."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "To some degree, an alien on another planet would not use degrees, especially if the degrees are motivated by these astronomical phenomena, but they might use what we're going to define as a radian. There's a certain degree of purity here. Radians. Let's just cut to the chase and define what a radian is. Let me draw a circle here. My best attempt at drawing a circle. Not bad."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let's just cut to the chase and define what a radian is. Let me draw a circle here. My best attempt at drawing a circle. Not bad. Let me draw the center of the circle. Now let me draw this radius. You might already notice the word radius is very close to the word radians, and that's not a coincidence."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Not bad. Let me draw the center of the circle. Now let me draw this radius. You might already notice the word radius is very close to the word radians, and that's not a coincidence. Let's say that this circle has a radius of length r. Let's construct an angle. I'll call that angle theta. Let's construct an angle theta."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "You might already notice the word radius is very close to the word radians, and that's not a coincidence. Let's say that this circle has a radius of length r. Let's construct an angle. I'll call that angle theta. Let's construct an angle theta. Let's call this angle right over here theta. Let's just say, for the sake of argument, that this angle is just the exact right measure. If you look at the arc that subtends this angle, that seems like a very fancy word."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let's construct an angle theta. Let's call this angle right over here theta. Let's just say, for the sake of argument, that this angle is just the exact right measure. If you look at the arc that subtends this angle, that seems like a very fancy word. Let me draw the angle. If you look at the arc that subtends the angle, that's a fancy word, but that's really just talking about the arc along the circle that intersects the two sides of the angles. This arc right over here subtends the angle theta."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "If you look at the arc that subtends this angle, that seems like a very fancy word. Let me draw the angle. If you look at the arc that subtends the angle, that's a fancy word, but that's really just talking about the arc along the circle that intersects the two sides of the angles. This arc right over here subtends the angle theta. Let me write that down. Subtends this arc. Subtends angle theta."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "This arc right over here subtends the angle theta. Let me write that down. Subtends this arc. Subtends angle theta. Let's say theta is the exact right size. This arc is also the same length as the radius of the circle. This arc is also of length r. Given that, if you were defining a new type of angle measurement, and you wanted to call it a radian, which is very close to a radius, how many radians would you define this angle to be?"}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Subtends angle theta. Let's say theta is the exact right size. This arc is also the same length as the radius of the circle. This arc is also of length r. Given that, if you were defining a new type of angle measurement, and you wanted to call it a radian, which is very close to a radius, how many radians would you define this angle to be? The most obvious one, if you view a radian as another way of saying radiuses or radii, you say, look, this is subtended by an arc of one radius, so why don't we call this right over here one radian? Which is exactly how a radian is defined. When you have a circle and you have an angle of one radian, the arc that subtends it is exactly one radius long, which you can imagine might be a little bit useful as we start to interpret more and more types of circles."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "This arc is also of length r. Given that, if you were defining a new type of angle measurement, and you wanted to call it a radian, which is very close to a radius, how many radians would you define this angle to be? The most obvious one, if you view a radian as another way of saying radiuses or radii, you say, look, this is subtended by an arc of one radius, so why don't we call this right over here one radian? Which is exactly how a radian is defined. When you have a circle and you have an angle of one radian, the arc that subtends it is exactly one radius long, which you can imagine might be a little bit useful as we start to interpret more and more types of circles. When you give a degree, you really have to do a little bit of math and think about the circumference and all of that to think about how many radiuses are subtending that angle. Here, the angle in radians tells you exactly how many arc lengths that is subtending the angle. Let's do a couple of thought experiments here."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "When you have a circle and you have an angle of one radian, the arc that subtends it is exactly one radius long, which you can imagine might be a little bit useful as we start to interpret more and more types of circles. When you give a degree, you really have to do a little bit of math and think about the circumference and all of that to think about how many radiuses are subtending that angle. Here, the angle in radians tells you exactly how many arc lengths that is subtending the angle. Let's do a couple of thought experiments here. Given that, what would be the angle in radians if we were to go... Let me draw another circle here. Let me draw another circle here. That's the center."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let's do a couple of thought experiments here. Given that, what would be the angle in radians if we were to go... Let me draw another circle here. Let me draw another circle here. That's the center. We'll start right over there. What would happen if I had an angle? What angle, if I wanted to measure in radians, what angle would this be in radians?"}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "That's the center. We'll start right over there. What would happen if I had an angle? What angle, if I wanted to measure in radians, what angle would this be in radians? You can almost think of it as radiuses. What would that angle be? Going one full revolution in degrees, that would be 360 degrees."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "What angle, if I wanted to measure in radians, what angle would this be in radians? You can almost think of it as radiuses. What would that angle be? Going one full revolution in degrees, that would be 360 degrees. Based on this definition, what would this be in radians? Let's think about the arc that subtends this angle. The arc that subtends this angle is the entire circumference of this circle."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Going one full revolution in degrees, that would be 360 degrees. Based on this definition, what would this be in radians? Let's think about the arc that subtends this angle. The arc that subtends this angle is the entire circumference of this circle. It's the entire circumference of this circle. What's the circumference of a circle in terms of radiuses? If this has length r, if the radius is length r, what's the circumference of the circle in terms of r?"}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "The arc that subtends this angle is the entire circumference of this circle. It's the entire circumference of this circle. What's the circumference of a circle in terms of radiuses? If this has length r, if the radius is length r, what's the circumference of the circle in terms of r? We know that. That's going to be 2 pi r. Going back to this angle, the length of the arc that subtends this angle is how many radiuses this is? What's 2 pi radiuses this is?"}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "If this has length r, if the radius is length r, what's the circumference of the circle in terms of r? We know that. That's going to be 2 pi r. Going back to this angle, the length of the arc that subtends this angle is how many radiuses this is? What's 2 pi radiuses this is? It's 2 pi times r. This angle right over here, I'll call this a different angle, x. x in this case is going to be 2 pi radians. It is subtended by an arc length of 2 pi radiuses. If the radius was one unit, then this would be 2 pi times 1, 2 pi radiuses."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "What's 2 pi radiuses this is? It's 2 pi times r. This angle right over here, I'll call this a different angle, x. x in this case is going to be 2 pi radians. It is subtended by an arc length of 2 pi radiuses. If the radius was one unit, then this would be 2 pi times 1, 2 pi radiuses. Given that, let's start to think about how we can convert between radians and degrees and vice versa. If I were to have, and we can just follow up over here, if we do one full revolution, that is 2 pi radians, how many degrees is this going to be equal to? We already know this."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "If the radius was one unit, then this would be 2 pi times 1, 2 pi radiuses. Given that, let's start to think about how we can convert between radians and degrees and vice versa. If I were to have, and we can just follow up over here, if we do one full revolution, that is 2 pi radians, how many degrees is this going to be equal to? We already know this. A full revolution in degrees is 360 degrees. I could either write out the word degrees or I can use this little degree notation there. Actually, let me write out the word degrees."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We already know this. A full revolution in degrees is 360 degrees. I could either write out the word degrees or I can use this little degree notation there. Actually, let me write out the word degrees. It might make things a little bit clearer that we're using units in both cases. If we wanted to simplify this a little bit, we could divide both sides by 2, in which case we would get on the left-hand side, we would get pi radians would be equal to how many degrees? It would be equal to 180 degrees."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Actually, let me write out the word degrees. It might make things a little bit clearer that we're using units in both cases. If we wanted to simplify this a little bit, we could divide both sides by 2, in which case we would get on the left-hand side, we would get pi radians would be equal to how many degrees? It would be equal to 180 degrees. 180 degrees. I could write it that way or I could write it that way. You see over here, this is 180 degrees."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It would be equal to 180 degrees. 180 degrees. I could write it that way or I could write it that way. You see over here, this is 180 degrees. You also see if you were to draw a circle around here, we've gone halfway around the circle. The arc length or the arc that subtends the angle is half the circumference. Half the circumference are pi radiuses."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "You see over here, this is 180 degrees. You also see if you were to draw a circle around here, we've gone halfway around the circle. The arc length or the arc that subtends the angle is half the circumference. Half the circumference are pi radiuses. We call this pi radians. Pi radians is 180 degrees. From this, we can come up with conversions."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Half the circumference are pi radiuses. We call this pi radians. Pi radians is 180 degrees. From this, we can come up with conversions. One radian would be how many degrees? To do that, we would just have to divide both sides by pi. On the left-hand side, you'd be left with 1."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "From this, we can come up with conversions. One radian would be how many degrees? To do that, we would just have to divide both sides by pi. On the left-hand side, you'd be left with 1. I'll just write it singular now. One radian is equal to, I'm just dividing both sides. Let me make it clear what I'm doing here just to show you."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "On the left-hand side, you'd be left with 1. I'll just write it singular now. One radian is equal to, I'm just dividing both sides. Let me make it clear what I'm doing here just to show you. This isn't some voodoo. I'm just dividing both sides by pi here. On the left-hand side, you're left with 1."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let me make it clear what I'm doing here just to show you. This isn't some voodoo. I'm just dividing both sides by pi here. On the left-hand side, you're left with 1. On the right-hand side, you're left with 180 over pi degrees. One radian is equal to 180 over pi degrees, which is starting to make it an interesting way to convert them. Let's think about it the other way."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "On the left-hand side, you're left with 1. On the right-hand side, you're left with 180 over pi degrees. One radian is equal to 180 over pi degrees, which is starting to make it an interesting way to convert them. Let's think about it the other way. If I were to have one degree, how many radians is that? Let's start off with, let me rewrite this thing over here. We said pi radians is equal to 180 degrees."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let's think about it the other way. If I were to have one degree, how many radians is that? Let's start off with, let me rewrite this thing over here. We said pi radians is equal to 180 degrees. Now we want to think about one degree. Let's solve for one degree. One degree, we can divide both sides by 180."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We said pi radians is equal to 180 degrees. Now we want to think about one degree. Let's solve for one degree. One degree, we can divide both sides by 180. We are left with pi over 180 radians is equal to one degree. Pi over 180 radians is equal to one degree. This might seem confusing and daunting, and it was for me the first time I was exposed to it, especially because we're not exposed to this in our everyday life."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "One degree, we can divide both sides by 180. We are left with pi over 180 radians is equal to one degree. Pi over 180 radians is equal to one degree. This might seem confusing and daunting, and it was for me the first time I was exposed to it, especially because we're not exposed to this in our everyday life. What we're going to see over the next few examples is that as long as we keep in mind this whole idea that 2 pi radians is equal to 360 degrees or that pi radians is equal to 180 degrees, which is the two things that I do keep in my mind, we can always re-derive these two things. You might say, hey, how do I remember if it's pi over 180 or 180 over pi to convert the two things? Well, just remember, which is hopefully intuitive, that 2 pi radians is equal to 360 degrees."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "So over here I have triangle BDC, it's inside of triangle AEC. They both share this angle right over there, so that gives us one angle. We need two to get to angle-angle, which gives us similarity. And we know that these two lines are parallel. We know if two lines are parallel and we have a transversal, that corresponding angles are going to be congruent. So that angle is going to correspond to that angle right over there. And we're done."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "And we know that these two lines are parallel. We know if two lines are parallel and we have a transversal, that corresponding angles are going to be congruent. So that angle is going to correspond to that angle right over there. And we're done. We have one angle in triangle AEC that is congruent to another angle in BDC. And then we have this angle that's obviously congruent to itself that's in both triangles. So both triangles have a pair of corresponding angles that are congruent, so they must be similar."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "And we're done. We have one angle in triangle AEC that is congruent to another angle in BDC. And then we have this angle that's obviously congruent to itself that's in both triangles. So both triangles have a pair of corresponding angles that are congruent, so they must be similar. So we can write triangle ACE is going to be similar to triangle, and we want to get the letters in the right order. So where the blue angle is here, the blue angle there is vertex B. Then we go to the white angle, C. And then we go to the unlabeled angle right over there, BCD."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "So both triangles have a pair of corresponding angles that are congruent, so they must be similar. So we can write triangle ACE is going to be similar to triangle, and we want to get the letters in the right order. So where the blue angle is here, the blue angle there is vertex B. Then we go to the white angle, C. And then we go to the unlabeled angle right over there, BCD. So we did that first one. Now let's do this one right over here. This is kind of similar, but at least it looks just superficially looking at it, that YZ is definitely not parallel to ST."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "Then we go to the white angle, C. And then we go to the unlabeled angle right over there, BCD. So we did that first one. Now let's do this one right over here. This is kind of similar, but at least it looks just superficially looking at it, that YZ is definitely not parallel to ST. So we won't be able to do this corresponding angle argument, especially because they didn't even label it as parallel. And so you don't want to look at things just by the way they look. You definitely want to say, what am I given and what am I not given?"}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "This is kind of similar, but at least it looks just superficially looking at it, that YZ is definitely not parallel to ST. So we won't be able to do this corresponding angle argument, especially because they didn't even label it as parallel. And so you don't want to look at things just by the way they look. You definitely want to say, what am I given and what am I not given? If these weren't labeled parallel, we wouldn't be able to make the statement, even if they looked parallel. One thing we do have is that we have this angle right here that's common to the inner triangle and to the outer triangle. And they've given us a bunch of sides."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "You definitely want to say, what am I given and what am I not given? If these weren't labeled parallel, we wouldn't be able to make the statement, even if they looked parallel. One thing we do have is that we have this angle right here that's common to the inner triangle and to the outer triangle. And they've given us a bunch of sides. So maybe we can use SAS for similarity, meaning if we can show the ratio of the sides on either side of this angle, if they have the same ratio from the smaller triangle to the larger triangle, then we can show similarity. So let's go, and we have to go on either side of this angle right over here. Let's look at the shorter side on either side of this angle."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "And they've given us a bunch of sides. So maybe we can use SAS for similarity, meaning if we can show the ratio of the sides on either side of this angle, if they have the same ratio from the smaller triangle to the larger triangle, then we can show similarity. So let's go, and we have to go on either side of this angle right over here. Let's look at the shorter side on either side of this angle. So the shorter side is 2. And let's look at the shorter side on either side of the angle for the larger triangle. Well, then the shorter side is on the right-hand side, and that's going to be xt."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "Let's look at the shorter side on either side of this angle. So the shorter side is 2. And let's look at the shorter side on either side of the angle for the larger triangle. Well, then the shorter side is on the right-hand side, and that's going to be xt. So what we want to compare is the ratio between, let me write it this way, we want to see is xy over xt, is that equal to the ratio of the longer side, or if we're looking relative to this angle, the longer of the two, not necessarily the longest of the triangle, although it looks like that as well, is that equal to the ratio of xz over the longer of the two sides when you're looking at this angle right here on either side of that angle for the larger triangle over xs. And it's a little confusing because we've kind of flipped which side, but I'm just thinking about the shorter side on either side of this angle in between and then the longer side on either side of this angle. So these are the shorter sides for the smaller triangle and the larger triangle."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "Well, then the shorter side is on the right-hand side, and that's going to be xt. So what we want to compare is the ratio between, let me write it this way, we want to see is xy over xt, is that equal to the ratio of the longer side, or if we're looking relative to this angle, the longer of the two, not necessarily the longest of the triangle, although it looks like that as well, is that equal to the ratio of xz over the longer of the two sides when you're looking at this angle right here on either side of that angle for the larger triangle over xs. And it's a little confusing because we've kind of flipped which side, but I'm just thinking about the shorter side on either side of this angle in between and then the longer side on either side of this angle. So these are the shorter sides for the smaller triangle and the larger triangle. These are the longer sides for the smaller triangle and the larger triangle. And we see xy, this is 2, xt is 3 plus 1 is 4, xz is 3, and xs is 6. So you have 2 over 4, which is 1 half, which is the same thing as 3, 6."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "So these are the shorter sides for the smaller triangle and the larger triangle. These are the longer sides for the smaller triangle and the larger triangle. And we see xy, this is 2, xt is 3 plus 1 is 4, xz is 3, and xs is 6. So you have 2 over 4, which is 1 half, which is the same thing as 3, 6. So the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle for both triangles, the ratio is the same. So by SAS we know that the two triangles are congruent. But we have to be careful on how we state the triangles."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "So you have 2 over 4, which is 1 half, which is the same thing as 3, 6. So the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle for both triangles, the ratio is the same. So by SAS we know that the two triangles are congruent. But we have to be careful on how we state the triangles. We want to make sure we get the corresponding sides. So we could say that triangle, and I'm running out of space here, let me write it right above here. We can write that triangle xyz is similar to triangle."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "But we have to be careful on how we state the triangles. We want to make sure we get the corresponding sides. So we could say that triangle, and I'm running out of space here, let me write it right above here. We can write that triangle xyz is similar to triangle. So we started up at x, which is the vertex at the angle, and we went to the shorter side first. So now we want to start at x and go to the shorter side of the large triangle. So you go to xts."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "We can write that triangle xyz is similar to triangle. So we started up at x, which is the vertex at the angle, and we went to the shorter side first. So now we want to start at x and go to the shorter side of the large triangle. So you go to xts. xyz is similar to xts. Now let's look at this right over here. So in our larger triangle we have a right angle here, but we really know nothing about what's going on with any of these smaller triangles in terms of their actual angles."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "So you go to xts. xyz is similar to xts. Now let's look at this right over here. So in our larger triangle we have a right angle here, but we really know nothing about what's going on with any of these smaller triangles in terms of their actual angles. Even though this looks like a right angle, we cannot assume it. And it shares, if we look at this smaller triangle right over here, it shares one side with the larger triangle, but that's not enough to do anything. And then this triangle over here also shares another side, but that also doesn't do anything."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "So in our larger triangle we have a right angle here, but we really know nothing about what's going on with any of these smaller triangles in terms of their actual angles. Even though this looks like a right angle, we cannot assume it. And it shares, if we look at this smaller triangle right over here, it shares one side with the larger triangle, but that's not enough to do anything. And then this triangle over here also shares another side, but that also doesn't do anything. So we really can't make any statement here about any kind of similarity. So there's no similarity going on here. If they gave us, well, there are some shared angles."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "And then this triangle over here also shares another side, but that also doesn't do anything. So we really can't make any statement here about any kind of similarity. So there's no similarity going on here. If they gave us, well, there are some shared angles. This guy, they both share that angle. The larger triangle and the smaller triangle. So there could be a statement of similarity we could make if we knew that this definitely was a right angle."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "If they gave us, well, there are some shared angles. This guy, they both share that angle. The larger triangle and the smaller triangle. So there could be a statement of similarity we could make if we knew that this definitely was a right angle. Then we could make some interesting statements about similarity. But right now we can't really do anything as is. Let's try this one out, or this pair right over here."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "So there could be a statement of similarity we could make if we knew that this definitely was a right angle. Then we could make some interesting statements about similarity. But right now we can't really do anything as is. Let's try this one out, or this pair right over here. So these are the first ones that we've actually separated out the triangles. So they've given us the three sides of both triangles. So let's just figure out if the ratios between corresponding sides are a constant."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "Let's try this one out, or this pair right over here. So these are the first ones that we've actually separated out the triangles. So they've given us the three sides of both triangles. So let's just figure out if the ratios between corresponding sides are a constant. So let's start with the short side. So the short side here is 3. The shortest side here is 9 square roots of 3."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "So let's just figure out if the ratios between corresponding sides are a constant. So let's start with the short side. So the short side here is 3. The shortest side here is 9 square roots of 3. So we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here is 3 square roots of 3, is equal to 3 square roots of 3, over the next longest side over here, which is 27. And then see if that's going to be equal to the ratio of the longest side. So the longest side here is 6."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "The shortest side here is 9 square roots of 3. So we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here is 3 square roots of 3, is equal to 3 square roots of 3, over the next longest side over here, which is 27. And then see if that's going to be equal to the ratio of the longest side. So the longest side here is 6. And then the longest side over here is 18 square roots of 3. So this is going to give us, let's see, this is 3. Let me do this in a neutral color."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "So the longest side here is 6. And then the longest side over here is 18 square roots of 3. So this is going to give us, let's see, this is 3. Let me do this in a neutral color. So this becomes 1 over 3 square roots of 3. This becomes 1 over square roots of 3, root 3 over 9, which seems like a different number, but we want to be careful here. And then this right over here, this becomes, this is a, if you divide the numerator and denominator by 6, this becomes a 1 and this becomes 3 square roots of 3."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "Let me do this in a neutral color. So this becomes 1 over 3 square roots of 3. This becomes 1 over square roots of 3, root 3 over 9, which seems like a different number, but we want to be careful here. And then this right over here, this becomes, this is a, if you divide the numerator and denominator by 6, this becomes a 1 and this becomes 3 square roots of 3. So you get 1 square roots of 3, 1 over 3 root 3 needs to be equal to 1, needs to be equal to square root of 3 over 9, which needs to be equal to 1 over 3 square roots of 3. At first they don't look equal, but we can actually rationalize this denominator right over here. We can show that 1 over 3 square roots of 3, if you multiply it by square root of 3 over square root of 3, this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 times 3 is 9."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "And then this right over here, this becomes, this is a, if you divide the numerator and denominator by 6, this becomes a 1 and this becomes 3 square roots of 3. So you get 1 square roots of 3, 1 over 3 root 3 needs to be equal to 1, needs to be equal to square root of 3 over 9, which needs to be equal to 1 over 3 square roots of 3. At first they don't look equal, but we can actually rationalize this denominator right over here. We can show that 1 over 3 square roots of 3, if you multiply it by square root of 3 over square root of 3, this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 times 3 is 9. So these actually are all the same. This is actually saying, this is 1 over 3 root 3, which is the same thing as square root of 3 over 9, which is this right over here, which is the same thing as 1 over 3 root 3. So actually these are similar triangles."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "We can show that 1 over 3 square roots of 3, if you multiply it by square root of 3 over square root of 3, this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 times 3 is 9. So these actually are all the same. This is actually saying, this is 1 over 3 root 3, which is the same thing as square root of 3 over 9, which is this right over here, which is the same thing as 1 over 3 root 3. So actually these are similar triangles. So we can actually say it, and I'll make sure I get the order right. So I'll start with E, which is between the blue and the magenta side. So that's between the blue and the magenta side, that is H right over here."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "So actually these are similar triangles. So we can actually say it, and I'll make sure I get the order right. So I'll start with E, which is between the blue and the magenta side. So that's between the blue and the magenta side, that is H right over here. So triangle E, I'll do it like this. Triangle E, and then I'll go along the blue side, F. Then I'll go along the blue side over here. Oh, sorry, let me do it this way."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "So that's between the blue and the magenta side, that is H right over here. So triangle E, I'll do it like this. Triangle E, and then I'll go along the blue side, F. Then I'll go along the blue side over here. Oh, sorry, let me do it this way. Actually, let me just write it this way. E, triangle E, F, G, we know is similar to triangle. So E is between the blue and the magenta side."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "Oh, sorry, let me do it this way. Actually, let me just write it this way. E, triangle E, F, G, we know is similar to triangle. So E is between the blue and the magenta side. Blue and magenta side, that is H. And then we go along the blue side to F, go along the blue side to I. And then you went along the orange side to G. And then you go along the orange side to J. So triangle E, F, G is similar to triangle H, I, J by side, side, side similarity."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "So E is between the blue and the magenta side. Blue and magenta side, that is H. And then we go along the blue side to F, go along the blue side to I. And then you went along the orange side to G. And then you go along the orange side to J. So triangle E, F, G is similar to triangle H, I, J by side, side, side similarity. They're not congruent sides. They all have just the same ratio or the same scaling factor. Now let's do this last one right over here."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "So triangle E, F, G is similar to triangle H, I, J by side, side, side similarity. They're not congruent sides. They all have just the same ratio or the same scaling factor. Now let's do this last one right over here. So we have an angle that's congruent to another angle right over there. And we have two sides. And so it might be tempting to use side, angle, side, because we have side, angle, side here."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "Now let's do this last one right over here. So we have an angle that's congruent to another angle right over there. And we have two sides. And so it might be tempting to use side, angle, side, because we have side, angle, side here. And even the ratios look kind of tempting, because 4 times 2 is 8, 5 times 2 is 10. But it's tricky here, because they aren't the same corresponding sides. In order to use side, angle, side, the two sides that have the same corresponding ratios, they have to be on either side of the angle."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "And so it might be tempting to use side, angle, side, because we have side, angle, side here. And even the ratios look kind of tempting, because 4 times 2 is 8, 5 times 2 is 10. But it's tricky here, because they aren't the same corresponding sides. In order to use side, angle, side, the two sides that have the same corresponding ratios, they have to be on either side of the angle. So in this case, they are on either side of the angle. In this case, the 4 is on one side of the angle, but the 5 is not. So because if this 5 was over here, then we could make an argument for similarity."}, {"video_title": "Similar triangle example problems Similarity Geometry Khan Academy.mp3", "Sentence": "In order to use side, angle, side, the two sides that have the same corresponding ratios, they have to be on either side of the angle. So in this case, they are on either side of the angle. In this case, the 4 is on one side of the angle, but the 5 is not. So because if this 5 was over here, then we could make an argument for similarity. But with this 5 not being on the other side of the angle, it's not sandwiching the angle with the 4, we can't use side, angle, side. And frankly, there's nothing that we can do over here. So we can't make some strong statement about similarity for this last one."}, {"video_title": "Another example using compass and straightedge for tangent line Geometry Khan Academy.mp3", "Sentence": "And in this example, P doesn't sit on this circle. P is outside of the circle. So we want to draw something like, and actually let me get my straight edge out. So my straight edge, and I have my controls up here. It is on the Khan Academy exercise, constructing a line tangent to a circle. So I have that up there. So let me add a straight edge."}, {"video_title": "Another example using compass and straightedge for tangent line Geometry Khan Academy.mp3", "Sentence": "So my straight edge, and I have my controls up here. It is on the Khan Academy exercise, constructing a line tangent to a circle. So I have that up there. So let me add a straight edge. Once again, you could try to eyeball it. I'm going to go through P. You want to be tangent to the circle. So hey, yeah, maybe something like that."}, {"video_title": "Another example using compass and straightedge for tangent line Geometry Khan Academy.mp3", "Sentence": "So let me add a straight edge. Once again, you could try to eyeball it. I'm going to go through P. You want to be tangent to the circle. So hey, yeah, maybe something like that. That looks pretty good. But like we've said in other geometric constructions videos, that's just eyeballing it. We don't know how precise that actually is."}, {"video_title": "Another example using compass and straightedge for tangent line Geometry Khan Academy.mp3", "Sentence": "So hey, yeah, maybe something like that. That looks pretty good. But like we've said in other geometric constructions videos, that's just eyeballing it. We don't know how precise that actually is. So what if we had a compass and a straight edge? How do we make something more precise? And actually, in the process of doing it, we'll make really fun patterns."}, {"video_title": "Another example using compass and straightedge for tangent line Geometry Khan Academy.mp3", "Sentence": "We don't know how precise that actually is. So what if we had a compass and a straight edge? How do we make something more precise? And actually, in the process of doing it, we'll make really fun patterns. So how do we do that? Well, what I'm going to do is I'm going to try to construct a circle that has the segment PC as a diameter. So let me draw that."}, {"video_title": "Another example using compass and straightedge for tangent line Geometry Khan Academy.mp3", "Sentence": "And actually, in the process of doing it, we'll make really fun patterns. So how do we do that? Well, what I'm going to do is I'm going to try to construct a circle that has the segment PC as a diameter. So let me draw that. So I want to construct a circle that has segment PC as a diameter. And in order to do that, I need to figure out where the center of that circle is. And you'll see in a second why it's useful to have this circle that has PC as a diameter."}, {"video_title": "Another example using compass and straightedge for tangent line Geometry Khan Academy.mp3", "Sentence": "So let me draw that. So I want to construct a circle that has segment PC as a diameter. And in order to do that, I need to figure out where the center of that circle is. And you'll see in a second why it's useful to have this circle that has PC as a diameter. So where is the center? It looks like it's roughly here. But how do we actually construct it?"}, {"video_title": "Another example using compass and straightedge for tangent line Geometry Khan Academy.mp3", "Sentence": "And you'll see in a second why it's useful to have this circle that has PC as a diameter. So where is the center? It looks like it's roughly here. But how do we actually construct it? To do that, we're going to have to figure out the midpoint of segment PC or CP. And to do that, what I'm going to do is construct two circles. I'll just make them a little bit larger."}, {"video_title": "Another example using compass and straightedge for tangent line Geometry Khan Academy.mp3", "Sentence": "But how do we actually construct it? To do that, we're going to have to figure out the midpoint of segment PC or CP. And to do that, what I'm going to do is construct two circles. I'll just make them a little bit larger. So one centered at C. I'll make it reasonably large, maybe that big. And then I'm going to construct another circle of the same radius. It's that large one that I just constructed."}, {"video_title": "Another example using compass and straightedge for tangent line Geometry Khan Academy.mp3", "Sentence": "I'll just make them a little bit larger. So one centered at C. I'll make it reasonably large, maybe that big. And then I'm going to construct another circle of the same radius. It's that large one that I just constructed. So it's that same radius. But I'm now going to center it at P. Now why is this interesting? Why is what I just did interesting?"}, {"video_title": "Another example using compass and straightedge for tangent line Geometry Khan Academy.mp3", "Sentence": "It's that large one that I just constructed. So it's that same radius. But I'm now going to center it at P. Now why is this interesting? Why is what I just did interesting? Well, where these two larger circles intersect are going to be equidistant to P and C. How do we know that? Well, all the points on this circle are equidistant to C. All the points on this circle are equidistant to P. And these circles have the same radius. And at this point right over here, they are equidistant to both of them because it sits on both circles."}, {"video_title": "Another example using compass and straightedge for tangent line Geometry Khan Academy.mp3", "Sentence": "Why is what I just did interesting? Well, where these two larger circles intersect are going to be equidistant to P and C. How do we know that? Well, all the points on this circle are equidistant to C. All the points on this circle are equidistant to P. And these circles have the same radius. And at this point right over here, they are equidistant to both of them because it sits on both circles. So this point is equidistant to both of them. And so is this point right over here. And so they both sit on the perpendicular bisector of this segment, of the segment CP."}, {"video_title": "Another example using compass and straightedge for tangent line Geometry Khan Academy.mp3", "Sentence": "And at this point right over here, they are equidistant to both of them because it sits on both circles. So this point is equidistant to both of them. And so is this point right over here. And so they both sit on the perpendicular bisector of this segment, of the segment CP. So actually, let me draw that. So if I draw a line that looks something like this, this right over here is a perpendicular bisector of line of segment CP. Now what we really just care about is that it is bisecting it because we wanted to find the midpoint."}, {"video_title": "Another example using compass and straightedge for tangent line Geometry Khan Academy.mp3", "Sentence": "And so they both sit on the perpendicular bisector of this segment, of the segment CP. So actually, let me draw that. So if I draw a line that looks something like this, this right over here is a perpendicular bisector of line of segment CP. Now what we really just care about is that it is bisecting it because we wanted to find the midpoint. And now that we have found the midpoint of our segment, we're ready to construct that circle I talked about. A circle centered at the midpoint. And it has a diameter."}, {"video_title": "Another example using compass and straightedge for tangent line Geometry Khan Academy.mp3", "Sentence": "Now what we really just care about is that it is bisecting it because we wanted to find the midpoint. And now that we have found the midpoint of our segment, we're ready to construct that circle I talked about. A circle centered at the midpoint. And it has a diameter. It has CP as a diameter. So I got that far. But why did I go through all of this trouble to do that?"}, {"video_title": "Another example using compass and straightedge for tangent line Geometry Khan Academy.mp3", "Sentence": "And it has a diameter. It has CP as a diameter. So I got that far. But why did I go through all of this trouble to do that? Well, now we're going to use the idea that if you have a triangle embedded in a circle where one side of the triangle is a diameter, then you're going to have a right triangle. Well, what am I talking about? Well, let me actually just draw the triangle."}, {"video_title": "Another example using compass and straightedge for tangent line Geometry Khan Academy.mp3", "Sentence": "But why did I go through all of this trouble to do that? Well, now we're going to use the idea that if you have a triangle embedded in a circle where one side of the triangle is a diameter, then you're going to have a right triangle. Well, what am I talking about? Well, let me actually just draw the triangle. So let me add a straight edge here. So I'm going to draw a triangle. So it has one side as a diameter."}, {"video_title": "Another example using compass and straightedge for tangent line Geometry Khan Academy.mp3", "Sentence": "Well, let me actually just draw the triangle. So let me add a straight edge here. So I'm going to draw a triangle. So it has one side as a diameter. So we're going to embed it in this circle that I just constructed. The circle centered at this midpoint, this circle right over here. CP is clearly a diameter."}, {"video_title": "Another example using compass and straightedge for tangent line Geometry Khan Academy.mp3", "Sentence": "So it has one side as a diameter. So we're going to embed it in this circle that I just constructed. The circle centered at this midpoint, this circle right over here. CP is clearly a diameter. And I'm going to embed it in CP. But I'm going to put it at this point right over here. Because this point sits on this circle centered at C. And I'm going to draw another line."}, {"video_title": "Another example using compass and straightedge for tangent line Geometry Khan Academy.mp3", "Sentence": "CP is clearly a diameter. And I'm going to embed it in CP. But I'm going to put it at this point right over here. Because this point sits on this circle centered at C. And I'm going to draw another line. So this other line, let me put that there and this there. So my claim is that this is a right triangle. How do I make this claim?"}, {"video_title": "Another example using compass and straightedge for tangent line Geometry Khan Academy.mp3", "Sentence": "Because this point sits on this circle centered at C. And I'm going to draw another line. So this other line, let me put that there and this there. So my claim is that this is a right triangle. How do I make this claim? And I've proved it in other videos. Is it's embedded in a circle. It's embedded in this circle right over here that's highlighted in yellow."}, {"video_title": "Another example using compass and straightedge for tangent line Geometry Khan Academy.mp3", "Sentence": "How do I make this claim? And I've proved it in other videos. Is it's embedded in a circle. It's embedded in this circle right over here that's highlighted in yellow. It has a diameter. And it has a diameter for one of its sides. And that side is actually its hypotenuse."}, {"video_title": "Another example using compass and straightedge for tangent line Geometry Khan Academy.mp3", "Sentence": "It's embedded in this circle right over here that's highlighted in yellow. It has a diameter. And it has a diameter for one of its sides. And that side is actually its hypotenuse. And we prove it in other videos. So this is a right triangle. Well, why is this useful for proving, or why is this useful for constructing a tangent line to this circle over here, to circle centered at C?"}, {"video_title": "Another example using compass and straightedge for tangent line Geometry Khan Academy.mp3", "Sentence": "And that side is actually its hypotenuse. And we prove it in other videos. So this is a right triangle. Well, why is this useful for proving, or why is this useful for constructing a tangent line to this circle over here, to circle centered at C? Well, this side right over here that I have in orange, that I'm highlighting right now, that's a radius of C. And if that forms a right angle with this line right over here, then this line right over here must be tangent. To really make it look tangent, I'll elongate it a little bit just like that. And there you go."}, {"video_title": "Another example using compass and straightedge for tangent line Geometry Khan Academy.mp3", "Sentence": "Well, why is this useful for proving, or why is this useful for constructing a tangent line to this circle over here, to circle centered at C? Well, this side right over here that I have in orange, that I'm highlighting right now, that's a radius of C. And if that forms a right angle with this line right over here, then this line right over here must be tangent. To really make it look tangent, I'll elongate it a little bit just like that. And there you go. You should feel good that this looks like it's truly intersecting at a right angle, and that this segment that I'm highlighting in orange right now is truly tangent. So once again, a lot more trouble than just eyeballing it and trying to draw it. And actually, a lot of you all probably could have eyeballed this segment right over here."}, {"video_title": "Another example using compass and straightedge for tangent line Geometry Khan Academy.mp3", "Sentence": "And there you go. You should feel good that this looks like it's truly intersecting at a right angle, and that this segment that I'm highlighting in orange right now is truly tangent. So once again, a lot more trouble than just eyeballing it and trying to draw it. And actually, a lot of you all probably could have eyeballed this segment right over here. But if you're doing something on a larger scale, you want to be more precise, it's useful to be able to do these constructions. And frankly, in the process here, while you're dealing with the compass and the ruler or the straight edge, you get a new appreciation for what these tools can do. And you're also making some pretty fun patterns and designs."}, {"video_title": "Finding measures using rigid transformations.mp3", "Sentence": "Fair enough. So based on that, they're going to ask us some questions, and I encourage you to pause this video and see if you can figure out the answers to these questions on your own before I work through them. So the first question they say is, well, what's A prime, C prime? This is really, what's the length of segment A prime, C prime? So they want the length of this right over here. How do we figure that out? Well, the key realization here is a reflection is a rigid transformation, rigid transformation, which is a very fancy word, but it's really just saying that it's a transformation where the length between corresponding points don't change."}, {"video_title": "Finding measures using rigid transformations.mp3", "Sentence": "This is really, what's the length of segment A prime, C prime? So they want the length of this right over here. How do we figure that out? Well, the key realization here is a reflection is a rigid transformation, rigid transformation, which is a very fancy word, but it's really just saying that it's a transformation where the length between corresponding points don't change. So if we're talking about a shape like a triangle, the angle measures won't change, the perimeter won't change, and the area won't change. So we're gonna use the fact that the length between corresponding points won't change, so the length between A prime and C prime is gonna be the same as the length between A and C. So A prime, C prime is going to be equal to AC, which is equal to, they tell us, right over there. That's this corresponding side of the triangle."}, {"video_title": "Finding measures using rigid transformations.mp3", "Sentence": "Well, the key realization here is a reflection is a rigid transformation, rigid transformation, which is a very fancy word, but it's really just saying that it's a transformation where the length between corresponding points don't change. So if we're talking about a shape like a triangle, the angle measures won't change, the perimeter won't change, and the area won't change. So we're gonna use the fact that the length between corresponding points won't change, so the length between A prime and C prime is gonna be the same as the length between A and C. So A prime, C prime is going to be equal to AC, which is equal to, they tell us, right over there. That's this corresponding side of the triangle. That has a length of three. So we answered the first question, and maybe that gave you a good clue, and so I encourage you to keep pausing the video when you feel like you can have a go at it. All right, the next question is, what is the measure of angle B prime?"}, {"video_title": "Finding measures using rigid transformations.mp3", "Sentence": "That's this corresponding side of the triangle. That has a length of three. So we answered the first question, and maybe that gave you a good clue, and so I encourage you to keep pausing the video when you feel like you can have a go at it. All right, the next question is, what is the measure of angle B prime? So that's this angle right over here, and we're gonna use the exact same property. Angle B prime corresponds to angle B. It underwent a rigid transformation of a reflection."}, {"video_title": "Finding measures using rigid transformations.mp3", "Sentence": "All right, the next question is, what is the measure of angle B prime? So that's this angle right over here, and we're gonna use the exact same property. Angle B prime corresponds to angle B. It underwent a rigid transformation of a reflection. This would also be true if we had a translation or if we had a rotation, and so right over here, the measure of angle B prime would be the same as the measure of angle B, but what is that going to be equal to? Well, we can use the fact that if we call that measure, let's just call that X, X plus 53 degrees, we'll do it all in degrees, plus 90 degrees, this right angle here, well, the sum of the interior angles of a triangle add up to 180 degrees, and so what do we have? We could subtract, let's see, 53 plus 90 is X, plus 143 degrees is equal to 180 degrees, and so subtract 143 degrees from both sides, you get X is equal to, let's see, 180, 80 minus 40 would be 40, 80 minus 43 would be 37 degrees."}, {"video_title": "Finding measures using rigid transformations.mp3", "Sentence": "It underwent a rigid transformation of a reflection. This would also be true if we had a translation or if we had a rotation, and so right over here, the measure of angle B prime would be the same as the measure of angle B, but what is that going to be equal to? Well, we can use the fact that if we call that measure, let's just call that X, X plus 53 degrees, we'll do it all in degrees, plus 90 degrees, this right angle here, well, the sum of the interior angles of a triangle add up to 180 degrees, and so what do we have? We could subtract, let's see, 53 plus 90 is X, plus 143 degrees is equal to 180 degrees, and so subtract 143 degrees from both sides, you get X is equal to, let's see, 180, 80 minus 40 would be 40, 80 minus 43 would be 37 degrees. X is equal to 37 degrees, so that is 37 degrees. If that's 37 degrees, then this is also going to be 37 degrees. Next, they ask us, what is the area of triangle ABC, ABC?"}, {"video_title": "Finding measures using rigid transformations.mp3", "Sentence": "We could subtract, let's see, 53 plus 90 is X, plus 143 degrees is equal to 180 degrees, and so subtract 143 degrees from both sides, you get X is equal to, let's see, 180, 80 minus 40 would be 40, 80 minus 43 would be 37 degrees. X is equal to 37 degrees, so that is 37 degrees. If that's 37 degrees, then this is also going to be 37 degrees. Next, they ask us, what is the area of triangle ABC, ABC? Well, it's gonna have the same area as A prime, B prime, C prime, and so a couple of ways we could think about it. We could try to find the area of A prime, B prime, C prime based on the fact that we already know that this length is three and this is a right triangle, or we can use the fact that this length right over here, four, from A prime to B prime, is gonna be the same thing as this length right over here, from A prime to B prime, which is four. And so the area of this triangle, especially this is a right triangle, it's quite straightforward, it's the base times the height times 1 1\u20442."}, {"video_title": "Finding measures using rigid transformations.mp3", "Sentence": "Next, they ask us, what is the area of triangle ABC, ABC? Well, it's gonna have the same area as A prime, B prime, C prime, and so a couple of ways we could think about it. We could try to find the area of A prime, B prime, C prime based on the fact that we already know that this length is three and this is a right triangle, or we can use the fact that this length right over here, four, from A prime to B prime, is gonna be the same thing as this length right over here, from A prime to B prime, which is four. And so the area of this triangle, especially this is a right triangle, it's quite straightforward, it's the base times the height times 1 1\u20442. So this area is gonna be 1 1\u20442 times the base, four, times the height, three, which is equal to 1\u20442, which is equal to six square units. And then last but not least, what's the perimeter of triangle A prime, B prime, C prime? Well, here we just use the Pythagorean theorem to figure out the length of this hypotenuse, and we know that this is a length of three based on the whole rigid transformation and lengths are preserved."}, {"video_title": "Finding measures using rigid transformations.mp3", "Sentence": "And so the area of this triangle, especially this is a right triangle, it's quite straightforward, it's the base times the height times 1 1\u20442. So this area is gonna be 1 1\u20442 times the base, four, times the height, three, which is equal to 1\u20442, which is equal to six square units. And then last but not least, what's the perimeter of triangle A prime, B prime, C prime? Well, here we just use the Pythagorean theorem to figure out the length of this hypotenuse, and we know that this is a length of three based on the whole rigid transformation and lengths are preserved. And so you might immediately recognize that if you have a right triangle where one side is three and another side is four, that the hypotenuse is five, three, four, five triangles, or you could just use the Pythagorean theorem. You say three squared plus four squared, four squared is equal to, let's just say, the hypotenuse, the hypotenuse squared. Well, three squared plus four squared, that's nine plus 16."}, {"video_title": "Finding measures using rigid transformations.mp3", "Sentence": "Well, here we just use the Pythagorean theorem to figure out the length of this hypotenuse, and we know that this is a length of three based on the whole rigid transformation and lengths are preserved. And so you might immediately recognize that if you have a right triangle where one side is three and another side is four, that the hypotenuse is five, three, four, five triangles, or you could just use the Pythagorean theorem. You say three squared plus four squared, four squared is equal to, let's just say, the hypotenuse, the hypotenuse squared. Well, three squared plus four squared, that's nine plus 16. 25 is equal to the hypotenuse squared, and so the hypotenuse right over here will be equal to five. And so they're not asking us the length of the hypotenuse, they wanna know the perimeter. So it's gonna be four plus three plus five, which is equal to 12, the perimeter of either of those triangles, because it's just one's the image of the other under a rigid transformation, they're gonna have the same perimeter, the same area."}, {"video_title": "Proving triangle congruence Congruence High school geometry Khan Academy.mp3", "Sentence": "Pause this video and see if you can figure that out on your own. All right, now let's work through this together. So let's see what we can figure out. We see that segment DC is parallel to segment AB, that's what these little arrows tell us. And so you can view this segment AC as something of a transversal across those parallel lines. And we know that alternate interior angles would be congruent. So we know, for example, that the measure of this angle is the same as the measure of this angle, or that those angles are congruent."}, {"video_title": "Proving triangle congruence Congruence High school geometry Khan Academy.mp3", "Sentence": "We see that segment DC is parallel to segment AB, that's what these little arrows tell us. And so you can view this segment AC as something of a transversal across those parallel lines. And we know that alternate interior angles would be congruent. So we know, for example, that the measure of this angle is the same as the measure of this angle, or that those angles are congruent. We also know that both of these triangles, both triangle DCA and triangle BAC, they share this side, which by reflexivity is going to be congruent to itself. So in both triangles, we have an angle and a side that are congruent, but can we figure out anything else? Well, you might be tempted to make a similar argument thinking that this is parallel to that because it looks parallel, but you can't make that assumption just based on how it looks."}, {"video_title": "Proving triangle congruence Congruence High school geometry Khan Academy.mp3", "Sentence": "So we know, for example, that the measure of this angle is the same as the measure of this angle, or that those angles are congruent. We also know that both of these triangles, both triangle DCA and triangle BAC, they share this side, which by reflexivity is going to be congruent to itself. So in both triangles, we have an angle and a side that are congruent, but can we figure out anything else? Well, you might be tempted to make a similar argument thinking that this is parallel to that because it looks parallel, but you can't make that assumption just based on how it looks. If you did know that, then you would be able to make some other assumptions about some other angles here and maybe prove congruency. But it turns out given the information that we have, we can't just assume that because something looks parallel, or because something looks congruent that they are. And so based on just the information given, we actually can't prove congruency."}, {"video_title": "Proving triangle congruence Congruence High school geometry Khan Academy.mp3", "Sentence": "Well, you might be tempted to make a similar argument thinking that this is parallel to that because it looks parallel, but you can't make that assumption just based on how it looks. If you did know that, then you would be able to make some other assumptions about some other angles here and maybe prove congruency. But it turns out given the information that we have, we can't just assume that because something looks parallel, or because something looks congruent that they are. And so based on just the information given, we actually can't prove congruency. Now, let me ask you a slightly different question. Let's say that we did give you a little bit more information. Let's say we told you that the measure of this angle right over here is 31 degrees, and the measure of this angle right over here is 31 degrees."}, {"video_title": "Proving triangle congruence Congruence High school geometry Khan Academy.mp3", "Sentence": "And so based on just the information given, we actually can't prove congruency. Now, let me ask you a slightly different question. Let's say that we did give you a little bit more information. Let's say we told you that the measure of this angle right over here is 31 degrees, and the measure of this angle right over here is 31 degrees. Can you now prove that triangle DCA is congruent to triangle BAC? So let's see what we can deduce now. Well, we know that AC is in both triangles, so it's going to be congruent to itself."}, {"video_title": "Proving triangle congruence Congruence High school geometry Khan Academy.mp3", "Sentence": "Let's say we told you that the measure of this angle right over here is 31 degrees, and the measure of this angle right over here is 31 degrees. Can you now prove that triangle DCA is congruent to triangle BAC? So let's see what we can deduce now. Well, we know that AC is in both triangles, so it's going to be congruent to itself. And let me write that down. We know that segment AC is congruent to segment AC. It sits in both triangles, and this is by reflexivity, which is a fancy way of saying that something is going to be congruent to itself."}, {"video_title": "Proving triangle congruence Congruence High school geometry Khan Academy.mp3", "Sentence": "Well, we know that AC is in both triangles, so it's going to be congruent to itself. And let me write that down. We know that segment AC is congruent to segment AC. It sits in both triangles, and this is by reflexivity, which is a fancy way of saying that something is going to be congruent to itself. Now, we also see that AB is parallel to DC just like before, and AC can be viewed as part of a transversal. So we can deduce that angle CAB, let me write this down, actually I'm gonna do it in a different color. We can deduce that angle CAB, CAB is congruent to angle ACD, angle ACD, because they are alternate interior angles where a transversal intersects two parallel lines."}, {"video_title": "Proving triangle congruence Congruence High school geometry Khan Academy.mp3", "Sentence": "It sits in both triangles, and this is by reflexivity, which is a fancy way of saying that something is going to be congruent to itself. Now, we also see that AB is parallel to DC just like before, and AC can be viewed as part of a transversal. So we can deduce that angle CAB, let me write this down, actually I'm gonna do it in a different color. We can deduce that angle CAB, CAB is congruent to angle ACD, angle ACD, because they are alternate interior angles where a transversal intersects two parallel lines. So just to be clear, this angle, which is CAB, is congruent to this angle, which is ACD. And so now we have two angles and a side, two angles and a side that are congruent. So we can now deduce by angle-angle-side postulate that the triangles are indeed congruent."}, {"video_title": "Proving triangle congruence Congruence High school geometry Khan Academy.mp3", "Sentence": "We can deduce that angle CAB, CAB is congruent to angle ACD, angle ACD, because they are alternate interior angles where a transversal intersects two parallel lines. So just to be clear, this angle, which is CAB, is congruent to this angle, which is ACD. And so now we have two angles and a side, two angles and a side that are congruent. So we can now deduce by angle-angle-side postulate that the triangles are indeed congruent. So we now know that triangle DCA is indeed congruent to triangle BAC because of angle-angle-side congruency, which we've talked about in previous videos. And just to be clear, sometimes people like the two-column proofs, I can make this look a little bit more like a two-column proof by saying these are my statements, statement, and this is my rationale right over here. And we're done."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And like how all of these proofs start, let's construct ourselves a right triangle. So what I'm going to do, I'm going to construct it so this hypotenuse sits on the bottom. So that's the hypotenuse of my right triangle. Try to draw it as big as possible so that we have space to work with. So that's going to be my hypotenuse. And then let's say that this is the longer side that's not the hypotenuse. We could have two sides that are equal, but I'll just draw it so it looks a little bit longer."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "Try to draw it as big as possible so that we have space to work with. So that's going to be my hypotenuse. And then let's say that this is the longer side that's not the hypotenuse. We could have two sides that are equal, but I'll just draw it so it looks a little bit longer. Let's call that side length a. And then let's draw this side right over here. It has to be a right triangle, so maybe it goes right over there, that side of length b."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "We could have two sides that are equal, but I'll just draw it so it looks a little bit longer. Let's call that side length a. And then let's draw this side right over here. It has to be a right triangle, so maybe it goes right over there, that side of length b. Let me extend the length a a little bit. So it definitely looks like a right triangle. And this is our 90 degree angle."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "It has to be a right triangle, so maybe it goes right over there, that side of length b. Let me extend the length a a little bit. So it definitely looks like a right triangle. And this is our 90 degree angle. So the first thing that I'm going to do is take this triangle and then rotate it counterclockwise by 90 degrees. So if I rotate it counterclockwise by 90 degrees, I'm literally just going to rotate it like that and draw another completely congruent version of this one. So I'm going to rotate it by 90 degrees."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And this is our 90 degree angle. So the first thing that I'm going to do is take this triangle and then rotate it counterclockwise by 90 degrees. So if I rotate it counterclockwise by 90 degrees, I'm literally just going to rotate it like that and draw another completely congruent version of this one. So I'm going to rotate it by 90 degrees. And if I did that, the hypotenuse would then sit straight up. So I'm going to do my best attempt to draw it almost to scale, as much as I can eyeball it. This side of length a will now come out, will now look something like this."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So I'm going to rotate it by 90 degrees. And if I did that, the hypotenuse would then sit straight up. So I'm going to do my best attempt to draw it almost to scale, as much as I can eyeball it. This side of length a will now come out, will now look something like this. It'll actually be parallel to this over here. So let's see. Let me see how well I could draw it."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "This side of length a will now come out, will now look something like this. It'll actually be parallel to this over here. So let's see. Let me see how well I could draw it. So this is the side of length a. And if we care, this would be 90 degrees. The rotation between the corresponding sides are just going to be 90 degrees in every case."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "Let me see how well I could draw it. So this is the side of length a. And if we care, this would be 90 degrees. The rotation between the corresponding sides are just going to be 90 degrees in every case. This is going to be 90 degrees. That's going to be 90 degrees. Now let me draw side b."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "The rotation between the corresponding sides are just going to be 90 degrees in every case. This is going to be 90 degrees. That's going to be 90 degrees. Now let me draw side b. So it's going to look something like that, or the side that's length b. And the right angle is now here. So all I did is I rotated this by 90 degrees counterclockwise."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "Now let me draw side b. So it's going to look something like that, or the side that's length b. And the right angle is now here. So all I did is I rotated this by 90 degrees counterclockwise. Now what I want to do is construct a parallelogram. I'm going to construct a parallelogram by essentially, and let me label this. So this is height c right over here."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So all I did is I rotated this by 90 degrees counterclockwise. Now what I want to do is construct a parallelogram. I'm going to construct a parallelogram by essentially, and let me label this. So this is height c right over here. Let me do that white color. This is height c. Now what I want to do is go from this point and go up c as well. Now, so this is height c as well."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So this is height c right over here. Let me do that white color. This is height c. Now what I want to do is go from this point and go up c as well. Now, so this is height c as well. And what is this length? What is the length over here from this point to this point going to be? What is this length going to be?"}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "Now, so this is height c as well. And what is this length? What is the length over here from this point to this point going to be? What is this length going to be? Well, a little clue is this is a parallelogram. This line right over here is going to be parallel to this line. It's maintained the same distance."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "What is this length going to be? Well, a little clue is this is a parallelogram. This line right over here is going to be parallel to this line. It's maintained the same distance. And since it's traveling the same distance in the x direction or in the horizontal direction and the vertical direction, this is going to be the same length. So this is going to be of length a. Now the next question I have for you is what is the area of this parallelogram that I have just constructed?"}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "It's maintained the same distance. And since it's traveling the same distance in the x direction or in the horizontal direction and the vertical direction, this is going to be the same length. So this is going to be of length a. Now the next question I have for you is what is the area of this parallelogram that I have just constructed? Well, to think about that, let's redraw this part of the diagram so that the parallelogram is kind of sitting on the ground. So this is length a. This is length a."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "Now the next question I have for you is what is the area of this parallelogram that I have just constructed? Well, to think about that, let's redraw this part of the diagram so that the parallelogram is kind of sitting on the ground. So this is length a. This is length a. This is length c. This is length c. And if you look at this part right over here, it gives you a clue. The height of the parallelogram is given right over here. This side is perpendicular to the base."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "This is length a. This is length c. This is length c. And if you look at this part right over here, it gives you a clue. The height of the parallelogram is given right over here. This side is perpendicular to the base. So the height of the parallelogram is a as well. So what's the area? Well, the area of a parallelogram is just the base times the height."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "This side is perpendicular to the base. So the height of the parallelogram is a as well. So what's the area? Well, the area of a parallelogram is just the base times the height. So the area of this parallelogram right over here is going to be a squared. Now let's do the same thing, but let's rotate our original right triangle. Let's rotate it the other way."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "Well, the area of a parallelogram is just the base times the height. So the area of this parallelogram right over here is going to be a squared. Now let's do the same thing, but let's rotate our original right triangle. Let's rotate it the other way. So let's rotate it 90 degrees clockwise. And this time, instead of pivoting on this point, we're going to pivot on that point right over there. So what are we going to get?"}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "Let's rotate it the other way. So let's rotate it 90 degrees clockwise. And this time, instead of pivoting on this point, we're going to pivot on that point right over there. So what are we going to get? So this side of length c, if we rotate it like that, it's going to end up right over here. Try to draw it as close to scale as possible. So that side has length c. Now the side of length b is going to pop out, look something like this."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So what are we going to get? So this side of length c, if we rotate it like that, it's going to end up right over here. Try to draw it as close to scale as possible. So that side has length c. Now the side of length b is going to pop out, look something like this. It's going to be parallel to that. This is going to be a right angle. So let me draw it like that."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So that side has length c. Now the side of length b is going to pop out, look something like this. It's going to be parallel to that. This is going to be a right angle. So let me draw it like that. That looks pretty good. And then the side of length a is going to be out here. So that's a."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So let me draw it like that. That looks pretty good. And then the side of length a is going to be out here. So that's a. And then this right over here is b. And I want to do that b in blue. So let me do the b in blue."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So that's a. And then this right over here is b. And I want to do that b in blue. So let me do the b in blue. And then this right angle, once we've rotated it, is just sitting right over here. Now let's do the same exercise. Let's construct a parallelogram right over here."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So let me do the b in blue. And then this right angle, once we've rotated it, is just sitting right over here. Now let's do the same exercise. Let's construct a parallelogram right over here. So this is height c. This is height c as well. So by the same logic we used over here, if this length is b, this length is b as well. These are parallel lines."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "Let's construct a parallelogram right over here. So this is height c. This is height c as well. So by the same logic we used over here, if this length is b, this length is b as well. These are parallel lines. We're going the same distance in the horizontal direction. We're rising the same in the vertical direction. We know that because they're parallel."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "These are parallel lines. We're going the same distance in the horizontal direction. We're rising the same in the vertical direction. We know that because they're parallel. So this is length b down here. This is length b up there. Now what is the area of this parallelogram right over there?"}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "We know that because they're parallel. So this is length b down here. This is length b up there. Now what is the area of this parallelogram right over there? What is the area of that parallelogram going to be? Well, once again, to help us visualize it, we can draw it kind of sitting flat. So this is that side."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "Now what is the area of this parallelogram right over there? What is the area of that parallelogram going to be? Well, once again, to help us visualize it, we can draw it kind of sitting flat. So this is that side. Then you have another side right over here. They both have length b. And you have the sides of length c. So that's c. What is its height?"}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So this is that side. Then you have another side right over here. They both have length b. And you have the sides of length c. So that's c. What is its height? Well, you see it right over here. Its height is length b as well. It gives right there."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And you have the sides of length c. So that's c. What is its height? Well, you see it right over here. Its height is length b as well. It gives right there. We know that this is 90 degrees. We did a 90 degree rotation. So this is how we constructed the thing."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "It gives right there. We know that this is 90 degrees. We did a 90 degree rotation. So this is how we constructed the thing. So given that, the area of a parallelogram is just the base times the height. The area of this parallelogram is b squared. So now things are starting to get interesting."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So this is how we constructed the thing. So given that, the area of a parallelogram is just the base times the height. The area of this parallelogram is b squared. So now things are starting to get interesting. And what I'm going to do is I'm going to copy and paste this part right over here. Because this is, in my mind, the most interesting part of our diagram. Let me see how well I can select it."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So now things are starting to get interesting. And what I'm going to do is I'm going to copy and paste this part right over here. Because this is, in my mind, the most interesting part of our diagram. Let me see how well I can select it. So let me select this part right over here. So let me copy. And then I am going to scroll down."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "Let me see how well I can select it. So let me select this part right over here. So let me copy. And then I am going to scroll down. And then let me paste it. So this diagram that we've constructed right over here, it's pretty clear what the area of it is, the combined diagram. And let me delete a little few parts of it just so that it cleans it up."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And then I am going to scroll down. And then let me paste it. So this diagram that we've constructed right over here, it's pretty clear what the area of it is, the combined diagram. And let me delete a little few parts of it just so that it cleans it up. So let me clean this thing up so we really get the part that we want to focus on. So cleaning that up, cleaning this up, right over there. So what is, and actually let me delete this right down here as well."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And let me delete a little few parts of it just so that it cleans it up. So let me clean this thing up so we really get the part that we want to focus on. So cleaning that up, cleaning this up, right over there. So what is, and actually let me delete this right down here as well. Let me delete this right over here. Although we know that this length was c. And actually I'll draw it right over here. This is from our original construction."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So what is, and actually let me delete this right down here as well. Let me delete this right over here. Although we know that this length was c. And actually I'll draw it right over here. This is from our original construction. We know that this length is c. We know this height is c. We know this down here is c. But my question for you is, what is the area of this combined shape? Well, it's just a squared plus b squared. Let me write that down."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "This is from our original construction. We know that this length is c. We know this height is c. We know this down here is c. But my question for you is, what is the area of this combined shape? Well, it's just a squared plus b squared. Let me write that down. The area is just a squared plus b squared, the area of those two parallelograms. Now how can we maybe rearrange pieces of this shape so that we can express it in terms of c? Well, it might have jumped out at you when I drew this line right over here."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "Let me write that down. The area is just a squared plus b squared, the area of those two parallelograms. Now how can we maybe rearrange pieces of this shape so that we can express it in terms of c? Well, it might have jumped out at you when I drew this line right over here. We know that this is of length, I want to do that in white. We know that this part right over here is of length c. This comes from our original construction. Whoops, lost my diagram."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "Well, it might have jumped out at you when I drew this line right over here. We know that this is of length, I want to do that in white. We know that this part right over here is of length c. This comes from our original construction. Whoops, lost my diagram. This is of length c, that's of length c, and then this right over here is of length c. And so what we could do is take this top right triangle, which is completely congruent to our original right triangle, and shift it down. So remember, the entire area, including this top right triangle, is a squared plus b squared. And we're excluding this part down here, which is our original triangle."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "Whoops, lost my diagram. This is of length c, that's of length c, and then this right over here is of length c. And so what we could do is take this top right triangle, which is completely congruent to our original right triangle, and shift it down. So remember, the entire area, including this top right triangle, is a squared plus b squared. And we're excluding this part down here, which is our original triangle. But what happens if we take that? So let me actually cut, and then let me paste it. And all I'm doing is I'm moving that triangle down here."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And we're excluding this part down here, which is our original triangle. But what happens if we take that? So let me actually cut, and then let me paste it. And all I'm doing is I'm moving that triangle down here. So now it looks like this. So I've just rearranged the area that was a squared b squared. So this entire area of this entire square is still a squared plus b squared."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And all I'm doing is I'm moving that triangle down here. So now it looks like this. So I've just rearranged the area that was a squared b squared. So this entire area of this entire square is still a squared plus b squared. a squared is this entire area right over here. It was before a parallelogram. I just shifted that top part of the parallelogram down."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So this entire area of this entire square is still a squared plus b squared. a squared is this entire area right over here. It was before a parallelogram. I just shifted that top part of the parallelogram down. b squared is this entire area right over here. Well, what's this going to be in terms of c? Well, we know that this entire thing is a c by c squared."}, {"video_title": "Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "I just shifted that top part of the parallelogram down. b squared is this entire area right over here. Well, what's this going to be in terms of c? Well, we know that this entire thing is a c by c squared. So the area in terms of c is just c squared. So a squared plus b squared is equal to c squared. And we have once again proven the Pythagorean theorem."}, {"video_title": "Dilating points example.mp3", "Sentence": "So however far A is from point P, it's going to be three times further under the dilation, three times further in the same direction. So how do we think about that? Well one way to think about it is to go from P to A, you have to go one down and two to the left. So minus one and minus two. And so if you dilate it with a factor of three, then you're gonna want to go three times as far down, so minus three, minus three, and three times as far to the left, so you'll go minus six. So one, or that's, let me do this. So negative one, negative two, negative three, negative four, negative five, negative six."}, {"video_title": "Dilating points example.mp3", "Sentence": "So minus one and minus two. And so if you dilate it with a factor of three, then you're gonna want to go three times as far down, so minus three, minus three, and three times as far to the left, so you'll go minus six. So one, or that's, let me do this. So negative one, negative two, negative three, negative four, negative five, negative six. So you will end up right over there. And you can even see it, that this is indeed three times as far from P in the same direction. And so we could call the image of point A, maybe we call that A prime."}, {"video_title": "Dilating points example.mp3", "Sentence": "So negative one, negative two, negative three, negative four, negative five, negative six. So you will end up right over there. And you can even see it, that this is indeed three times as far from P in the same direction. And so we could call the image of point A, maybe we call that A prime. And so there you have it, it has been dilated with a scale factor of three. And so you might be saying wait, I'm used to dilation being stretching or scaling, how have I stretched or scaled something? Well imagine a bunch of points here that represents some type of picture."}, {"video_title": "Dilating points example.mp3", "Sentence": "And so we could call the image of point A, maybe we call that A prime. And so there you have it, it has been dilated with a scale factor of three. And so you might be saying wait, I'm used to dilation being stretching or scaling, how have I stretched or scaled something? Well imagine a bunch of points here that represents some type of picture. And if you push them all three times further from point P, which you could use your center of dilation, then you would expand the size of your picture by a scale factor of three. Let's do another example with a point. So here we're told, plot the image of point A under a dilation about the origin with a scale factor of 1 3rd."}, {"video_title": "Dilating points example.mp3", "Sentence": "Well imagine a bunch of points here that represents some type of picture. And if you push them all three times further from point P, which you could use your center of dilation, then you would expand the size of your picture by a scale factor of three. Let's do another example with a point. So here we're told, plot the image of point A under a dilation about the origin with a scale factor of 1 3rd. So first of all, we don't even see the point A here, so it's actually below the fold. So let's see, there we go, that's our point A. We want it to be about the origin, so about the point zero, zero."}, {"video_title": "Dilating points example.mp3", "Sentence": "So here we're told, plot the image of point A under a dilation about the origin with a scale factor of 1 3rd. So first of all, we don't even see the point A here, so it's actually below the fold. So let's see, there we go, that's our point A. We want it to be about the origin, so about the point zero, zero. This is what we want to, the dilation about the origin with a scale factor of 1 3rd. Scale is 1 3rd, scale factor I should say. So how do we do this?"}, {"video_title": "Dilating points example.mp3", "Sentence": "We want it to be about the origin, so about the point zero, zero. This is what we want to, the dilation about the origin with a scale factor of 1 3rd. Scale is 1 3rd, scale factor I should say. So how do we do this? Well here, however far A is from the origin, we now want to be in the same direction, but 1 3rd as far. So one way to think about it, to go from the origin to A, you have to go six down and three to the right. So 1 3rd of that would be two down and one to the right."}, {"video_title": "Dilating points example.mp3", "Sentence": "So how do we do this? Well here, however far A is from the origin, we now want to be in the same direction, but 1 3rd as far. So one way to think about it, to go from the origin to A, you have to go six down and three to the right. So 1 3rd of that would be two down and one to the right. Two is 1 3rd of six and one is 1 3rd of three. So you will end up right over here. That would be our A prime."}, {"video_title": "Parallel and perpendicular lines intro Analytic geometry Geometry Khan Academy.mp3", "Sentence": "One is the idea of things being perpendicular. And usually people are talking about perpendicular, actually let me, I'm misspelling it, perpendicular, perpendicular lines, normally perpendicular lines, and the idea of parallel lines, parallel, parallel lines. So perpendicular lines are two lines that intersect at a right angle. So what am I talking about? So let's say that this is one line right over here, and that this is another line right over here. We would say these two lines are perpendicular if they intersect at a right angle. So they clearly intersect."}, {"video_title": "Parallel and perpendicular lines intro Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So what am I talking about? So let's say that this is one line right over here, and that this is another line right over here. We would say these two lines are perpendicular if they intersect at a right angle. So they clearly intersect. In order for them to intersect at a right angle, the angle formed between these two lines needs to be 90 degrees. And if any one of these angles is 90 degrees, the rest of them are going to be 90 degrees. So if this is 90 degrees then these are perpendicular lines."}, {"video_title": "Parallel and perpendicular lines intro Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So they clearly intersect. In order for them to intersect at a right angle, the angle formed between these two lines needs to be 90 degrees. And if any one of these angles is 90 degrees, the rest of them are going to be 90 degrees. So if this is 90 degrees then these are perpendicular lines. And if that's 90 degrees, then that's going to be 90 degrees, that's going to be 90 degrees, and that's going to be 90 degrees. So if any of them are 90 degrees, the rest of them are 90 degrees, and we have perpendicular lines. If you have two lines that on a two-dimensional surface like your paper or like the screen never intersect, they stay the same distance apart, then we are talking about parallel lines."}, {"video_title": "Parallel and perpendicular lines intro Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So if this is 90 degrees then these are perpendicular lines. And if that's 90 degrees, then that's going to be 90 degrees, that's going to be 90 degrees, and that's going to be 90 degrees. So if any of them are 90 degrees, the rest of them are 90 degrees, and we have perpendicular lines. If you have two lines that on a two-dimensional surface like your paper or like the screen never intersect, they stay the same distance apart, then we are talking about parallel lines. So this line right over here, and this line right over here, the way I've drawn them, are parallel lines, they aren't intersecting. They're both kind of going in the same direction, but they're kind of shifted versions of each other. They will never intersect with each other."}, {"video_title": "Parallel and perpendicular lines intro Analytic geometry Geometry Khan Academy.mp3", "Sentence": "If you have two lines that on a two-dimensional surface like your paper or like the screen never intersect, they stay the same distance apart, then we are talking about parallel lines. So this line right over here, and this line right over here, the way I've drawn them, are parallel lines, they aren't intersecting. They're both kind of going in the same direction, but they're kind of shifted versions of each other. They will never intersect with each other. So these two are parallel. If we have two lines that, let's say they intersect, but they don't intersect at a right angle, so let's say we have that line, and we have this line right over here, and they're clearly not intersecting at a right angle, then we call these neither perpendicular nor parallel lines. These lines just intersect."}, {"video_title": "Finding central angle measure given arc length Circles Geometry Khan Academy.mp3", "Sentence": "What is the central angle of the arc in degrees? So they're asking for this one. So this is the arc that they're talking about. That's 221 over 18 pi long. And they want to know this angle that it subtends, this central angle right over here. So we just have to remind ourselves that the ratio of this arc length to the entire circumference, let me write that down, the ratio of this arc length, which is 221 over 18 pi, to the entire circumference, which is 20 pi, is going to be equal to the ratio of this central angle, which we can call theta, the ratio of theta to 360 degrees if we were to go all the way around the circle, to 360 degrees. This will give us our theta in degrees."}, {"video_title": "Finding central angle measure given arc length Circles Geometry Khan Academy.mp3", "Sentence": "That's 221 over 18 pi long. And they want to know this angle that it subtends, this central angle right over here. So we just have to remind ourselves that the ratio of this arc length to the entire circumference, let me write that down, the ratio of this arc length, which is 221 over 18 pi, to the entire circumference, which is 20 pi, is going to be equal to the ratio of this central angle, which we can call theta, the ratio of theta to 360 degrees if we were to go all the way around the circle, to 360 degrees. This will give us our theta in degrees. If we wanted it in radians, we would think of it in terms of 2 pi radians around the circle. But it was 360 degrees since we're in degrees. Now we just have to simplify."}, {"video_title": "Finding central angle measure given arc length Circles Geometry Khan Academy.mp3", "Sentence": "This will give us our theta in degrees. If we wanted it in radians, we would think of it in terms of 2 pi radians around the circle. But it was 360 degrees since we're in degrees. Now we just have to simplify. Now the easiest thing is just to multiply both sides times 360 degrees. So let's do that. So this is 300."}, {"video_title": "Finding central angle measure given arc length Circles Geometry Khan Academy.mp3", "Sentence": "Now we just have to simplify. Now the easiest thing is just to multiply both sides times 360 degrees. So let's do that. So this is 300. If we multiply the left-hand side by 360 degrees, we get 360 times 221 times pi over, see we have 18 times 20 times 20 times pi times pi. And on the right-hand side, if we multiply it by 360, we are just left with theta. So we really just have to simplify this now."}, {"video_title": "Finding central angle measure given arc length Circles Geometry Khan Academy.mp3", "Sentence": "So this is 300. If we multiply the left-hand side by 360 degrees, we get 360 times 221 times pi over, see we have 18 times 20 times 20 times pi times pi. And on the right-hand side, if we multiply it by 360, we are just left with theta. So we really just have to simplify this now. Let's see, pi divided by pi is going to be 1. 360 divided by 20, well, it's going to be the same thing as 36 over 2, which is the same thing as 18. And 18 divided by 18 is 1."}, {"video_title": "Example identifying the center of dilation.mp3", "Sentence": "We are told the triangle N prime is the image of triangle N under a dilation. So this is N prime in this red color and then the original N is in this blue color. What is the center of dilation? And they give us some choices here. Choice A, B, C, or D is the center of dilation. So pause this video and see if you can figure it out on your own. So there's a couple of ways to think about it."}, {"video_title": "Example identifying the center of dilation.mp3", "Sentence": "And they give us some choices here. Choice A, B, C, or D is the center of dilation. So pause this video and see if you can figure it out on your own. So there's a couple of ways to think about it. One way, I like to just first think about well what is the scale factor here? So in our original N, we have this side here, it has a length of two. And then once we dilated it by, and used that scale factor, that corresponding side has a length of four."}, {"video_title": "Example identifying the center of dilation.mp3", "Sentence": "So there's a couple of ways to think about it. One way, I like to just first think about well what is the scale factor here? So in our original N, we have this side here, it has a length of two. And then once we dilated it by, and used that scale factor, that corresponding side has a length of four. So we went from two to four. So we can figure out our scale factor. Scale factor is equal to two."}, {"video_title": "Example identifying the center of dilation.mp3", "Sentence": "And then once we dilated it by, and used that scale factor, that corresponding side has a length of four. So we went from two to four. So we can figure out our scale factor. Scale factor is equal to two. Two times two is equal to four. Now what about our center of dilation? So one way to think about it is pick two corresponding points."}, {"video_title": "Example identifying the center of dilation.mp3", "Sentence": "Scale factor is equal to two. Two times two is equal to four. Now what about our center of dilation? So one way to think about it is pick two corresponding points. So let's say we were to pick this point and this point. So the image, the corresponding point on N prime is going to be the scale factor as far away from our center of dilation as the original point. So in this example, we know the scale factor is two."}, {"video_title": "Example identifying the center of dilation.mp3", "Sentence": "So one way to think about it is pick two corresponding points. So let's say we were to pick this point and this point. So the image, the corresponding point on N prime is going to be the scale factor as far away from our center of dilation as the original point. So in this example, we know the scale factor is two. So this is going to be twice as far from our center of dilation as the corresponding point. Well you can immediately see, and it's going to be in the same direction. So actually, if you just draw a line connecting these two, there's actually only one choice that sits on that line, and that is choice D right over here as being the center of dilation."}, {"video_title": "Example identifying the center of dilation.mp3", "Sentence": "So in this example, we know the scale factor is two. So this is going to be twice as far from our center of dilation as the corresponding point. Well you can immediately see, and it's going to be in the same direction. So actually, if you just draw a line connecting these two, there's actually only one choice that sits on that line, and that is choice D right over here as being the center of dilation. And you can also verify that notice, this first point on the original triangle, its change in X is two and its change in Y is three, two, three, to go from point D to point to that point. And then if you want to go to point D to its image, well now you gotta go twice as far. Your change in X is four and your change in Y is six."}, {"video_title": "Example identifying the center of dilation.mp3", "Sentence": "So actually, if you just draw a line connecting these two, there's actually only one choice that sits on that line, and that is choice D right over here as being the center of dilation. And you can also verify that notice, this first point on the original triangle, its change in X is two and its change in Y is three, two, three, to go from point D to point to that point. And then if you want to go to point D to its image, well now you gotta go twice as far. Your change in X is four and your change in Y is six. You could use the Pythagorean theorem to calculate this distance and then the longer distance. But what you see is is that the corresponding point is now twice as far from your center of dilation. So there's a couple of ways to think about it."}, {"video_title": "Writing equations of perpendicular lines (example 2) High School Math Khan Academy.mp3", "Sentence": "Find the equation of a line perpendicular to this line that passes through the point 2, 8. So this first piece of information, that it's perpendicular to that line right over there. What does that tell us? Well, if it's perpendicular to this line, its slope has to be the negative inverse of 2 5ths. So its slope, the negative inverse of 2 5ths, the inverse of 2 5ths is 5. Let me do it in a better color, a nicer green. If this line's slope is negative 2 5ths, the equation of the line we have to figure out that's perpendicular, its slope is going to be the inverse."}, {"video_title": "Writing equations of perpendicular lines (example 2) High School Math Khan Academy.mp3", "Sentence": "Well, if it's perpendicular to this line, its slope has to be the negative inverse of 2 5ths. So its slope, the negative inverse of 2 5ths, the inverse of 2 5ths is 5. Let me do it in a better color, a nicer green. If this line's slope is negative 2 5ths, the equation of the line we have to figure out that's perpendicular, its slope is going to be the inverse. So instead of 2 5ths, it's going to be 5 halves. Instead of being a negative, it's going to be a positive. So this is the negative inverse of negative 2 5ths."}, {"video_title": "Writing equations of perpendicular lines (example 2) High School Math Khan Academy.mp3", "Sentence": "If this line's slope is negative 2 5ths, the equation of the line we have to figure out that's perpendicular, its slope is going to be the inverse. So instead of 2 5ths, it's going to be 5 halves. Instead of being a negative, it's going to be a positive. So this is the negative inverse of negative 2 5ths. You take the negative sign, it becomes positive. You swap the 5 and the 2, you get 5 halves. So that is going to have to be our slope."}, {"video_title": "Writing equations of perpendicular lines (example 2) High School Math Khan Academy.mp3", "Sentence": "So this is the negative inverse of negative 2 5ths. You take the negative sign, it becomes positive. You swap the 5 and the 2, you get 5 halves. So that is going to have to be our slope. And we can actually use the point slope form right here. It goes through this point right there. So let's use point slope form."}, {"video_title": "Writing equations of perpendicular lines (example 2) High School Math Khan Academy.mp3", "Sentence": "So that is going to have to be our slope. And we can actually use the point slope form right here. It goes through this point right there. So let's use point slope form. y minus this y value, which has to be on the line, is equal to our slope, 5 halves, times x minus this x value, the x value when y is equal to 8. And this is the equation of the line in point slope form. If you want to put it in slope intercept form, you can just do a little bit of algebraic manipulation."}, {"video_title": "Writing equations of perpendicular lines (example 2) High School Math Khan Academy.mp3", "Sentence": "So let's use point slope form. y minus this y value, which has to be on the line, is equal to our slope, 5 halves, times x minus this x value, the x value when y is equal to 8. And this is the equation of the line in point slope form. If you want to put it in slope intercept form, you can just do a little bit of algebraic manipulation. y minus 8 is equal to, let's distribute the 5 halves. So 5 halves x minus 5 halves times 2 is just 5. And then add 8 to both sides."}, {"video_title": "Writing equations of perpendicular lines (example 2) High School Math Khan Academy.mp3", "Sentence": "If you want to put it in slope intercept form, you can just do a little bit of algebraic manipulation. y minus 8 is equal to, let's distribute the 5 halves. So 5 halves x minus 5 halves times 2 is just 5. And then add 8 to both sides. You get y is equal to 5 halves x. Add 8 to negative 5, so plus 3. And we are done."}, {"video_title": "Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3", "Sentence": "So pause this video and try to think about this on your own before we work through it together. All right, now let's work through it together. So in general, if you know that something is already a parallelogram and you wanna determine whether it's a rectangle, it's really a question of whether the adjacent sides intersect at a right angle. So for example, a parallelogram might look something like this. What we know about a parallelogram is that the opposite sides are parallel. So this side is parallel to that side and that this side is parallel to this side. And all rectangles are parallelograms, but not all parallelograms are rectangles."}, {"video_title": "Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3", "Sentence": "So for example, a parallelogram might look something like this. What we know about a parallelogram is that the opposite sides are parallel. So this side is parallel to that side and that this side is parallel to this side. And all rectangles are parallelograms, but not all parallelograms are rectangles. In order for a parallelogram to be a rectangle, these sides need to intersect at right angles. And clearly the way I drew this one, it doesn't look like that. But let's see if we can figure that out based on the coordinates that they have given us."}, {"video_title": "Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3", "Sentence": "And all rectangles are parallelograms, but not all parallelograms are rectangles. In order for a parallelogram to be a rectangle, these sides need to intersect at right angles. And clearly the way I drew this one, it doesn't look like that. But let's see if we can figure that out based on the coordinates that they have given us. And to help us visualize, let me just put some coordinates. Let me draw some axes here. So that's my X axis, and then this is my Y axis."}, {"video_title": "Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3", "Sentence": "But let's see if we can figure that out based on the coordinates that they have given us. And to help us visualize, let me just put some coordinates. Let me draw some axes here. So that's my X axis, and then this is my Y axis. Let's see the coordinates. Let's see, we have twos, fours, sixes. Let me actually count by an eight."}, {"video_title": "Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3", "Sentence": "So that's my X axis, and then this is my Y axis. Let's see the coordinates. Let's see, we have twos, fours, sixes. Let me actually count by an eight. So let me count by twos here. So we have two, four, six, and eight. And then we have negative two, negative four, negative six, negative eight."}, {"video_title": "Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3", "Sentence": "Let me actually count by an eight. So let me count by twos here. So we have two, four, six, and eight. And then we have negative two, negative four, negative six, negative eight. We have two, four, six, and eight. And then we'd have negative two, negative four, negative six, and negative eight. So each hash mark is another two."}, {"video_title": "Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3", "Sentence": "And then we have negative two, negative four, negative six, negative eight. We have two, four, six, and eight. And then we'd have negative two, negative four, negative six, and negative eight. So each hash mark is another two. I'm counting by twos here. And so let's plot these points. And I'll do it in different colors so we can keep track."}, {"video_title": "Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3", "Sentence": "So each hash mark is another two. I'm counting by twos here. And so let's plot these points. And I'll do it in different colors so we can keep track. So A is negative six, negative four. So negative two, negative four, negative six, and then negative four would go right over here. That is point A."}, {"video_title": "Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3", "Sentence": "And I'll do it in different colors so we can keep track. So A is negative six, negative four. So negative two, negative four, negative six, and then negative four would go right over here. That is point A. Then we have point B, which is negative two, six. So negative two, six. So that's going to go up to four and six."}, {"video_title": "Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3", "Sentence": "That is point A. Then we have point B, which is negative two, six. So negative two, six. So that's going to go up to four and six. So that is point B right over there. Then we have point C, which is at eight, two. So eight, comma, two right over there."}, {"video_title": "Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3", "Sentence": "So that's going to go up to four and six. So that is point B right over there. Then we have point C, which is at eight, two. So eight, comma, two right over there. That is point C. And then last but not least, we have point D, which is at four, comma, negative eight. Four, comma, negative eight right over there, point D. And so our quadrilateral, or we actually know it's a parallelogram, looks like this. So you have segment AB like that."}, {"video_title": "Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3", "Sentence": "So eight, comma, two right over there. That is point C. And then last but not least, we have point D, which is at four, comma, negative eight. Four, comma, negative eight right over there, point D. And so our quadrilateral, or we actually know it's a parallelogram, looks like this. So you have segment AB like that. You have segment BC that looks like that. Segment CD looks like this. And segment AD looks like this."}, {"video_title": "Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3", "Sentence": "So you have segment AB like that. You have segment BC that looks like that. Segment CD looks like this. And segment AD looks like this. And we know already that it's a parallelogram. So we know that segment AB is parallel to segment DC, and segment BC is parallel to segment AD. But what we really need to do is figure out whether they are intersecting at right angles."}, {"video_title": "Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3", "Sentence": "And segment AD looks like this. And we know already that it's a parallelogram. So we know that segment AB is parallel to segment DC, and segment BC is parallel to segment AD. But what we really need to do is figure out whether they are intersecting at right angles. And to do that, using the coordinates to figure that out, we have to figure out the slopes of these different line segments. And so let's figure out first the slope of AB. So the slope of segment AB is going to be equal to our change in y over change in x."}, {"video_title": "Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3", "Sentence": "But what we really need to do is figure out whether they are intersecting at right angles. And to do that, using the coordinates to figure that out, we have to figure out the slopes of these different line segments. And so let's figure out first the slope of AB. So the slope of segment AB is going to be equal to our change in y over change in x. So our change in y is going to be six minus negative four, six minus negative four, over negative two minus negative six. Negative two minus negative six. And so this is going to be equal to six plus four, which is 10, over negative two minus negative six."}, {"video_title": "Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3", "Sentence": "So the slope of segment AB is going to be equal to our change in y over change in x. So our change in y is going to be six minus negative four, six minus negative four, over negative two minus negative six. Negative two minus negative six. And so this is going to be equal to six plus four, which is 10, over negative two minus negative six. That's the same thing as negative two plus six. So that's going to be over four, which is the same thing as 5 1\u20442. All right, that's interesting."}, {"video_title": "Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3", "Sentence": "And so this is going to be equal to six plus four, which is 10, over negative two minus negative six. That's the same thing as negative two plus six. So that's going to be over four, which is the same thing as 5 1\u20442. All right, that's interesting. What is the slope of segment BC? The slope of segment BC is going to be equal to, once again, change in y over change in x. Our y-coordinates, change in y is two minus six, two minus six, over eight minus negative two."}, {"video_title": "Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3", "Sentence": "All right, that's interesting. What is the slope of segment BC? The slope of segment BC is going to be equal to, once again, change in y over change in x. Our y-coordinates, change in y is two minus six, two minus six, over eight minus negative two. Eight minus negative two, which is equal to negative four, over, and then eight minus negative two is the same thing as eight plus two, over 10, which is the same thing as negative 2\u2075. Now, in other videos in your algebra class, you might have learned that the slopes of lines that intersect at right angles, or the slopes of lines that form a right angle at their point of intersection, that they are going to be the opposite reciprocals. And you can actually see that right over here."}, {"video_title": "Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3", "Sentence": "Our y-coordinates, change in y is two minus six, two minus six, over eight minus negative two. Eight minus negative two, which is equal to negative four, over, and then eight minus negative two is the same thing as eight plus two, over 10, which is the same thing as negative 2\u2075. Now, in other videos in your algebra class, you might have learned that the slopes of lines that intersect at right angles, or the slopes of lines that form a right angle at their point of intersection, that they are going to be the opposite reciprocals. And you can actually see that right over here. These are opposite reciprocals. If you take the reciprocal of this top slope, you'd get 2\u2075, and then you take the opposite of it, or in this case, the negative of it, you are going to get negative 2\u2075. So these are actually, these are perpendicular lines."}, {"video_title": "Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3", "Sentence": "And you can actually see that right over here. These are opposite reciprocals. If you take the reciprocal of this top slope, you'd get 2\u2075, and then you take the opposite of it, or in this case, the negative of it, you are going to get negative 2\u2075. So these are actually, these are perpendicular lines. So this lets us know that AB is perpendicular, segment AB is perpendicular to segment BC. So we know that this is the case. And we could keep on to doing that, but in a parallelogram, if one set of segments intersect at a right angle, all of them are going to intersect at a right angle, and we could show that more rigorously in other places, but this is enough evidence for me to know that this is indeed going to be a rectangle."}, {"video_title": "Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3", "Sentence": "So these are actually, these are perpendicular lines. So this lets us know that AB is perpendicular, segment AB is perpendicular to segment BC. So we know that this is the case. And we could keep on to doing that, but in a parallelogram, if one set of segments intersect at a right angle, all of them are going to intersect at a right angle, and we could show that more rigorously in other places, but this is enough evidence for me to know that this is indeed going to be a rectangle. If you want, you could continue to do this analysis, and you will see that this is perpendicular, this is perpendicular, and that is perpendicular as well. But let's see which of these choices match up to what we just deduced. So choice A says yes, and yes would be, it is a rectangle because AB is equal."}, {"video_title": "Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3", "Sentence": "And we could keep on to doing that, but in a parallelogram, if one set of segments intersect at a right angle, all of them are going to intersect at a right angle, and we could show that more rigorously in other places, but this is enough evidence for me to know that this is indeed going to be a rectangle. If you want, you could continue to do this analysis, and you will see that this is perpendicular, this is perpendicular, and that is perpendicular as well. But let's see which of these choices match up to what we just deduced. So choice A says yes, and yes would be, it is a rectangle because AB is equal. So the length of segment AB is equal to the length of segment AD, and the length of segment BC is equal to the length of segment CD. So that might be true, I haven't validated it, but just because this is true, and because we do know that ABCD is a parallelogram, that wouldn't let me know that we are actually dealing with a rectangle. For example, you can have a parallelogram where even all the sides are congruent."}, {"video_title": "Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3", "Sentence": "So choice A says yes, and yes would be, it is a rectangle because AB is equal. So the length of segment AB is equal to the length of segment AD, and the length of segment BC is equal to the length of segment CD. So that might be true, I haven't validated it, but just because this is true, and because we do know that ABCD is a parallelogram, that wouldn't let me know that we are actually dealing with a rectangle. For example, you can have a parallelogram where even all the sides are congruent. So you could have a parallelogram that looks like this. And obviously, if all of the sides are congruent, you're dealing with a rhombus, but a rhombus is still not necessarily going to be a rectangle. And so I would rule this top one out."}, {"video_title": "Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3", "Sentence": "For example, you can have a parallelogram where even all the sides are congruent. So you could have a parallelogram that looks like this. And obviously, if all of the sides are congruent, you're dealing with a rhombus, but a rhombus is still not necessarily going to be a rectangle. And so I would rule this top one out. The second choice says yes, and it says because BC is perpendicular to AB. Yeah, we saw that by seeing that their slopes are the opposite reciprocals of each other. And of course, we know that ABCD is a parallelogram."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "In this first problem over here, we're asked to find out the length of this segment, segment CE. And we have these two parallel lines. AB is parallel to DE. And then we have these two essentially transversals that form these two triangles. So let's see what we can do here. So the first thing that might jump out at you is that this angle and this angle are vertical angles, so they are going to be congruent. The other thing that might jump out at you is that angle CDE is an alternate interior angle with CBA."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "And then we have these two essentially transversals that form these two triangles. So let's see what we can do here. So the first thing that might jump out at you is that this angle and this angle are vertical angles, so they are going to be congruent. The other thing that might jump out at you is that angle CDE is an alternate interior angle with CBA. So we have this transversal right over here. And these are alternate interior angles. And they are going to be congruent."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "The other thing that might jump out at you is that angle CDE is an alternate interior angle with CBA. So we have this transversal right over here. And these are alternate interior angles. And they are going to be congruent. Or you could say that if you continue this transversal, you would have a corresponding angle with CDE right up here. And this one's just vertical. Either way, this angle and this angle are going to be congruent."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "And they are going to be congruent. Or you could say that if you continue this transversal, you would have a corresponding angle with CDE right up here. And this one's just vertical. Either way, this angle and this angle are going to be congruent. So we've established that we have two triangles. And two of the corresponding angles are the same. And that by itself is enough to establish similarity."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "Either way, this angle and this angle are going to be congruent. So we've established that we have two triangles. And two of the corresponding angles are the same. And that by itself is enough to establish similarity. You can actually, we actually could show that this angle and this angle are also congruent by alternate interior angles. But we don't have to. So we already know that they are similar."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "And that by itself is enough to establish similarity. You can actually, we actually could show that this angle and this angle are also congruent by alternate interior angles. But we don't have to. So we already know that they are similar. Actually, we could just say it just by alternate interior angles, these are also going to be congruent. But we already know enough to say that they are similar even before doing that. So we already know that triangle, I'll try to write it, I'll color code it so that we have the same corresponding vertices."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "So we already know that they are similar. Actually, we could just say it just by alternate interior angles, these are also going to be congruent. But we already know enough to say that they are similar even before doing that. So we already know that triangle, I'll try to write it, I'll color code it so that we have the same corresponding vertices. And that's really important to know what angles and what sides correspond to what side so that you don't mess up your ratios. So that you do know what's corresponding to what. So we know triangle AB, triangle ABC, is similar to triangle."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "So we already know that triangle, I'll try to write it, I'll color code it so that we have the same corresponding vertices. And that's really important to know what angles and what sides correspond to what side so that you don't mess up your ratios. So that you do know what's corresponding to what. So we know triangle AB, triangle ABC, is similar to triangle. So A, this vertex A corresponds to vertex E over here, is similar to vertex E. And then vertex B right over here corresponds to vertex D, EDC. Now what does that do for us? Well, that tells us that the ratio of corresponding sides are going to be the same."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "So we know triangle AB, triangle ABC, is similar to triangle. So A, this vertex A corresponds to vertex E over here, is similar to vertex E. And then vertex B right over here corresponds to vertex D, EDC. Now what does that do for us? Well, that tells us that the ratio of corresponding sides are going to be the same. They're going to be some constant value. So we have corresponding side. So the ratio, for example, the corresponding side for BC is going to be DC."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "Well, that tells us that the ratio of corresponding sides are going to be the same. They're going to be some constant value. So we have corresponding side. So the ratio, for example, the corresponding side for BC is going to be DC. We can see it just the way that we've written down the similarity. This is true. Then BC is the corresponding side to DC."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "So the ratio, for example, the corresponding side for BC is going to be DC. We can see it just the way that we've written down the similarity. This is true. Then BC is the corresponding side to DC. So we know that the length of BC over DC is going to be equal to the length of, well, we want to figure out what CE is. That's what we care about. And I'm using BC and DC because we know those values."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "Then BC is the corresponding side to DC. So we know that the length of BC over DC is going to be equal to the length of, well, we want to figure out what CE is. That's what we care about. And I'm using BC and DC because we know those values. So BC over DC is going to be equal to what's the corresponding side to CE? The corresponding side over here is CA. It's going to be equal to CA over CE."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "And I'm using BC and DC because we know those values. So BC over DC is going to be equal to what's the corresponding side to CE? The corresponding side over here is CA. It's going to be equal to CA over CE. Corresponding sides. This is the last and the first, last and the first. CA over CE."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "It's going to be equal to CA over CE. Corresponding sides. This is the last and the first, last and the first. CA over CE. And we know what BC is. BC right over here is 5. We know what DC is."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "CA over CE. And we know what BC is. BC right over here is 5. We know what DC is. It is 3. We know what CA or AC is right over here. CA is 4."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "We know what DC is. It is 3. We know what CA or AC is right over here. CA is 4. And now we can just solve for CE. So we can, well, there's multiple ways that you could think about this. You could cross multiply, which is really just multiplying both sides by both denominators."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "CA is 4. And now we can just solve for CE. So we can, well, there's multiple ways that you could think about this. You could cross multiply, which is really just multiplying both sides by both denominators. So you get 5 times the length of CE is equal to 3 times 4, which is just going to be equal to 12. And then we get CE is equal to 12 over 5, which is the same thing as 2 and 2 fifths or 2.4. So this is going to be 2 and 2 fifths."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "You could cross multiply, which is really just multiplying both sides by both denominators. So you get 5 times the length of CE is equal to 3 times 4, which is just going to be equal to 12. And then we get CE is equal to 12 over 5, which is the same thing as 2 and 2 fifths or 2.4. So this is going to be 2 and 2 fifths. And we're done. We were able to use similarity to figure out this side, just knowing that the ratio between the corresponding sides are going to be the same. Now let's do this problem right over here."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "So this is going to be 2 and 2 fifths. And we're done. We were able to use similarity to figure out this side, just knowing that the ratio between the corresponding sides are going to be the same. Now let's do this problem right over here. Let me draw a little line here to show that this is a different problem now. This is a different problem. So in this problem, we need to figure out what DE is."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "Now let's do this problem right over here. Let me draw a little line here to show that this is a different problem now. This is a different problem. So in this problem, we need to figure out what DE is. And we all, once again, have these two parallel lines like this. And so we know corresponding angles are congruent. So we know that angle is going to be congruent to that angle, because you could view this as a transversal."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "So in this problem, we need to figure out what DE is. And we all, once again, have these two parallel lines like this. And so we know corresponding angles are congruent. So we know that angle is going to be congruent to that angle, because you could view this as a transversal. We also know that this angle right over here is going to be congruent to that angle right over there. Once again, corresponding angles for transversal. And also in both triangles."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "So we know that angle is going to be congruent to that angle, because you could view this as a transversal. We also know that this angle right over here is going to be congruent to that angle right over there. Once again, corresponding angles for transversal. And also in both triangles. So I'm looking at triangle CBD and triangle CAE. They both share this angle up here. Once again, we could have stopped at two angles."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "And also in both triangles. So I'm looking at triangle CBD and triangle CAE. They both share this angle up here. Once again, we could have stopped at two angles. But we've actually shown that all three angles of these two triangles, all three of the corresponding angles, are congruent to each other. So we now know, and once again, this is an important thing to do, is to make sure that you write it in the right order when you write your similarity. We now know that triangle CBD is similar."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "Once again, we could have stopped at two angles. But we've actually shown that all three angles of these two triangles, all three of the corresponding angles, are congruent to each other. So we now know, and once again, this is an important thing to do, is to make sure that you write it in the right order when you write your similarity. We now know that triangle CBD is similar. Not congruent. It is similar to triangle CCAE, which means that the ratio of corresponding sides are going to be constant. So we know, for example, that the ratio of CB to CA, let's write this down."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "We now know that triangle CBD is similar. Not congruent. It is similar to triangle CCAE, which means that the ratio of corresponding sides are going to be constant. So we know, for example, that the ratio of CB to CA, let's write this down. We know that the ratio of CB over CA is going to be equal to the ratio of CD over CE. And we know what CB is. CB over here is 5."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "So we know, for example, that the ratio of CB to CA, let's write this down. We know that the ratio of CB over CA is going to be equal to the ratio of CD over CE. And we know what CB is. CB over here is 5. We know what CA is. And we have to be careful here. It's not 3."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "CB over here is 5. We know what CA is. And we have to be careful here. It's not 3. CA, this entire side, is going to be 5 plus 3. So this is going to be 8. And we know what CD is."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "It's not 3. CA, this entire side, is going to be 5 plus 3. So this is going to be 8. And we know what CD is. CD is going to be 4. And so once again, we can cross multiply. We have 5 times CE is equal to 8 times 4."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "And we know what CD is. CD is going to be 4. And so once again, we can cross multiply. We have 5 times CE is equal to 8 times 4. 8 times 4 is 32. And so CE is equal to 32 over 5. Or this is another way to think about it."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "We have 5 times CE is equal to 8 times 4. 8 times 4 is 32. And so CE is equal to 32 over 5. Or this is another way to think about it. That's 6 and 2 fifths. Now we're not done, because they didn't ask for what CE is. They're asking for just this part right over here."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "Or this is another way to think about it. That's 6 and 2 fifths. Now we're not done, because they didn't ask for what CE is. They're asking for just this part right over here. They're asking for DE. So we know that this entire length, CE right over here, this is 6 and 2 fifths. And so DE right over here, what we actually have to figure out, it's going to be this entire length, 6 and 2 fifths, minus 4, minus CD right over here."}, {"video_title": "Similarity example problems Similarity Geometry Khan Academy.mp3", "Sentence": "They're asking for just this part right over here. They're asking for DE. So we know that this entire length, CE right over here, this is 6 and 2 fifths. And so DE right over here, what we actually have to figure out, it's going to be this entire length, 6 and 2 fifths, minus 4, minus CD right over here. So it's going to be 2 and 2 fifths. 6 and 2 fifths minus 4 and 2 fifths is 2 and 2 fifths. So we're done."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "So that is our inscribed angle. I'll denote it by psi. I'll use psi for inscribed angle and angles in this video, that psi, the inscribed angle, is going to be exactly 1 half of the central angle that subtends the same arc. So I just used a lot of fancy words, but I think you'll get what I'm saying. So this is psi. It is an inscribed angle. It sits, its vertex sits on the circumference."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "So I just used a lot of fancy words, but I think you'll get what I'm saying. So this is psi. It is an inscribed angle. It sits, its vertex sits on the circumference. And if you draw out the two rays that come out from this angle, or the two cords that define this angle, it intersects the circle at the other end. And if you look at the part of the circumference of the circle that's inside of it, that is the arc that is subtended by psi. So it's all very fancy words, but I think the idea is pretty straightforward."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "It sits, its vertex sits on the circumference. And if you draw out the two rays that come out from this angle, or the two cords that define this angle, it intersects the circle at the other end. And if you look at the part of the circumference of the circle that's inside of it, that is the arc that is subtended by psi. So it's all very fancy words, but I think the idea is pretty straightforward. This right here is the arc subtended by psi, where psi is that inscribed angle right over there, the vertex sitting on the circumference. Now, a central angle is an angle where the vertex is sitting at the center of the circle. So let's say that this right here, I'll try to eyeball it, that right there is the center of the circle."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "So it's all very fancy words, but I think the idea is pretty straightforward. This right here is the arc subtended by psi, where psi is that inscribed angle right over there, the vertex sitting on the circumference. Now, a central angle is an angle where the vertex is sitting at the center of the circle. So let's say that this right here, I'll try to eyeball it, that right there is the center of the circle. And so let me draw a central angle that subtends this same arc. So that looks like a central angle subtending that same arc, just like that. Let's call this theta."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "So let's say that this right here, I'll try to eyeball it, that right there is the center of the circle. And so let me draw a central angle that subtends this same arc. So that looks like a central angle subtending that same arc, just like that. Let's call this theta. So this angle is psi. This angle right here is theta. And what I'm going to prove in this video is that psi is always going to be equal to 1 half of theta."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "Let's call this theta. So this angle is psi. This angle right here is theta. And what I'm going to prove in this video is that psi is always going to be equal to 1 half of theta. So if I were to tell you that psi is equal to, I don't know, 25 degrees, then you would immediately know that theta must be equal to 50 degrees. Or if I told you that theta was 80 degrees, then you'd immediately know that psi was 40 degrees. So let's actually prove this."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "And what I'm going to prove in this video is that psi is always going to be equal to 1 half of theta. So if I were to tell you that psi is equal to, I don't know, 25 degrees, then you would immediately know that theta must be equal to 50 degrees. Or if I told you that theta was 80 degrees, then you'd immediately know that psi was 40 degrees. So let's actually prove this. So let me clear this. So a good place to start, or the place I'm going to start, is a special case. I'm going to draw a inscribed angle."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "So let's actually prove this. So let me clear this. So a good place to start, or the place I'm going to start, is a special case. I'm going to draw a inscribed angle. But one of the chords that define it is going to be the diameter of the circle. So this isn't going to be the general case. This is just going to be a special case."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "I'm going to draw a inscribed angle. But one of the chords that define it is going to be the diameter of the circle. So this isn't going to be the general case. This is just going to be a special case. So let me see. This is the center right here of my circle. Trying to eyeball it."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "This is just going to be a special case. So let me see. This is the center right here of my circle. Trying to eyeball it. Let me draw it a little bit better. That looks, center looks like that. So let me draw a diameter."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "Trying to eyeball it. Let me draw it a little bit better. That looks, center looks like that. So let me draw a diameter. So the diameter looks like that. And then let me define my inscribed angle. This diameter is one side of it."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "So let me draw a diameter. So the diameter looks like that. And then let me define my inscribed angle. This diameter is one side of it. And then the other side maybe is just like that. So let me call this right here psi. If that's psi, this length right here is a radius."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "This diameter is one side of it. And then the other side maybe is just like that. So let me call this right here psi. If that's psi, this length right here is a radius. That's our radius of our circle. And then this length right here is also going to be the radius of our circle. We're going from the center to the circumference."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "If that's psi, this length right here is a radius. That's our radius of our circle. And then this length right here is also going to be the radius of our circle. We're going from the center to the circumference. The circumference is defined by all of the points that are exactly a radius away from the center. So that's also a radius. Now, this triangle right here is an isosceles triangle."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "We're going from the center to the circumference. The circumference is defined by all of the points that are exactly a radius away from the center. So that's also a radius. Now, this triangle right here is an isosceles triangle. It has two sides that are equal. Two sides that are definitely equal. And we know that when we have two sides being equal, their base angles are also equal."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "Now, this triangle right here is an isosceles triangle. It has two sides that are equal. Two sides that are definitely equal. And we know that when we have two sides being equal, their base angles are also equal. So this will also be equal to psi. You might not recognize it because it's tilted up like that. But I think many of us, when we see a triangle that looks like this, if I told you this is r and that is r, that these two sides are equal, and if this is psi, then you would also know that this angle is also going to be psi."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "And we know that when we have two sides being equal, their base angles are also equal. So this will also be equal to psi. You might not recognize it because it's tilted up like that. But I think many of us, when we see a triangle that looks like this, if I told you this is r and that is r, that these two sides are equal, and if this is psi, then you would also know that this angle is also going to be psi. Base angles are equivalent on an isosceles triangle. So if this is psi, that is also psi. Now, let me look at the central angle."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "But I think many of us, when we see a triangle that looks like this, if I told you this is r and that is r, that these two sides are equal, and if this is psi, then you would also know that this angle is also going to be psi. Base angles are equivalent on an isosceles triangle. So if this is psi, that is also psi. Now, let me look at the central angle. This is the central angle subtending the same arc. Let's highlight the arc that they're both subtending. This right here is the arc that they're both going to subtend."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "Now, let me look at the central angle. This is the central angle subtending the same arc. Let's highlight the arc that they're both subtending. This right here is the arc that they're both going to subtend. So this is my central angle right there, theta. Now, if this angle is theta, what's this angle going to be? This angle right here."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "This right here is the arc that they're both going to subtend. So this is my central angle right there, theta. Now, if this angle is theta, what's this angle going to be? This angle right here. Well, this angle is supplementary to theta. So it's 180 minus theta. When you add these two angles together, you go 180 degrees around, or they kind of form a line."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "This angle right here. Well, this angle is supplementary to theta. So it's 180 minus theta. When you add these two angles together, you go 180 degrees around, or they kind of form a line. They're supplementary to each other. Now, we also know that these three angles are sitting inside of the same triangle. So they must add up to 180 degrees."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "When you add these two angles together, you go 180 degrees around, or they kind of form a line. They're supplementary to each other. Now, we also know that these three angles are sitting inside of the same triangle. So they must add up to 180 degrees. So we get psi, this psi, plus that psi, plus psi, plus this angle, which is 180 minus theta, plus 180 minus theta. These three angles must add up to 180 degrees. They're the three angles of a triangle."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "So they must add up to 180 degrees. So we get psi, this psi, plus that psi, plus psi, plus this angle, which is 180 minus theta, plus 180 minus theta. These three angles must add up to 180 degrees. They're the three angles of a triangle. Now, we could subtract 180 from both sides. Psi plus psi is 2 psi minus theta is equal to 0. Add theta to both sides, you get 2 psi is equal to theta."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "They're the three angles of a triangle. Now, we could subtract 180 from both sides. Psi plus psi is 2 psi minus theta is equal to 0. Add theta to both sides, you get 2 psi is equal to theta. Multiply both sides by 1 half, or divide both sides by 2. You get psi is equal to 1 half of theta. So we just proved what we set out to prove for the special case where our inscribed angle is defined where one of the rays, if you want to view these lines as rays, where one of the rays that defines this inscribed angle is along the diameter, or the diameter forms part of that ray."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "Add theta to both sides, you get 2 psi is equal to theta. Multiply both sides by 1 half, or divide both sides by 2. You get psi is equal to 1 half of theta. So we just proved what we set out to prove for the special case where our inscribed angle is defined where one of the rays, if you want to view these lines as rays, where one of the rays that defines this inscribed angle is along the diameter, or the diameter forms part of that ray. So this is a special case where one edge is sitting on the diameter. So this could apply, so already we could generalize this. So now that we know that if this is 50, that this is going to be 100 degrees and likewise, whatever psi is, or whatever theta is, psi is going to be 1 half of that, or whatever psi is, theta is going to be 2 times that."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "So we just proved what we set out to prove for the special case where our inscribed angle is defined where one of the rays, if you want to view these lines as rays, where one of the rays that defines this inscribed angle is along the diameter, or the diameter forms part of that ray. So this is a special case where one edge is sitting on the diameter. So this could apply, so already we could generalize this. So now that we know that if this is 50, that this is going to be 100 degrees and likewise, whatever psi is, or whatever theta is, psi is going to be 1 half of that, or whatever psi is, theta is going to be 2 times that. And now this will apply for any time. We can use this notion any time, let me clear this, any time that, so just using that result we just got, we can now generalize it a little bit, although this won't apply to all inscribed angles. Let's have an inscribed angle that looks like this."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "So now that we know that if this is 50, that this is going to be 100 degrees and likewise, whatever psi is, or whatever theta is, psi is going to be 1 half of that, or whatever psi is, theta is going to be 2 times that. And now this will apply for any time. We can use this notion any time, let me clear this, any time that, so just using that result we just got, we can now generalize it a little bit, although this won't apply to all inscribed angles. Let's have an inscribed angle that looks like this. So in this situation, the center, you can kind of view it as it's inside of the angle. That's my inscribed angle. And I want to find a relationship between this inscribed angle and the central angle that's subtending the same arc."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "Let's have an inscribed angle that looks like this. So in this situation, the center, you can kind of view it as it's inside of the angle. That's my inscribed angle. And I want to find a relationship between this inscribed angle and the central angle that's subtending the same arc. So that's my central angle subtending the same arc. Well, you might say, hey Gene, none of these ends or these cords that define this angle, neither of these are diameters, but what we can do is we can draw a diameter. If the center is within these two cords, we can draw a diameter just like that."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "And I want to find a relationship between this inscribed angle and the central angle that's subtending the same arc. So that's my central angle subtending the same arc. Well, you might say, hey Gene, none of these ends or these cords that define this angle, neither of these are diameters, but what we can do is we can draw a diameter. If the center is within these two cords, we can draw a diameter just like that. If we define this angle as psi 1, that angle as psi 2, clearly psi is the sum of those two angles, and we call this angle theta 1 and this angle theta 2, we immediately know that, just using the result I just got, since we have one side of our angles in both cases being a diameter now, we know that psi 1 is going to be equal to 1 half theta 1, and we know that psi 2 is going to be 1 half theta 2. Psi 2 is going to be 1 half theta 2. And so psi, which is psi 1 plus psi 2, so psi 1 plus, let me write the psi a little better, psi 1 plus psi 2 is going to be equal to these two things, 1 half theta 1 plus 1 half theta 2."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "If the center is within these two cords, we can draw a diameter just like that. If we define this angle as psi 1, that angle as psi 2, clearly psi is the sum of those two angles, and we call this angle theta 1 and this angle theta 2, we immediately know that, just using the result I just got, since we have one side of our angles in both cases being a diameter now, we know that psi 1 is going to be equal to 1 half theta 1, and we know that psi 2 is going to be 1 half theta 2. Psi 2 is going to be 1 half theta 2. And so psi, which is psi 1 plus psi 2, so psi 1 plus, let me write the psi a little better, psi 1 plus psi 2 is going to be equal to these two things, 1 half theta 1 plus 1 half theta 2. Psi 1 plus psi 2, this is equal to the first inscribed angle that we want to deal with, just regular psi. That's psi, and this right here, this is equal to 1 half times theta 1 plus theta 2. What's theta 1 plus theta 2?"}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "And so psi, which is psi 1 plus psi 2, so psi 1 plus, let me write the psi a little better, psi 1 plus psi 2 is going to be equal to these two things, 1 half theta 1 plus 1 half theta 2. Psi 1 plus psi 2, this is equal to the first inscribed angle that we want to deal with, just regular psi. That's psi, and this right here, this is equal to 1 half times theta 1 plus theta 2. What's theta 1 plus theta 2? Well, that's just our original theta that we were dealing with. So now we see that psi is equal to 1 half theta. So now we've proved it for a slightly more general case where our center is inside of the two rays that define that angle."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "What's theta 1 plus theta 2? Well, that's just our original theta that we were dealing with. So now we see that psi is equal to 1 half theta. So now we've proved it for a slightly more general case where our center is inside of the two rays that define that angle. Now, we still haven't addressed a slightly harder situation or a more general situation where if this is the center of our circle, and I have an inscribed angle where the center isn't sitting inside of the two chords. So let me draw that. So let's say I have, that's going to be my vertex, and I'll switch colors."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "So now we've proved it for a slightly more general case where our center is inside of the two rays that define that angle. Now, we still haven't addressed a slightly harder situation or a more general situation where if this is the center of our circle, and I have an inscribed angle where the center isn't sitting inside of the two chords. So let me draw that. So let's say I have, that's going to be my vertex, and I'll switch colors. So let's say that is one of the chords that defines the angle, just like that. And let's say that is the other chord that defines the angle, just like that. So how do we find the relationship between, let's call this angle right here, let's call it psi 1."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "So let's say I have, that's going to be my vertex, and I'll switch colors. So let's say that is one of the chords that defines the angle, just like that. And let's say that is the other chord that defines the angle, just like that. So how do we find the relationship between, let's call this angle right here, let's call it psi 1. How do we find the relationship between psi 1 and the central angle that subtends this same arc? So when I talk about the same arc, that's that right there. So the central angle that subtends the same arc will look like this."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "So how do we find the relationship between, let's call this angle right here, let's call it psi 1. How do we find the relationship between psi 1 and the central angle that subtends this same arc? So when I talk about the same arc, that's that right there. So the central angle that subtends the same arc will look like this. It will look like this. Let's call that theta 1. What we could do is use what we just learned when one side of our inscribed angle is a diameter."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "So the central angle that subtends the same arc will look like this. It will look like this. Let's call that theta 1. What we could do is use what we just learned when one side of our inscribed angle is a diameter. So let's construct that. So let me draw a diameter here. The result we want still is that this should be 1 half of this."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "What we could do is use what we just learned when one side of our inscribed angle is a diameter. So let's construct that. So let me draw a diameter here. The result we want still is that this should be 1 half of this. But let's prove it. Let's draw a diameter. I have to draw a straighter diameter than that."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "The result we want still is that this should be 1 half of this. But let's prove it. Let's draw a diameter. I have to draw a straighter diameter than that. So let's draw a diameter just like that. And let me call this angle right here, this angle right there. Let me call that psi 2."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "I have to draw a straighter diameter than that. So let's draw a diameter just like that. And let me call this angle right here, this angle right there. Let me call that psi 2. And it is subtending this arc right there. Let me do that in a darker color. It is subtending this arc right there."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "Let me call that psi 2. And it is subtending this arc right there. Let me do that in a darker color. It is subtending this arc right there. And so the central angle that subtends that same arc, let me call that theta 2. Now, we know from the last, from really just the earlier part of this video, that psi 2 is going to be equal to 1 half theta 2. They share the diameter is right there."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "It is subtending this arc right there. And so the central angle that subtends that same arc, let me call that theta 2. Now, we know from the last, from really just the earlier part of this video, that psi 2 is going to be equal to 1 half theta 2. They share the diameter is right there. The diameter is one of the chords that forms the angle. So psi 2 is going to be equal to 1 half theta 2. This is exactly what we've been doing in the last video, right?"}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "They share the diameter is right there. The diameter is one of the chords that forms the angle. So psi 2 is going to be equal to 1 half theta 2. This is exactly what we've been doing in the last video, right? This is an inscribed angle. One of the chords that define it is sitting on the diameter. So this is going to be 1 half of this angle, of the central angle that subtends the same arc."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "This is exactly what we've been doing in the last video, right? This is an inscribed angle. One of the chords that define it is sitting on the diameter. So this is going to be 1 half of this angle, of the central angle that subtends the same arc. Now, let's look at this larger angle right here. Psi 1 plus psi 2. That larger angle is psi 1 plus psi 2."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "So this is going to be 1 half of this angle, of the central angle that subtends the same arc. Now, let's look at this larger angle right here. Psi 1 plus psi 2. That larger angle is psi 1 plus psi 2. Once again, this subtends this entire arc right here. So it's going to be, and it has a diameter as one of the chords that defines this huge angle. So this is going to be 1 half of the central angle that subtends the same arc."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "That larger angle is psi 1 plus psi 2. Once again, this subtends this entire arc right here. So it's going to be, and it has a diameter as one of the chords that defines this huge angle. So this is going to be 1 half of the central angle that subtends the same arc. We're just using what we've already shown in this video. So this is going to be equal to 1 half of this huge central angle of theta 1 plus theta 2. So far we've just used everything that we've learned earlier in this video."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "So this is going to be 1 half of the central angle that subtends the same arc. We're just using what we've already shown in this video. So this is going to be equal to 1 half of this huge central angle of theta 1 plus theta 2. So far we've just used everything that we've learned earlier in this video. Now, we already know that psi 2 is equal to 1 half theta 2. So let me make that substitution. This is equal to that."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "So far we've just used everything that we've learned earlier in this video. Now, we already know that psi 2 is equal to 1 half theta 2. So let me make that substitution. This is equal to that. So we can say that psi 1 plus, instead of psi 2, I'll write 1 half theta 2 is equal to 1 half theta 1 plus 1 half theta 2. We can subtract 1 half theta 2 from both sides, and we get our result, psi 1 is equal to 1 half theta 1. And now we're done."}, {"video_title": "Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3", "Sentence": "This is equal to that. So we can say that psi 1 plus, instead of psi 2, I'll write 1 half theta 2 is equal to 1 half theta 1 plus 1 half theta 2. We can subtract 1 half theta 2 from both sides, and we get our result, psi 1 is equal to 1 half theta 1. And now we're done. We've proven the situation that the inscribed angle is always 1 half of the central angle that subtends the same arc, regardless of whether the center of the circle is inside of the angle, outside of the angle, whether we have a diameter on one side. And so any situation can be constructed as, any other angle can be constructed as a sum of any or all of these that we've already done. But hopefully you found this useful."}, {"video_title": "Example translating points.mp3", "Sentence": "So for example, they say plot the image of point P under a translation by five units to the left and three units up. So let's just do that at first, and then we're gonna think about other ways of describing this. So we wanna go five units to the left. So we start right over here, we're gonna go one, two, three, four, five units to the left, and then we're gonna go three units up. So that's going to be one, two, three. And so the image of point P, I guess, would show up right over here after this translation described this way. Now there are other ways that you could describe this translation."}, {"video_title": "Example translating points.mp3", "Sentence": "So we start right over here, we're gonna go one, two, three, four, five units to the left, and then we're gonna go three units up. So that's going to be one, two, three. And so the image of point P, I guess, would show up right over here after this translation described this way. Now there are other ways that you could describe this translation. Here we described it just in plain English by five units to the left and three units up, but you could, and this will look fancy, but as we'll see, it's hopefully a pretty intuitive way to describe a translation. You could say, look, I'm gonna take some point with the coordinates x comma y, and the x coordinate tells me what's my coordinate in the horizontal direction to the left or the right, and so I want that to be five less. So I would say x minus five comma y, and what do we do to the y coordinate?"}, {"video_title": "Example translating points.mp3", "Sentence": "Now there are other ways that you could describe this translation. Here we described it just in plain English by five units to the left and three units up, but you could, and this will look fancy, but as we'll see, it's hopefully a pretty intuitive way to describe a translation. You could say, look, I'm gonna take some point with the coordinates x comma y, and the x coordinate tells me what's my coordinate in the horizontal direction to the left or the right, and so I want that to be five less. So I would say x minus five comma y, and what do we do to the y coordinate? Well, we're going to increase it by three. We're gonna translate three units up. So y plus three."}, {"video_title": "Example translating points.mp3", "Sentence": "So I would say x minus five comma y, and what do we do to the y coordinate? Well, we're going to increase it by three. We're gonna translate three units up. So y plus three. So all this is saying is whatever x and y coordinates you have, this translation will make you take five from the x. That's what meaning this right over here is five units to the left, five units to the left, and then this right over here is saying three units up. Increase your y coordinate by three, decrease your x coordinate by five, and so let's just test this out with this particular coordinate, with this particular point."}, {"video_title": "Example translating points.mp3", "Sentence": "So y plus three. So all this is saying is whatever x and y coordinates you have, this translation will make you take five from the x. That's what meaning this right over here is five units to the left, five units to the left, and then this right over here is saying three units up. Increase your y coordinate by three, decrease your x coordinate by five, and so let's just test this out with this particular coordinate, with this particular point. So this point right over, p has the coordinates, its x coordinate is three, and its y coordinate is negative four. So let's see how that works. So I have three comma negative four, and I want to apply this translation."}, {"video_title": "Example translating points.mp3", "Sentence": "Increase your y coordinate by three, decrease your x coordinate by five, and so let's just test this out with this particular coordinate, with this particular point. So this point right over, p has the coordinates, its x coordinate is three, and its y coordinate is negative four. So let's see how that works. So I have three comma negative four, and I want to apply this translation. What happens? Well, let me just do my coordinates, and so I started off with three and negative four, and I'm going to subtract five from the three. So subtract five here."}, {"video_title": "Example translating points.mp3", "Sentence": "So I have three comma negative four, and I want to apply this translation. What happens? Well, let me just do my coordinates, and so I started off with three and negative four, and I'm going to subtract five from the three. So subtract five here. We see that right over there, and we're going to add three to the y. So notice, instead of an x, now I have a three. Instead of an x, now I have a three."}, {"video_title": "Example translating points.mp3", "Sentence": "So subtract five here. We see that right over there, and we're going to add three to the y. So notice, instead of an x, now I have a three. Instead of an x, now I have a three. Instead of a y, now I have a negative four. Instead of a y, now I have a negative four. And so another way of writing this, we're going from three comma negative four to three minus five is negative two, and negative four plus three is negative one."}, {"video_title": "Example translating points.mp3", "Sentence": "Instead of an x, now I have a three. Instead of a y, now I have a negative four. Instead of a y, now I have a negative four. And so another way of writing this, we're going from three comma negative four to three minus five is negative two, and negative four plus three is negative one. So what are the coordinates right over here? Well, the coordinate of this point is indeed negative two comma negative one. So notice how this, as you can see, this formula, the algebraic formula that shows how we map our coordinates, how it's able to draw the connection between the coordinates."}, {"video_title": "Example translating points.mp3", "Sentence": "And so another way of writing this, we're going from three comma negative four to three minus five is negative two, and negative four plus three is negative one. So what are the coordinates right over here? Well, the coordinate of this point is indeed negative two comma negative one. So notice how this, as you can see, this formula, the algebraic formula that shows how we map our coordinates, how it's able to draw the connection between the coordinates. And so you'll see questions where they'll tell you, hey, plot the image and they'll describe it like this. Translate x units to the left or the right, or three units up or down. You'll sometimes see it like this, but just recognize."}, {"video_title": "Example translating points.mp3", "Sentence": "So notice how this, as you can see, this formula, the algebraic formula that shows how we map our coordinates, how it's able to draw the connection between the coordinates. And so you'll see questions where they'll tell you, hey, plot the image and they'll describe it like this. Translate x units to the left or the right, or three units up or down. You'll sometimes see it like this, but just recognize. This is just saying take your x and subtract five from it, which means move five to the left. And this just means take your y coordinate and add three to it, which means move three up. And sometimes they'll ask you, hey, what's the new coordinate?"}, {"video_title": "Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "And what we have right here is a fairly classic problem. And what I want to do is I want to figure out, just given the information here, so obviously I have a triangle here. I have another triangle over here. We were given some of the angles inside of these triangles. Given the information over here, I want to figure out what the measure of this angle is right over there. I need to figure out what that question mark is. And so you might want to give a go at it, just knowing what you know about the sums of the measures of the angles inside of a triangle and maybe a little bit of what you know about supplementary angles."}, {"video_title": "Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "We were given some of the angles inside of these triangles. Given the information over here, I want to figure out what the measure of this angle is right over there. I need to figure out what that question mark is. And so you might want to give a go at it, just knowing what you know about the sums of the measures of the angles inside of a triangle and maybe a little bit of what you know about supplementary angles. So you might want to pause it and give it a try yourself, because I'm about to give you the solution. So the first thing you might say, and this is a general way to think about a lot of these problems, where they give you some angles and you have to figure out some other angles based on the sum of angles in a triangle equaling 180, or this one doesn't have parallel lines on it, but you might see some with parallel lines and supplementary lines and complementary lines, is to just fill in everything that you can figure out. And one way or another, you'll probably be able to figure out what this question mark is."}, {"video_title": "Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "And so you might want to give a go at it, just knowing what you know about the sums of the measures of the angles inside of a triangle and maybe a little bit of what you know about supplementary angles. So you might want to pause it and give it a try yourself, because I'm about to give you the solution. So the first thing you might say, and this is a general way to think about a lot of these problems, where they give you some angles and you have to figure out some other angles based on the sum of angles in a triangle equaling 180, or this one doesn't have parallel lines on it, but you might see some with parallel lines and supplementary lines and complementary lines, is to just fill in everything that you can figure out. And one way or another, you'll probably be able to figure out what this question mark is. So the first thing that kind of pops out to me is we have one triangle right over here. We have this triangle on the left. And on this triangle on the left, we're given two of the angles."}, {"video_title": "Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "And one way or another, you'll probably be able to figure out what this question mark is. So the first thing that kind of pops out to me is we have one triangle right over here. We have this triangle on the left. And on this triangle on the left, we're given two of the angles. And if you have two of the angles in a triangle, you can always figure out the third angle, because they're going to add up to 180 degrees. So if you call that x, we know that x plus 50 plus 64 is going to be equal to 180 degrees. Or we could say x plus, what is this, 114 is equal to 180 degrees."}, {"video_title": "Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "And on this triangle on the left, we're given two of the angles. And if you have two of the angles in a triangle, you can always figure out the third angle, because they're going to add up to 180 degrees. So if you call that x, we know that x plus 50 plus 64 is going to be equal to 180 degrees. Or we could say x plus, what is this, 114 is equal to 180 degrees. We can subtract 114 from both sides of this equation. And we get x is equal to 180 minus 114. So 80 minus 14, 80 minus 10 would be 70, minus another 4 is 66."}, {"video_title": "Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "Or we could say x plus, what is this, 114 is equal to 180 degrees. We can subtract 114 from both sides of this equation. And we get x is equal to 180 minus 114. So 80 minus 14, 80 minus 10 would be 70, minus another 4 is 66. So x is 66 degrees. Now, if x is 66 degrees, I think you might find that there's another angle that's not too hard to figure out. So let me write it like this."}, {"video_title": "Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "So 80 minus 14, 80 minus 10 would be 70, minus another 4 is 66. So x is 66 degrees. Now, if x is 66 degrees, I think you might find that there's another angle that's not too hard to figure out. So let me write it like this. So x is equal to 66 degrees. Well, if we know this angle right over here, if we know the measure of this angle is 66 degrees, we know that that angle is supplementary with this angle right over here. Their outer sides form a straight angle, and they are adjacent."}, {"video_title": "Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "So let me write it like this. So x is equal to 66 degrees. Well, if we know this angle right over here, if we know the measure of this angle is 66 degrees, we know that that angle is supplementary with this angle right over here. Their outer sides form a straight angle, and they are adjacent. So if we call this angle right over here y, we know that y plus x is going to be equal to 180 degrees. And we know x is equal to 66 degrees. So this is 66."}, {"video_title": "Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "Their outer sides form a straight angle, and they are adjacent. So if we call this angle right over here y, we know that y plus x is going to be equal to 180 degrees. And we know x is equal to 66 degrees. So this is 66. And so we can subtract 66 from both sides. And we get y is equal to, these cancel out, 180 minus 66 is 114. And that number might look a little familiar to you."}, {"video_title": "Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "So this is 66. And so we can subtract 66 from both sides. And we get y is equal to, these cancel out, 180 minus 66 is 114. And that number might look a little familiar to you. Notice this 114 was the exact same sum of these two angles over here. And that's actually a general idea. And I'll do it on the side here just to prove it to you."}, {"video_title": "Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "And that number might look a little familiar to you. Notice this 114 was the exact same sum of these two angles over here. And that's actually a general idea. And I'll do it on the side here just to prove it to you. If I have, let's say that these two angles, let's say that the measure of that angle is a, the measure of that angle is b, the measure of this angle we know is going to be 180 minus a minus b. That's this angle right over here. And then this angle, which is considered to be an exterior angle, so in this example, y is an exterior angle."}, {"video_title": "Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "And I'll do it on the side here just to prove it to you. If I have, let's say that these two angles, let's say that the measure of that angle is a, the measure of that angle is b, the measure of this angle we know is going to be 180 minus a minus b. That's this angle right over here. And then this angle, which is considered to be an exterior angle, so in this example, y is an exterior angle. In this example, that is our exterior angle. That is going to be supplementary to 180 minus a minus b. So this angle plus 180 minus a minus b is going to be equal to 180."}, {"video_title": "Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "And then this angle, which is considered to be an exterior angle, so in this example, y is an exterior angle. In this example, that is our exterior angle. That is going to be supplementary to 180 minus a minus b. So this angle plus 180 minus a minus b is going to be equal to 180. So if you call this angle y, you would have y plus 180 minus a minus b is equal to 180. You can subtract 180 from both sides. You can add a plus b to both sides."}, {"video_title": "Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "So this angle plus 180 minus a minus b is going to be equal to 180. So if you call this angle y, you would have y plus 180 minus a minus b is equal to 180. You can subtract 180 from both sides. You can add a plus b to both sides. So plus a plus b, running out of space on the right-hand side. And then you're left with, these cancel out. On the left-hand side, you're left with y."}, {"video_title": "Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "You can add a plus b to both sides. So plus a plus b, running out of space on the right-hand side. And then you're left with, these cancel out. On the left-hand side, you're left with y. On the right-hand side, is equal to a plus b. So this is just a general property. You can just reason it through yourself just with the sum of the measures of the angles inside of a triangle add up to 180 degrees."}, {"video_title": "Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "On the left-hand side, you're left with y. On the right-hand side, is equal to a plus b. So this is just a general property. You can just reason it through yourself just with the sum of the measures of the angles inside of a triangle add up to 180 degrees. And then you have the supplementary angles right over here. Or you could just say, look, if I have the exterior angles right over here, it's equal to the sum of the remote interior angles. That's just a little terminology you'd see there."}, {"video_title": "Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "You can just reason it through yourself just with the sum of the measures of the angles inside of a triangle add up to 180 degrees. And then you have the supplementary angles right over here. Or you could just say, look, if I have the exterior angles right over here, it's equal to the sum of the remote interior angles. That's just a little terminology you'd see there. So y is equal to a plus b. 114 degrees, we've already shown to ourselves, is equal to 64 plus 50 degrees. But anyway, regardless of how we do it, if we just reason it out step by step, or if we just knew this property from the get-go, if we know that y is equal to 114 degrees, and I like to reason it out every time just to make sure I'm not jumping to conclusions."}, {"video_title": "Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "That's just a little terminology you'd see there. So y is equal to a plus b. 114 degrees, we've already shown to ourselves, is equal to 64 plus 50 degrees. But anyway, regardless of how we do it, if we just reason it out step by step, or if we just knew this property from the get-go, if we know that y is equal to 114 degrees, and I like to reason it out every time just to make sure I'm not jumping to conclusions. So if y is 114 degrees, now we know this angle. We were given this angle in the beginning. Now we just have to figure out this third angle in this triangle."}, {"video_title": "Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "But anyway, regardless of how we do it, if we just reason it out step by step, or if we just knew this property from the get-go, if we know that y is equal to 114 degrees, and I like to reason it out every time just to make sure I'm not jumping to conclusions. So if y is 114 degrees, now we know this angle. We were given this angle in the beginning. Now we just have to figure out this third angle in this triangle. So if we call this z, if we call this question mark is equal to z, we know that z plus 114 plus 31 is equal to 180 degrees. The sums of the measures of the angle inside of a triangle add up to 180 degrees. That's the only property we're using in this step."}, {"video_title": "Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "Now we just have to figure out this third angle in this triangle. So if we call this z, if we call this question mark is equal to z, we know that z plus 114 plus 31 is equal to 180 degrees. The sums of the measures of the angle inside of a triangle add up to 180 degrees. That's the only property we're using in this step. So we get z plus, what is this? 145 is equal to 180. Did I do that right?"}, {"video_title": "Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "That's the only property we're using in this step. So we get z plus, what is this? 145 is equal to 180. Did I do that right? We have a 15 and then a 30. Yep, 145 is equal to 180. Subtract 145 from both sides of this equation, and we are left with z is equal to 80 minus 45 is equal to 35."}, {"video_title": "Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3", "Sentence": "We're asked which of these lines are parallel. So they give us three equations of three different lines. And if they're parallel, then they have to have the same slope. So all we have to do over here is figure out the slopes of each of these lines. And if any of them are equal, they're parallel. So let's do line A. Line A, it's 2y is equal to 12x plus 10."}, {"video_title": "Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3", "Sentence": "So all we have to do over here is figure out the slopes of each of these lines. And if any of them are equal, they're parallel. So let's do line A. Line A, it's 2y is equal to 12x plus 10. We're almost in slope-intercept form. We can just divide both sides of this equation by 2. Let's divide all the terms by 2."}, {"video_title": "Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3", "Sentence": "Line A, it's 2y is equal to 12x plus 10. We're almost in slope-intercept form. We can just divide both sides of this equation by 2. Let's divide all the terms by 2. We get y is equal to 6x. 12 divided by 2. 6x plus 5."}, {"video_title": "Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3", "Sentence": "Let's divide all the terms by 2. We get y is equal to 6x. 12 divided by 2. 6x plus 5. So our slope in this case, we have it in slope-intercept form, our slope in this case is equal to 6. Let's try line B. Line B is y is equal to 6."}, {"video_title": "Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3", "Sentence": "6x plus 5. So our slope in this case, we have it in slope-intercept form, our slope in this case is equal to 6. Let's try line B. Line B is y is equal to 6. Now this might be, you might say, hey, this is a bizarre character. How do I get this into slope-intercept form? Where is the x?"}, {"video_title": "Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3", "Sentence": "Line B is y is equal to 6. Now this might be, you might say, hey, this is a bizarre character. How do I get this into slope-intercept form? Where is the x? And my answer to you is that it already is in slope-intercept form. I could just rewrite it as y is equal to 0x plus 6. The x term is being multiplied by 0, because the slope here is 0. y is going to be equal to 6 no matter how much you change x."}, {"video_title": "Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3", "Sentence": "Where is the x? And my answer to you is that it already is in slope-intercept form. I could just rewrite it as y is equal to 0x plus 6. The x term is being multiplied by 0, because the slope here is 0. y is going to be equal to 6 no matter how much you change x. Change in y is always going to be 0. It's always going to be 6. So here, our slope is 0."}, {"video_title": "Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3", "Sentence": "The x term is being multiplied by 0, because the slope here is 0. y is going to be equal to 6 no matter how much you change x. Change in y is always going to be 0. It's always going to be 6. So here, our slope is 0. Our slope is equal to 0. So these two lines are definitely not parallel. They have different slopes."}, {"video_title": "Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3", "Sentence": "So here, our slope is 0. Our slope is equal to 0. So these two lines are definitely not parallel. They have different slopes. Let's try line C. I'll do it down here. So that's y minus 2 is equal to 6 times x plus 2. And this is actually in point-slope form, where the point x is equal to negative 2, y is equal to 2."}, {"video_title": "Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3", "Sentence": "They have different slopes. Let's try line C. I'll do it down here. So that's y minus 2 is equal to 6 times x plus 2. And this is actually in point-slope form, where the point x is equal to negative 2, y is equal to 2. So it's a point negative 2, 2 is being represented here, because you're subtracting the points. And the slope is 6. So we already know that the slope is equal to 6."}, {"video_title": "Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3", "Sentence": "And this is actually in point-slope form, where the point x is equal to negative 2, y is equal to 2. So it's a point negative 2, 2 is being represented here, because you're subtracting the points. And the slope is 6. So we already know that the slope is equal to 6. And sometimes people are more comfortable with slope-intercept form. So let's put it in slope-intercept form just to confirm that if we put it in this form, the slope will still be equal to 6. So if we distribute this 6, we get y minus 2 is equal to 6 times x, 6x plus 6 times 2 is 12."}, {"video_title": "Parallel lines from equation (example 3) Mathematics I High School Math Khan Academy.mp3", "Sentence": "So we already know that the slope is equal to 6. And sometimes people are more comfortable with slope-intercept form. So let's put it in slope-intercept form just to confirm that if we put it in this form, the slope will still be equal to 6. So if we distribute this 6, we get y minus 2 is equal to 6 times x, 6x plus 6 times 2 is 12. And then if you add this 2, if you add 2 to both sides of the equation, you get y, because these guys cancel out, is equal to 6x plus 14. So you see once again, the slope is 6. So line A and line C have the same slope."}, {"video_title": "Area of a sector given a central angle Circles Geometry Khan Academy.mp3", "Sentence": "A circle with area 81 pi has a sector with a 350 degree central angle. So this whole sector right over here that's shaded in this kind of pale orange yellowish color, that has a 350 degree central angle. So you see the central angle is a very large angle. It's going all the way around, all the way around like that. And they ask us, what is the area of the sector? So we just need to realize that the ratio between the area of the sector, area of sector, the ratio between the area of the sector and the total area of the circle. And they tell us what the total area is."}, {"video_title": "Area of a sector given a central angle Circles Geometry Khan Academy.mp3", "Sentence": "It's going all the way around, all the way around like that. And they ask us, what is the area of the sector? So we just need to realize that the ratio between the area of the sector, area of sector, the ratio between the area of the sector and the total area of the circle. And they tell us what the total area is. It's 81 pi. And 81 pi is going to be equal to the ratio of its central angle, which is 350 degrees, over the total number of degrees in a circle, over 360. So the area of the sector over the total area is equal to the degrees in the central angle over the total degrees in a circle."}, {"video_title": "Area of a sector given a central angle Circles Geometry Khan Academy.mp3", "Sentence": "And they tell us what the total area is. It's 81 pi. And 81 pi is going to be equal to the ratio of its central angle, which is 350 degrees, over the total number of degrees in a circle, over 360. So the area of the sector over the total area is equal to the degrees in the central angle over the total degrees in a circle. And then we just can solve for area of the sector by multiplying both sides by 81 pi. 81 pi, 81 pi. So these cancel out."}, {"video_title": "Area of a sector given a central angle Circles Geometry Khan Academy.mp3", "Sentence": "So the area of the sector over the total area is equal to the degrees in the central angle over the total degrees in a circle. And then we just can solve for area of the sector by multiplying both sides by 81 pi. 81 pi, 81 pi. So these cancel out. 350 divided by 360 is 35 over 36. And so our area, our sector area, is equal to, see in the numerator, we have 35 times, instead of 81, I'm just going to write that as, let's see, that's going to be 9 times 9 pi. And the denominator, I have 36."}, {"video_title": "Area of a sector given a central angle Circles Geometry Khan Academy.mp3", "Sentence": "So these cancel out. 350 divided by 360 is 35 over 36. And so our area, our sector area, is equal to, see in the numerator, we have 35 times, instead of 81, I'm just going to write that as, let's see, that's going to be 9 times 9 pi. And the denominator, I have 36. Well, that's the same thing as 9 times 4. And so we can divide the numerator and the denominator both by 9. And so we are left with 35 times 9."}, {"video_title": "Area of a sector given a central angle Circles Geometry Khan Academy.mp3", "Sentence": "And the denominator, I have 36. Well, that's the same thing as 9 times 4. And so we can divide the numerator and the denominator both by 9. And so we are left with 35 times 9. And neither of these are divisible by 4. So that's about as simplified as we can get it. So let's think about what 35 times 9 is."}, {"video_title": "Area of a sector given a central angle Circles Geometry Khan Academy.mp3", "Sentence": "And so we are left with 35 times 9. And neither of these are divisible by 4. So that's about as simplified as we can get it. So let's think about what 35 times 9 is. 35 times 9. It's going to be 350 minus 35, which would be 315, I guess. 315."}, {"video_title": "Area of a sector given a central angle Circles Geometry Khan Academy.mp3", "Sentence": "So let's think about what 35 times 9 is. 35 times 9. It's going to be 350 minus 35, which would be 315, I guess. 315. Did I do that right? Yeah, it's going to be 270 plus 45, which is 315 pi over 4. 315 pi over 4 is the area of the sector."}, {"video_title": "Points on line of reflection Transformations Geometry Khan Academy.mp3", "Sentence": "Let's see what they expect from us if we want to add a reflection. So if I click on this, it says reflection over the line from, and then we have two coordinate pairs. So they want us to define the line that we're going to reflect over with two points on that line. So let's see if we can do that. And to do that, I think I need to write something down. So let me get my scratch pad out, and I copied and pasted the same diagram. And the line of reflection, one way to think about it, we want to map point E to this point right over here."}, {"video_title": "Points on line of reflection Transformations Geometry Khan Academy.mp3", "Sentence": "So let's see if we can do that. And to do that, I think I need to write something down. So let me get my scratch pad out, and I copied and pasted the same diagram. And the line of reflection, one way to think about it, we want to map point E to this point right over here. We want to map point M to this point over here. And so between any point and its corresponding point on the image after the reflection, these should be equidistant from the line of reflection. This and this should be equidistant from the line of reflection."}, {"video_title": "Points on line of reflection Transformations Geometry Khan Academy.mp3", "Sentence": "And the line of reflection, one way to think about it, we want to map point E to this point right over here. We want to map point M to this point over here. And so between any point and its corresponding point on the image after the reflection, these should be equidistant from the line of reflection. This and this should be equidistant from the line of reflection. This and this should be, E and this point should be equidistant from the line of reflection. Or another way of thinking about it, that line of reflection should contain the midpoint between these two magenta points, and it should contain the midpoint between these two deep navy blue points. So let's just calculate the midpoints."}, {"video_title": "Points on line of reflection Transformations Geometry Khan Academy.mp3", "Sentence": "This and this should be equidistant from the line of reflection. This and this should be, E and this point should be equidistant from the line of reflection. Or another way of thinking about it, that line of reflection should contain the midpoint between these two magenta points, and it should contain the midpoint between these two deep navy blue points. So let's just calculate the midpoints. So we could do that with a little bit of mathematics. The coordinates for E right over here, that is, let's see, that is X equals negative four, Y is equal to negative four. And the coordinates for the corresponding point to E in the image, this is X is equal to two, X is equal to two, and Y is equal to negative six."}, {"video_title": "Points on line of reflection Transformations Geometry Khan Academy.mp3", "Sentence": "So let's just calculate the midpoints. So we could do that with a little bit of mathematics. The coordinates for E right over here, that is, let's see, that is X equals negative four, Y is equal to negative four. And the coordinates for the corresponding point to E in the image, this is X is equal to two, X is equal to two, and Y is equal to negative six. So what's the midpoint between negative four, negative four, and two comma negative six? Well, you just have to take the average of the Xs and take the average of the Ys. Let me do that, actually I'll do it over here."}, {"video_title": "Points on line of reflection Transformations Geometry Khan Academy.mp3", "Sentence": "And the coordinates for the corresponding point to E in the image, this is X is equal to two, X is equal to two, and Y is equal to negative six. So what's the midpoint between negative four, negative four, and two comma negative six? Well, you just have to take the average of the Xs and take the average of the Ys. Let me do that, actually I'll do it over here. So if I take the average of the Xs, it's going to be negative four, negative four, plus two, plus two, over two, that's the average of the Xs. And then the average of the Ys is gonna be negative four plus negative six over two. Negative four plus negative six over two."}, {"video_title": "Points on line of reflection Transformations Geometry Khan Academy.mp3", "Sentence": "Let me do that, actually I'll do it over here. So if I take the average of the Xs, it's going to be negative four, negative four, plus two, plus two, over two, that's the average of the Xs. And then the average of the Ys is gonna be negative four plus negative six over two. Negative four plus negative six over two. And then, close the parentheses. Let's see, negative four plus two is negative two. Divided by two is negative one, so it's gonna be negative one comma, negative four plus negative six, that's the same thing as negative four minus six, which is gonna be negative 10, divided by two is negative five."}, {"video_title": "Points on line of reflection Transformations Geometry Khan Academy.mp3", "Sentence": "Negative four plus negative six over two. And then, close the parentheses. Let's see, negative four plus two is negative two. Divided by two is negative one, so it's gonna be negative one comma, negative four plus negative six, that's the same thing as negative four minus six, which is gonna be negative 10, divided by two is negative five. Is, let me do that in a blue color so you see where it came from, is going to be negative five. So there you have it. That's going to be the midpoint between E and the corresponding point on its image."}, {"video_title": "Points on line of reflection Transformations Geometry Khan Academy.mp3", "Sentence": "Divided by two is negative one, so it's gonna be negative one comma, negative four plus negative six, that's the same thing as negative four minus six, which is gonna be negative 10, divided by two is negative five. Is, let me do that in a blue color so you see where it came from, is going to be negative five. So there you have it. That's going to be the midpoint between E and the corresponding point on its image. So let's see if I can plot that. So this is going to be, this point right over here is going to be negative one comma negative five. So X is negative one, Y is negative five."}, {"video_title": "Points on line of reflection Transformations Geometry Khan Academy.mp3", "Sentence": "That's going to be the midpoint between E and the corresponding point on its image. So let's see if I can plot that. So this is going to be, this point right over here is going to be negative one comma negative five. So X is negative one, Y is negative five. So it's this point right over here. And it does indeed look like the midpoint. It looks like it's equidistant between E and this point right over here."}, {"video_title": "Points on line of reflection Transformations Geometry Khan Academy.mp3", "Sentence": "So X is negative one, Y is negative five. So it's this point right over here. And it does indeed look like the midpoint. It looks like it's equidistant between E and this point right over here. And so this should sit on the line of reflection. So now let's find the midpoint between M and this point right over here. The coordinates of M are X is negative five and Y is equal to three."}, {"video_title": "Points on line of reflection Transformations Geometry Khan Academy.mp3", "Sentence": "It looks like it's equidistant between E and this point right over here. And so this should sit on the line of reflection. So now let's find the midpoint between M and this point right over here. The coordinates of M are X is negative five and Y is equal to three. The coordinates here are X is equal to seven and Y is equal to negative one. So the midpoint, the X coordinate of the midpoint is going to be the average of the X's here. So let's see, it's going to be negative five plus seven over two."}, {"video_title": "Points on line of reflection Transformations Geometry Khan Academy.mp3", "Sentence": "The coordinates of M are X is negative five and Y is equal to three. The coordinates here are X is equal to seven and Y is equal to negative one. So the midpoint, the X coordinate of the midpoint is going to be the average of the X's here. So let's see, it's going to be negative five plus seven over two. And the Y coordinate of the midpoint is going to be the average of the Y coordinates. So three plus negative one over two. Let's see, negative five plus seven is positive two over two is one."}, {"video_title": "Points on line of reflection Transformations Geometry Khan Academy.mp3", "Sentence": "So let's see, it's going to be negative five plus seven over two. And the Y coordinate of the midpoint is going to be the average of the Y coordinates. So three plus negative one over two. Let's see, negative five plus seven is positive two over two is one. Three minus one, three plus negative one, that's positive two over two is one. So the point one comma one is the midpoint between these two. So one comma one, just like that."}, {"video_title": "Points on line of reflection Transformations Geometry Khan Academy.mp3", "Sentence": "Let's see, negative five plus seven is positive two over two is one. Three minus one, three plus negative one, that's positive two over two is one. So the point one comma one is the midpoint between these two. So one comma one, just like that. So the line of reflection is going to contain these two points and two points define a line. In fact, I could, let me draw the line of reflection just because we did all of this work. The line of reflection is going to look something like, is going to look something like, I want to draw this a little bit straighter than that."}, {"video_title": "Points on line of reflection Transformations Geometry Khan Academy.mp3", "Sentence": "So one comma one, just like that. So the line of reflection is going to contain these two points and two points define a line. In fact, I could, let me draw the line of reflection just because we did all of this work. The line of reflection is going to look something like, is going to look something like, I want to draw this a little bit straighter than that. It's going to look something like, it's going to look something like, something like this. And this makes sense that this is a line of reflection. I missed that magenta point a little bit."}, {"video_title": "Points on line of reflection Transformations Geometry Khan Academy.mp3", "Sentence": "The line of reflection is going to look something like, is going to look something like, I want to draw this a little bit straighter than that. It's going to look something like, it's going to look something like, something like this. And this makes sense that this is a line of reflection. I missed that magenta point a little bit. So let me go through the magenta point. Okay, there you go. This makes sense that this is a line of reflection because you see that every, you pick an arbitrary point on segment ME, say that point, and if you reflect it over this line, it's this, this is its shortest distance from the line."}, {"video_title": "Points on line of reflection Transformations Geometry Khan Academy.mp3", "Sentence": "I missed that magenta point a little bit. So let me go through the magenta point. Okay, there you go. This makes sense that this is a line of reflection because you see that every, you pick an arbitrary point on segment ME, say that point, and if you reflect it over this line, it's this, this is its shortest distance from the line. You just go onto the other side of the line and equal distance and you get to its corresponding point on the image. So it makes a lot of sense that these are mirror images if this is kind of the mirror here. You can imagine that this is, you know, this is kind of the surface of the water if you're looking at it at an angle."}, {"video_title": "Points on line of reflection Transformations Geometry Khan Academy.mp3", "Sentence": "This makes sense that this is a line of reflection because you see that every, you pick an arbitrary point on segment ME, say that point, and if you reflect it over this line, it's this, this is its shortest distance from the line. You just go onto the other side of the line and equal distance and you get to its corresponding point on the image. So it makes a lot of sense that these are mirror images if this is kind of the mirror here. You can imagine that this is, you know, this is kind of the surface of the water if you're looking at it at an angle. I don't know if that helps you or not. But anyway, we found two points. We found two points that define that line of reflection."}, {"video_title": "Points on line of reflection Transformations Geometry Khan Academy.mp3", "Sentence": "You can imagine that this is, you know, this is kind of the surface of the water if you're looking at it at an angle. I don't know if that helps you or not. But anyway, we found two points. We found two points that define that line of reflection. So now let's use the tool to type them in. One is negative one, negative five. The other one is one comma one."}, {"video_title": "Points on line of reflection Transformations Geometry Khan Academy.mp3", "Sentence": "We found two points that define that line of reflection. So now let's use the tool to type them in. One is negative one, negative five. The other one is one comma one. So let me see if I can remember that. I have a bad memory. But so one is negative one comma negative five."}, {"video_title": "Points on line of reflection Transformations Geometry Khan Academy.mp3", "Sentence": "The other one is one comma one. So let me see if I can remember that. I have a bad memory. But so one is negative one comma negative five. And then the other one is one comma one. And we see it worked. We see it worked."}, {"video_title": "Points on line of reflection Transformations Geometry Khan Academy.mp3", "Sentence": "But so one is negative one comma negative five. And then the other one is one comma one. And we see it worked. We see it worked. By when I did that, it actually made the reflection happen. And notice, it completely went from this point and now our blue is over the image that we wanted to get to. So we are done."}, {"video_title": "Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3", "Sentence": "And then on top of that, on top of that we have what you could call a right pyramid, where the height of this right pyramid, so if you start at the center of its base right over here, and you go to the top, this height right over here is one unit. And this hasn't been drawn completely to scale, and kind of the perspective skews it a little bit, but our goal here, our goal here is to figure out what is the length, what is the length of one of these, of one of these edges right over here? So either that one, or this one right over here. What is that length? And we will call that, we will call that x. And so I encourage you to pause this video, and try to think about it on your own. Remember, this is a right, this is a right pyramid, so what that tells us is that this red line, that's one unit long, it is perpendicular, it is perpendicular to this entire plane."}, {"video_title": "Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3", "Sentence": "What is that length? And we will call that, we will call that x. And so I encourage you to pause this video, and try to think about it on your own. Remember, this is a right, this is a right pyramid, so what that tells us is that this red line, that's one unit long, it is perpendicular, it is perpendicular to this entire plane. It's perpendicular to the top of the rectangular, of the rectangular prism. So with that in mind, I encourage you to pause the video, and see if you can figure it out. And I will give you a hint."}, {"video_title": "Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3", "Sentence": "Remember, this is a right, this is a right pyramid, so what that tells us is that this red line, that's one unit long, it is perpendicular, it is perpendicular to this entire plane. It's perpendicular to the top of the rectangular, of the rectangular prism. So with that in mind, I encourage you to pause the video, and see if you can figure it out. And I will give you a hint. You will have to use the Pythagorean theorem, maybe more than once. All right, so I'm assuming you've at least given it a shot. So let's work through it together."}, {"video_title": "Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3", "Sentence": "And I will give you a hint. You will have to use the Pythagorean theorem, maybe more than once. All right, so I'm assuming you've at least given it a shot. So let's work through it together. So the key here is to realize, well okay, this point, this base right over here, this point right over here, it's halfway in this direction, and halfway in this direction. So we can figure out, well this entire, this entire length right over here, is length four, so halfway, this is going to be, I'll write it with perspective, that's going to be two, and that's going to be two, just like that. And then the other thing we can figure out, we can figure out what this length is going to be, because once again, it's halfway in that direction."}, {"video_title": "Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3", "Sentence": "So let's work through it together. So the key here is to realize, well okay, this point, this base right over here, this point right over here, it's halfway in this direction, and halfway in this direction. So we can figure out, well this entire, this entire length right over here, is length four, so halfway, this is going to be, I'll write it with perspective, that's going to be two, and that's going to be two, just like that. And then the other thing we can figure out, we can figure out what this length is going to be, because once again, it's halfway in that direction. So if this whole thing is two, and we see it right over here, this is a rectangular prism, so this length is going to be the same thing as this length. So if this whole thing is two, then each of these, this is going to be one, and this is going to be one right over there. Well how does that help us?"}, {"video_title": "Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3", "Sentence": "And then the other thing we can figure out, we can figure out what this length is going to be, because once again, it's halfway in that direction. So if this whole thing is two, and we see it right over here, this is a rectangular prism, so this length is going to be the same thing as this length. So if this whole thing is two, then each of these, this is going to be one, and this is going to be one right over there. Well how does that help us? Well using that information, we should be able to figure out, we should be able to figure out this length. Actually let me, actually I'll keep it in this color, because this color is easy to see. We should be able to figure out this length."}, {"video_title": "Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3", "Sentence": "Well how does that help us? Well using that information, we should be able to figure out, we should be able to figure out this length. Actually let me, actually I'll keep it in this color, because this color is easy to see. We should be able to figure out this length. Well why is this length interesting? Well if we know that length, that length forms a right triangle. That length and the one are the two non-hypotenuse sides of a right triangle, and then the x would be the hypotenuse."}, {"video_title": "Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3", "Sentence": "We should be able to figure out this length. Well why is this length interesting? Well if we know that length, that length forms a right triangle. That length and the one are the two non-hypotenuse sides of a right triangle, and then the x would be the hypotenuse. So we could just apply the Pythagorean theorem. So if we can figure out this, we can figure out x. So let's do one step at a time."}, {"video_title": "Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3", "Sentence": "That length and the one are the two non-hypotenuse sides of a right triangle, and then the x would be the hypotenuse. So we could just apply the Pythagorean theorem. So if we can figure out this, we can figure out x. So let's do one step at a time. How do we figure out, how do we figure out, I don't know, let me call this length, how do we call that, how do we figure out length a? Well let's just take it out, and look at it in two dimensions. So if we look at it in two dimensions, if we look at it in two dimensions, it would look something like this."}, {"video_title": "Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3", "Sentence": "So let's do one step at a time. How do we figure out, how do we figure out, I don't know, let me call this length, how do we call that, how do we figure out length a? Well let's just take it out, and look at it in two dimensions. So if we look at it in two dimensions, if we look at it in two dimensions, it would look something like this. So that's our length a. We know that this length is half of this side right over here, so that's going to be one. And actually let me do it with the same colors."}, {"video_title": "Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3", "Sentence": "So if we look at it in two dimensions, if we look at it in two dimensions, it would look something like this. So that's our length a. We know that this length is half of this side right over here, so that's going to be one. And actually let me do it with the same colors. So this right over here is the same thing as this right over here, and it is going to be of length one. And then this right over here is going to be the same as this right over here, which is going to be of length two. And so we can just use the Pythagorean theorem here."}, {"video_title": "Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3", "Sentence": "And actually let me do it with the same colors. So this right over here is the same thing as this right over here, and it is going to be of length one. And then this right over here is going to be the same as this right over here, which is going to be of length two. And so we can just use the Pythagorean theorem here. We know that the hypotenuse squared is going to be equal to one squared, one squared plus two squared. One squared plus two squared, which of course is equal to five, sorry, is equal to this one plus four, which is equal to five. So we could write a, let me do this in the magenta color, we can write a squared is equal to five, or we could say that a is equal to the principal root of five."}, {"video_title": "Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3", "Sentence": "And so we can just use the Pythagorean theorem here. We know that the hypotenuse squared is going to be equal to one squared, one squared plus two squared. One squared plus two squared, which of course is equal to five, sorry, is equal to this one plus four, which is equal to five. So we could write a, let me do this in the magenta color, we can write a squared is equal to five, or we could say that a is equal to the principal root of five. So this length right over here is the square root of five, the principal root of five. And now we can use that information to solve for x. And let's take this right triangle, and it takes a little bit of visualization practice to visualize this, right?"}, {"video_title": "Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3", "Sentence": "So we could write a, let me do this in the magenta color, we can write a squared is equal to five, or we could say that a is equal to the principal root of five. So this length right over here is the square root of five, the principal root of five. And now we can use that information to solve for x. And let's take this right triangle, and it takes a little bit of visualization practice to visualize this, right? But notice this is a right triangle. This height right here of length one is perpendicular to this entire plane. So let's see if I can draw it."}, {"video_title": "Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3", "Sentence": "And let's take this right triangle, and it takes a little bit of visualization practice to visualize this, right? But notice this is a right triangle. This height right here of length one is perpendicular to this entire plane. So let's see if I can draw it. So this, so we have this side is the square root of five, and then we have a height. Let me do it, it looks like it's in a, well, let's say we're talking about this height right over here. This height right over here is length one."}, {"video_title": "Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3", "Sentence": "So let's see if I can draw it. So this, so we have this side is the square root of five, and then we have a height. Let me do it, it looks like it's in a, well, let's say we're talking about this height right over here. This height right over here is length one. So I'll draw that. That is of length one. And we are trying to figure out x."}, {"video_title": "Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3", "Sentence": "This height right over here is length one. So I'll draw that. That is of length one. And we are trying to figure out x. We are trying to figure out x in orange. So we're trying to figure out, we're trying to figure out this x. And once again, we know this is a right triangle."}, {"video_title": "Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3", "Sentence": "And we are trying to figure out x. We are trying to figure out x in orange. So we're trying to figure out, we're trying to figure out this x. And once again, we know this is a right triangle. This is a right triangle, so we can apply the Pythagorean Theorem again. So we will have x squared is equal to one squared, which is just, I'll just write it one squared plus square root of five squared, plus square root of five squared. Well, that gets us x squared is equal to one plus five, right, square root of five squared is just five."}, {"video_title": "Pythagorean theorem in 3D Geometry 8th grade Khan Academy.mp3", "Sentence": "And once again, we know this is a right triangle. This is a right triangle, so we can apply the Pythagorean Theorem again. So we will have x squared is equal to one squared, which is just, I'll just write it one squared plus square root of five squared, plus square root of five squared. Well, that gets us x squared is equal to one plus five, right, square root of five squared is just five. So one plus five is equal to six. So we get x is equal to the square root of six. And we're done."}, {"video_title": "Drawing image of translation Transformations Geometry Khan Academy.mp3", "Sentence": "We're gonna move, it's kinda small, I hope you can see it on your video screen. We're gonna move positive eight. Every point here is gonna move positive eight in the x direction. It's x coordinate is going to increase by eight, and, or the corresponding point in the image is x coordinate is going to increase by eight, and the corresponding point in the image's y coordinate is going to decrease by one. So let's do that. And I'll focus on the vertices. Whoops, let me drag that to the trash."}, {"video_title": "Drawing image of translation Transformations Geometry Khan Academy.mp3", "Sentence": "It's x coordinate is going to increase by eight, and, or the corresponding point in the image is x coordinate is going to increase by eight, and the corresponding point in the image's y coordinate is going to decrease by one. So let's do that. And I'll focus on the vertices. Whoops, let me drag that to the trash. I didn't mean to do that. I'm gonna focus on the vertices, because that's, well, that's the easiest thing for my brain to work with. So, and actually this is what the tool expects as well."}, {"video_title": "Drawing image of translation Transformations Geometry Khan Academy.mp3", "Sentence": "Whoops, let me drag that to the trash. I didn't mean to do that. I'm gonna focus on the vertices, because that's, well, that's the easiest thing for my brain to work with. So, and actually this is what the tool expects as well. So the point B is gonna move eight to the right, or its corresponding point in the image is gonna have an x coordinate eight larger. So right now the x coordinate is negative four. If you added eight to that, it would be positive four."}, {"video_title": "Drawing image of translation Transformations Geometry Khan Academy.mp3", "Sentence": "So, and actually this is what the tool expects as well. So the point B is gonna move eight to the right, or its corresponding point in the image is gonna have an x coordinate eight larger. So right now the x coordinate is negative four. If you added eight to that, it would be positive four. And its y coordinate is gonna be one lower. Right now point B's y coordinate is eight. One lower than that is seven."}, {"video_title": "Drawing image of translation Transformations Geometry Khan Academy.mp3", "Sentence": "If you added eight to that, it would be positive four. And its y coordinate is gonna be one lower. Right now point B's y coordinate is eight. One lower than that is seven. So in the image, the corresponding point in the image would be right over there. And you see we moved eight to the right and one down. Let's do that with point C. It's at x equals negative seven."}, {"video_title": "Drawing image of translation Transformations Geometry Khan Academy.mp3", "Sentence": "One lower than that is seven. So in the image, the corresponding point in the image would be right over there. And you see we moved eight to the right and one down. Let's do that with point C. It's at x equals negative seven. If you move eight to the right, if you increase your x coordinate by eight, you're gonna move to x equals one. And then if you change your y coordinate by negative one, you're gonna move down one, and you're gonna get to that point right over there. Now let's do it with point A."}, {"video_title": "Drawing image of translation Transformations Geometry Khan Academy.mp3", "Sentence": "Let's do that with point C. It's at x equals negative seven. If you move eight to the right, if you increase your x coordinate by eight, you're gonna move to x equals one. And then if you change your y coordinate by negative one, you're gonna move down one, and you're gonna get to that point right over there. Now let's do it with point A. So point A's x coordinate is negative one. If you add eight to it, it's going to be positive seven. And its current y coordinate is two."}, {"video_title": "Drawing image of translation Transformations Geometry Khan Academy.mp3", "Sentence": "Now let's do it with point A. So point A's x coordinate is negative one. If you add eight to it, it's going to be positive seven. And its current y coordinate is two. If you take one away from it, you're gonna get to a y coordinate of one. And so there you have it. Let's see, how do I connect these two?"}, {"video_title": "Drawing image of translation Transformations Geometry Khan Academy.mp3", "Sentence": "And its current y coordinate is two. If you take one away from it, you're gonna get to a y coordinate of one. And so there you have it. Let's see, how do I connect these two? Oh, there you go. And we can check our answer. And we got it right."}, {"video_title": "Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3", "Sentence": "So pause this video and see if you can work this out on your own before we work through this together. All right, now let's work through this together, and it looks like for every one of these, or actually almost every one of these, they've given us two angles and they've given us a side. This triangle IJH, they've only given us two angles. So what I'd like to do is if I know two angles of a triangle, I can figure out the third angle because the sum of the angles of a triangle have to add up to 180 degrees. And then I can use that information, maybe with the sides that they give us, in order to judge which of these triangles are congruent. So first of all, what is going to be the measure of this angle right over here, the measure of angle ACB? Pause the video and try to think about that."}, {"video_title": "Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3", "Sentence": "So what I'd like to do is if I know two angles of a triangle, I can figure out the third angle because the sum of the angles of a triangle have to add up to 180 degrees. And then I can use that information, maybe with the sides that they give us, in order to judge which of these triangles are congruent. So first of all, what is going to be the measure of this angle right over here, the measure of angle ACB? Pause the video and try to think about that. Well, one way to think about it, if we call the measure of that angle X, we know that X plus 36 plus 82 needs to be equal to 180. I'm just giving their measures in degrees here. And so you could say X plus, let's see, 36 plus 82 is 118."}, {"video_title": "Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3", "Sentence": "Pause the video and try to think about that. Well, one way to think about it, if we call the measure of that angle X, we know that X plus 36 plus 82 needs to be equal to 180. I'm just giving their measures in degrees here. And so you could say X plus, let's see, 36 plus 82 is 118. Did I do that right? Six plus two is eight, and then three plus eight is 11. Yep, that's right."}, {"video_title": "Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3", "Sentence": "And so you could say X plus, let's see, 36 plus 82 is 118. Did I do that right? Six plus two is eight, and then three plus eight is 11. Yep, that's right. So that's going to be equal to 180. And then if I subtract 118 from both sides, I'm going to get X is equal to, 180 minus 18 is 62. So this is X is equal to 62, or this is a 62-degree angle, I guess is another way of thinking about it."}, {"video_title": "Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3", "Sentence": "Yep, that's right. So that's going to be equal to 180. And then if I subtract 118 from both sides, I'm going to get X is equal to, 180 minus 18 is 62. So this is X is equal to 62, or this is a 62-degree angle, I guess is another way of thinking about it. I could put everything in terms of degrees if you like. All right, now let's do the same thing with this one right over here. Well, this one has an 82-degree angle and a 62-degree angle, just like this triangle over here."}, {"video_title": "Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3", "Sentence": "So this is X is equal to 62, or this is a 62-degree angle, I guess is another way of thinking about it. I could put everything in terms of degrees if you like. All right, now let's do the same thing with this one right over here. Well, this one has an 82-degree angle and a 62-degree angle, just like this triangle over here. So we know that the third angle needs to be 36 degrees. 36 degrees, because we know 82 and 62, if you need to get to 180, it has to be 36. We just figured that out from this first triangle over here."}, {"video_title": "Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3", "Sentence": "Well, this one has an 82-degree angle and a 62-degree angle, just like this triangle over here. So we know that the third angle needs to be 36 degrees. 36 degrees, because we know 82 and 62, if you need to get to 180, it has to be 36. We just figured that out from this first triangle over here. Now if we look over here, 36 degrees and 59, this definitely looks like it has different angles, but let's figure out what this angle would have to be. So if we call that Y degrees, we know, I'll do it over here, Y plus 36 plus 59 is equal to 180. And I'm just thinking in terms of degrees here."}, {"video_title": "Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3", "Sentence": "We just figured that out from this first triangle over here. Now if we look over here, 36 degrees and 59, this definitely looks like it has different angles, but let's figure out what this angle would have to be. So if we call that Y degrees, we know, I'll do it over here, Y plus 36 plus 59 is equal to 180. And I'm just thinking in terms of degrees here. So Y plus, this is going to be equal to, what is this? This is going to be equal to 95 is equal to 180. Did I do that right?"}, {"video_title": "Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3", "Sentence": "And I'm just thinking in terms of degrees here. So Y plus, this is going to be equal to, what is this? This is going to be equal to 95 is equal to 180. Did I do that right? Yep, that's 80 plus 15, yep, 95. And then if I subtract 95 from both sides, what am I left with? I'm left with Y is equal to 85 degrees."}, {"video_title": "Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3", "Sentence": "Did I do that right? Yep, that's 80 plus 15, yep, 95. And then if I subtract 95 from both sides, what am I left with? I'm left with Y is equal to 85 degrees. And so this is going to be equal to 85 degrees. And then this last triangle right over here, I have an angle that has measure 36, another one that's 59. So by the same logic, this one over here has to be 85 degrees."}, {"video_title": "Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3", "Sentence": "I'm left with Y is equal to 85 degrees. And so this is going to be equal to 85 degrees. And then this last triangle right over here, I have an angle that has measure 36, another one that's 59. So by the same logic, this one over here has to be 85 degrees. So let's ask ourselves, now that we've figured out a little bit more about these triangles, which of these two must be congruent? So you might be tempted to look at these bottom two triangles and say, hey, look, all of their angles are the same. You have angle, angle, angle, and angle, angle, angle."}, {"video_title": "Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3", "Sentence": "So by the same logic, this one over here has to be 85 degrees. So let's ask ourselves, now that we've figured out a little bit more about these triangles, which of these two must be congruent? So you might be tempted to look at these bottom two triangles and say, hey, look, all of their angles are the same. You have angle, angle, angle, and angle, angle, angle. Well, they would be similar. If you have three angles that are the same, you definitely have similar triangles. But we don't have any length information for triangle IJH."}, {"video_title": "Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3", "Sentence": "You have angle, angle, angle, and angle, angle, angle. Well, they would be similar. If you have three angles that are the same, you definitely have similar triangles. But we don't have any length information for triangle IJH. You need to know at least one of the lengths of one of the sides in order to even think, start to think about congruence. And so we can't make any conclusion that IJH and LMK, triangles IJH and triangles LMK are congruent to each other. Now let's look at these candidates up here."}, {"video_title": "Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3", "Sentence": "But we don't have any length information for triangle IJH. You need to know at least one of the lengths of one of the sides in order to even think, start to think about congruence. And so we can't make any conclusion that IJH and LMK, triangles IJH and triangles LMK are congruent to each other. Now let's look at these candidates up here. We know that their angles are all the same. And so we could apply, we could apply angle, I'll do this in a different color, angle, side, angle. 36 degrees, length six, 82 degrees."}, {"video_title": "Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "A tiny but horrible alien is standing at the top of the Eiffel Tower, so this is where the tiny but horrible alien is, which is 324 meters tall, and they label that the height of the Eiffel Tower, and threatening to destroy the city of Paris. A men in black, or I guess a men in black agent, I was about to say maybe it should be a man in black, a men in black agent is standing at ground level, 54 meters across the Eiffel Square, so 54 meters from, I guess you could say the center of the base of the Eiffel Tower, aiming his laser gun at the alien, so this is him aiming the laser gun. At what angle should the agent shoot his laser gun? Round your answer if necessary to two decimal places. So if we construct a right triangle here, and we can, so the height of this right triangle is 324 meters, this width right over here is 54 meters, it is a right triangle, what they're really asking us is what is the angle, what is this angle, what is this angle right over here? And they've given us two pieces of information, they gave us the side that is opposite the angle, and they've given us the side that is adjacent to the angle. So what trig function deals with opposite and adjacent?"}, {"video_title": "Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Round your answer if necessary to two decimal places. So if we construct a right triangle here, and we can, so the height of this right triangle is 324 meters, this width right over here is 54 meters, it is a right triangle, what they're really asking us is what is the angle, what is this angle, what is this angle right over here? And they've given us two pieces of information, they gave us the side that is opposite the angle, and they've given us the side that is adjacent to the angle. So what trig function deals with opposite and adjacent? And to remind ourselves, we can write like I always like to do, SOH CAH TOA. And these are really by definition, so you just kind of have to know this, and SOH CAH TOA helps us. Sine is opposite over hypotenuse."}, {"video_title": "Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So what trig function deals with opposite and adjacent? And to remind ourselves, we can write like I always like to do, SOH CAH TOA. And these are really by definition, so you just kind of have to know this, and SOH CAH TOA helps us. Sine is opposite over hypotenuse. Cosine is adjacent over hypotenuse. Tangent is opposite over adjacent. So we can say that the tangent, we can write that the tangent of theta, the tangent of theta is equal to the length of the opposite side, 324 meters, 324 meters over the length of the adjacent side, over 54 meters."}, {"video_title": "Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Sine is opposite over hypotenuse. Cosine is adjacent over hypotenuse. Tangent is opposite over adjacent. So we can say that the tangent, we can write that the tangent of theta, the tangent of theta is equal to the length of the opposite side, 324 meters, 324 meters over the length of the adjacent side, over 54 meters. Now you might say, well, okay, that's fine, so what angle gives, when I take its tangent, gives me 324 over 54? Well, for this, it will probably be useful to use a calculator. And the way that we'd use a calculator is we would use the inverse tan function."}, {"video_title": "Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So we can say that the tangent, we can write that the tangent of theta, the tangent of theta is equal to the length of the opposite side, 324 meters, 324 meters over the length of the adjacent side, over 54 meters. Now you might say, well, okay, that's fine, so what angle gives, when I take its tangent, gives me 324 over 54? Well, for this, it will probably be useful to use a calculator. And the way that we'd use a calculator is we would use the inverse tan function. So we could rewrite this as we're going to take the inverse tangent, and sometimes it's written as tangent, kind of this negative 1 superscript. So the inverse tangent of tan of theta is going to be equal to the inverse tangent of 324 over 54. And just to be clear, what is this inverse tangent?"}, {"video_title": "Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And the way that we'd use a calculator is we would use the inverse tan function. So we could rewrite this as we're going to take the inverse tangent, and sometimes it's written as tangent, kind of this negative 1 superscript. So the inverse tangent of tan of theta is going to be equal to the inverse tangent of 324 over 54. And just to be clear, what is this inverse tangent? This literally says, this will return what is the angle that when I take the tangent of it, gives me 324 over 54. This says, what is the angle that when I take the tangent of it, gives me tangent of theta? So this right over here, this just simplifies to theta."}, {"video_title": "Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And just to be clear, what is this inverse tangent? This literally says, this will return what is the angle that when I take the tangent of it, gives me 324 over 54. This says, what is the angle that when I take the tangent of it, gives me tangent of theta? So this right over here, this just simplifies to theta. Theta is the angle that when you get the tangent of it, gets you tangent of theta. And so we get theta is equal to tangent, inverse tangent of 324 over 54. Once again, this inverse tangent thing, you might find confusing, but all this is saying is, over here it's saying tangent of some angle is 324 over 54."}, {"video_title": "Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So this right over here, this just simplifies to theta. Theta is the angle that when you get the tangent of it, gets you tangent of theta. And so we get theta is equal to tangent, inverse tangent of 324 over 54. Once again, this inverse tangent thing, you might find confusing, but all this is saying is, over here it's saying tangent of some angle is 324 over 54. This is just saying is my angle is whatever angle I need, so that when I take the tangent of it, I get 324 over 54. It's how we will solve for theta. So let's get our calculator out, and let's say that we want our answer in degrees."}, {"video_title": "Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Once again, this inverse tangent thing, you might find confusing, but all this is saying is, over here it's saying tangent of some angle is 324 over 54. This is just saying is my angle is whatever angle I need, so that when I take the tangent of it, I get 324 over 54. It's how we will solve for theta. So let's get our calculator out, and let's say that we want our answer in degrees. And actually they should, well, I'm just going to assume that they want our answer in degrees. So let me make sure my calculator is actually in degree mode. So I'll go to the second mode right over here, and actually it's in radian mode right now."}, {"video_title": "Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So let's get our calculator out, and let's say that we want our answer in degrees. And actually they should, well, I'm just going to assume that they want our answer in degrees. So let me make sure my calculator is actually in degree mode. So I'll go to the second mode right over here, and actually it's in radian mode right now. So let me make sure I'm in degree mode to get my answer in degrees. Now let me exit out of here, and let me just type in the inverse tangent. So it's in this yellow color right here."}, {"video_title": "Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So I'll go to the second mode right over here, and actually it's in radian mode right now. So let me make sure I'm in degree mode to get my answer in degrees. Now let me exit out of here, and let me just type in the inverse tangent. So it's in this yellow color right here. Inverse tangent of 324 divided by 54 is going to be 80 point, and they told us to round to two decimal places, 80.54 degrees. So theta is equal to 80.54 degrees. That's the angle at which you should shoot the gun to help defeat this horrible alien."}, {"video_title": "Pythagorean theorem with right triangle.mp3", "Sentence": "So pause this video and see if you can figure that out. Well, the key realization to solve this is to realize that this altitude that they dropped, this is going to form a right angle here and a right angle here. And notice, both of these triangles, because this whole thing is an isosceles triangle, we are going to have two angles that are the same. This angle is the same as that angle because it's an isosceles triangle. This 90 degrees is the same as that 90 degrees. And so the third angle needs to be the same. So that is going to be the same as that right over there."}, {"video_title": "Pythagorean theorem with right triangle.mp3", "Sentence": "This angle is the same as that angle because it's an isosceles triangle. This 90 degrees is the same as that 90 degrees. And so the third angle needs to be the same. So that is going to be the same as that right over there. And since you have two angles that are the same and you have a side between them that is the same, this side, this altitude of length 12 is on both triangles, we know that both of these triangles are congruent. So they're both gonna have 13, they're gonna have one side that's 13, one side that is 12. And so this and this side are going to be the same."}, {"video_title": "Pythagorean theorem with right triangle.mp3", "Sentence": "So that is going to be the same as that right over there. And since you have two angles that are the same and you have a side between them that is the same, this side, this altitude of length 12 is on both triangles, we know that both of these triangles are congruent. So they're both gonna have 13, they're gonna have one side that's 13, one side that is 12. And so this and this side are going to be the same. So this is going to be x over two and this is going to be x over two. And so now we can use that information and the fact and the Pythagorean theorem to solve for x. Let's use the Pythagorean theorem on this right triangle on the right-hand side."}, {"video_title": "Pythagorean theorem with right triangle.mp3", "Sentence": "And so this and this side are going to be the same. So this is going to be x over two and this is going to be x over two. And so now we can use that information and the fact and the Pythagorean theorem to solve for x. Let's use the Pythagorean theorem on this right triangle on the right-hand side. We can say that x over two squared, that's the base right over here, this side right over here, we could write that x over two squared plus the other side, plus 12 squared, is going to be equal to our hypotenuse squared, is going to be equal to 13 squared. This is just the Pythagorean theorem now. And so we can simplify."}, {"video_title": "Pythagorean theorem with right triangle.mp3", "Sentence": "Let's use the Pythagorean theorem on this right triangle on the right-hand side. We can say that x over two squared, that's the base right over here, this side right over here, we could write that x over two squared plus the other side, plus 12 squared, is going to be equal to our hypotenuse squared, is going to be equal to 13 squared. This is just the Pythagorean theorem now. And so we can simplify. This is going to be x, let me do that same color, this is going to be x squared over four, that's just x squared over two squared, plus 144, 144 is equal to, 13 squared is 169. Now I can subtract 144 from both sides. I'm gonna try to solve for x, that's the whole goal here."}, {"video_title": "Pythagorean theorem with right triangle.mp3", "Sentence": "And so we can simplify. This is going to be x, let me do that same color, this is going to be x squared over four, that's just x squared over two squared, plus 144, 144 is equal to, 13 squared is 169. Now I can subtract 144 from both sides. I'm gonna try to solve for x, that's the whole goal here. So subtracting 144 from both sides. And what do we get? On the left-hand side, we have x squared over four is equal to 169 minus 144, let's see, 69 minus 44 is 25."}, {"video_title": "Pythagorean theorem with right triangle.mp3", "Sentence": "I'm gonna try to solve for x, that's the whole goal here. So subtracting 144 from both sides. And what do we get? On the left-hand side, we have x squared over four is equal to 169 minus 144, let's see, 69 minus 44 is 25. So this is going to be equal to 25. We can multiply both sides by four to isolate the x squared. And so we get x squared is equal to 25 times four is equal to 100."}, {"video_title": "Pythagorean theorem with right triangle.mp3", "Sentence": "On the left-hand side, we have x squared over four is equal to 169 minus 144, let's see, 69 minus 44 is 25. So this is going to be equal to 25. We can multiply both sides by four to isolate the x squared. And so we get x squared is equal to 25 times four is equal to 100. Now if we were just looking at this purely mathematically, you'd say, oh, x could be positive or negative 10. But since we're dealing with distances, we know that we want the positive value of it. So x is equal to the principal root of 100, which is equal to positive 10."}, {"video_title": "Pythagorean theorem with right triangle.mp3", "Sentence": "And so we get x squared is equal to 25 times four is equal to 100. Now if we were just looking at this purely mathematically, you'd say, oh, x could be positive or negative 10. But since we're dealing with distances, we know that we want the positive value of it. So x is equal to the principal root of 100, which is equal to positive 10. So there you have it, we have solved for x. This distance right here, the whole thing, the whole thing is going to be equal to 10. Half of that is going to be five."}, {"video_title": "Pythagorean theorem with right triangle.mp3", "Sentence": "So x is equal to the principal root of 100, which is equal to positive 10. So there you have it, we have solved for x. This distance right here, the whole thing, the whole thing is going to be equal to 10. Half of that is going to be five. So if we just looked at this length right over here, I'm doing that in the same color, let me see. So this length right over here, that's going to be five, and indeed, five squared plus 12 squared, that's 25 plus 144, is 169, 13 squared. So the key realization here is, isosceles triangle, the altitude splits it into two congruent right triangles."}, {"video_title": "Pythagorean theorem with right triangle.mp3", "Sentence": "Half of that is going to be five. So if we just looked at this length right over here, I'm doing that in the same color, let me see. So this length right over here, that's going to be five, and indeed, five squared plus 12 squared, that's 25 plus 144, is 169, 13 squared. So the key realization here is, isosceles triangle, the altitude splits it into two congruent right triangles. And so it also splits this base into two. So this is x over two, and this is x over two. And we use that information and the Pythagorean theorem to solve for x."}, {"video_title": "Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3", "Sentence": "So this is a diameter, that's what they're telling us. That's SE, and they say, what is the measure of angle ISE? So what we care about, the measure of angle ISE, I-S-E. So we're trying to figure out this angle right over there. And like always, I encourage you to pause the video and see if you can work through it yourself. So there's a bunch of ways that we can actually tackle this problem. The first one that jumps out at me is, you know, there's a bunch of triangles here, and we can use the fact that the angles, the interior angles of a triangle add up to 180 degrees."}, {"video_title": "Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3", "Sentence": "So we're trying to figure out this angle right over there. And like always, I encourage you to pause the video and see if you can work through it yourself. So there's a bunch of ways that we can actually tackle this problem. The first one that jumps out at me is, you know, there's a bunch of triangles here, and we can use the fact that the angles, the interior angles of a triangle add up to 180 degrees. So we could look at this triangle right over here. And so we know one of the angles already. We know this angle has a measure of 27 degrees."}, {"video_title": "Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3", "Sentence": "The first one that jumps out at me is, you know, there's a bunch of triangles here, and we can use the fact that the angles, the interior angles of a triangle add up to 180 degrees. So we could look at this triangle right over here. And so we know one of the angles already. We know this angle has a measure of 27 degrees. If we could figure out this angle right over here, this is what would be angle SIE, then if we know two angles, two interior angles of a triangle, we can figure out the third. And this one, SIE, we can figure it out because it's supplementary to this 61 degree angle. So this angle right over here is just going to be this angle plus the 61 degree angle is going to be equal to 180 degrees because they are supplementary."}, {"video_title": "Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3", "Sentence": "We know this angle has a measure of 27 degrees. If we could figure out this angle right over here, this is what would be angle SIE, then if we know two angles, two interior angles of a triangle, we can figure out the third. And this one, SIE, we can figure it out because it's supplementary to this 61 degree angle. So this angle right over here is just going to be this angle plus the 61 degree angle is going to be equal to 180 degrees because they are supplementary. Or we could say that this angle right over here is going to be 180 minus 61. So what is that going to be? 180 minus 60 would be 120, and then minus one would be, this is 119 degrees."}, {"video_title": "Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3", "Sentence": "So this angle right over here is just going to be this angle plus the 61 degree angle is going to be equal to 180 degrees because they are supplementary. Or we could say that this angle right over here is going to be 180 minus 61. So what is that going to be? 180 minus 60 would be 120, and then minus one would be, this is 119 degrees. And so this angle that we're trying to figure out, this angle plus the 119 degrees, plus the 27 degrees, is going to be equal to 180 degrees. Or we could say that this angle is going to be 180 minus 119 minus 27, which is going to be equal to, so let's see, 180 minus 119 is 61, and then 61 minus 27 is going to be 34. So there you have it."}, {"video_title": "Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3", "Sentence": "180 minus 60 would be 120, and then minus one would be, this is 119 degrees. And so this angle that we're trying to figure out, this angle plus the 119 degrees, plus the 27 degrees, is going to be equal to 180 degrees. Or we could say that this angle is going to be 180 minus 119 minus 27, which is going to be equal to, so let's see, 180 minus 119 is 61, and then 61 minus 27 is going to be 34. So there you have it. Measure of angle ISE is 34 degrees. Now I mentioned that there is multiple ways that we could figure this out. Let me do it one more way."}, {"video_title": "Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3", "Sentence": "So there you have it. Measure of angle ISE is 34 degrees. Now I mentioned that there is multiple ways that we could figure this out. Let me do it one more way. So let me unwind everything that I just wrote. We already figured out the answer, but I want to show you that there's multiple ways that we can tackle this. So ISE is still the thing that we want to figure out."}, {"video_title": "Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3", "Sentence": "Let me do it one more way. So let me unwind everything that I just wrote. We already figured out the answer, but I want to show you that there's multiple ways that we can tackle this. So ISE is still the thing that we want to figure out. Another way that we could approach it is, well, we know we have some angles, inscribed angles on this circle, and we know that if an inscribed angle intercepts the diameter, then it's going to be a right angle. It's going to be a 90 degree angle. So this angle right over here is a 90 degree angle, and we can use that information to figure out this angle, and we could also use that information if we look at a different, so if we look at this triangle, we could use 90 plus 61 plus this angle is going to be equal to 180 degrees."}, {"video_title": "Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3", "Sentence": "So ISE is still the thing that we want to figure out. Another way that we could approach it is, well, we know we have some angles, inscribed angles on this circle, and we know that if an inscribed angle intercepts the diameter, then it's going to be a right angle. It's going to be a 90 degree angle. So this angle right over here is a 90 degree angle, and we can use that information to figure out this angle, and we could also use that information if we look at a different, so if we look at this triangle, we could use 90 plus 61 plus this angle is going to be equal to 180 degrees. So this angle right over here, another way to think about it, it's going to be 180 minus 90 minus 61, which is equal to, 180 minus 90 is 90, minus 61 is 29 degrees. So this one right over here is 29 degrees. 29 degrees."}, {"video_title": "Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3", "Sentence": "So this angle right over here is a 90 degree angle, and we can use that information to figure out this angle, and we could also use that information if we look at a different, so if we look at this triangle, we could use 90 plus 61 plus this angle is going to be equal to 180 degrees. So this angle right over here, another way to think about it, it's going to be 180 minus 90 minus 61, which is equal to, 180 minus 90 is 90, minus 61 is 29 degrees. So this one right over here is 29 degrees. 29 degrees. And then we could look at this larger triangle. We could look at this larger triangle right over here to figure out this entire angle. To figure out this entire angle."}, {"video_title": "Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3", "Sentence": "29 degrees. And then we could look at this larger triangle. We could look at this larger triangle right over here to figure out this entire angle. To figure out this entire angle. If we know this entire angle, you subtract 29, then you figure out angle ISE. And so this large, or what I've depicted as this kind of, this magenta, this measure right over here, of that angle, plus 90 degrees, plus 27 degrees is going to be equal to 180, because they're the interior angles of triangle SLE. So this angle right over here is going to be 180 minus 61 minus 27."}, {"video_title": "Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3", "Sentence": "To figure out this entire angle. If we know this entire angle, you subtract 29, then you figure out angle ISE. And so this large, or what I've depicted as this kind of, this magenta, this measure right over here, of that angle, plus 90 degrees, plus 27 degrees is going to be equal to 180, because they're the interior angles of triangle SLE. So this angle right over here is going to be 180 minus 61 minus 27. Sorry, not minus 61, minus 90. Minus 90. It's 180 minus 90 minus 27 is going to give us this angle right over here, because the three angles add up to 180."}, {"video_title": "Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3", "Sentence": "So this angle right over here is going to be 180 minus 61 minus 27. Sorry, not minus 61, minus 90. Minus 90. It's 180 minus 90 minus 27 is going to give us this angle right over here, because the three angles add up to 180. So minus 90 minus 27, which is equal to, so 180 minus 90 is 90. 90 minus 27, 90 minus 27 is 63. 63 degrees."}, {"video_title": "Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3", "Sentence": "It's 180 minus 90 minus 27 is going to give us this angle right over here, because the three angles add up to 180. So minus 90 minus 27, which is equal to, so 180 minus 90 is 90. 90 minus 27, 90 minus 27 is 63. 63 degrees. So this large one over here is 63 degrees, and the smaller one is 29 degrees. And so angle ISE, which we set out to figure out, is going to be 63 degrees minus the 29 degrees. So 63 minus 29 is once again equal to 34 degrees."}, {"video_title": "Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3", "Sentence": "63 degrees. So this large one over here is 63 degrees, and the smaller one is 29 degrees. And so angle ISE, which we set out to figure out, is going to be 63 degrees minus the 29 degrees. So 63 minus 29 is once again equal to 34 degrees. So the way I did it just now, a little bit harder. It really depends what jumps out at you. The first way I tackled it, it does seem a little bit easier, a little bit clearer, but it's good to see these different things, and at least here, we use this idea of an inscribed angle that intercepts a diameter."}, {"video_title": "Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3", "Sentence": "So 63 minus 29 is once again equal to 34 degrees. So the way I did it just now, a little bit harder. It really depends what jumps out at you. The first way I tackled it, it does seem a little bit easier, a little bit clearer, but it's good to see these different things, and at least here, we use this idea of an inscribed angle that intercepts a diameter. And if you say, hey, how do we know? I mean, we prove it in other videos, but it comes straight out of the idea that an inscribed angle, the measure of an inscribed angle is going to be half of the measure of the arc that it intercepts. And notice, it's intercepting an arc that has a measure of 180 degrees."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "Now, we call the longest side of a right triangle, we call that side, and you can either view it as the longest side of the right triangle, or the side opposite the 90 degree angle, it is called the hypotenuse. It's a very fancy word for a fairly simple idea, just the longest side of a right triangle, or the side opposite the 90 degree angle. And it's just good to know that because someone might say hypotenuse, like, oh, they're just talking about this side right here, the side longest, the side opposite the 90 degree angle. Now, what I want to do in this video is prove a relationship, a very famous relationship, and you might see where this is going, a very famous relationship between the lengths of the sides of a right triangle. So let's say that the length of AC, so uppercase A, uppercase C, let's call that length lowercase a, let's call the length of BC, lowercase b right over here, I'll use uppercases for points, lowercases for lengths, and let's call the length of the hypotenuse, the length of AB, let's call that C. And let's see if we can come up with a relationship between A, B, and C. And to do that, I'm first going to construct another line, or another segment, I should say, between C and the hypotenuse, and I'm going to construct it so that they intersect at a right angle. And you can always do that, and we'll call this point right over here, we'll call this point capital D. And if you're wondering, how can you always do that, you can imagine rotating this entire triangle like this, and this isn't a rigorous proof, but it just kind of gives you the general idea of how you can always construct a point like this. So if I've rotated it around, so now our hypotenuse, we're now sitting on our hypotenuse, this is now point B, this is point A, and if I rotated the whole thing all the way around, this is point C, you can imagine just dropping a rock from point C, maybe with a string attached, and it would hit the hypotenuse at a right angle."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "Now, what I want to do in this video is prove a relationship, a very famous relationship, and you might see where this is going, a very famous relationship between the lengths of the sides of a right triangle. So let's say that the length of AC, so uppercase A, uppercase C, let's call that length lowercase a, let's call the length of BC, lowercase b right over here, I'll use uppercases for points, lowercases for lengths, and let's call the length of the hypotenuse, the length of AB, let's call that C. And let's see if we can come up with a relationship between A, B, and C. And to do that, I'm first going to construct another line, or another segment, I should say, between C and the hypotenuse, and I'm going to construct it so that they intersect at a right angle. And you can always do that, and we'll call this point right over here, we'll call this point capital D. And if you're wondering, how can you always do that, you can imagine rotating this entire triangle like this, and this isn't a rigorous proof, but it just kind of gives you the general idea of how you can always construct a point like this. So if I've rotated it around, so now our hypotenuse, we're now sitting on our hypotenuse, this is now point B, this is point A, and if I rotated the whole thing all the way around, this is point C, you can imagine just dropping a rock from point C, maybe with a string attached, and it would hit the hypotenuse at a right angle. So that's all we did here to establish segment CD, where we put our point D right over there. And the reason why I did that is now we can do all sorts of interesting relationships between similar triangles, because we have three triangles here. We have triangle ADC, we have triangle DBC, and then we have the larger original triangle, and we can hopefully establish similarity between those triangles."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "So if I've rotated it around, so now our hypotenuse, we're now sitting on our hypotenuse, this is now point B, this is point A, and if I rotated the whole thing all the way around, this is point C, you can imagine just dropping a rock from point C, maybe with a string attached, and it would hit the hypotenuse at a right angle. So that's all we did here to establish segment CD, where we put our point D right over there. And the reason why I did that is now we can do all sorts of interesting relationships between similar triangles, because we have three triangles here. We have triangle ADC, we have triangle DBC, and then we have the larger original triangle, and we can hopefully establish similarity between those triangles. And first I'll show you that ADC is similar to the larger one, because both of them have a right angle. ADC has a right angle right over here. Clearly if this angle is 90 degrees, then this angle is going to be 90 degrees as well."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "We have triangle ADC, we have triangle DBC, and then we have the larger original triangle, and we can hopefully establish similarity between those triangles. And first I'll show you that ADC is similar to the larger one, because both of them have a right angle. ADC has a right angle right over here. Clearly if this angle is 90 degrees, then this angle is going to be 90 degrees as well. They're supplementary. They have to add up to 180. And so they both have a right angle in them."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "Clearly if this angle is 90 degrees, then this angle is going to be 90 degrees as well. They're supplementary. They have to add up to 180. And so they both have a right angle in them. So the smaller one has a right angle, the larger one clearly has a right angle. That's where we started from. And they also both share this angle right over here."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "And so they both have a right angle in them. So the smaller one has a right angle, the larger one clearly has a right angle. That's where we started from. And they also both share this angle right over here. Angle DAC or BAC, however you want to refer to it. So we can actually write down that triangle. I'm going to start with the smaller one."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "And they also both share this angle right over here. Angle DAC or BAC, however you want to refer to it. So we can actually write down that triangle. I'm going to start with the smaller one. ADC, and maybe I'll shade it in right over here. So this is the triangle we're talking about. Triangle ADC, and I went from the blue angle to the right angle to the unlabeled angle from the point of view of triangle ADC."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "I'm going to start with the smaller one. ADC, and maybe I'll shade it in right over here. So this is the triangle we're talking about. Triangle ADC, and I went from the blue angle to the right angle to the unlabeled angle from the point of view of triangle ADC. This right angle isn't applying to that right over there. It's applying to the larger triangle. So we can say triangle ADC is similar to triangle."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "Triangle ADC, and I went from the blue angle to the right angle to the unlabeled angle from the point of view of triangle ADC. This right angle isn't applying to that right over there. It's applying to the larger triangle. So we can say triangle ADC is similar to triangle. Once again, you want to start at the blue angle, A. Then we went to the right angle, so we want to go to the right angle again. So it's ACB."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "So we can say triangle ADC is similar to triangle. Once again, you want to start at the blue angle, A. Then we went to the right angle, so we want to go to the right angle again. So it's ACB. And because they're similar, we can set up a relationship between the ratios of their sides. For example, we know the ratio of corresponding sides are going to, well, in general for a similar triangle, we know that the ratio of the corresponding sides are going to be a constant. So we could take the ratio of the hypotenuse of this side, of the smaller triangle."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "So it's ACB. And because they're similar, we can set up a relationship between the ratios of their sides. For example, we know the ratio of corresponding sides are going to, well, in general for a similar triangle, we know that the ratio of the corresponding sides are going to be a constant. So we could take the ratio of the hypotenuse of this side, of the smaller triangle. So the hypotenuse is AC. So AC over the hypotenuse over the larger one, which is AB. AC over AB is going to be the same thing as AD, as one of the legs."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "So we could take the ratio of the hypotenuse of this side, of the smaller triangle. So the hypotenuse is AC. So AC over the hypotenuse over the larger one, which is AB. AC over AB is going to be the same thing as AD, as one of the legs. AD, just to show that I'm just taking corresponding points on both similar triangles. This is AD over AC. You can look at these triangles yourself and show, look, AD, point AD is between the blue angle and the right angle."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "AC over AB is going to be the same thing as AD, as one of the legs. AD, just to show that I'm just taking corresponding points on both similar triangles. This is AD over AC. You can look at these triangles yourself and show, look, AD, point AD is between the blue angle and the right angle. Sorry, side AD is between the blue angle and the right angle. Side AC is between the blue angle and the right angle on the larger triangle. So both of these are from the larger triangle."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "You can look at these triangles yourself and show, look, AD, point AD is between the blue angle and the right angle. Sorry, side AD is between the blue angle and the right angle. Side AC is between the blue angle and the right angle on the larger triangle. So both of these are from the larger triangle. These are the corresponding sides on the smaller triangle. And if that is confusing looking at them visually, as long as we wrote our similarity statement correctly, you can just find the corresponding points. AC corresponds to AB on the larger triangle."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "So both of these are from the larger triangle. These are the corresponding sides on the smaller triangle. And if that is confusing looking at them visually, as long as we wrote our similarity statement correctly, you can just find the corresponding points. AC corresponds to AB on the larger triangle. AD on the smaller triangle corresponds to AC on the larger triangle. And we know that AC, we can rewrite that as lowercase a. AC is lowercase a. We don't have any label for AD or for AB."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "AC corresponds to AB on the larger triangle. AD on the smaller triangle corresponds to AC on the larger triangle. And we know that AC, we can rewrite that as lowercase a. AC is lowercase a. We don't have any label for AD or for AB. Sorry, we do have a label for AB. That is C right over here. We don't have a label for AD."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "We don't have any label for AD or for AB. Sorry, we do have a label for AB. That is C right over here. We don't have a label for AD. So let's just call that lowercase d. So lowercase d applies to that part right over there. C applies to that entire part right over there. And then we'll call DB, let's call that length E. That will just make things a little bit simpler for us."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "We don't have a label for AD. So let's just call that lowercase d. So lowercase d applies to that part right over there. C applies to that entire part right over there. And then we'll call DB, let's call that length E. That will just make things a little bit simpler for us. So AD we'll just call d. And so we have a over c is equal to d over a. If we cross multiply, you have a times a, which is a squared, is equal to c times d, which is cd. So that's a little bit of an interesting result."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "And then we'll call DB, let's call that length E. That will just make things a little bit simpler for us. So AD we'll just call d. And so we have a over c is equal to d over a. If we cross multiply, you have a times a, which is a squared, is equal to c times d, which is cd. So that's a little bit of an interesting result. Let's see what we can do with the other triangle right over here. So this triangle right over here. So once again, it has a right angle."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "So that's a little bit of an interesting result. Let's see what we can do with the other triangle right over here. So this triangle right over here. So once again, it has a right angle. The larger one has a right angle. And they both share this angle right over here. So by angle-angle similarity, the two triangles are going to be similar."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "So once again, it has a right angle. The larger one has a right angle. And they both share this angle right over here. So by angle-angle similarity, the two triangles are going to be similar. So we can say triangle BDC. We went from pink to right to not labeled. So triangle BDC is similar to triangle."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "So by angle-angle similarity, the two triangles are going to be similar. So we can say triangle BDC. We went from pink to right to not labeled. So triangle BDC is similar to triangle. Now we're going to look at the larger triangle. We're going to start at the pink angle, B. Now we go to the right angle, CA."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "So triangle BDC is similar to triangle. Now we're going to look at the larger triangle. We're going to start at the pink angle, B. Now we go to the right angle, CA. From pink angle to right angle to non-labeled angle, at least from the point of view here. We labeled it before with that blue. So now let's set up some type of a relationship here."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "Now we go to the right angle, CA. From pink angle to right angle to non-labeled angle, at least from the point of view here. We labeled it before with that blue. So now let's set up some type of a relationship here. We can say that the ratio on the smaller triangle, BC over BA. Once again, we're taking the hypotenuses of both of them. So BC over BA is going to be equal to BD."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "So now let's set up some type of a relationship here. We can say that the ratio on the smaller triangle, BC over BA. Once again, we're taking the hypotenuses of both of them. So BC over BA is going to be equal to BD. Let me do this in another color. So this is one of the legs. The way I drew it is the shorter legs."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "So BC over BA is going to be equal to BD. Let me do this in another color. So this is one of the legs. The way I drew it is the shorter legs. BD over BC. I'm just taking the corresponding vertices. Over BC."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "The way I drew it is the shorter legs. BD over BC. I'm just taking the corresponding vertices. Over BC. And once again, we know BC is the same thing as lowercase b. BA is lowercase c. And then BD we defined as lowercase e. This is lowercase e. We can cross multiply here and we get B times B. I've mentioned this in many videos. Cross multiplying is really the same thing as multiplying both sides by both denominators. B times B is B squared is equal to CE."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "Over BC. And once again, we know BC is the same thing as lowercase b. BA is lowercase c. And then BD we defined as lowercase e. This is lowercase e. We can cross multiply here and we get B times B. I've mentioned this in many videos. Cross multiplying is really the same thing as multiplying both sides by both denominators. B times B is B squared is equal to CE. And now we can do something kind of interesting. We can add these two statements. Let me rewrite the statement down here."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "B times B is B squared is equal to CE. And now we can do something kind of interesting. We can add these two statements. Let me rewrite the statement down here. So B squared is equal to CE. So if we add the left-hand sides, we get A squared plus B squared is equal to CD plus CE. And then we have a C in both of these terms so we can factor it out."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "Let me rewrite the statement down here. So B squared is equal to CE. So if we add the left-hand sides, we get A squared plus B squared is equal to CD plus CE. And then we have a C in both of these terms so we can factor it out. This is going to be equal to C times D plus E. Close the parentheses. Now what is D plus E? D is this length."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "And then we have a C in both of these terms so we can factor it out. This is going to be equal to C times D plus E. Close the parentheses. Now what is D plus E? D is this length. E is this length. So D plus E is actually going to be C as well. So this is going to be C. So you have C times C which is just the same thing as C squared."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "D is this length. E is this length. So D plus E is actually going to be C as well. So this is going to be C. So you have C times C which is just the same thing as C squared. So now we have an interesting relationship. We have that A squared plus B squared is equal to C squared. Let me rewrite that."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "So this is going to be C. So you have C times C which is just the same thing as C squared. So now we have an interesting relationship. We have that A squared plus B squared is equal to C squared. Let me rewrite that. A squared. I'll do that in an arbitrary new color. I deleted that by accident."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "Let me rewrite that. A squared. I'll do that in an arbitrary new color. I deleted that by accident. So let me rewrite it. So we've just established that A squared plus B squared is equal to C squared. And this is just an arbitrary right triangle."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "I deleted that by accident. So let me rewrite it. So we've just established that A squared plus B squared is equal to C squared. And this is just an arbitrary right triangle. This is true for any two right triangles. We've just established that the sum of the squares of each of the legs is equal to the square of the hypotenuse. And this is probably what's easily one of the most famous theorems in mathematics named for Pythagoras."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy.mp3", "Sentence": "And this is just an arbitrary right triangle. This is true for any two right triangles. We've just established that the sum of the squares of each of the legs is equal to the square of the hypotenuse. And this is probably what's easily one of the most famous theorems in mathematics named for Pythagoras. Not clear if he's the first person to establish this, but it's called the Pythagorean theorem. And it's really the basis of, well, not all of geometry, but a lot of the geometry that we're going to do. And it forms the basis of a lot of the trigonometry we're going to do."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "And I thought we would use this to really just get some practice with line and angle proofs. And what's neat about this, this even uses translations and transformations as ways to actually prove things. So let's look at what they're telling us. So it says line AB and line DE are parallel lines. All right. Perform a translation that proves corresponding angles are always equal and select the option which explains the proof. All right."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "So it says line AB and line DE are parallel lines. All right. Perform a translation that proves corresponding angles are always equal and select the option which explains the proof. All right. So let's see what they have down here. So they say perform a translation that proves corresponding angles are always equal. And then select the option that explains the proof."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "All right. So let's see what they have down here. So they say perform a translation that proves corresponding angles are always equal. And then select the option that explains the proof. So they've picked two corresponding angles here. And so you see this is kind of the bottom left angle, this phi. And then you have theta right here, is the bottom left angle down here."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "And then select the option that explains the proof. So they've picked two corresponding angles here. And so you see this is kind of the bottom left angle, this phi. And then you have theta right here, is the bottom left angle down here. So these are corresponding angles. Line FB is a transversal. And they already told us that line AB, and what did they call it, did they call it DE?"}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "And then you have theta right here, is the bottom left angle down here. So these are corresponding angles. Line FB is a transversal. And they already told us that line AB, and what did they call it, did they call it DE? And line DE are indeed parallel lines. So we want to prove that these two things, that the measure of these two angles are equal. So there's many ways that you can do this geometrically."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "And they already told us that line AB, and what did they call it, did they call it DE? And line DE are indeed parallel lines. So we want to prove that these two things, that the measure of these two angles are equal. So there's many ways that you can do this geometrically. And we do that in many Khan Academy videos. But this one, they offer us the option of translating, of doing a translation. So let's see what that is."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "So there's many ways that you can do this geometrically. And we do that in many Khan Academy videos. But this one, they offer us the option of translating, of doing a translation. So let's see what that is. So I press the Translate button. And when I move this around, notice it essentially translates these four points, which has the effect of translating this entire intersection here. So this point where my mouse is right now, that is the point D. And I'm translating it around."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "So let's see what that is. So I press the Translate button. And when I move this around, notice it essentially translates these four points, which has the effect of translating this entire intersection here. So this point where my mouse is right now, that is the point D. And I'm translating it around. If I move that over to B, what it shows is, because under a translation, the angle measures shouldn't change. So when I did that, this angle, so it's down here. Theta is the measure of angle CDF."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "So this point where my mouse is right now, that is the point D. And I'm translating it around. If I move that over to B, what it shows is, because under a translation, the angle measures shouldn't change. So when I did that, this angle, so it's down here. Theta is the measure of angle CDF. And so when you move it over here, this right over here should be the same. This angle's measure is the same as CDF. I'm just translating it."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "Theta is the measure of angle CDF. And so when you move it over here, this right over here should be the same. This angle's measure is the same as CDF. I'm just translating it. And when you move it over here, you see, look, that's the same exact measure as phi. So this is one way to think about it. I just translated the point D to B."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "I'm just translating it. And when you move it over here, you see, look, that's the same exact measure as phi. So this is one way to think about it. I just translated the point D to B. And then it really just translated angle CDF over angle ABD to show that these have the same measure, or at least to feel good about the idea of them having the same measure. So let's see which choices describe that. So let me, I'm having trouble operating my mouse."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "I just translated the point D to B. And then it really just translated angle CDF over angle ABD to show that these have the same measure, or at least to feel good about the idea of them having the same measure. So let's see which choices describe that. So let me, I'm having trouble operating my mouse. All right. The translation mapping point F to point D. So point F to point D. We didn't map point F to point D. So this is already looking suspect. Produces a new line, which is a bisector of segment DB."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "So let me, I'm having trouble operating my mouse. All right. The translation mapping point F to point D. So point F to point D. We didn't map point F to point D. So this is already looking suspect. Produces a new line, which is a bisector of segment DB. OK, this doesn't seem anything like what I just did. So I'm just going to move on to the next one. Since the image of a line under translation is parallel to the original line, that's true."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "Produces a new line, which is a bisector of segment DB. OK, this doesn't seem anything like what I just did. So I'm just going to move on to the next one. Since the image of a line under translation is parallel to the original line, that's true. The translation that mapped point D to point B. That's what I did right over here. Maps angle CDF to ABD."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "Since the image of a line under translation is parallel to the original line, that's true. The translation that mapped point D to point B. That's what I did right over here. Maps angle CDF to ABD. And that's what I did. I mapped angle CDF to angle ABD. That's exactly what I did right over there."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "Maps angle CDF to ABD. And that's what I did. I mapped angle CDF to angle ABD. That's exactly what I did right over there. So this is to ABD. Translations preserve angle measures, so theta is equal to phi. Yeah, that looks pretty good."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "That's exactly what I did right over there. So this is to ABD. Translations preserve angle measures, so theta is equal to phi. Yeah, that looks pretty good. The translation that maps point D to E. I didn't do that. I didn't take point D and move it over to E like that. That didn't really help me."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "Yeah, that looks pretty good. The translation that maps point D to E. I didn't do that. I didn't take point D and move it over to E like that. That didn't really help me. Let's just keep reading it just to make sure. Produces a parallelogram. That actually is true."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "That didn't really help me. Let's just keep reading it just to make sure. Produces a parallelogram. That actually is true. If I translate point D to point E, I have this parallelogram constructed. But it really doesn't help us establishing that phi is equal to theta. So that one I also don't feel good about."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "That actually is true. If I translate point D to point E, I have this parallelogram constructed. But it really doesn't help us establishing that phi is equal to theta. So that one I also don't feel good about. And it's good, because we felt good about the middle choice. Let's do one more of these. So they're telling us that line AOB."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "So that one I also don't feel good about. And it's good, because we felt good about the middle choice. Let's do one more of these. So they're telling us that line AOB. And they could have just said line AB, but I guess they wanted to put the O in there to show that point O is on that line, that AOB are collinear. And COD is our straight lines. All right, fair enough."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "So they're telling us that line AOB. And they could have just said line AB, but I guess they wanted to put the O in there to show that point O is on that line, that AOB are collinear. And COD is our straight lines. All right, fair enough. Which of these statements prove vertical angles are always equal? So vertical angles would be the angles on the opposite sides of an intersection. So in order to prove that vertical, so for example, angle AOC and angle DOB are vertical angles."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "All right, fair enough. Which of these statements prove vertical angles are always equal? So vertical angles would be the angles on the opposite sides of an intersection. So in order to prove that vertical, so for example, angle AOC and angle DOB are vertical angles. And if we wanted to prove that they are equal, we would say, well, their measures are going to be equal. So theta should be equal to phi. So let's see which of these statements actually does that."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "So in order to prove that vertical, so for example, angle AOC and angle DOB are vertical angles. And if we wanted to prove that they are equal, we would say, well, their measures are going to be equal. So theta should be equal to phi. So let's see which of these statements actually does that. So this one says segment OA is congruent to OD. We don't know that. They never even told us that."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "So let's see which of these statements actually does that. So this one says segment OA is congruent to OD. We don't know that. They never even told us that. So I don't even have to read the rest of it. This is already saying, I don't know how far D is away from O. I don't know if it's the same distance as A is from O. So we can just rule this first choice out."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "They never even told us that. So I don't even have to read the rest of it. This is already saying, I don't know how far D is away from O. I don't know if it's the same distance as A is from O. So we can just rule this first choice out. I can just stop reading. They started with a statement that we don't know based on the information they gave us. So let's look at the second choice."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "So we can just rule this first choice out. I can just stop reading. They started with a statement that we don't know based on the information they gave us. So let's look at the second choice. If ray OA and ray OC are each rotated 180 degrees about point O, they must map to OB and OD respectively. If two rays are rotated by the same amount, the angle between them will not change. So phi must be equal to theta."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "So let's look at the second choice. If ray OA and ray OC are each rotated 180 degrees about point O, they must map to OB and OD respectively. If two rays are rotated by the same amount, the angle between them will not change. So phi must be equal to theta. So this is interesting. So let's just slow down and think about what they're saying. If ray OA and OC are each rotated 180 degrees, so if you take ray OA, this ray right over here, if you rotate it 180 degrees, it's going to go all the way around and point in the other direction."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "So phi must be equal to theta. So this is interesting. So let's just slow down and think about what they're saying. If ray OA and OC are each rotated 180 degrees, so if you take ray OA, this ray right over here, if you rotate it 180 degrees, it's going to go all the way around and point in the other direction. So it's going to map to ray OB. So I definitely believe that. OA is going to map to ray OB."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "If ray OA and OC are each rotated 180 degrees, so if you take ray OA, this ray right over here, if you rotate it 180 degrees, it's going to go all the way around and point in the other direction. So it's going to map to ray OB. So I definitely believe that. OA is going to map to ray OB. And OC, ray OC, if you rotate 180 degrees, is going to map to ray OD. And so this first statement is true. If ray OA and ray OC are each rotated 180 degrees about point O, they must map to ray OB and OD respectively."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "OA is going to map to ray OB. And OC, ray OC, if you rotate 180 degrees, is going to map to ray OD. And so this first statement is true. If ray OA and ray OC are each rotated 180 degrees about point O, they must map to ray OB and OD respectively. And when people say respectively, they're saying in the same order, that ray OA maps to ray OB and that ray OC maps to ray OD. And we saw that. Ray OA maps, if you rotate it all the way around 180 degrees, it'll map to OB."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "If ray OA and ray OC are each rotated 180 degrees about point O, they must map to ray OB and OD respectively. And when people say respectively, they're saying in the same order, that ray OA maps to ray OB and that ray OC maps to ray OD. And we saw that. Ray OA maps, if you rotate it all the way around 180 degrees, it'll map to OB. And OC, if you rotate 180 degrees, will map to OD. So I'm feeling good about that first sentence. If two rays are rotated by the same amount, the angle between them will not change."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "Ray OA maps, if you rotate it all the way around 180 degrees, it'll map to OB. And OC, if you rotate 180 degrees, will map to OD. So I'm feeling good about that first sentence. If two rays are rotated by the same amount, the angle between them will not change. Especially if they are rotated around, yeah, I'll go with that. If two rays are rotated by the same amount, the angle between them will not change. So if we rotate both of these rays by 180 degrees, then we've essentially mapped to OB and OD."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "If two rays are rotated by the same amount, the angle between them will not change. Especially if they are rotated around, yeah, I'll go with that. If two rays are rotated by the same amount, the angle between them will not change. So if we rotate both of these rays by 180 degrees, then we've essentially mapped to OB and OD. Or another way to think about it, this angle, angle AOC, is going to map to angle BOD. And so the measure of those angles are going to be the same. So phi must be equal to theta."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "So if we rotate both of these rays by 180 degrees, then we've essentially mapped to OB and OD. Or another way to think about it, this angle, angle AOC, is going to map to angle BOD. And so the measure of those angles are going to be the same. So phi must be equal to theta. So I actually like this second statement a lot. So let's see this last statement. Rotations preserve lengths and angles."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "So phi must be equal to theta. So I actually like this second statement a lot. So let's see this last statement. Rotations preserve lengths and angles. AB is congruent to CD. Actually, we don't know whether segment AB is congruent to CD. They never told us that."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "Rotations preserve lengths and angles. AB is congruent to CD. Actually, we don't know whether segment AB is congruent to CD. They never told us that. We don't know how far apart these things are. So we know that phi is equal to theta. So this statement right over here is just suspect."}, {"video_title": "Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3", "Sentence": "They never told us that. We don't know how far apart these things are. So we know that phi is equal to theta. So this statement right over here is just suspect. And so actually, I don't like that one. So I'm going to go with the first one, which is it takes a little bit of visualization going on. But if you took angle AOC and you rotated it 180 degrees, which means take the corresponding rays, or the rays that make it up, and rotate them 180 degrees, you get to angle BOD."}, {"video_title": "Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3", "Sentence": "So we have this larger triangle here, and inside of that we have these other triangles. And we're given this information right over here, that triangle BCD is congruent to triangle BCA, which is congruent to triangle ECD. And given just this information, what I want to do in this drawing, I want to figure out what every angle on this drawing is. What's the measure of every angle? So let's see what we can do here. So let's just start with the information that they've actually given us. So we know that triangle BCD is congruent to, well, we know all of these three triangles are congruent to each other."}, {"video_title": "Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3", "Sentence": "What's the measure of every angle? So let's see what we can do here. So let's just start with the information that they've actually given us. So we know that triangle BCD is congruent to, well, we know all of these three triangles are congruent to each other. So for example, BCD is congruent to ECD. And so their corresponding sides and corresponding angles will also be congruent. So just looking at the order in which they're written, vertex B corresponds in this triangle, in BCD, corresponds to vertex B in BCA, so this is the B vertex in BCA, which corresponds to the E vertex in ECD."}, {"video_title": "Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3", "Sentence": "So we know that triangle BCD is congruent to, well, we know all of these three triangles are congruent to each other. So for example, BCD is congruent to ECD. And so their corresponding sides and corresponding angles will also be congruent. So just looking at the order in which they're written, vertex B corresponds in this triangle, in BCD, corresponds to vertex B in BCA, so this is the B vertex in BCA, which corresponds to the E vertex in ECD. So everything that I've done in magenta, all of these angles are congruent. And then we also know that the C angle, so in BCA, sorry, BCD, this angle, this angle right over here, is congruent to the C angle in BCA. BCA, the C angle is right over here, or C is the vertex for that angle in BCA."}, {"video_title": "Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3", "Sentence": "So just looking at the order in which they're written, vertex B corresponds in this triangle, in BCD, corresponds to vertex B in BCA, so this is the B vertex in BCA, which corresponds to the E vertex in ECD. So everything that I've done in magenta, all of these angles are congruent. And then we also know that the C angle, so in BCA, sorry, BCD, this angle, this angle right over here, is congruent to the C angle in BCA. BCA, the C angle is right over here, or C is the vertex for that angle in BCA. And that is also the C angle, I guess we could call it, in ECD. But in ECD, we're talking about this angle right over here. So these three angles are going to be congruent."}, {"video_title": "Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3", "Sentence": "BCA, the C angle is right over here, or C is the vertex for that angle in BCA. And that is also the C angle, I guess we could call it, in ECD. But in ECD, we're talking about this angle right over here. So these three angles are going to be congruent. And I think you could already guess a way to come up with the values of those three angles. But let's just keep looking at everything else that they're telling us. Finally, we have vertex D over here."}, {"video_title": "Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3", "Sentence": "So these three angles are going to be congruent. And I think you could already guess a way to come up with the values of those three angles. But let's just keep looking at everything else that they're telling us. Finally, we have vertex D over here. So angle, so this is the last one where we listed B. So in triangle BCD, this angle right over here corresponds to the A vertex angle in BCA. So BCA, that's going to correspond to this angle right over here."}, {"video_title": "Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3", "Sentence": "Finally, we have vertex D over here. So angle, so this is the last one where we listed B. So in triangle BCD, this angle right over here corresponds to the A vertex angle in BCA. So BCA, that's going to correspond to this angle right over here. It's really the only one that we haven't labeled yet. And that corresponds to this angle, this vertex right over here, that angle right over there. And just to make it consistent, this C should also be circled in yellow."}, {"video_title": "Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3", "Sentence": "So BCA, that's going to correspond to this angle right over here. It's really the only one that we haven't labeled yet. And that corresponds to this angle, this vertex right over here, that angle right over there. And just to make it consistent, this C should also be circled in yellow. And so we have all these congruencies, and now we can come up with some interesting things about them. First of all, here, angle BCA, angle BCD, and angle DCE, they're all congruent. And when you add them up together, you get to 180 degrees."}, {"video_title": "Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3", "Sentence": "And just to make it consistent, this C should also be circled in yellow. And so we have all these congruencies, and now we can come up with some interesting things about them. First of all, here, angle BCA, angle BCD, and angle DCE, they're all congruent. And when you add them up together, you get to 180 degrees. If you put them all adjacent, as they all are right here, they end up with a straight angle if you look at their outer sides. So if these are each x, you have three of them added together have to be 180 degrees, which tells us that each of these have to be 60 degrees. That's the only way you have three of the same thing adding up to 180 degrees."}, {"video_title": "Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3", "Sentence": "And when you add them up together, you get to 180 degrees. If you put them all adjacent, as they all are right here, they end up with a straight angle if you look at their outer sides. So if these are each x, you have three of them added together have to be 180 degrees, which tells us that each of these have to be 60 degrees. That's the only way you have three of the same thing adding up to 180 degrees. Fair enough. What else can we do? Well, we have these two characters up here."}, {"video_title": "Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3", "Sentence": "That's the only way you have three of the same thing adding up to 180 degrees. Fair enough. What else can we do? Well, we have these two characters up here. They are both equal, and they add up to 180 degrees. They are supplementary. And the only way you can have two equal things that add up to 180 is if they're both 90 degrees."}, {"video_title": "Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3", "Sentence": "Well, we have these two characters up here. They are both equal, and they add up to 180 degrees. They are supplementary. And the only way you can have two equal things that add up to 180 is if they're both 90 degrees. So these two characters are both 90 degrees, or we could say this is a right angle, that's a right angle. And this is congruent to both of those, so that is also 90 degrees. And then we're left with these magenta parts of the angle."}, {"video_title": "Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3", "Sentence": "And the only way you can have two equal things that add up to 180 is if they're both 90 degrees. So these two characters are both 90 degrees, or we could say this is a right angle, that's a right angle. And this is congruent to both of those, so that is also 90 degrees. And then we're left with these magenta parts of the angle. And here we can just say, well, 90 plus 60 plus something is going to add up to 180. 90 plus 60 is 150, so this has to be 30 degrees to add up to 180. And if that's 30 degrees, then this is 30 degrees."}, {"video_title": "Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3", "Sentence": "And then we're left with these magenta parts of the angle. And here we can just say, well, 90 plus 60 plus something is going to add up to 180. 90 plus 60 is 150, so this has to be 30 degrees to add up to 180. And if that's 30 degrees, then this is 30 degrees. And then this thing right over here is 30 degrees. And then the last thing, we've actually done what we said we would do. We found out all of the angles."}, {"video_title": "Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3", "Sentence": "And if that's 30 degrees, then this is 30 degrees. And then this thing right over here is 30 degrees. And then the last thing, we've actually done what we said we would do. We found out all of the angles. We can also think about these outer angles. Or not the outer angles, but these combined angles. So angle, say, angle ABE."}, {"video_title": "Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3", "Sentence": "We found out all of the angles. We can also think about these outer angles. Or not the outer angles, but these combined angles. So angle, say, angle ABE. So this whole angle we see is 60 degrees. This angle is 90 degrees. And this angle here is 30."}, {"video_title": "Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3", "Sentence": "So angle, say, angle ABE. So this whole angle we see is 60 degrees. This angle is 90 degrees. And this angle here is 30. So what's interesting is these three smaller triangles, they all have the exact same angles, 30, 60, 90, and the exact same side lengths. And we know that because they're congruent. And what's interesting is when you put them together this way, they construct this larger triangle, triangle ABE, that's clearly not congruent."}, {"video_title": "Figuring out all the angles for congruent triangles example Congruence Geometry Khan Academy.mp3", "Sentence": "And this angle here is 30. So what's interesting is these three smaller triangles, they all have the exact same angles, 30, 60, 90, and the exact same side lengths. And we know that because they're congruent. And what's interesting is when you put them together this way, they construct this larger triangle, triangle ABE, that's clearly not congruent. It's a larger triangle. It has different measures for its length. But it has the same angles, 30, 60, and then 90."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "I know what you're thinking, what's all of the silliness on the right-hand side? This is actually just the view we use when we're trying to debug things on Khan Academy, but we can still do the exercise. So it says drag the center point and perimeter of the circle to graph the equation. So the first thing we want to think about is, well, what's the center of this equation? Well, a standard form of a circle is x minus the x-coordinate of the center squared plus y minus the y-coordinate of the center squared is equal to the radius squared. So x minus the x-coordinate of the center, so the x-coordinate of the center must be negative five. Cause the way we can get a positive 5 here is by subtracting a negative 5."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "So the first thing we want to think about is, well, what's the center of this equation? Well, a standard form of a circle is x minus the x-coordinate of the center squared plus y minus the y-coordinate of the center squared is equal to the radius squared. So x minus the x-coordinate of the center, so the x-coordinate of the center must be negative five. Cause the way we can get a positive 5 here is by subtracting a negative 5. So the x-coordinate must be negative 5 and the y-coordinate must be positive 5 cause y minus the y-coordinate of the center. So y-coordinate is positive five. And then the radius squared is going to be equal to four."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "Cause the way we can get a positive 5 here is by subtracting a negative 5. So the x-coordinate must be negative 5 and the y-coordinate must be positive 5 cause y minus the y-coordinate of the center. So y-coordinate is positive five. And then the radius squared is going to be equal to four. So that means that the radius is equal to two. And the way it's drawn right now, I mean we could drag this out like this, but this, the way it's drawn, the radius is indeed equal to two. And so we are done."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "And then the radius squared is going to be equal to four. So that means that the radius is equal to two. And the way it's drawn right now, I mean we could drag this out like this, but this, the way it's drawn, the radius is indeed equal to two. And so we are done. And I really want to hit the point home of what I just did. So let me get my little, get my scratch pad out. So this, sorry for knocking the microphone just now."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "And so we are done. And I really want to hit the point home of what I just did. So let me get my little, get my scratch pad out. So this, sorry for knocking the microphone just now. So that equation was x plus five squared plus y minus five squared is equal to four squared. And so I want to rewrite this as, this is x minus negative five, x minus negative five squared plus y minus positive five, positive five squared is equal to, and instead of writing it as four, I'll write it as two squared. And so this right over here tells us that the center of the circle is going to be, x equals negative five, y equals five, and the radius is going to be equal to two."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "So this, sorry for knocking the microphone just now. So that equation was x plus five squared plus y minus five squared is equal to four squared. And so I want to rewrite this as, this is x minus negative five, x minus negative five squared plus y minus positive five, positive five squared is equal to, and instead of writing it as four, I'll write it as two squared. And so this right over here tells us that the center of the circle is going to be, x equals negative five, y equals five, and the radius is going to be equal to two. And once again, this is no magic here. This is not just, I don't want you to just memorize this formula. I want you to appreciate that this formula comes straight out of the Pythagorean theorem, straight out of the distance formula, which comes out of the Pythagorean theorem."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "And so this right over here tells us that the center of the circle is going to be, x equals negative five, y equals five, and the radius is going to be equal to two. And once again, this is no magic here. This is not just, I don't want you to just memorize this formula. I want you to appreciate that this formula comes straight out of the Pythagorean theorem, straight out of the distance formula, which comes out of the Pythagorean theorem. Remember, if you have some center, in this case it's the point negative five comma five. So negative five comma five. And you want to find all of the x's and y's that are two away from it."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "I want you to appreciate that this formula comes straight out of the Pythagorean theorem, straight out of the distance formula, which comes out of the Pythagorean theorem. Remember, if you have some center, in this case it's the point negative five comma five. So negative five comma five. And you want to find all of the x's and y's that are two away from it. So you want to find all the x's and y's that are two away from it. So that would be one of them. X comma y."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "And you want to find all of the x's and y's that are two away from it. So you want to find all the x's and y's that are two away from it. So that would be one of them. X comma y. This distance is two. And there's going to be a bunch of them. And when you plot all of them together, you're going to get a circle with a radius two around that center."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "X comma y. This distance is two. And there's going to be a bunch of them. And when you plot all of them together, you're going to get a circle with a radius two around that center. But let's think about how we got that actual formula. Well, the distance between that coordinate, between any of these x's and y's, it could be an x and y here, it could be an x and y here, and this is going to be two. So we could have our change in x."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "And when you plot all of them together, you're going to get a circle with a radius two around that center. But let's think about how we got that actual formula. Well, the distance between that coordinate, between any of these x's and y's, it could be an x and y here, it could be an x and y here, and this is going to be two. So we could have our change in x. So we have x minus negative five. So that's our change in x between any point x comma y and negative five comma five. So our change in x squared plus our change in y squared, so that's going to be y minus the y coordinate over here, squared, is going to be equal to the radius squared."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "So we could have our change in x. So we have x minus negative five. So that's our change in x between any point x comma y and negative five comma five. So our change in x squared plus our change in y squared, so that's going to be y minus the y coordinate over here, squared, is going to be equal to the radius squared. So the change in y, it's going to be from this y to that y. If this is the end point, it'd be the end minus the beginning, y minus five. Y minus five squared."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "So our change in x squared plus our change in y squared, so that's going to be y minus the y coordinate over here, squared, is going to be equal to the radius squared. So the change in y, it's going to be from this y to that y. If this is the end point, it'd be the end minus the beginning, y minus five. Y minus five squared. And so this shows for any xy that is two away from the center, this equation will hold. And it becomes, I'll just write this in neutral color, x plus five squared plus y minus five squared is equal to the radius squared, is equal to two, or let me just write that, is equal to four. And let me make it, I really want to, you know, I dislike it when formulas are just memorized and you don't see the connection to other things."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "Y minus five squared. And so this shows for any xy that is two away from the center, this equation will hold. And it becomes, I'll just write this in neutral color, x plus five squared plus y minus five squared is equal to the radius squared, is equal to two, or let me just write that, is equal to four. And let me make it, I really want to, you know, I dislike it when formulas are just memorized and you don't see the connection to other things. Notice, we can construct, we can construct a nice little right triangle here. So our change in x is that right over there. So that is our change in x, change in x."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "And let me make it, I really want to, you know, I dislike it when formulas are just memorized and you don't see the connection to other things. Notice, we can construct, we can construct a nice little right triangle here. So our change in x is that right over there. So that is our change in x, change in x. And our change in y, our change in y, not the change in y squared, but our change in y is that right over there. Change in y, our change in y, you could view that as y minus, so this is change in y is going to be y minus five. And our change in x is x minus negative five."}, {"video_title": "Graphing a circle from its standard equation Mathematics II High School Math Khan Academy.mp3", "Sentence": "So that is our change in x, change in x. And our change in y, our change in y, not the change in y squared, but our change in y is that right over there. Change in y, our change in y, you could view that as y minus, so this is change in y is going to be y minus five. And our change in x is x minus negative five. X minus negative five. So this is just change in x squared plus change in y squared is equal to the hypotenuse squared, which is the length of, which is this radius. So once again, comes straight out of the Pythagorean theorem."}, {"video_title": "30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And I think you know why they're called this. The measures of its angles are 30 degrees, 60 degrees, and 90 degrees. And what we're going to prove in this video, and this tends to be a very useful result, at least for a lot of what you see in a geometry class and then later on in a trigonometry class, is the ratios between the sides of a 30-60-90 triangle. That if the hypotenuse has length x, so remember the hypotenuse is opposite the 90-degree side, if the hypotenuse has length x, what we're going to prove is that the shortest side, which is opposite the 30-degree side, has length x over 2, and that the 60-degree side, or the side that's opposite the 60-degree angle, I should say, is going to be square root of 3 times the shortest side. So square root of 3 times x over 2. That's going to be its length. So that's what we're going to prove in this video."}, {"video_title": "30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "That if the hypotenuse has length x, so remember the hypotenuse is opposite the 90-degree side, if the hypotenuse has length x, what we're going to prove is that the shortest side, which is opposite the 30-degree side, has length x over 2, and that the 60-degree side, or the side that's opposite the 60-degree angle, I should say, is going to be square root of 3 times the shortest side. So square root of 3 times x over 2. That's going to be its length. So that's what we're going to prove in this video. And then in other videos we're going to apply this. We're going to show that this is actually a pretty useful result. Now let's start with a triangle that we're very familiar with."}, {"video_title": "30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So that's what we're going to prove in this video. And then in other videos we're going to apply this. We're going to show that this is actually a pretty useful result. Now let's start with a triangle that we're very familiar with. So let me draw ourselves an equilateral triangle. So drawing the triangles is always the hard part. So this is my best shot at an equilateral triangle."}, {"video_title": "30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "Now let's start with a triangle that we're very familiar with. So let me draw ourselves an equilateral triangle. So drawing the triangles is always the hard part. So this is my best shot at an equilateral triangle. And let's call this ABC. I'm just going to assume that I've constructed an equilateral triangle. So triangle ABC is equilateral."}, {"video_title": "30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So this is my best shot at an equilateral triangle. And let's call this ABC. I'm just going to assume that I've constructed an equilateral triangle. So triangle ABC is equilateral. And if it's equilateral, that means all of its sides are equal. And let's say that it has length, and let's say equilateral with sides of length x. So this is going to be x, this is going to be x, and this is going to be x."}, {"video_title": "30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So triangle ABC is equilateral. And if it's equilateral, that means all of its sides are equal. And let's say that it has length, and let's say equilateral with sides of length x. So this is going to be x, this is going to be x, and this is going to be x. We also know, based on what we've seen from equilateral triangles before, that the measures of all of these angles are going to be 60 degrees. So this is going to be 60 degrees, this is going to be 60 degrees, and then this is going to be 60 degrees. Now what I'm going to do is I'm going to drop an altitude from this top point right over here."}, {"video_title": "30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So this is going to be x, this is going to be x, and this is going to be x. We also know, based on what we've seen from equilateral triangles before, that the measures of all of these angles are going to be 60 degrees. So this is going to be 60 degrees, this is going to be 60 degrees, and then this is going to be 60 degrees. Now what I'm going to do is I'm going to drop an altitude from this top point right over here. So I'm going to drop an altitude right down, and by definition when I'm constructing an altitude, it's going to intersect the base right here at a right angle. So that's going to be a right angle, and then this is going to be a right angle. And it's a pretty straightforward proof to show that this is not only an altitude, not only is it perpendicular to this base, but it's a pretty straightforward proof to show that it bisects the base."}, {"video_title": "30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "Now what I'm going to do is I'm going to drop an altitude from this top point right over here. So I'm going to drop an altitude right down, and by definition when I'm constructing an altitude, it's going to intersect the base right here at a right angle. So that's going to be a right angle, and then this is going to be a right angle. And it's a pretty straightforward proof to show that this is not only an altitude, not only is it perpendicular to this base, but it's a pretty straightforward proof to show that it bisects the base. And you could pause it if you like and prove it yourself. But it really comes out of the fact that it's easy to prove that these two triangles are congruent. So let me prove it for you."}, {"video_title": "30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And it's a pretty straightforward proof to show that this is not only an altitude, not only is it perpendicular to this base, but it's a pretty straightforward proof to show that it bisects the base. And you could pause it if you like and prove it yourself. But it really comes out of the fact that it's easy to prove that these two triangles are congruent. So let me prove it for you. So let's call this point D right over here. So triangles ABD and BDC, they clearly both share this side. So this side is common to both of them right over here."}, {"video_title": "30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So let me prove it for you. So let's call this point D right over here. So triangles ABD and BDC, they clearly both share this side. So this side is common to both of them right over here. And then we could do, we have this angle right over here is congruent to this angle over there. This angle right over here is congruent to this angle over here. And so if these two are congruent to each other, then the third angle has to be congruent to each other."}, {"video_title": "30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So this side is common to both of them right over here. And then we could do, we have this angle right over here is congruent to this angle over there. This angle right over here is congruent to this angle over here. And so if these two are congruent to each other, then the third angle has to be congruent to each other. So this angle right over here needs to be congruent to that angle right over there. So these two are congruent. You can use actually a variety of our congruence postulates."}, {"video_title": "30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And so if these two are congruent to each other, then the third angle has to be congruent to each other. So this angle right over here needs to be congruent to that angle right over there. So these two are congruent. You can use actually a variety of our congruence postulates. We could say side, angle, side, side, angle, side congruence. We could use angle, side, angle, any of those to show that triangle A, B, D is congruent to triangle C, B, D. And what that does for us, and we could use, as I said, we could use angle, side, angle, or side, angle, side, whatever we like to use for this. What that does for us is it tells us that the corresponding sides of these triangles are going to be equal."}, {"video_title": "30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "You can use actually a variety of our congruence postulates. We could say side, angle, side, side, angle, side congruence. We could use angle, side, angle, any of those to show that triangle A, B, D is congruent to triangle C, B, D. And what that does for us, and we could use, as I said, we could use angle, side, angle, or side, angle, side, whatever we like to use for this. What that does for us is it tells us that the corresponding sides of these triangles are going to be equal. In particular, the length of, in particular, A, D is going to be equal to C, D. These are corresponding sides. So these are going to be equal to each other. And if we know that they're equal to each other and they add up to x, remember this was an equilateral triangle of length x, we know that this side right over here is going to be x over 2."}, {"video_title": "30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "What that does for us is it tells us that the corresponding sides of these triangles are going to be equal. In particular, the length of, in particular, A, D is going to be equal to C, D. These are corresponding sides. So these are going to be equal to each other. And if we know that they're equal to each other and they add up to x, remember this was an equilateral triangle of length x, we know that this side right over here is going to be x over 2. We know this is going to be x over 2. Not only do we know that, but we also knew when we dropped this altitude, we showed that this angle has to be congruent to that angle, and their measures have to add up to 60. So if two things are the same and they add up to 60, this is going to be at 30 degrees, and this is going to be 30 degrees."}, {"video_title": "30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And if we know that they're equal to each other and they add up to x, remember this was an equilateral triangle of length x, we know that this side right over here is going to be x over 2. We know this is going to be x over 2. Not only do we know that, but we also knew when we dropped this altitude, we showed that this angle has to be congruent to that angle, and their measures have to add up to 60. So if two things are the same and they add up to 60, this is going to be at 30 degrees, and this is going to be 30 degrees. So we've already shown one of the interesting parts of a 30-60-90 triangle that if the hypotenuse, notice, and I guess I didn't point this out, by dropping this altitude, but I've essentially split this equilateral triangle into two 30-60-90 triangles. And so we've already shown that if the side opposite the 90 degree side is x, that the side opposite the 30 degree side is going to be x over 2. That's what we showed right over here."}, {"video_title": "30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So if two things are the same and they add up to 60, this is going to be at 30 degrees, and this is going to be 30 degrees. So we've already shown one of the interesting parts of a 30-60-90 triangle that if the hypotenuse, notice, and I guess I didn't point this out, by dropping this altitude, but I've essentially split this equilateral triangle into two 30-60-90 triangles. And so we've already shown that if the side opposite the 90 degree side is x, that the side opposite the 30 degree side is going to be x over 2. That's what we showed right over here. Now we just have to come up with the third side, the side that is opposite the 60 degree side right over there. And let's call that length, well, I'll just use the letters that we already have here. This is BD."}, {"video_title": "30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "That's what we showed right over here. Now we just have to come up with the third side, the side that is opposite the 60 degree side right over there. And let's call that length, well, I'll just use the letters that we already have here. This is BD. And we can just use the Pythagorean theorem right here. BD squared plus this length right over here squared plus x over 2 squared is going to be equal to the hypotenuse squared. So we get BD squared plus x over 2 squared."}, {"video_title": "30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "This is BD. And we can just use the Pythagorean theorem right here. BD squared plus this length right over here squared plus x over 2 squared is going to be equal to the hypotenuse squared. So we get BD squared plus x over 2 squared. This is just straight out of the Pythagorean theorem. Plus x over 2 squared is going to equal this hypotenuse squared is going to equal x squared. And just to be clear, I'm looking at this triangle right here."}, {"video_title": "30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So we get BD squared plus x over 2 squared. This is just straight out of the Pythagorean theorem. Plus x over 2 squared is going to equal this hypotenuse squared is going to equal x squared. And just to be clear, I'm looking at this triangle right here. I'm looking at this triangle right over here on the right. And I'm just applying the Pythagorean theorem. This side squared plus this side squared is going to equal the hypotenuse squared."}, {"video_title": "30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And just to be clear, I'm looking at this triangle right here. I'm looking at this triangle right over here on the right. And I'm just applying the Pythagorean theorem. This side squared plus this side squared is going to equal the hypotenuse squared. And let's solve now for BD. You get BD squared plus x squared over 4 is equal to x squared. You could view this as 4x squared over 4."}, {"video_title": "30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "This side squared plus this side squared is going to equal the hypotenuse squared. And let's solve now for BD. You get BD squared plus x squared over 4 is equal to x squared. You could view this as 4x squared over 4. That's the same thing, obviously, as x squared. And then if you subtract 1 fourth x squared from both sides or x squared over 4 from both sides, you get BD squared is equal to 4x squared over 4 minus x squared over 4 is going to be 3x squared over 4. So we're just going to be 3x squared over 4."}, {"video_title": "30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "You could view this as 4x squared over 4. That's the same thing, obviously, as x squared. And then if you subtract 1 fourth x squared from both sides or x squared over 4 from both sides, you get BD squared is equal to 4x squared over 4 minus x squared over 4 is going to be 3x squared over 4. So we're just going to be 3x squared over 4. Take the principal root of both sides. You get BD is equal to the square root of 3 times x. Principal root of 3 is square root of 3."}, {"video_title": "30-60-90 triangle side ratios proof Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So we're just going to be 3x squared over 4. Take the principal root of both sides. You get BD is equal to the square root of 3 times x. Principal root of 3 is square root of 3. Principal root of x squared is just x over the principal root of 4, which is 2. And BD is the side opposite the 60 degree side. So we're done."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Point A is at negative 5 comma 5. So this is negative 5 right over here. This is 1, 2, 3, 4, 5. That's 5 right over there. So point A is right about there. So that is point A, just like that, at negative 5 comma 5. And then it's the center of circle A, which I won't draw just yet because I don't know the radius of circle A."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "That's 5 right over there. So point A is right about there. So that is point A, just like that, at negative 5 comma 5. And then it's the center of circle A, which I won't draw just yet because I don't know the radius of circle A. Point B is at, let me underline these in the appropriate color, point B is at 3 comma 1. So 1, 2, 3 comma 1. So that's point B right over there."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And then it's the center of circle A, which I won't draw just yet because I don't know the radius of circle A. Point B is at, let me underline these in the appropriate color, point B is at 3 comma 1. So 1, 2, 3 comma 1. So that's point B right over there. It's the center of circle B. Point P is at 0, 0. So it's right over there at the origin."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So that's point B right over there. It's the center of circle B. Point P is at 0, 0. So it's right over there at the origin. And it is on circles A and B. Well, that's a big piece of information because that tells us if this is on both circles, then that means that this is B's radius away from the point B, from the center. And this tells us that it is circle A's radius away from its center, which is at point A."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So it's right over there at the origin. And it is on circles A and B. Well, that's a big piece of information because that tells us if this is on both circles, then that means that this is B's radius away from the point B, from the center. And this tells us that it is circle A's radius away from its center, which is at point A. So let's figure out what those radii actually are. And so we can imagine, let me draw a radius, or the radius for circle A. We now know since P sits on it that this could be considered the radius for circle A."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And this tells us that it is circle A's radius away from its center, which is at point A. So let's figure out what those radii actually are. And so we can imagine, let me draw a radius, or the radius for circle A. We now know since P sits on it that this could be considered the radius for circle A. And you could use a distance formula. But what we'll see is that the distance formula is really just falling out of the Pythagorean theorem. So the distance formula tells us the radius right over here, this is just the distance between those two points."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "We now know since P sits on it that this could be considered the radius for circle A. And you could use a distance formula. But what we'll see is that the distance formula is really just falling out of the Pythagorean theorem. So the distance formula tells us the radius right over here, this is just the distance between those two points. So the radius, or the distance between those two points squared, is going to be equal to our change in x values between A and P. So our change in x values, we could write it as negative 5 minus 0 squared. Negative 5 minus 0 squared, that's our change in x. Negative 5 minus 0 squared plus our change in y."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So the distance formula tells us the radius right over here, this is just the distance between those two points. So the radius, or the distance between those two points squared, is going to be equal to our change in x values between A and P. So our change in x values, we could write it as negative 5 minus 0 squared. Negative 5 minus 0 squared, that's our change in x. Negative 5 minus 0 squared plus our change in y. 5 minus 0 squared, which gets us that our distance between these two points, which is the length of the radius, squared, is equal to negative 5 squared plus 5 squared. Or we could say that the radius is equal to the square root of, this is 25, this is 25, 50. 50 we can write as 25 times 2."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Negative 5 minus 0 squared plus our change in y. 5 minus 0 squared, which gets us that our distance between these two points, which is the length of the radius, squared, is equal to negative 5 squared plus 5 squared. Or we could say that the radius is equal to the square root of, this is 25, this is 25, 50. 50 we can write as 25 times 2. So this is equal to the square root of 25 times the square root of 2, which is 5 times the square root of 2. So this distance right over here is 5 times the square root of 2. Now I said this is just the same thing as the Pythagorean theorem."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "50 we can write as 25 times 2. So this is equal to the square root of 25 times the square root of 2, which is 5 times the square root of 2. So this distance right over here is 5 times the square root of 2. Now I said this is just the same thing as the Pythagorean theorem. Why? Well, if we were to construct a right triangle right over here, then we can look at this distance. This distance would be the absolute value of negative 5 minus 0."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Now I said this is just the same thing as the Pythagorean theorem. Why? Well, if we were to construct a right triangle right over here, then we can look at this distance. This distance would be the absolute value of negative 5 minus 0. Or you could say it's 0 minus negative 5. This distance right over here is 5. This distance is the distance between 0 and 5 in the y direction."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "This distance would be the absolute value of negative 5 minus 0. Or you could say it's 0 minus negative 5. This distance right over here is 5. This distance is the distance between 0 and 5 in the y direction. That's 5. Pythagorean theorem tells us that 5 squared, which is 25, plus 5 squared, another 25, is going to be equal to your hypotenuse squared. And that's exactly what we have here."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "This distance is the distance between 0 and 5 in the y direction. That's 5. Pythagorean theorem tells us that 5 squared, which is 25, plus 5 squared, another 25, is going to be equal to your hypotenuse squared. And that's exactly what we have here. Now you might be saying, wait, wait, wait. This thing had a negative 5 squared here, while here you had a positive 5. But the reason why we could do this is when you square it, the negative disappears."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And that's exactly what we have here. Now you might be saying, wait, wait, wait. This thing had a negative 5 squared here, while here you had a positive 5. But the reason why we could do this is when you square it, the negative disappears. The distance formula, you could write it this way, where you're taking the absolute value. And then it becomes very clear that this really is just the Pythagorean theorem. This would be 5 squared plus 5 squared."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "But the reason why we could do this is when you square it, the negative disappears. The distance formula, you could write it this way, where you're taking the absolute value. And then it becomes very clear that this really is just the Pythagorean theorem. This would be 5 squared plus 5 squared. 5 squared plus 5 squared. The reason why you don't have to do this is because a sign doesn't matter when you square it. It will become a positive value."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "This would be 5 squared plus 5 squared. 5 squared plus 5 squared. The reason why you don't have to do this is because a sign doesn't matter when you square it. It will become a positive value. But either way, we figured out this radius. Now let's figure out the radius of circle B. The radius of circle B."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "It will become a positive value. But either way, we figured out this radius. Now let's figure out the radius of circle B. The radius of circle B. The same exact idea. The radius of circle B squared is equal to our change in x. So we could write it as 3 minus 0 or 0 minus 3."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "The radius of circle B. The same exact idea. The radius of circle B squared is equal to our change in x. So we could write it as 3 minus 0 or 0 minus 3. But we'll just write it as 3 minus 0. 3 minus 0 squared plus 1 minus 0 squared. Or the radius, or the distance between these two points, is equal to the square root of, let's see, this is 3 squared plus 1 squared."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So we could write it as 3 minus 0 or 0 minus 3. But we'll just write it as 3 minus 0. 3 minus 0 squared plus 1 minus 0 squared. Or the radius, or the distance between these two points, is equal to the square root of, let's see, this is 3 squared plus 1 squared. This is 9 plus 1. This is the square root of 10. The radius of B is the square root of 10."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Or the radius, or the distance between these two points, is equal to the square root of, let's see, this is 3 squared plus 1 squared. This is 9 plus 1. This is the square root of 10. The radius of B is the square root of 10. Now they ask us, which of the following points are on circle A, circle B, or both circles? So all we have to do now is look at these points. If this point is a square root of 10 away from point B, then it's on the circle."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "The radius of B is the square root of 10. Now they ask us, which of the following points are on circle A, circle B, or both circles? So all we have to do now is look at these points. If this point is a square root of 10 away from point B, then it's on the circle. It's a radius away. A circle is a locus of all points that are a radius away from the center. If it's 5 square roots of 2 from this point, then it's on circle A."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "If this point is a square root of 10 away from point B, then it's on the circle. It's a radius away. A circle is a locus of all points that are a radius away from the center. If it's 5 square roots of 2 from this point, then it's on circle A. If it's neither, then it's neither. Or it could be both. So let's try these out one by one."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "If it's 5 square roots of 2 from this point, then it's on circle A. If it's neither, then it's neither. Or it could be both. So let's try these out one by one. So point C is at 4, negative 2. So let me color this in a new color. So point C, let me do it in orange."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So let's try these out one by one. So point C is at 4, negative 2. So let me color this in a new color. So point C, let me do it in orange. Point C is at 1, 2, 3, 4, negative 2. Point C is right over there. Now it looks pretty close."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So point C, let me do it in orange. Point C is at 1, 2, 3, 4, negative 2. Point C is right over there. Now it looks pretty close. Just this is a hand-drawn drawing, so it's not perfect. So point C is there. It looks pretty close."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Now it looks pretty close. Just this is a hand-drawn drawing, so it's not perfect. So point C is there. It looks pretty close. But let's actually verify it. The distance between point C and point D, so the distance squared is going to be equal to the change in x's. So we could say 4 minus, so we're trying the distance between C and B."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "It looks pretty close. But let's actually verify it. The distance between point C and point D, so the distance squared is going to be equal to the change in x's. So we could say 4 minus, so we're trying the distance between C and B. It's 4 minus 3 squared. Plus negative 2 minus 1 squared, which is equal to 1 squared plus negative 3 squared. And so our distance squared is equal to 10, or our distance is equal to the square root of 10."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So we could say 4 minus, so we're trying the distance between C and B. It's 4 minus 3 squared. Plus negative 2 minus 1 squared, which is equal to 1 squared plus negative 3 squared. And so our distance squared is equal to 10, or our distance is equal to the square root of 10. So this is also, the distance right over here is the square root of 10. So this is on the circle. If we wanted to draw circle B, it would look something like this."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And so our distance squared is equal to 10, or our distance is equal to the square root of 10. So this is also, the distance right over here is the square root of 10. So this is on the circle. If we wanted to draw circle B, it would look something like this. And once again, I'm hand-drawing it, so it's not perfect. But it would look something, I'm going to draw part of it, looks something like this. This is exactly a radius away."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "If we wanted to draw circle B, it would look something like this. And once again, I'm hand-drawing it, so it's not perfect. But it would look something, I'm going to draw part of it, looks something like this. This is exactly a radius away. So let me write, this is on circle B. Now let's look at this point, the point 5, 3. So I'll do that in pink."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "This is exactly a radius away. So let me write, this is on circle B. Now let's look at this point, the point 5, 3. So I'll do that in pink. So 1, 2, 3, 4, 5, 3. So this looks close, but let's verify just in case. So now our distance is equal to, let me just write it this way."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So I'll do that in pink. So 1, 2, 3, 4, 5, 3. So this looks close, but let's verify just in case. So now our distance is equal to, let me just write it this way. Our distance squared is going to be our change in x squared. So 5 minus 3 squared plus 3 minus 1 squared. Change in y."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So now our distance is equal to, let me just write it this way. Our distance squared is going to be our change in x squared. So 5 minus 3 squared plus 3 minus 1 squared. Change in y. 3 minus 1 squared. And so our distance is going to be equal to, actually I don't want to skip too many steps. You see this is 2 squared, which is 4, plus 2 squared, which is another 4."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Change in y. 3 minus 1 squared. And so our distance is going to be equal to, actually I don't want to skip too many steps. You see this is 2 squared, which is 4, plus 2 squared, which is another 4. So our distance is going to be equal to the square root of 8, which is the same thing as the square root of 2 times 4, which is the same thing as 2 times the square root of 2. Square root of 4 is 2, and then of course you just have the 2 left in the radical. So this is a different distance away than square root of 10."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "You see this is 2 squared, which is 4, plus 2 squared, which is another 4. So our distance is going to be equal to the square root of 8, which is the same thing as the square root of 2 times 4, which is the same thing as 2 times the square root of 2. Square root of 4 is 2, and then of course you just have the 2 left in the radical. So this is a different distance away than square root of 10. So this one right over here is definitely not on circle B. And just eyeballing it, you can see that it's not going to be on circle A. This distance, just eyeballing it, is much further than 5 square roots of 2."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So this is a different distance away than square root of 10. So this one right over here is definitely not on circle B. And just eyeballing it, you can see that it's not going to be on circle A. This distance, just eyeballing it, is much further than 5 square roots of 2. And that's also true for point C. Point C is much further than 5 square roots of 2. You can just look at that visually. They're much further than a radius away from A."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "This distance, just eyeballing it, is much further than 5 square roots of 2. And that's also true for point C. Point C is much further than 5 square roots of 2. You can just look at that visually. They're much further than a radius away from A. So this point right over here, this is neither. This is on neither circle. Now finally, we have the point negative 2 comma 8."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "They're much further than a radius away from A. So this point right over here, this is neither. This is on neither circle. Now finally, we have the point negative 2 comma 8. So let me find, I'm running out of colors. Let me see, I could use yellow again. Negative 2 comma 8."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Now finally, we have the point negative 2 comma 8. So let me find, I'm running out of colors. Let me see, I could use yellow again. Negative 2 comma 8. So that's negative 2 comma 1, 2, 3, 4, 5, 6, 7, 8. So it's right over here. That is point E. Just eyeballing it, this distance, so it's clearly way too far."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Negative 2 comma 8. So that's negative 2 comma 1, 2, 3, 4, 5, 6, 7, 8. So it's right over here. That is point E. Just eyeballing it, this distance, so it's clearly way too far. Just looking at it, just eyeballing it, it's clearly more than a radius away from B. So this isn't going to be on circle B. And also looking at it relative to point A, it looks much closer to point A."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "That is point E. Just eyeballing it, this distance, so it's clearly way too far. Just looking at it, just eyeballing it, it's clearly more than a radius away from B. So this isn't going to be on circle B. And also looking at it relative to point A, it looks much closer to point A. It doesn't even seem close than point P is. So it looks, just inspecting it, you could rule this one out that this is going to be neither. But we can verify this on our own if we like."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And also looking at it relative to point A, it looks much closer to point A. It doesn't even seem close than point P is. So it looks, just inspecting it, you could rule this one out that this is going to be neither. But we can verify this on our own if we like. We can just find the distance between these two points. Our distance squared is going to be our change in x's. So negative 2 minus negative 5 squared plus our change in y."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "But we can verify this on our own if we like. We can just find the distance between these two points. Our distance squared is going to be our change in x's. So negative 2 minus negative 5 squared plus our change in y. So it's 8 minus 5 squared. And so this is our distance squared is going to be equal to negative 2 minus negative 5. That's negative 2 plus 5."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So negative 2 minus negative 5 squared plus our change in y. So it's 8 minus 5 squared. And so this is our distance squared is going to be equal to negative 2 minus negative 5. That's negative 2 plus 5. So that's going to be 3 squared plus 3 squared. And you see that right over here, Pythagorean theorem. This distance right over here is 3."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "That's negative 2 plus 5. So that's going to be 3 squared plus 3 squared. And you see that right over here, Pythagorean theorem. This distance right over here is 3. This distance right over here, this is your change in x, is 3. Change in y is 3. 3 squared plus 3 squared is going to be the distance squared, the hypotenuse squared."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "This distance right over here is 3. This distance right over here, this is your change in x, is 3. Change in y is 3. 3 squared plus 3 squared is going to be the distance squared, the hypotenuse squared. So our distance squared is going to be, or I could say our distance, skip a few steps, is equal to the square root of, we could write this as 9 times 2, or the distance is equal to 3 times the square root of 2. The radius of circle A is 5 times the square root of 2, not 3 times the square root of 2. So this is actually going to be inside the circle."}, {"video_title": "Recognizing points on a circle Analytic geometry Geometry Khan Academy.mp3", "Sentence": "3 squared plus 3 squared is going to be the distance squared, the hypotenuse squared. So our distance squared is going to be, or I could say our distance, skip a few steps, is equal to the square root of, we could write this as 9 times 2, or the distance is equal to 3 times the square root of 2. The radius of circle A is 5 times the square root of 2, not 3 times the square root of 2. So this is actually going to be inside the circle. So if we want to draw circle A, it's going to look something like this. And point E is on the inside. Point D and point C are on the outside of circle A."}, {"video_title": "Proof Rhombus diagonals are perpendicular bisectors Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "What they want us to prove is that their diagonals are perpendicular, that AC is perpendicular to BD. Now let's think about everything we know about a rhombus. First of all, a rhombus is a special case of a parallelogram. In a parallelogram, the opposite sides are parallel, so that side is parallel to that side, these two sides are parallel. And in a rhombus, not only are the opposite sides parallel, it's a parallelogram, but also all of the sides have equal length. So this side is equal to this side, which is equal to that side, which is equal to that side right over there. Now there's other interesting things we know about the diagonals of a parallelogram, which we know all rhombi are parallelograms."}, {"video_title": "Proof Rhombus diagonals are perpendicular bisectors Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "In a parallelogram, the opposite sides are parallel, so that side is parallel to that side, these two sides are parallel. And in a rhombus, not only are the opposite sides parallel, it's a parallelogram, but also all of the sides have equal length. So this side is equal to this side, which is equal to that side, which is equal to that side right over there. Now there's other interesting things we know about the diagonals of a parallelogram, which we know all rhombi are parallelograms. The other way around is not necessarily true. We know that for any parallelogram, and a rhombus is a parallelogram, that the diagonals bisect each other. So for example, let me label this point in the center, let me label it point E. We know that AE is going to be equal to EC, I'll put two slashes right over there, and we also know that EB is going to be equal to ED."}, {"video_title": "Proof Rhombus diagonals are perpendicular bisectors Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "Now there's other interesting things we know about the diagonals of a parallelogram, which we know all rhombi are parallelograms. The other way around is not necessarily true. We know that for any parallelogram, and a rhombus is a parallelogram, that the diagonals bisect each other. So for example, let me label this point in the center, let me label it point E. We know that AE is going to be equal to EC, I'll put two slashes right over there, and we also know that EB is going to be equal to ED. So this is all of what we know when someone just says that ABCD is a rhombus, based on other things that we've proven to ourselves. Now we need to prove that AC is perpendicular to BD. So an interesting way to prove it, and you can kind of look at it just by eyeballing, is if we can show that this triangle is congruent to this triangle, and that these two angles right over here correspond to each other, then they have to be the same, and they'll be supplementary, and then they'll be 90 degrees."}, {"video_title": "Proof Rhombus diagonals are perpendicular bisectors Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So for example, let me label this point in the center, let me label it point E. We know that AE is going to be equal to EC, I'll put two slashes right over there, and we also know that EB is going to be equal to ED. So this is all of what we know when someone just says that ABCD is a rhombus, based on other things that we've proven to ourselves. Now we need to prove that AC is perpendicular to BD. So an interesting way to prove it, and you can kind of look at it just by eyeballing, is if we can show that this triangle is congruent to this triangle, and that these two angles right over here correspond to each other, then they have to be the same, and they'll be supplementary, and then they'll be 90 degrees. So let's just prove it to ourselves. So the first thing we see is we have a side, a side, and a side. So we can see that triangle, let me write it here, triangle, let me just add a new color, we see that triangle ABE is congruent to triangle CBE, and we know that by side, side, side congruency."}, {"video_title": "Proof Rhombus diagonals are perpendicular bisectors Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So an interesting way to prove it, and you can kind of look at it just by eyeballing, is if we can show that this triangle is congruent to this triangle, and that these two angles right over here correspond to each other, then they have to be the same, and they'll be supplementary, and then they'll be 90 degrees. So let's just prove it to ourselves. So the first thing we see is we have a side, a side, and a side. So we can see that triangle, let me write it here, triangle, let me just add a new color, we see that triangle ABE is congruent to triangle CBE, and we know that by side, side, side congruency. We have a side, a side, and a side, a side, a side, and a side. And then once we know that, we know that all the corresponding angles are congruent, and in particular we know that AEB, we know that angle AEB is going to be congruent to angle, so AEB to CEB, to angle CEB, because they are corresponding angles of congruent, corresponding angles of congruent triangles. So this angle right over here is going to be equal to that angle over there."}, {"video_title": "Proof Rhombus diagonals are perpendicular bisectors Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So we can see that triangle, let me write it here, triangle, let me just add a new color, we see that triangle ABE is congruent to triangle CBE, and we know that by side, side, side congruency. We have a side, a side, and a side, a side, a side, and a side. And then once we know that, we know that all the corresponding angles are congruent, and in particular we know that AEB, we know that angle AEB is going to be congruent to angle, so AEB to CEB, to angle CEB, because they are corresponding angles of congruent, corresponding angles of congruent triangles. So this angle right over here is going to be equal to that angle over there. And we also know that they are supplementary, and so they're both supplementary, so we also know, and let me write it this way, they're congruent and they are supplementary. So we have these two are going to have the same measure, and they need to add up to 180 degrees. So if I have two things that are the same thing, and they add up to 180 degrees, what does that tell me?"}, {"video_title": "Proof Rhombus diagonals are perpendicular bisectors Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So this angle right over here is going to be equal to that angle over there. And we also know that they are supplementary, and so they're both supplementary, so we also know, and let me write it this way, they're congruent and they are supplementary. So we have these two are going to have the same measure, and they need to add up to 180 degrees. So if I have two things that are the same thing, and they add up to 180 degrees, what does that tell me? So that tells me that angle, the measure of angle AEB is equal to the measure of angle CEB, which is equal to, which must be equal to 90 degrees. They're the same measure and they are supplementary. So this is a right angle, and then this is a right angle."}, {"video_title": "Proof Rhombus diagonals are perpendicular bisectors Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So if I have two things that are the same thing, and they add up to 180 degrees, what does that tell me? So that tells me that angle, the measure of angle AEB is equal to the measure of angle CEB, which is equal to, which must be equal to 90 degrees. They're the same measure and they are supplementary. So this is a right angle, and then this is a right angle. And obviously if this is a right angle, this angle down here is a vertical angle, that's going to be a right angle. If this is a right angle, this over here is going to be a vertical angle. And you see the diagonals intersect at a 90 degree angle."}, {"video_title": "Proof Rhombus diagonals are perpendicular bisectors Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So this is a right angle, and then this is a right angle. And obviously if this is a right angle, this angle down here is a vertical angle, that's going to be a right angle. If this is a right angle, this over here is going to be a vertical angle. And you see the diagonals intersect at a 90 degree angle. So we've just proved, so this is interesting. In a parallelogram, the diagonals bisect each other. For a rhombus, where all the sides are equal, we've shown that not only do they bisect each other, but they're perpendicular bisectors of each other."}, {"video_title": "Finding a point part way between two points Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So let's think about what they're asking. So if that's point C, I'm just going to redraw this line segment just to conceptualize what they're asking for. And that's point A. They're asking us to find some point B that the distance between C and B, so that's this distance right over here. So if this distance is x, then the distance between B and A is going to be 3 times that. It's going to be 3 times that. So this will be 3x."}, {"video_title": "Finding a point part way between two points Analytic geometry Geometry Khan Academy.mp3", "Sentence": "They're asking us to find some point B that the distance between C and B, so that's this distance right over here. So if this distance is x, then the distance between B and A is going to be 3 times that. It's going to be 3 times that. So this will be 3x. That the ratio of AB to BC is 3 to 1. So that would be the ratio. Let me write this down."}, {"video_title": "Finding a point part way between two points Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So this will be 3x. That the ratio of AB to BC is 3 to 1. So that would be the ratio. Let me write this down. It would be AB, that looks like an HB. It would be AB to BC is going to be equal to 3x to x, which is the same thing as 3 to 1. Which is the same thing as 3 to 1, if we wanted to write it a slightly different way."}, {"video_title": "Finding a point part way between two points Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Let me write this down. It would be AB, that looks like an HB. It would be AB to BC is going to be equal to 3x to x, which is the same thing as 3 to 1. Which is the same thing as 3 to 1, if we wanted to write it a slightly different way. So how can we think about it? You might be tempted to say, oh, well, you could use the distance formula, find the distance, which by itself isn't completely uncomplicated. And then this will be one-fourth of the way."}, {"video_title": "Finding a point part way between two points Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Which is the same thing as 3 to 1, if we wanted to write it a slightly different way. So how can we think about it? You might be tempted to say, oh, well, you could use the distance formula, find the distance, which by itself isn't completely uncomplicated. And then this will be one-fourth of the way. Because if you think about it, this entire distance is going to be 4x. This entire distance is going to be, let me draw that a little bit neater. This entire distance, if you have an x plus a 3x, is going to be 4x."}, {"video_title": "Finding a point part way between two points Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And then this will be one-fourth of the way. Because if you think about it, this entire distance is going to be 4x. This entire distance is going to be, let me draw that a little bit neater. This entire distance, if you have an x plus a 3x, is going to be 4x. So you'd say, well, this is one out of the 4x's along the way. This is going to be one-fourth of the distance between the two points. So this is, let me write that down."}, {"video_title": "Finding a point part way between two points Analytic geometry Geometry Khan Academy.mp3", "Sentence": "This entire distance, if you have an x plus a 3x, is going to be 4x. So you'd say, well, this is one out of the 4x's along the way. This is going to be one-fourth of the distance between the two points. So this is, let me write that down. This is one-fourth of the way. One-fourth of the way between C and B. Going from C, sorry, going from C to A."}, {"video_title": "Finding a point part way between two points Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So this is, let me write that down. This is one-fourth of the way. One-fourth of the way between C and B. Going from C, sorry, going from C to A. B is going to be one-fourth of the way. So maybe you try to find the distance. And you say, oh, well, what are all the points that are one-fourth away?"}, {"video_title": "Finding a point part way between two points Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Going from C, sorry, going from C to A. B is going to be one-fourth of the way. So maybe you try to find the distance. And you say, oh, well, what are all the points that are one-fourth away? But it has to be one-fourth of that distance away. But then it has to be on that line. But that makes it complicated."}, {"video_title": "Finding a point part way between two points Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And you say, oh, well, what are all the points that are one-fourth away? But it has to be one-fourth of that distance away. But then it has to be on that line. But that makes it complicated. Because this line is at an incline. It's not just horizontal. It's not just vertical."}, {"video_title": "Finding a point part way between two points Analytic geometry Geometry Khan Academy.mp3", "Sentence": "But that makes it complicated. Because this line is at an incline. It's not just horizontal. It's not just vertical. What we can do, however, is break this problem down into the vertical change between A and C and the horizontal change between A and C. So for example, the horizontal change between A and C, A is at 9 right over here. And C is at negative 7. So this distance right over here is 9 minus negative 7, which is equal to 9 plus 7, which is equal to 16."}, {"video_title": "Finding a point part way between two points Analytic geometry Geometry Khan Academy.mp3", "Sentence": "It's not just vertical. What we can do, however, is break this problem down into the vertical change between A and C and the horizontal change between A and C. So for example, the horizontal change between A and C, A is at 9 right over here. And C is at negative 7. So this distance right over here is 9 minus negative 7, which is equal to 9 plus 7, which is equal to 16. And you see that here. 9 plus 7, this total distance is 16. That's the horizontal change going from A to C or going from C to A."}, {"video_title": "Finding a point part way between two points Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So this distance right over here is 9 minus negative 7, which is equal to 9 plus 7, which is equal to 16. And you see that here. 9 plus 7, this total distance is 16. That's the horizontal change going from A to C or going from C to A. And the vertical change, and you could even just count that, that's going to be 4. C is at 1. A is at 5."}, {"video_title": "Finding a point part way between two points Analytic geometry Geometry Khan Academy.mp3", "Sentence": "That's the horizontal change going from A to C or going from C to A. And the vertical change, and you could even just count that, that's going to be 4. C is at 1. A is at 5. Going from 1 to 5, you've changed vertically 4. So what we can say, going from C to B in each direction, in the vertical direction and the horizontal direction, we're going to go one-fourth of the way. So if we go one-fourth in the vertical direction, we're going to end up at y is equal to 2."}, {"video_title": "Finding a point part way between two points Analytic geometry Geometry Khan Academy.mp3", "Sentence": "A is at 5. Going from 1 to 5, you've changed vertically 4. So what we can say, going from C to B in each direction, in the vertical direction and the horizontal direction, we're going to go one-fourth of the way. So if we go one-fourth in the vertical direction, we're going to end up at y is equal to 2. So I'm just going, starting at C, one-fourth of the way. One-fourth of 4 is 1. So I've just moved up 1."}, {"video_title": "Finding a point part way between two points Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So if we go one-fourth in the vertical direction, we're going to end up at y is equal to 2. So I'm just going, starting at C, one-fourth of the way. One-fourth of 4 is 1. So I've just moved up 1. So our y is going to be equal to 2. And if we go one-fourth in the horizontal direction, one-fourth of 16 is 4. So we go 1, 2, 3, 4."}, {"video_title": "Finding a point part way between two points Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So I've just moved up 1. So our y is going to be equal to 2. And if we go one-fourth in the horizontal direction, one-fourth of 16 is 4. So we go 1, 2, 3, 4. So we end up right over here. Our x is negative 3. So we end up at that point right over there."}, {"video_title": "Finding a point part way between two points Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So we go 1, 2, 3, 4. So we end up right over here. Our x is negative 3. So we end up at that point right over there. We end up at this point. This is the point negative 3, 2. And if you were really careful with your drawing, you could have actually just drawn."}, {"video_title": "Finding a point part way between two points Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So we end up at that point right over there. We end up at this point. This is the point negative 3, 2. And if you were really careful with your drawing, you could have actually just drawn. Actually, you don't have to be that careful, since this is graph paper. You actually could have just said, hey, we're going to go one-fourth this way. And where does that intersect the line?"}, {"video_title": "Finding a point part way between two points Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And if you were really careful with your drawing, you could have actually just drawn. Actually, you don't have to be that careful, since this is graph paper. You actually could have just said, hey, we're going to go one-fourth this way. And where does that intersect the line? Hey, it intersects the line right over there. Or you could have said, we're going to go one-fourth this way. Where does that intersect the line?"}, {"video_title": "Finding a point part way between two points Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And where does that intersect the line? Hey, it intersects the line right over there. Or you could have said, we're going to go one-fourth this way. Where does that intersect the line? And that would have let you figure it out either way. So this point right over here is B. It is one-fourth of the way between C and A."}, {"video_title": "Finding a point part way between two points Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Where does that intersect the line? And that would have let you figure it out either way. So this point right over here is B. It is one-fourth of the way between C and A. Or another way of thinking about it, the distance between C and B, which we haven't even figured out, we could do that using the distance formula or the Pythagorean theorem, which it really is. This distance, the distance CB, is one-third the distance BA. Or BA, the ratio of AB to BC is 3 to 1."}, {"video_title": "Formal translation tool example Transformations Geometry Khan Academy.mp3", "Sentence": "So I'm gonna use the translation tool. So I'm gonna start with a translation. So it says translate by, and this is gonna say how much do we translate the X coordinates and how much do we translate the Y coordinates. Let's see, if I wanna map, if I wanna get point W to correspond to this point right over here, which it seems like it should, I would have to go from X equals two to X equals negative five. So my X would have to decrease by seven. So let me type that in, negative seven. And then, let's see, on the Y side, and we saw that."}, {"video_title": "Formal translation tool example Transformations Geometry Khan Academy.mp3", "Sentence": "Let's see, if I wanna map, if I wanna get point W to correspond to this point right over here, which it seems like it should, I would have to go from X equals two to X equals negative five. So my X would have to decrease by seven. So let me type that in, negative seven. And then, let's see, on the Y side, and we saw that. So by just typing in negative seven here, we've moved it to the left by seven. And now in the Y, in the Y axis, I need to move it down by, let's see, one, two, three, four. So my Y coordinates, I need to move it down by four."}, {"video_title": "Formal translation tool example Transformations Geometry Khan Academy.mp3", "Sentence": "And then, let's see, on the Y side, and we saw that. So by just typing in negative seven here, we've moved it to the left by seven. And now in the Y, in the Y axis, I need to move it down by, let's see, one, two, three, four. So my Y coordinates, I need to move it down by four. And let's see what happens. And it looks like I was able to successfully translate it. By translating X by negative seven, every point here, every point on this has been shifted to the left by seven and has been shifted down by four, and I was able to get onto this triangle."}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "The laws of nature are but the mathematical thoughts of God. And this is a quote by Euclid of Alexandria, who was a Greek mathematician and philosopher who lived about 300 years before Christ. And the reason why I include this quote is because Euclid is considered to be the father, the father of geometry. And it is a neat quote. Regardless of your views of God, whether or not God exists or the nature of God, it says something very fundamental about nature. The laws of nature are but the mathematical thoughts of God, that math underpins all of the laws of nature. And the word geometry itself has Greek roots."}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And it is a neat quote. Regardless of your views of God, whether or not God exists or the nature of God, it says something very fundamental about nature. The laws of nature are but the mathematical thoughts of God, that math underpins all of the laws of nature. And the word geometry itself has Greek roots. Geo comes from Greek for earth. Metry comes from Greek for measurement. You're probably used to something like the metric system."}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And the word geometry itself has Greek roots. Geo comes from Greek for earth. Metry comes from Greek for measurement. You're probably used to something like the metric system. And Euclid is considered to be the father of geometry, not because he was the first person who studied geometry. You could imagine the very first humans might have studied geometry. They might have looked at two twigs on the ground that looked something like that."}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "You're probably used to something like the metric system. And Euclid is considered to be the father of geometry, not because he was the first person who studied geometry. You could imagine the very first humans might have studied geometry. They might have looked at two twigs on the ground that looked something like that. And they might have looked at another pair of twigs that looked like that and said, this is a bigger opening. What is the relationship here? Or they might have looked at a tree that had a branch that came off like that."}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "They might have looked at two twigs on the ground that looked something like that. And they might have looked at another pair of twigs that looked like that and said, this is a bigger opening. What is the relationship here? Or they might have looked at a tree that had a branch that came off like that. And they said, oh, there's something similar about this opening here and this opening here. Or they might have asked themselves, what is the ratio? Or what is the relationship between the distance around a circle and the distance across it?"}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Or they might have looked at a tree that had a branch that came off like that. And they said, oh, there's something similar about this opening here and this opening here. Or they might have asked themselves, what is the ratio? Or what is the relationship between the distance around a circle and the distance across it? And is that the same for all circles? And is there a way for us to feel really good that that is definitely true? And then once you got to the early Greeks, they started to get even more thoughtful, essentially, about geometric things when you talk about Greek mathematicians like Pythagoras, who came before Euclid."}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Or what is the relationship between the distance around a circle and the distance across it? And is that the same for all circles? And is there a way for us to feel really good that that is definitely true? And then once you got to the early Greeks, they started to get even more thoughtful, essentially, about geometric things when you talk about Greek mathematicians like Pythagoras, who came before Euclid. But the reason why Euclid is considered to be the father of geometry and why we often talk about Euclidean geometry is around 300 BC. And this right over here is a picture of Euclid painted by Raphael. And no one really knows what Euclid looked like even when he was born or when he died."}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And then once you got to the early Greeks, they started to get even more thoughtful, essentially, about geometric things when you talk about Greek mathematicians like Pythagoras, who came before Euclid. But the reason why Euclid is considered to be the father of geometry and why we often talk about Euclidean geometry is around 300 BC. And this right over here is a picture of Euclid painted by Raphael. And no one really knows what Euclid looked like even when he was born or when he died. So this is just Raphael's impression of what Euclid might have looked like while he was teaching in Alexandria. But what made Euclid the father of geometry is really his writing of Euclid's elements. And what the elements were were essentially a 13-volume textbook, and arguably the most famous textbook of all time."}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And no one really knows what Euclid looked like even when he was born or when he died. So this is just Raphael's impression of what Euclid might have looked like while he was teaching in Alexandria. But what made Euclid the father of geometry is really his writing of Euclid's elements. And what the elements were were essentially a 13-volume textbook, and arguably the most famous textbook of all time. And what he did in those 13 volumes is he essentially did a rigorous, thoughtful, logical march through geometry and number theory, and then also solid geometry, so geometry in three dimensions. And this right over here is the frontispiece for the English version of, or the first translation of the English version of Euclid's elements. And this was done in 1570."}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And what the elements were were essentially a 13-volume textbook, and arguably the most famous textbook of all time. And what he did in those 13 volumes is he essentially did a rigorous, thoughtful, logical march through geometry and number theory, and then also solid geometry, so geometry in three dimensions. And this right over here is the frontispiece for the English version of, or the first translation of the English version of Euclid's elements. And this was done in 1570. But it was obviously first written in Greek. And then during much of the Middle Ages, that knowledge was shepherded by the Arabs. And it was translated into Arabic."}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And this was done in 1570. But it was obviously first written in Greek. And then during much of the Middle Ages, that knowledge was shepherded by the Arabs. And it was translated into Arabic. And then eventually in the late Middle Ages, translated into Latin, and then obviously eventually English. And when I say that he did a rigorous march, what Euclid did is he didn't just say, well, I think if you take the length of one side of a right triangle and the length of the other side of the right triangle, it's going to be the same as the square of the hypotenuse, all of these other things. And we'll go into depth about what all of these things are meant."}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And it was translated into Arabic. And then eventually in the late Middle Ages, translated into Latin, and then obviously eventually English. And when I say that he did a rigorous march, what Euclid did is he didn't just say, well, I think if you take the length of one side of a right triangle and the length of the other side of the right triangle, it's going to be the same as the square of the hypotenuse, all of these other things. And we'll go into depth about what all of these things are meant. He says, I don't want to just feel good that it's probably true. I want to prove to myself that it is true. And so what he did in Elements, especially the six books that are concerned with planar geometry, in fact, he did all of them, but from a geometrical point of view, he started with basic assumptions."}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And we'll go into depth about what all of these things are meant. He says, I don't want to just feel good that it's probably true. I want to prove to myself that it is true. And so what he did in Elements, especially the six books that are concerned with planar geometry, in fact, he did all of them, but from a geometrical point of view, he started with basic assumptions. So he started with basic assumptions. And those basic assumptions in geometric speak are called axioms or postulates. And from them, he proved, he deduced other statements or propositions, or these are sometimes called theorems."}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And so what he did in Elements, especially the six books that are concerned with planar geometry, in fact, he did all of them, but from a geometrical point of view, he started with basic assumptions. So he started with basic assumptions. And those basic assumptions in geometric speak are called axioms or postulates. And from them, he proved, he deduced other statements or propositions, or these are sometimes called theorems. And then he says, now I know if this is true and this is true, this must be true. And he could also prove that other things cannot be true. So then he could prove that this is not going to be the true."}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And from them, he proved, he deduced other statements or propositions, or these are sometimes called theorems. And then he says, now I know if this is true and this is true, this must be true. And he could also prove that other things cannot be true. So then he could prove that this is not going to be the true. He didn't just say, well, every circle I've said has this property. He says, I've now proven that this is true. And then from there, we can go and deduce other propositions or theorems."}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "So then he could prove that this is not going to be the true. He didn't just say, well, every circle I've said has this property. He says, I've now proven that this is true. And then from there, we can go and deduce other propositions or theorems. And we could use some of our original axioms to do that. And what's special about that is no one had really done that before. Rigorously proven beyond a shadow of a doubt, across a whole broad sweep of knowledge."}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And then from there, we can go and deduce other propositions or theorems. And we could use some of our original axioms to do that. And what's special about that is no one had really done that before. Rigorously proven beyond a shadow of a doubt, across a whole broad sweep of knowledge. So not just one proof here or there. He did it for an entire set of knowledge that we're talking about, a rigorous march through a subject so that he could build this scaffold of axioms and postulates and theorems and propositions. And theorems and propositions are essentially the same thing."}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "Rigorously proven beyond a shadow of a doubt, across a whole broad sweep of knowledge. So not just one proof here or there. He did it for an entire set of knowledge that we're talking about, a rigorous march through a subject so that he could build this scaffold of axioms and postulates and theorems and propositions. And theorems and propositions are essentially the same thing. And essentially, for about 2,000 years after Euclid, so this is unbelievable shelf life for a textbook, people didn't view you as educated if you did not read and understand Euclid's elements. And Euclid's elements, the book itself, was the second most printed book in the Western world after the Bible. This is a math textbook."}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "And theorems and propositions are essentially the same thing. And essentially, for about 2,000 years after Euclid, so this is unbelievable shelf life for a textbook, people didn't view you as educated if you did not read and understand Euclid's elements. And Euclid's elements, the book itself, was the second most printed book in the Western world after the Bible. This is a math textbook. It was second only to the Bible. When the first printing presses came out, they said, OK, let's print the Bible. What do we print next?"}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "This is a math textbook. It was second only to the Bible. When the first printing presses came out, they said, OK, let's print the Bible. What do we print next? Let's print Euclid's elements. And to show that this is relevant into the fairly recent past, although some would, whether or not you argue that about 150, 160 years ago is a recent past, this right here is a direct quote from Abraham Lincoln, obviously one of the great American presidents. I like this picture of Abraham Lincoln."}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "What do we print next? Let's print Euclid's elements. And to show that this is relevant into the fairly recent past, although some would, whether or not you argue that about 150, 160 years ago is a recent past, this right here is a direct quote from Abraham Lincoln, obviously one of the great American presidents. I like this picture of Abraham Lincoln. This is actually a photograph of Lincoln in his late 30s. But he was a huge fan of Euclid's elements. He would actually use it to fine tune his mind."}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "I like this picture of Abraham Lincoln. This is actually a photograph of Lincoln in his late 30s. But he was a huge fan of Euclid's elements. He would actually use it to fine tune his mind. While he was riding his horse, he would read Euclid's element. While he was in the White House, he would read Euclid's element. But this is a direct quote from Lincoln."}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "He would actually use it to fine tune his mind. While he was riding his horse, he would read Euclid's element. While he was in the White House, he would read Euclid's element. But this is a direct quote from Lincoln. In the course of my law reading, I constantly came upon the word demonstrate. I thought at first that I understood its meaning, but soon became satisfied that I did not. I said to myself, what do I do when I demonstrate more than when I reason or prove?"}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "But this is a direct quote from Lincoln. In the course of my law reading, I constantly came upon the word demonstrate. I thought at first that I understood its meaning, but soon became satisfied that I did not. I said to myself, what do I do when I demonstrate more than when I reason or prove? How does demonstration differ from any other proof? So in Lincoln's thing, there's this word demonstration that kind of means something more, proving beyond doubt, something more rigorous, more than just simple kind of feeling good about something or reasoning through it. I consulted Webster's Dictionary."}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "I said to myself, what do I do when I demonstrate more than when I reason or prove? How does demonstration differ from any other proof? So in Lincoln's thing, there's this word demonstration that kind of means something more, proving beyond doubt, something more rigorous, more than just simple kind of feeling good about something or reasoning through it. I consulted Webster's Dictionary. So Webster's Dictionary was around even around when Lincoln was around. They told of certain proof, proof beyond the possibility of doubt. But I could form no idea of what sort of proof that was."}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "I consulted Webster's Dictionary. So Webster's Dictionary was around even around when Lincoln was around. They told of certain proof, proof beyond the possibility of doubt. But I could form no idea of what sort of proof that was. I thought a great many things were proved beyond the possibility of doubt without recourse to any such extraordinary process of reasoning as I understood demonstration to be. I consulted all the dictionaries and books of reference I could find, but with no better results. You might as well have defined blue to a blind man."}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "But I could form no idea of what sort of proof that was. I thought a great many things were proved beyond the possibility of doubt without recourse to any such extraordinary process of reasoning as I understood demonstration to be. I consulted all the dictionaries and books of reference I could find, but with no better results. You might as well have defined blue to a blind man. At last I said, Lincoln, he's talking to himself, at last I said, Lincoln, you never can make a lawyer if you do not understand what demonstrate means. And I left my situation in Springfield, went home to my father's house, and stayed there till I could give any proposition in the six books of Euclid at site, so the six books concerned with planar geometry. I then found out what demonstrate means and went back to my law studies."}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "You might as well have defined blue to a blind man. At last I said, Lincoln, he's talking to himself, at last I said, Lincoln, you never can make a lawyer if you do not understand what demonstrate means. And I left my situation in Springfield, went home to my father's house, and stayed there till I could give any proposition in the six books of Euclid at site, so the six books concerned with planar geometry. I then found out what demonstrate means and went back to my law studies. So one of the greatest American presidents of all time felt that in order to be a great lawyer, he had to understand, be able to prove any proposition in the six books of Euclid's elements at site. And also, once he was in the White House, he continued to do this to make him, in his mind, to fine tune his mind to become a great president. And so what we're going to be doing in the geometry playlist is essentially that."}, {"video_title": "Euclid as the father of geometry Introduction to Euclidean geometry Geometry Khan Academy.mp3", "Sentence": "I then found out what demonstrate means and went back to my law studies. So one of the greatest American presidents of all time felt that in order to be a great lawyer, he had to understand, be able to prove any proposition in the six books of Euclid's elements at site. And also, once he was in the White House, he continued to do this to make him, in his mind, to fine tune his mind to become a great president. And so what we're going to be doing in the geometry playlist is essentially that. What we're going to study is we're going to think about how do we really tightly, rigorously prove things. We're essentially going to be, in a slightly more modern form, be studying what Euclid studied 2,300 years ago to really tighten our reasoning of different statements and being able to make sure that when we say something, we can really prove what we're saying. This is really some of the most fundamental real mathematics that you will do."}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "So we have a circle here, and they specified some points for us. This little orangish, or I guess maroonish red point right over here is the center of the circle, and then this blue point is a point that happens to sit on the circle. And so with that information, I want you to pause the video and see if you can figure out the equation for this circle. All right, let's work through this together. So let's first think about the center of the circle, and the center of the circle is just going to be the coordinates of that point. So the x-coordinate is negative one, and then the y-coordinate is one. So center is negative one, comma, one."}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "All right, let's work through this together. So let's first think about the center of the circle, and the center of the circle is just going to be the coordinates of that point. So the x-coordinate is negative one, and then the y-coordinate is one. So center is negative one, comma, one. And now let's think about what the radius of the circle is. Well, the radius is going to be the distance between the center and any point on the circle. So for example, this distance, the distance of that line, let's see, I can do a thicker version of that."}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "So center is negative one, comma, one. And now let's think about what the radius of the circle is. Well, the radius is going to be the distance between the center and any point on the circle. So for example, this distance, the distance of that line, let's see, I can do a thicker version of that. This line right over there, something strange about my, something strange about my pen tool is making that very thin. Let me do it one more time. Okay, that's better."}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "So for example, this distance, the distance of that line, let's see, I can do a thicker version of that. This line right over there, something strange about my, something strange about my pen tool is making that very thin. Let me do it one more time. Okay, that's better. The distance of that line right over there, that is going to be the radius. So how can we figure that out? Well, we can set up a right triangle and essentially use the distance formula which comes from the Pythagorean theorem."}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "Okay, that's better. The distance of that line right over there, that is going to be the radius. So how can we figure that out? Well, we can set up a right triangle and essentially use the distance formula which comes from the Pythagorean theorem. To figure out the length of that line, so this is the radius, we could figure out a change in x. So if we look at our change in x right over here, our change in x as we go from the center to this point, so this is our change in x, and then we could say that this is our change in y. That right over there is our change in y."}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, we can set up a right triangle and essentially use the distance formula which comes from the Pythagorean theorem. To figure out the length of that line, so this is the radius, we could figure out a change in x. So if we look at our change in x right over here, our change in x as we go from the center to this point, so this is our change in x, and then we could say that this is our change in y. That right over there is our change in y. And so our change in x squared plus our change in y squared is going to be our radius squared. That comes straight out of the Pythagorean theorem. This is a right triangle."}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "That right over there is our change in y. And so our change in x squared plus our change in y squared is going to be our radius squared. That comes straight out of the Pythagorean theorem. This is a right triangle. And so we can say that r squared is going to be equal to our change in x squared plus our change in y squared. Plus our change in y squared. Now what is our change in x going to be?"}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "This is a right triangle. And so we can say that r squared is going to be equal to our change in x squared plus our change in y squared. Plus our change in y squared. Now what is our change in x going to be? Our change in x is going to be equal to, well when we go from the radius to this point over here, our x goes from negative one to six. So you could view it as our ending x minus our starting x. So negative one minus, sorry, six minus negative one is equal to seven."}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "Now what is our change in x going to be? Our change in x is going to be equal to, well when we go from the radius to this point over here, our x goes from negative one to six. So you could view it as our ending x minus our starting x. So negative one minus, sorry, six minus negative one is equal to seven. So let me, so we have our change in x, this right over here, is equal to seven. If we viewed this as the start point and this as the end point, it would be negative seven, but we really care about the absolute value of the change in x, and once you square it, it all becomes a positive anyway. So our change in x right over here is going to be positive seven."}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "So negative one minus, sorry, six minus negative one is equal to seven. So let me, so we have our change in x, this right over here, is equal to seven. If we viewed this as the start point and this as the end point, it would be negative seven, but we really care about the absolute value of the change in x, and once you square it, it all becomes a positive anyway. So our change in x right over here is going to be positive seven. And our change in y, well, we are starting at, we are starting at y is equal to one, and we are going to y is equal to negative four, so it would be negative four minus one, which is equal to negative five. And so our change in y is negative five. You could view this distance right over here as the absolute value of our change in y, which of course would be the absolute value of five."}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "So our change in x right over here is going to be positive seven. And our change in y, well, we are starting at, we are starting at y is equal to one, and we are going to y is equal to negative four, so it would be negative four minus one, which is equal to negative five. And so our change in y is negative five. You could view this distance right over here as the absolute value of our change in y, which of course would be the absolute value of five. But once you square it, it doesn't matter, the negative sign goes away. And so this is going to simplify to seven squared, change in x squared is 49. Change in y squared, negative five squared is 25."}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "You could view this distance right over here as the absolute value of our change in y, which of course would be the absolute value of five. But once you square it, it doesn't matter, the negative sign goes away. And so this is going to simplify to seven squared, change in x squared is 49. Change in y squared, negative five squared is 25. So we get r squared, we get r squared is equal to 49 plus 25. So what's 49 plus 25? Let's see, that's going to be 54, was it 74?"}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "Change in y squared, negative five squared is 25. So we get r squared, we get r squared is equal to 49 plus 25. So what's 49 plus 25? Let's see, that's going to be 54, was it 74? R squared is equal to 74. Did I do that right? Yep, 74."}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "Let's see, that's going to be 54, was it 74? R squared is equal to 74. Did I do that right? Yep, 74. And so now we can write the equation for the circle. The circle is going to be all of the points that are, well, let me write, so if r squared is equal to 74, r is equal to the square root of 74. And so the equation of the circle is going to be all points x comma y that are this far away from the center."}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "Yep, 74. And so now we can write the equation for the circle. The circle is going to be all of the points that are, well, let me write, so if r squared is equal to 74, r is equal to the square root of 74. And so the equation of the circle is going to be all points x comma y that are this far away from the center. And so what are those points going to be? Well, the distance is going to be x minus the x coordinate of the center, x minus negative one squared. Let me do that in blue color."}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "And so the equation of the circle is going to be all points x comma y that are this far away from the center. And so what are those points going to be? Well, the distance is going to be x minus the x coordinate of the center, x minus negative one squared. Let me do that in blue color. Minus negative one squared plus y minus, y minus the y coordinate of the center, y minus one squared, is equal, is going to be equal to r squared, is going to be equal to the length of the radius squared. Well, r squared we already know is going to be 74. 74."}, {"video_title": "Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3", "Sentence": "Let me do that in blue color. Minus negative one squared plus y minus, y minus the y coordinate of the center, y minus one squared, is equal, is going to be equal to r squared, is going to be equal to the length of the radius squared. Well, r squared we already know is going to be 74. 74. And then if we want to simplify a little bit, you subtract a negative, this becomes a positive. So it simplifies to x plus one squared plus y minus one squared is equal to 74. Is equal to 74."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "Now, we call the longest side of a right triangle, we call that side, and you can either view it as the longest side of the right triangle, or the side opposite the 90 degree angle, it is called the hypotenuse. It's a very fancy word for a fairly simple idea, just the longest side of a right triangle, or the side opposite the 90 degree angle. And it's just good to know that because someone might say hypotenuse, like, oh, they're just talking about this side right here, the side longest, the side opposite the 90 degree angle. Now, what I want to do in this video is prove a relationship, a very famous relationship, and you might see where this is going, a very famous relationship between the lengths of the sides of a right triangle. So let's say that the length of AC, so uppercase A, uppercase C, let's call that length lowercase a, let's call the length of BC, lowercase b right over here, I'll use uppercases for points, lowercases for lengths, and let's call the length of the hypotenuse, the length of AB, let's call that C. And let's see if we can come up with a relationship between A, B, and C. And to do that, I'm first going to construct another line, or another segment, I should say, between C and the hypotenuse, and I'm going to construct it so that they intersect at a right angle. And you can always do that, and we'll call this point right over here, we'll call this point capital D. And if you're wondering, how can you always do that, you can imagine rotating this entire triangle like this, and this isn't a rigorous proof, but it just kind of gives you the general idea of how you can always construct a point like this. So if I've rotated it around, so now our hypotenuse, we're now sitting on our hypotenuse, this is now point B, this is point A, and if I rotated the whole thing all the way around, this is point C, you can imagine just dropping a rock from point C, maybe with a string attached, and it would hit the hypotenuse at a right angle."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "Now, what I want to do in this video is prove a relationship, a very famous relationship, and you might see where this is going, a very famous relationship between the lengths of the sides of a right triangle. So let's say that the length of AC, so uppercase A, uppercase C, let's call that length lowercase a, let's call the length of BC, lowercase b right over here, I'll use uppercases for points, lowercases for lengths, and let's call the length of the hypotenuse, the length of AB, let's call that C. And let's see if we can come up with a relationship between A, B, and C. And to do that, I'm first going to construct another line, or another segment, I should say, between C and the hypotenuse, and I'm going to construct it so that they intersect at a right angle. And you can always do that, and we'll call this point right over here, we'll call this point capital D. And if you're wondering, how can you always do that, you can imagine rotating this entire triangle like this, and this isn't a rigorous proof, but it just kind of gives you the general idea of how you can always construct a point like this. So if I've rotated it around, so now our hypotenuse, we're now sitting on our hypotenuse, this is now point B, this is point A, and if I rotated the whole thing all the way around, this is point C, you can imagine just dropping a rock from point C, maybe with a string attached, and it would hit the hypotenuse at a right angle. So that's all we did here to establish segment CD, where we put our point D right over there. And the reason why I did that is now we can do all sorts of interesting relationships between similar triangles, because we have three triangles here. We have triangle ADC, we have triangle DBC, and then we have the larger original triangle, and we can hopefully establish similarity between those triangles."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "So if I've rotated it around, so now our hypotenuse, we're now sitting on our hypotenuse, this is now point B, this is point A, and if I rotated the whole thing all the way around, this is point C, you can imagine just dropping a rock from point C, maybe with a string attached, and it would hit the hypotenuse at a right angle. So that's all we did here to establish segment CD, where we put our point D right over there. And the reason why I did that is now we can do all sorts of interesting relationships between similar triangles, because we have three triangles here. We have triangle ADC, we have triangle DBC, and then we have the larger original triangle, and we can hopefully establish similarity between those triangles. And first I'll show you that ADC is similar to the larger one, because both of them have a right angle. ADC has a right angle right over here. Clearly if this angle is 90 degrees, then this angle is going to be 90 degrees as well."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "We have triangle ADC, we have triangle DBC, and then we have the larger original triangle, and we can hopefully establish similarity between those triangles. And first I'll show you that ADC is similar to the larger one, because both of them have a right angle. ADC has a right angle right over here. Clearly if this angle is 90 degrees, then this angle is going to be 90 degrees as well. They're supplementary. They have to add up to 180. And so they both have a right angle in them."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "Clearly if this angle is 90 degrees, then this angle is going to be 90 degrees as well. They're supplementary. They have to add up to 180. And so they both have a right angle in them. So the smaller one has a right angle, the larger one clearly has a right angle. That's where we started from. And they also both share this angle right over here."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "And so they both have a right angle in them. So the smaller one has a right angle, the larger one clearly has a right angle. That's where we started from. And they also both share this angle right over here. Angle DAC or BAC, however you want to refer to it. So we can actually write down that triangle. I'm going to start with the smaller one."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "And they also both share this angle right over here. Angle DAC or BAC, however you want to refer to it. So we can actually write down that triangle. I'm going to start with the smaller one. ADC, and maybe I'll shade it in right over here. So this is the triangle we're talking about. Triangle ADC, and I went from the blue angle to the right angle to the unlabeled angle from the point of view of triangle ADC."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "I'm going to start with the smaller one. ADC, and maybe I'll shade it in right over here. So this is the triangle we're talking about. Triangle ADC, and I went from the blue angle to the right angle to the unlabeled angle from the point of view of triangle ADC. This right angle isn't applying to that right over there. It's applying to the larger triangle. So we can say triangle ADC is similar to triangle."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "Triangle ADC, and I went from the blue angle to the right angle to the unlabeled angle from the point of view of triangle ADC. This right angle isn't applying to that right over there. It's applying to the larger triangle. So we can say triangle ADC is similar to triangle. Once again, you want to start at the blue angle, A. Then we went to the right angle, so we want to go to the right angle again. So it's ACB."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "So we can say triangle ADC is similar to triangle. Once again, you want to start at the blue angle, A. Then we went to the right angle, so we want to go to the right angle again. So it's ACB. And because they're similar, we can set up a relationship between the ratios of their sides. For example, we know the ratio of corresponding sides are going to, well, in general for a similar triangle, we know that the ratio of the corresponding sides are going to be a constant. So we could take the ratio of the hypotenuse of this side, of the smaller triangle."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "So it's ACB. And because they're similar, we can set up a relationship between the ratios of their sides. For example, we know the ratio of corresponding sides are going to, well, in general for a similar triangle, we know that the ratio of the corresponding sides are going to be a constant. So we could take the ratio of the hypotenuse of this side, of the smaller triangle. So the hypotenuse is AC. So AC over the hypotenuse over the larger one, which is AB. AC over AB is going to be the same thing as AD, as one of the legs."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "So we could take the ratio of the hypotenuse of this side, of the smaller triangle. So the hypotenuse is AC. So AC over the hypotenuse over the larger one, which is AB. AC over AB is going to be the same thing as AD, as one of the legs. AD, just to show that I'm just taking corresponding points on both similar triangles. This is AD over AC. You can look at these triangles yourself and show, look, AD, point AD is between the blue angle and the right angle."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "AC over AB is going to be the same thing as AD, as one of the legs. AD, just to show that I'm just taking corresponding points on both similar triangles. This is AD over AC. You can look at these triangles yourself and show, look, AD, point AD is between the blue angle and the right angle. Sorry, side AD is between the blue angle and the right angle. Side AC is between the blue angle and the right angle on the larger triangle. So both of these are from the larger triangle."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "You can look at these triangles yourself and show, look, AD, point AD is between the blue angle and the right angle. Sorry, side AD is between the blue angle and the right angle. Side AC is between the blue angle and the right angle on the larger triangle. So both of these are from the larger triangle. These are the corresponding sides on the smaller triangle. And if that is confusing looking at them visually, as long as we wrote our similarity statement correctly, you can just find the corresponding points. AC corresponds to AB on the larger triangle."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "So both of these are from the larger triangle. These are the corresponding sides on the smaller triangle. And if that is confusing looking at them visually, as long as we wrote our similarity statement correctly, you can just find the corresponding points. AC corresponds to AB on the larger triangle. AD on the smaller triangle corresponds to AC on the larger triangle. And we know that AC, we can rewrite that as lowercase a. AC is lowercase a. We don't have any label for AD or for AB."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "AC corresponds to AB on the larger triangle. AD on the smaller triangle corresponds to AC on the larger triangle. And we know that AC, we can rewrite that as lowercase a. AC is lowercase a. We don't have any label for AD or for AB. Sorry, we do have a label for AB. That is C right over here. We don't have a label for AD."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "We don't have any label for AD or for AB. Sorry, we do have a label for AB. That is C right over here. We don't have a label for AD. So let's just call that lowercase d. So lowercase d applies to that part right over there. C applies to that entire part right over there. And then we'll call DB, let's call that length E. That will just make things a little bit simpler for us."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "We don't have a label for AD. So let's just call that lowercase d. So lowercase d applies to that part right over there. C applies to that entire part right over there. And then we'll call DB, let's call that length E. That will just make things a little bit simpler for us. So AD we'll just call d. And so we have a over c is equal to d over a. If we cross multiply, you have a times a, which is a squared, is equal to c times d, which is cd. So that's a little bit of an interesting result."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "And then we'll call DB, let's call that length E. That will just make things a little bit simpler for us. So AD we'll just call d. And so we have a over c is equal to d over a. If we cross multiply, you have a times a, which is a squared, is equal to c times d, which is cd. So that's a little bit of an interesting result. Let's see what we can do with the other triangle right over here. So this triangle right over here. So once again, it has a right angle."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "So that's a little bit of an interesting result. Let's see what we can do with the other triangle right over here. So this triangle right over here. So once again, it has a right angle. The larger one has a right angle. And they both share this angle right over here. So by angle-angle similarity, the two triangles are going to be similar."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "So once again, it has a right angle. The larger one has a right angle. And they both share this angle right over here. So by angle-angle similarity, the two triangles are going to be similar. So we can say triangle BDC. We went from pink to right to not labeled. So triangle BDC is similar to triangle."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "So by angle-angle similarity, the two triangles are going to be similar. So we can say triangle BDC. We went from pink to right to not labeled. So triangle BDC is similar to triangle. Now we're going to look at the larger triangle. We're going to start at the pink angle, B. Now we go to the right angle, CA."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "So triangle BDC is similar to triangle. Now we're going to look at the larger triangle. We're going to start at the pink angle, B. Now we go to the right angle, CA. From pink angle to right angle to non-labeled angle, at least from the point of view here. We labeled it before with that blue. So now let's set up some type of a relationship here."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "Now we go to the right angle, CA. From pink angle to right angle to non-labeled angle, at least from the point of view here. We labeled it before with that blue. So now let's set up some type of a relationship here. We can say that the ratio on the smaller triangle, BC over BA. Once again, we're taking the hypotenuses of both of them. So BC over BA is going to be equal to BD."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "So now let's set up some type of a relationship here. We can say that the ratio on the smaller triangle, BC over BA. Once again, we're taking the hypotenuses of both of them. So BC over BA is going to be equal to BD. Let me do this in another color. So this is one of the legs. The way I drew it is the shorter legs."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "So BC over BA is going to be equal to BD. Let me do this in another color. So this is one of the legs. The way I drew it is the shorter legs. BD over BC. I'm just taking the corresponding vertices. Over BC."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "The way I drew it is the shorter legs. BD over BC. I'm just taking the corresponding vertices. Over BC. And once again, we know BC is the same thing as lowercase b. BA is lowercase c. And then BD we defined as lowercase e. This is lowercase e. We can cross multiply here and we get B times B. I've mentioned this in many videos. Cross multiplying is really the same thing as multiplying both sides by both denominators. B times B is B squared is equal to CE."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "Over BC. And once again, we know BC is the same thing as lowercase b. BA is lowercase c. And then BD we defined as lowercase e. This is lowercase e. We can cross multiply here and we get B times B. I've mentioned this in many videos. Cross multiplying is really the same thing as multiplying both sides by both denominators. B times B is B squared is equal to CE. And now we can do something kind of interesting. We can add these two statements. Let me rewrite the statement down here."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "B times B is B squared is equal to CE. And now we can do something kind of interesting. We can add these two statements. Let me rewrite the statement down here. So B squared is equal to CE. So if we add the left-hand sides, we get A squared plus B squared is equal to CD plus CE. And then we have a C in both of these terms so we can factor it out."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "Let me rewrite the statement down here. So B squared is equal to CE. So if we add the left-hand sides, we get A squared plus B squared is equal to CD plus CE. And then we have a C in both of these terms so we can factor it out. This is going to be equal to C times D plus E. Close the parentheses. Now what is D plus E? D is this length."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "And then we have a C in both of these terms so we can factor it out. This is going to be equal to C times D plus E. Close the parentheses. Now what is D plus E? D is this length. E is this length. So D plus E is actually going to be C as well. So this is going to be C. So you have C times C which is just the same thing as C squared."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "D is this length. E is this length. So D plus E is actually going to be C as well. So this is going to be C. So you have C times C which is just the same thing as C squared. So now we have an interesting relationship. We have that A squared plus B squared is equal to C squared. Let me rewrite that."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "So this is going to be C. So you have C times C which is just the same thing as C squared. So now we have an interesting relationship. We have that A squared plus B squared is equal to C squared. Let me rewrite that. A squared. I'll do that in an arbitrary new color. I deleted that by accident."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "Let me rewrite that. A squared. I'll do that in an arbitrary new color. I deleted that by accident. So let me rewrite it. So we've just established that A squared plus B squared is equal to C squared. And this is just an arbitrary right triangle."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "I deleted that by accident. So let me rewrite it. So we've just established that A squared plus B squared is equal to C squared. And this is just an arbitrary right triangle. This is true for any two right triangles. We've just established that the sum of the squares of each of the legs is equal to the square of the hypotenuse. And this is probably what's easily one of the most famous theorems in mathematics named for Pythagoras."}, {"video_title": "Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3", "Sentence": "And this is just an arbitrary right triangle. This is true for any two right triangles. We've just established that the sum of the squares of each of the legs is equal to the square of the hypotenuse. And this is probably what's easily one of the most famous theorems in mathematics named for Pythagoras. Not clear if he's the first person to establish this, but it's called the Pythagorean theorem. And it's really the basis of, well, not all of geometry, but a lot of the geometry that we're going to do. And it forms the basis of a lot of the trigonometry we're going to do."}, {"video_title": "Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Her character is on a quest to vanquish an evil sorcerer and his minions from the land. Her character is a wizard whose spells have a range of six meters. The locations of objects in the game are stored by the computer in terms of x and y coordinates. So 5, 4 is the location of Alyssa's wizard. 8, 7 is the location of minion A. 2, negative 1 is the location of minion B. 9, 0 is the location of minion C. So what I want to do, and I want you to pause this video, and I want you to think about, given that her wizard has a range of six meters, which of these minions can the wizard actually reach?"}, {"video_title": "Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So 5, 4 is the location of Alyssa's wizard. 8, 7 is the location of minion A. 2, negative 1 is the location of minion B. 9, 0 is the location of minion C. So what I want to do, and I want you to pause this video, and I want you to think about, given that her wizard has a range of six meters, which of these minions can the wizard actually reach? I'm assuming you've given a go at it. And we just have to remind ourselves, to figure out which of these minions are in reach, we have to say, well, which of these points are within six units? We're assuming that these units are in meters right over here."}, {"video_title": "Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3", "Sentence": "9, 0 is the location of minion C. So what I want to do, and I want you to pause this video, and I want you to think about, given that her wizard has a range of six meters, which of these minions can the wizard actually reach? I'm assuming you've given a go at it. And we just have to remind ourselves, to figure out which of these minions are in reach, we have to say, well, which of these points are within six units? We're assuming that these units are in meters right over here. Which of these points are within six units of 5, 4? And to think about that, we just have to calculate the distance between this point and this point, that point and that point, that point and that point, and see if they are greater than or less than six meters. And how do we calculate a distance between two points?"}, {"video_title": "Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3", "Sentence": "We're assuming that these units are in meters right over here. Which of these points are within six units of 5, 4? And to think about that, we just have to calculate the distance between this point and this point, that point and that point, that point and that point, and see if they are greater than or less than six meters. And how do we calculate a distance between two points? Well, if this is some point right over here, that's x1 comma y1. And then this is another point right over here, x2 comma y2. And we want to calculate this distance right over here."}, {"video_title": "Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And how do we calculate a distance between two points? Well, if this is some point right over here, that's x1 comma y1. And then this is another point right over here, x2 comma y2. And we want to calculate this distance right over here. The distance formula comes straight out of the Pythagorean theorem. The Pythagorean theorem tells us if this side right over here is our change in y, and let's actually just write that as the absolute value of our change in y. And let's say that this side right over here is the absolute value of our change in x."}, {"video_title": "Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And we want to calculate this distance right over here. The distance formula comes straight out of the Pythagorean theorem. The Pythagorean theorem tells us if this side right over here is our change in y, and let's actually just write that as the absolute value of our change in y. And let's say that this side right over here is the absolute value of our change in x. The Pythagorean theorem tells us that this one, the hypotenuse, is going to be the square root of the sum of the squares of the two sides. So change in x squared plus change in y squared. You might say, hey, what happened to the absolute value?"}, {"video_title": "Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And let's say that this side right over here is the absolute value of our change in x. The Pythagorean theorem tells us that this one, the hypotenuse, is going to be the square root of the sum of the squares of the two sides. So change in x squared plus change in y squared. You might say, hey, what happened to the absolute value? Well, when I square it, it's going to be positive anyway, so I don't have to write down the absolute value. So really, I just need to figure out between each of these two points, what is the change in x, what's the change in y, square them, add them together, and then take the square root. So for example, let's find, if I were to call this P1, if I were to call this P2, let me call this P3."}, {"video_title": "Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3", "Sentence": "You might say, hey, what happened to the absolute value? Well, when I square it, it's going to be positive anyway, so I don't have to write down the absolute value. So really, I just need to figure out between each of these two points, what is the change in x, what's the change in y, square them, add them together, and then take the square root. So for example, let's find, if I were to call this P1, if I were to call this P2, let me call this P3. I'm going to do them in different colors so you can keep track of what I'm doing. This is P3, and let's say this is P4. So let's first think about the distance."}, {"video_title": "Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So for example, let's find, if I were to call this P1, if I were to call this P2, let me call this P3. I'm going to do them in different colors so you can keep track of what I'm doing. This is P3, and let's say this is P4. So let's first think about the distance. The distance between P1 and P2, well, that's going to be equal to the square root of our change in x squared. So our change in x is 3. That squared is 9."}, {"video_title": "Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So let's first think about the distance. The distance between P1 and P2, well, that's going to be equal to the square root of our change in x squared. So our change in x is 3. That squared is 9. Plus our change in y squared. Our change in y is also 3. That squared is 9."}, {"video_title": "Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3", "Sentence": "That squared is 9. Plus our change in y squared. Our change in y is also 3. That squared is 9. So this is going to be square root of 18, which is the same thing as 3 square roots of 2. Now, is this more or less than 6? Well, 3 times 2 is equal to 6."}, {"video_title": "Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3", "Sentence": "That squared is 9. So this is going to be square root of 18, which is the same thing as 3 square roots of 2. Now, is this more or less than 6? Well, 3 times 2 is equal to 6. Square root of 2 is less than 2. It's 1 point something. So this right over here is going to be less than 6."}, {"video_title": "Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Well, 3 times 2 is equal to 6. Square root of 2 is less than 2. It's 1 point something. So this right over here is going to be less than 6. So P2 is in range. Alyssa's wizard can get minion A. Minion A she can attack. Now let's think about minion B."}, {"video_title": "Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So this right over here is going to be less than 6. So P2 is in range. Alyssa's wizard can get minion A. Minion A she can attack. Now let's think about minion B. So the distance between P1 and P3 is going to be equal to the square root of. So your change in x, it's negative 3. Negative 3 squared is positive 9."}, {"video_title": "Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Now let's think about minion B. So the distance between P1 and P3 is going to be equal to the square root of. So your change in x, it's negative 3. Negative 3 squared is positive 9. Our change in y, to go from 4 to negative 1, it's negative 5. That squared is 25. So 9 plus 25, which is equal to the square root of 34."}, {"video_title": "Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Negative 3 squared is positive 9. Our change in y, to go from 4 to negative 1, it's negative 5. That squared is 25. So 9 plus 25, which is equal to the square root of 34. Now, is this greater than or equal to 6? Well, the square root of 36 is 6. So this is a square root of a lower number."}, {"video_title": "Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So 9 plus 25, which is equal to the square root of 34. Now, is this greater than or equal to 6? Well, the square root of 36 is 6. So this is a square root of a lower number. So this is going to be less than 6 as well. So P minion B is also in reach. Now let's think about this last point, the distance."}, {"video_title": "Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So this is a square root of a lower number. So this is going to be less than 6 as well. So P minion B is also in reach. Now let's think about this last point, the distance. The distance between P1 and P4 is going to be equal to the square root of our change in x squared. Change in x is 4, squared is 16. Plus our change in y squared."}, {"video_title": "Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Now let's think about this last point, the distance. The distance between P1 and P4 is going to be equal to the square root of our change in x squared. Change in x is 4, squared is 16. Plus our change in y squared. Our change in y is negative 4, but you square that, you get another 16. So this is going to be square root of 32, which could be written as, actually, we could just leave it as square root of 32. Square root of 32 is clearly less than the square root of 36, which is 6."}, {"video_title": "Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Plus our change in y squared. Our change in y is negative 4, but you square that, you get another 16. So this is going to be square root of 32, which could be written as, actually, we could just leave it as square root of 32. Square root of 32 is clearly less than the square root of 36, which is 6. So this is also going to be less than 6. So she can get at all of the minions. They're all within 6 units of her."}, {"video_title": "Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Square root of 32 is clearly less than the square root of 36, which is 6. So this is also going to be less than 6. So she can get at all of the minions. They're all within 6 units of her. Now, which of these is the furthest away? Well, actually, if we were to write this, we simplified this, but we could write this as the square root of 18. Square root of 18 is clearly the smallest out of square root of 18, square root of 32, and square root of 34."}, {"video_title": "Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3", "Sentence": "They're all within 6 units of her. Now, which of these is the furthest away? Well, actually, if we were to write this, we simplified this, but we could write this as the square root of 18. Square root of 18 is clearly the smallest out of square root of 18, square root of 32, and square root of 34. So minion A is the closest, and minion B, square root of 34, is the furthest. Or I should say the farthest. is the farthest."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "1, 2, 3, 4. So that's 3, negative 4. And I also have the point 6, 1. So 1, 2, 3, 4, 5, 6, 1. So just like that. 6, 1. In the last video, we figured out that we could just use the Pythagorean theorem if we wanted to figure out the distance between these two points."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So 1, 2, 3, 4, 5, 6, 1. So just like that. 6, 1. In the last video, we figured out that we could just use the Pythagorean theorem if we wanted to figure out the distance between these two points. We just drew a triangle there and realized that this was the hypotenuse. In this video, we're going to try to figure out what is the coordinate of the point that is exactly halfway between this point and that point. So this right here is kind of the distance, the line that connects them."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "In the last video, we figured out that we could just use the Pythagorean theorem if we wanted to figure out the distance between these two points. We just drew a triangle there and realized that this was the hypotenuse. In this video, we're going to try to figure out what is the coordinate of the point that is exactly halfway between this point and that point. So this right here is kind of the distance, the line that connects them. Now what is the coordinate of the point that is exactly halfway in between the two? What is this coordinate right here? It's something comma something."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So this right here is kind of the distance, the line that connects them. Now what is the coordinate of the point that is exactly halfway in between the two? What is this coordinate right here? It's something comma something. And to do that, let me draw it really big here. Because I think you're going to find out that it's actually pretty straightforward. At first it seems like a really tough problem."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "It's something comma something. And to do that, let me draw it really big here. Because I think you're going to find out that it's actually pretty straightforward. At first it seems like a really tough problem. Gee, let me use the distance formula with some variables. But you're going to see it's actually one of the simplest things you'll learn in algebra and geometry. So let's say that this is my triangle right there."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "At first it seems like a really tough problem. Gee, let me use the distance formula with some variables. But you're going to see it's actually one of the simplest things you'll learn in algebra and geometry. So let's say that this is my triangle right there. That is my triangle right there. This right here is the point 6, 1. This down here is the point 3 comma negative 4."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So let's say that this is my triangle right there. That is my triangle right there. This right here is the point 6, 1. This down here is the point 3 comma negative 4. And we're looking for the point that is smack dab in between those two points. What are its coordinates? It seems very hard at first."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "This down here is the point 3 comma negative 4. And we're looking for the point that is smack dab in between those two points. What are its coordinates? It seems very hard at first. But it's easy when you think about it in terms of just the x and the y coordinates. What's this guy's x coordinate going to be? This line out here represents x is equal to 6."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "It seems very hard at first. But it's easy when you think about it in terms of just the x and the y coordinates. What's this guy's x coordinate going to be? This line out here represents x is equal to 6. This over here, let me do it in a little darker color, this over here represents x is equal to 6. This over here represents x is equal to 3. What will this guy's x coordinate be?"}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "This line out here represents x is equal to 6. This over here, let me do it in a little darker color, this over here represents x is equal to 6. This over here represents x is equal to 3. What will this guy's x coordinate be? Well, his x coordinate is going to be smack dab in between the two x coordinates. This is x is equal to 3. This is x equal to 6."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "What will this guy's x coordinate be? Well, his x coordinate is going to be smack dab in between the two x coordinates. This is x is equal to 3. This is x equal to 6. He's going to be right in between. This distance is going to be equal to that distance. His x coordinate is going to be right in between the 3 and the 6."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "This is x equal to 6. He's going to be right in between. This distance is going to be equal to that distance. His x coordinate is going to be right in between the 3 and the 6. So what do we call the number that's right in between the 3 and the 6? Well, we could even call that the midpoint. Or we could call it the mean or the average, or however you want to talk about it."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "His x coordinate is going to be right in between the 3 and the 6. So what do we call the number that's right in between the 3 and the 6? Well, we could even call that the midpoint. Or we could call it the mean or the average, or however you want to talk about it. We just want to know what's the average of 3 and 6. So to figure out this point, the point halfway between 3 and 6, you literally just figure out 3 plus 6 over 2, which is equal to 4.5. So this x coordinate is going to be 4.5."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Or we could call it the mean or the average, or however you want to talk about it. We just want to know what's the average of 3 and 6. So to figure out this point, the point halfway between 3 and 6, you literally just figure out 3 plus 6 over 2, which is equal to 4.5. So this x coordinate is going to be 4.5. Let me draw that on this graph. 1, 2, 3, 4.5. And you see it's smack dab in between."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So this x coordinate is going to be 4.5. Let me draw that on this graph. 1, 2, 3, 4.5. And you see it's smack dab in between. That's its x coordinate. Now, by the exact same logic, this guy's y coordinate is going to be smack dab between y is equal to negative 4 and y is equal to 1. So it's going to be right in between those."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And you see it's smack dab in between. That's its x coordinate. Now, by the exact same logic, this guy's y coordinate is going to be smack dab between y is equal to negative 4 and y is equal to 1. So it's going to be right in between those. So this is the x right there. The y coordinate is just going to be right in between y is equal to negative 4 and y is equal to 1. So you just take the average."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So it's going to be right in between those. So this is the x right there. The y coordinate is just going to be right in between y is equal to negative 4 and y is equal to 1. So you just take the average. 1 plus negative 4 over 2, that's equal to negative 3 over 2, or you could say negative 1.5. So you go down 1.5. It is literally right there."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So you just take the average. 1 plus negative 4 over 2, that's equal to negative 3 over 2, or you could say negative 1.5. So you go down 1.5. It is literally right there. So just like that, you literally take the average of the x's, take the average of the y's, or maybe I should say the mean to be a little bit more specific, a mean of only two points, and you will get the midpoint of those two points, the point that's equidistant from both of them. It's the midpoint of the line that connects them. So the coordinates are 4.5 comma negative 1.5."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "It is literally right there. So just like that, you literally take the average of the x's, take the average of the y's, or maybe I should say the mean to be a little bit more specific, a mean of only two points, and you will get the midpoint of those two points, the point that's equidistant from both of them. It's the midpoint of the line that connects them. So the coordinates are 4.5 comma negative 1.5. Let's do a couple more of these. These actually, you're going to find are very, very straightforward. But just to visualize it, let me graph it."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So the coordinates are 4.5 comma negative 1.5. Let's do a couple more of these. These actually, you're going to find are very, very straightforward. But just to visualize it, let me graph it. Let's say I have the point 4, negative 5. So 1, 2, 3, 4, and then go down 5. 1, 2, 3, 4, 5."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "But just to visualize it, let me graph it. Let's say I have the point 4, negative 5. So 1, 2, 3, 4, and then go down 5. 1, 2, 3, 4, 5. So that's 4, negative 5. And I have the point 8 comma 2. So 1, 2, 3, 4, 5, 6, 7, 8 comma 2."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "1, 2, 3, 4, 5. So that's 4, negative 5. And I have the point 8 comma 2. So 1, 2, 3, 4, 5, 6, 7, 8 comma 2. So what is the coordinate of the midpoint of these two points, the point that is smack dab in between them? Well, we just average the x's, average the y's. So the midpoint is going to be the x values are 8 and 4."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So 1, 2, 3, 4, 5, 6, 7, 8 comma 2. So what is the coordinate of the midpoint of these two points, the point that is smack dab in between them? Well, we just average the x's, average the y's. So the midpoint is going to be the x values are 8 and 4. So it's going to be 8 plus 4 over 2. And the y value is going to be, well, we have a 2 and a negative 5, so you get 2 plus negative 5 over 2. And what is this equal to?"}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So the midpoint is going to be the x values are 8 and 4. So it's going to be 8 plus 4 over 2. And the y value is going to be, well, we have a 2 and a negative 5, so you get 2 plus negative 5 over 2. And what is this equal to? This is 12 over 2, which is 6 comma 2 minus 5 is negative 3. Negative 3 over 2 is negative 1.5. So that right there is the midpoint."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And what is this equal to? This is 12 over 2, which is 6 comma 2 minus 5 is negative 3. Negative 3 over 2 is negative 1.5. So that right there is the midpoint. You literally just average the x's and average the y's or find their mean. So let's graph it just to make sure it looks like the midpoint. 6, negative 5."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So that right there is the midpoint. You literally just average the x's and average the y's or find their mean. So let's graph it just to make sure it looks like the midpoint. 6, negative 5. 1, 2, 3, 4, 5, 6. Negative 1.5. Negative 1, negative 1.5."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "6, negative 5. 1, 2, 3, 4, 5, 6. Negative 1.5. Negative 1, negative 1.5. Yep, looks pretty good. It looks like it's equidistant from this point and that point up there. Now, that's all you have to remember."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Negative 1, negative 1.5. Yep, looks pretty good. It looks like it's equidistant from this point and that point up there. Now, that's all you have to remember. Average the x or take the mean of the x or find the x that's right in between the two. Average the y's. You've got the midpoint."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Now, that's all you have to remember. Average the x or take the mean of the x or find the x that's right in between the two. Average the y's. You've got the midpoint. What I'm going to show you now is what's in many textbooks. They'll write, oh, if I have the point x1, y1, and then I have the point, actually I'm just sticking yellow. It's kind of painful to switch colors all the time."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "You've got the midpoint. What I'm going to show you now is what's in many textbooks. They'll write, oh, if I have the point x1, y1, and then I have the point, actually I'm just sticking yellow. It's kind of painful to switch colors all the time. And then I have the point x2, y2. Many books will give you something called the midpoint formula, which once again, I think is kind of silly to memorize. Just remember, you just average."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "It's kind of painful to switch colors all the time. And then I have the point x2, y2. Many books will give you something called the midpoint formula, which once again, I think is kind of silly to memorize. Just remember, you just average. Find the x in between, find the y in between. So midpoint formula. What they'll really say is the midpoint."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Just remember, you just average. Find the x in between, find the y in between. So midpoint formula. What they'll really say is the midpoint. So maybe we'll say the midpoint x, or maybe I'll call it this way. I'm just making up notation. The x midpoint and the y midpoint is going to be equal to, and they'll give you this formula, x1 plus x2 over 2."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "What they'll really say is the midpoint. So maybe we'll say the midpoint x, or maybe I'll call it this way. I'm just making up notation. The x midpoint and the y midpoint is going to be equal to, and they'll give you this formula, x1 plus x2 over 2. And then y1 plus y2 over 2. And it looks like something you have to memorize. But all you have to say is, look, that's just the average or the mean of these two numbers."}, {"video_title": "Midpoint formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "The x midpoint and the y midpoint is going to be equal to, and they'll give you this formula, x1 plus x2 over 2. And then y1 plus y2 over 2. And it looks like something you have to memorize. But all you have to say is, look, that's just the average or the mean of these two numbers. This is just the average or the mean of these two numbers. I'm just saying I'm adding the two together, dividing by 2. Adding these two together, dividing by 2."}, {"video_title": "More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3", "Sentence": "Several videos ago, I very quickly went through why side-side angle is not a valid postulate. What I want to do in this video is explore it a little bit more. It's not called angle-side-side for obvious reasons, because then the acronym would make people giggle in geometry class. I guess we don't want people giggling while they're doing mathematics. Let's just think about a triangle here. Let's say I have a triangle that looks something like this. If I have a triangle that looks something like that."}, {"video_title": "More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3", "Sentence": "I guess we don't want people giggling while they're doing mathematics. Let's just think about a triangle here. Let's say I have a triangle that looks something like this. If I have a triangle that looks something like that. Let's say that we've found another triangle that has a congruent side. A side that is congruent to this side right over here. That is next to any side on the triangle."}, {"video_title": "More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3", "Sentence": "If I have a triangle that looks something like that. Let's say that we've found another triangle that has a congruent side. A side that is congruent to this side right over here. That is next to any side on the triangle. Next to that is a side that is congruent to this side right over here. That side is one of the sides of an angle. It forms one of the parts of an angle right over here."}, {"video_title": "More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3", "Sentence": "That is next to any side on the triangle. Next to that is a side that is congruent to this side right over here. That side is one of the sides of an angle. It forms one of the parts of an angle right over here. That other triangle has a congruent angle right over here. This is the angle that that first side is not a part of. Only that second side is part of this angle."}, {"video_title": "More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3", "Sentence": "It forms one of the parts of an angle right over here. That other triangle has a congruent angle right over here. This is the angle that that first side is not a part of. Only that second side is part of this angle. This is side-side-angle, or you could call it angle-side-side. And giggle a little bit about it. How do we know that this doesn't by itself show that this is congruent?"}, {"video_title": "More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3", "Sentence": "Only that second side is part of this angle. This is side-side-angle, or you could call it angle-side-side. And giggle a little bit about it. How do we know that this doesn't by itself show that this is congruent? We'd have to show that this could actually imply two different triangles. To think about that, let's say we know that the angle, we know that this other triangle has that same yellow angle there. Which means that the blue side is going to have to look something like that."}, {"video_title": "More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3", "Sentence": "How do we know that this doesn't by itself show that this is congruent? We'd have to show that this could actually imply two different triangles. To think about that, let's say we know that the angle, we know that this other triangle has that same yellow angle there. Which means that the blue side is going to have to look something like that. Just the way we drew it over here. This side down here, I'll make it a green side. This green side down here, we know nothing about."}, {"video_title": "More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3", "Sentence": "Which means that the blue side is going to have to look something like that. Just the way we drew it over here. This side down here, I'll make it a green side. This green side down here, we know nothing about. We never said that this is congruent to anything. If we knew, then we could use side-side-side. We only know that this side is congruent, this side is congruent, and this angle is congruent."}, {"video_title": "More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3", "Sentence": "This green side down here, we know nothing about. We never said that this is congruent to anything. If we knew, then we could use side-side-side. We only know that this side is congruent, this side is congruent, and this angle is congruent. So this green side, and I'll draw it as a dotted line, it could be of any length. We don't know what the length is of that green side. Now we have this magenta side."}, {"video_title": "More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3", "Sentence": "We only know that this side is congruent, this side is congruent, and this angle is congruent. So this green side, and I'll draw it as a dotted line, it could be of any length. We don't know what the length is of that green side. Now we have this magenta side. We have another side that's congruent here. This thing could pivot over here. We know nothing about this angle."}, {"video_title": "More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3", "Sentence": "Now we have this magenta side. We have another side that's congruent here. This thing could pivot over here. We know nothing about this angle. It could form any angle, but it does have to get to this other side. One possibility is that maybe the triangles are congruent. Maybe this side does go down just like that."}, {"video_title": "More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3", "Sentence": "We know nothing about this angle. It could form any angle, but it does have to get to this other side. One possibility is that maybe the triangles are congruent. Maybe this side does go down just like that. In which case, we actually would have congruent triangles. The kind of aha moment here, or the reason why SSA isn't possible, is that this side could also come down like this. There's two ways to get down to this base, if you want to call it that way."}, {"video_title": "More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3", "Sentence": "Maybe this side does go down just like that. In which case, we actually would have congruent triangles. The kind of aha moment here, or the reason why SSA isn't possible, is that this side could also come down like this. There's two ways to get down to this base, if you want to call it that way. You can come out that way, or you can kind of come in this way. That's why SSA by itself, with no other information, is ambiguous. It does not give you enough information to say that those triangles are definitely the same."}, {"video_title": "More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3", "Sentence": "There's two ways to get down to this base, if you want to call it that way. You can come out that way, or you can kind of come in this way. That's why SSA by itself, with no other information, is ambiguous. It does not give you enough information to say that those triangles are definitely the same. Now, there are special cases. In this situation, right over here, our angle, the angle in our SSA, our angle was acute. This is an acute angle right over here."}, {"video_title": "More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3", "Sentence": "It does not give you enough information to say that those triangles are definitely the same. Now, there are special cases. In this situation, right over here, our angle, the angle in our SSA, our angle was acute. This is an acute angle right over here. When you have an acute angle as one of the sides of your triangle, the other sides of the triangle, you could still have an obtuse angle. Remember, acute means less than 90 degrees. Obtuse means greater than 90 degrees."}, {"video_title": "More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3", "Sentence": "This is an acute angle right over here. When you have an acute angle as one of the sides of your triangle, the other sides of the triangle, you could still have an obtuse angle. Remember, acute means less than 90 degrees. Obtuse means greater than 90 degrees. You could still have an obtuse angle. That's why this is an option. One option is that you have two other acute angles."}, {"video_title": "More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3", "Sentence": "Obtuse means greater than 90 degrees. You could still have an obtuse angle. That's why this is an option. One option is that you have two other acute angles. This could be acute. This is also acute. Also acute."}, {"video_title": "More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3", "Sentence": "One option is that you have two other acute angles. This could be acute. This is also acute. Also acute. Also acute. You had the option where this is even more acute, even narrower, and then this becomes an obtuse angle. That is an obtuse angle."}, {"video_title": "More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3", "Sentence": "Also acute. Also acute. You had the option where this is even more acute, even narrower, and then this becomes an obtuse angle. That is an obtuse angle. That's only possible if you don't have two obtuse angles in the same triangle. You can't have two things that have larger than 90 degree measure in the same triangle. That's why there is a possibility where if you have another triangle that looks like this, and if I were to tell you very clearly that this angle right over here is obtuse, and that is the A in the SSA, I would say I have another triangle where this angle is congruent to that other triangle, some angle of that other triangle, and then one of the sides adjacent to it is congruent, and then the next side over is also congruent, but that's not so ambiguous because we could try to draw that."}, {"video_title": "More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3", "Sentence": "That is an obtuse angle. That's only possible if you don't have two obtuse angles in the same triangle. You can't have two things that have larger than 90 degree measure in the same triangle. That's why there is a possibility where if you have another triangle that looks like this, and if I were to tell you very clearly that this angle right over here is obtuse, and that is the A in the SSA, I would say I have another triangle where this angle is congruent to that other triangle, some angle of that other triangle, and then one of the sides adjacent to it is congruent, and then the next side over is also congruent, but that's not so ambiguous because we could try to draw that. Let's draw that same congruent obtuse angle. We know nothing about this side down here because we haven't said that that's necessarily congruent, so that could be of any length. We do know that this triangle is going to have the same length for this side of the angle."}, {"video_title": "More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3", "Sentence": "That's why there is a possibility where if you have another triangle that looks like this, and if I were to tell you very clearly that this angle right over here is obtuse, and that is the A in the SSA, I would say I have another triangle where this angle is congruent to that other triangle, some angle of that other triangle, and then one of the sides adjacent to it is congruent, and then the next side over is also congruent, but that's not so ambiguous because we could try to draw that. Let's draw that same congruent obtuse angle. We know nothing about this side down here because we haven't said that that's necessarily congruent, so that could be of any length. We do know that this triangle is going to have the same length for this side of the angle. It looks like this. Then we know that this side is also going to be the same length. We haven't told you anything about this angle right over here, so this side could pivot over here."}, {"video_title": "More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3", "Sentence": "We do know that this triangle is going to have the same length for this side of the angle. It looks like this. Then we know that this side is also going to be the same length. We haven't told you anything about this angle right over here, so this side could pivot over here. We could kind of rotate it over there, but there's only one way now that this orange side can reach this green side. Now the only way is this way over here. We were more constrained, or this case isn't ambiguous, because we used up our obtuse angle here."}, {"video_title": "More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3", "Sentence": "We haven't told you anything about this angle right over here, so this side could pivot over here. We could kind of rotate it over there, but there's only one way now that this orange side can reach this green side. Now the only way is this way over here. We were more constrained, or this case isn't ambiguous, because we used up our obtuse angle here. The A here is an obtuse one, so then it constrains what the triangle can become. I don't want to make you say, maybe SSA in general, if SSA you do not want to use it as a postulate. I just wanted to make it clear that there is this special case where if you know that the A in the SSA is obtuse, then it becomes a little bit less ambiguous."}, {"video_title": "More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3", "Sentence": "We were more constrained, or this case isn't ambiguous, because we used up our obtuse angle here. The A here is an obtuse one, so then it constrains what the triangle can become. I don't want to make you say, maybe SSA in general, if SSA you do not want to use it as a postulate. I just wanted to make it clear that there is this special case where if you know that the A in the SSA is obtuse, then it becomes a little bit less ambiguous. Then finally there's a circumstance, so this is an acute angle where it would be ambiguous. You have the obtuse angle, and then you have something in between, which is the right angle, where you have the A in SSA is a right angle. If you had it like this, if you have a right angle, and you have some base of unknown length, but you fix this length right over here, if you know that this is fixed, because you're saying it's congruent to some other triangle, and if you know that the next length is fixed, and if you think about it, this next side is going to be the side opposite the right angle, it's going to have to be the hypotenuse of the right angle, if you know that the only way you can construct this, and similar to the obtuse case, and if you know the length of this, the only way you can do it is to bring it down over here."}, {"video_title": "More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3", "Sentence": "I just wanted to make it clear that there is this special case where if you know that the A in the SSA is obtuse, then it becomes a little bit less ambiguous. Then finally there's a circumstance, so this is an acute angle where it would be ambiguous. You have the obtuse angle, and then you have something in between, which is the right angle, where you have the A in SSA is a right angle. If you had it like this, if you have a right angle, and you have some base of unknown length, but you fix this length right over here, if you know that this is fixed, because you're saying it's congruent to some other triangle, and if you know that the next length is fixed, and if you think about it, this next side is going to be the side opposite the right angle, it's going to have to be the hypotenuse of the right angle, if you know that the only way you can construct this, and similar to the obtuse case, and if you know the length of this, the only way you can do it is to bring it down over here. That actually does lead to another postulate called the right angle side hypotenuse postulate, which is really just a special case of SSA where the angle is actually a right angle. Here they wrote the angle first. You can view this as angle, side, side, and they were able to do it because now they can write right angle, and so it doesn't form that embarrassing acronym."}, {"video_title": "More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3", "Sentence": "If you had it like this, if you have a right angle, and you have some base of unknown length, but you fix this length right over here, if you know that this is fixed, because you're saying it's congruent to some other triangle, and if you know that the next length is fixed, and if you think about it, this next side is going to be the side opposite the right angle, it's going to have to be the hypotenuse of the right angle, if you know that the only way you can construct this, and similar to the obtuse case, and if you know the length of this, the only way you can do it is to bring it down over here. That actually does lead to another postulate called the right angle side hypotenuse postulate, which is really just a special case of SSA where the angle is actually a right angle. Here they wrote the angle first. You can view this as angle, side, side, and they were able to do it because now they can write right angle, and so it doesn't form that embarrassing acronym. This would also be a little bit common sense, because if you know two sides of a right triangle, and we haven't gone into depth in this in the geometry playlist, but you might already be familiar with it, by Pythagorean theorem, you can always figure out the third side. If you have this information about any triangle, you can always figure out the third side, and then you can use side, side, side. I just want to show you this little special case, but in general, the important thing is that you can't just use SSA unless you have more information."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "Because at least it looks that triangle CFE is similar to ABE. And the intuition there is it's kind of embedded inside of it. And we're going to prove that to ourselves. And it also looks like triangle CFB is going to be similar to triangle DEB. But once again, we're going to have to prove that to ourselves. And then maybe we can deal with all the ratios of the different sides to CF right over here. And then actually figure out what CF is going to be."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "And it also looks like triangle CFB is going to be similar to triangle DEB. But once again, we're going to have to prove that to ourselves. And then maybe we can deal with all the ratios of the different sides to CF right over here. And then actually figure out what CF is going to be. So first, let's prove to ourselves that these definitely are similar triangles. So you have this 90 degree angle in ABE. And you have this 90 degree angle in CFE."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "And then actually figure out what CF is going to be. So first, let's prove to ourselves that these definitely are similar triangles. So you have this 90 degree angle in ABE. And you have this 90 degree angle in CFE. If we can prove just one other angle or one other set of corresponding angles is congruent in both, then we've proved that they're similar. And we can either show that, look, they both share this angle right over here. Angle CEF is the same as angle AEB."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "And you have this 90 degree angle in CFE. If we can prove just one other angle or one other set of corresponding angles is congruent in both, then we've proved that they're similar. And we can either show that, look, they both share this angle right over here. Angle CEF is the same as angle AEB. So we've shown two angles, two corresponding angles in these triangles. This is an angle in both triangles. They are congruent, so the triangles are going to be similar."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "Angle CEF is the same as angle AEB. So we've shown two angles, two corresponding angles in these triangles. This is an angle in both triangles. They are congruent, so the triangles are going to be similar. You could also show that this line is parallel to this line, because obviously these two angles are the same. And so these angles will also be the same. So they're definitely similar triangles."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "They are congruent, so the triangles are going to be similar. You could also show that this line is parallel to this line, because obviously these two angles are the same. And so these angles will also be the same. So they're definitely similar triangles. So let's just write that down, get that out of the way. We know that triangle ABE is similar to triangle CFE. And you want to make sure you get in the right order."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "So they're definitely similar triangles. So let's just write that down, get that out of the way. We know that triangle ABE is similar to triangle CFE. And you want to make sure you get in the right order. F is where the 90 degree angle is. B is where the 90 degree angle is. And then E is where this orange angle is."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "And you want to make sure you get in the right order. F is where the 90 degree angle is. B is where the 90 degree angle is. And then E is where this orange angle is. So CFE, it's similar to triangle CFE. Now let's see if we can figure out that same statement going the other way, looking at triangle DEB. So once again, you have a 90 degree angle here."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "And then E is where this orange angle is. So CFE, it's similar to triangle CFE. Now let's see if we can figure out that same statement going the other way, looking at triangle DEB. So once again, you have a 90 degree angle here. If this is 90, then this is definitely going to be 90 as well. You have a 90 degree angle here at CFB. You have a 90 degree angle at DEF or DEB, however you want to call it."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "So once again, you have a 90 degree angle here. If this is 90, then this is definitely going to be 90 as well. You have a 90 degree angle here at CFB. You have a 90 degree angle at DEF or DEB, however you want to call it. So they have one set of corresponding angles that are congruent. And then you'll also see that they both share this angle right over here on the smaller triangle. So I'm now looking at this triangle right over here as opposed to the one on the right."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "You have a 90 degree angle at DEF or DEB, however you want to call it. So they have one set of corresponding angles that are congruent. And then you'll also see that they both share this angle right over here on the smaller triangle. So I'm now looking at this triangle right over here as opposed to the one on the right. So they both share this angle right over here, DBE. Angle DBE is the same as angle CBF. So I've shown you already that we have this angle is congruent to this angle."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "So I'm now looking at this triangle right over here as opposed to the one on the right. So they both share this angle right over here, DBE. Angle DBE is the same as angle CBF. So I've shown you already that we have this angle is congruent to this angle. And we have this angle as a part of both. So it's obviously congruent to itself. So we have two corresponding angles that are congruent to each other."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "So I've shown you already that we have this angle is congruent to this angle. And we have this angle as a part of both. So it's obviously congruent to itself. So we have two corresponding angles that are congruent to each other. So we know that this larger triangle over here is similar to this smaller triangle over there. So let me write this down. So we also know that triangle DEB is similar to triangle CFB."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "So we have two corresponding angles that are congruent to each other. So we know that this larger triangle over here is similar to this smaller triangle over there. So let me write this down. So we also know that triangle DEB is similar to triangle CFB. Now what can we do from here? Well, we know that the ratios of corresponding sides for each of those similar triangles are going to have to be the same. But we only have one side of one of the triangles."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "So we also know that triangle DEB is similar to triangle CFB. Now what can we do from here? Well, we know that the ratios of corresponding sides for each of those similar triangles are going to have to be the same. But we only have one side of one of the triangles. So in the case of AB and CFB, we've only been given one side. In the case of DEB and CFB, we've only been given one side right over here. So there doesn't seem to be a lot to work with."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "But we only have one side of one of the triangles. So in the case of AB and CFB, we've only been given one side. In the case of DEB and CFB, we've only been given one side right over here. So there doesn't seem to be a lot to work with. And this is why this is a slightly more challenging problem here. Let's just go ahead and see if we can assume one of the sides, or maybe a side that's shared by both of these larger triangles, and then maybe things will work out from there. So let's just assume that this length right over here, let's just assume that BE is equal to Y."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "So there doesn't seem to be a lot to work with. And this is why this is a slightly more challenging problem here. Let's just go ahead and see if we can assume one of the sides, or maybe a side that's shared by both of these larger triangles, and then maybe things will work out from there. So let's just assume that this length right over here, let's just assume that BE is equal to Y. So let me just write this down. This whole length is going to be equal to Y, because this at least gives us something to work with. And Y is shared by both ABE and DEB, so that seems useful."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "So let's just assume that this length right over here, let's just assume that BE is equal to Y. So let me just write this down. This whole length is going to be equal to Y, because this at least gives us something to work with. And Y is shared by both ABE and DEB, so that seems useful. And then we're going to have to think about the smaller triangles right over there. So maybe we'll call this length BFx. And then let's call FE."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "And Y is shared by both ABE and DEB, so that seems useful. And then we're going to have to think about the smaller triangles right over there. So maybe we'll call this length BFx. And then let's call FE. Well, if this is X, then this is Y minus X. So we've introduced a bunch of variables here, but maybe with all the proportionalities and things, just maybe things will work out, or at least we'll have a little bit more sense of where we can go with this actual problem. But now we can start dealing with the similar triangles."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "And then let's call FE. Well, if this is X, then this is Y minus X. So we've introduced a bunch of variables here, but maybe with all the proportionalities and things, just maybe things will work out, or at least we'll have a little bit more sense of where we can go with this actual problem. But now we can start dealing with the similar triangles. For example, so we want to figure out what CF is. We now know that for these two triangles right here, the ratio of the corresponding sides are going to be constant. So for example, the ratio between CF and 9, their corresponding sides, the ratio between CF and 9 has got to be equal to the ratio between Y minus X, that's that side right there, Y minus X and the corresponding side of the larger triangle."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "But now we can start dealing with the similar triangles. For example, so we want to figure out what CF is. We now know that for these two triangles right here, the ratio of the corresponding sides are going to be constant. So for example, the ratio between CF and 9, their corresponding sides, the ratio between CF and 9 has got to be equal to the ratio between Y minus X, that's that side right there, Y minus X and the corresponding side of the larger triangle. Well, the corresponding side of the larger triangle is this entire length. And that entire length right over there is Y. So it's equal to Y minus X over Y."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "So for example, the ratio between CF and 9, their corresponding sides, the ratio between CF and 9 has got to be equal to the ratio between Y minus X, that's that side right there, Y minus X and the corresponding side of the larger triangle. Well, the corresponding side of the larger triangle is this entire length. And that entire length right over there is Y. So it's equal to Y minus X over Y. So we could simplify this a little bit. Well, I'll hold off for a second. Let's see if we can do something similar with this thing on the right."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "So it's equal to Y minus X over Y. So we could simplify this a little bit. Well, I'll hold off for a second. Let's see if we can do something similar with this thing on the right. So once again, we have CF, its corresponding side on DEB. So now we're looking at the triangle CFB, not looking at triangle CFE anymore. So now when we're looking at this triangle, CF corresponds to DE."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "Let's see if we can do something similar with this thing on the right. So once again, we have CF, its corresponding side on DEB. So now we're looking at the triangle CFB, not looking at triangle CFE anymore. So now when we're looking at this triangle, CF corresponds to DE. So we have CF over DE is going to be equal to, so CF over DE is going to be equal to X is going to be equal to, let me do that in a different color, it's going to be equal to, I'm using all my colors, it's going to be equal to X over this entire base right over here, so this entire B, which once again we know is Y, so over Y. And now this looks interesting because it looks like we have three unknowns. We have CF, sorry, we know what DE is already."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "So now when we're looking at this triangle, CF corresponds to DE. So we have CF over DE is going to be equal to, so CF over DE is going to be equal to X is going to be equal to, let me do that in a different color, it's going to be equal to, I'm using all my colors, it's going to be equal to X over this entire base right over here, so this entire B, which once again we know is Y, so over Y. And now this looks interesting because it looks like we have three unknowns. We have CF, sorry, we know what DE is already. This is 12. I could have written CF over 12. The ratio between CF and 12 is going to be the ratio between X and Y."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "We have CF, sorry, we know what DE is already. This is 12. I could have written CF over 12. The ratio between CF and 12 is going to be the ratio between X and Y. So we have three unknowns and only two equations, so it seems hard to solve it first because there's one unknown, another unknown, another unknown, another unknown, another unknown, and another unknown. But it looks like I can write this right here, this expression, in terms of X over Y, and then we could do a substitution. So that's why this was a little tricky."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "The ratio between CF and 12 is going to be the ratio between X and Y. So we have three unknowns and only two equations, so it seems hard to solve it first because there's one unknown, another unknown, another unknown, another unknown, another unknown, and another unknown. But it looks like I can write this right here, this expression, in terms of X over Y, and then we could do a substitution. So that's why this was a little tricky. So this one right here we can rewrite as CF, let me do it in that same green color, we can rewrite it as CF over 9 is equal to Y minus X over Y, the same thing as Y over Y minus X over Y, over, or 1 minus X over Y. All I did is I essentially, I guess you could say, distributed the 1 over Y times both of these terms. Y over Y minus X over Y, or 1 minus X minus Y."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "So that's why this was a little tricky. So this one right here we can rewrite as CF, let me do it in that same green color, we can rewrite it as CF over 9 is equal to Y minus X over Y, the same thing as Y over Y minus X over Y, over, or 1 minus X over Y. All I did is I essentially, I guess you could say, distributed the 1 over Y times both of these terms. Y over Y minus X over Y, or 1 minus X minus Y. And this is useful because we already know what X minus Y is equal to. Sorry, X over Y is equal to. We already know that X over Y is equal to CF over 12."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "Y over Y minus X over Y, or 1 minus X minus Y. And this is useful because we already know what X minus Y is equal to. Sorry, X over Y is equal to. We already know that X over Y is equal to CF over 12. So this right over here I can replace with this, CF over 12. So then we get, this is a home stretch here, CF, which is what we care about, CF over 9 is equal to 1 minus CF over 12. And now we have one equation with one unknown, and we should be able to solve this right over here."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "We already know that X over Y is equal to CF over 12. So this right over here I can replace with this, CF over 12. So then we get, this is a home stretch here, CF, which is what we care about, CF over 9 is equal to 1 minus CF over 12. And now we have one equation with one unknown, and we should be able to solve this right over here. So we could add CF over 12 to both sides. So you have CF over 9 plus CF over 12 is equal to 1. We just have to find a common denominator here, and I think 36 will do the trick."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "And now we have one equation with one unknown, and we should be able to solve this right over here. So we could add CF over 12 to both sides. So you have CF over 9 plus CF over 12 is equal to 1. We just have to find a common denominator here, and I think 36 will do the trick. So 9 times 4 is 36, so if you had to multiply 9 times 4, you have to multiply CF times 4. So you have 4 CF, 4 CF over 36 is the same thing as CF over 9, and then plus CF over 12 is the same thing as 3 CF over 36. And this is going to be equal to 1."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "We just have to find a common denominator here, and I think 36 will do the trick. So 9 times 4 is 36, so if you had to multiply 9 times 4, you have to multiply CF times 4. So you have 4 CF, 4 CF over 36 is the same thing as CF over 9, and then plus CF over 12 is the same thing as 3 CF over 36. And this is going to be equal to 1. And then we are left with 4 CF plus 3 CF is 7. CF over 36 is equal to 1. And then to solve for CF, we can multiply both sides times the reciprocal of 7 over 36."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "And this is going to be equal to 1. And then we are left with 4 CF plus 3 CF is 7. CF over 36 is equal to 1. And then to solve for CF, we can multiply both sides times the reciprocal of 7 over 36. So 36 over 7, multiply both sides times that, 36 over 7. This side things cancel out. And we are left with our final, we get our drum roll now."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy (2).mp3", "Sentence": "And then to solve for CF, we can multiply both sides times the reciprocal of 7 over 36. So 36 over 7, multiply both sides times that, 36 over 7. This side things cancel out. And we are left with our final, we get our drum roll now. CF is equal to CF, so all of this stuff cancels out. CF is equal to 1 times 36 over 7, or it's just 36 over 7. And this was a pretty cool problem, because what it shows you is if you have two things, let's see, this thing is some type of a pole or a stick or maybe the wall of a building or who knows what it is."}, {"video_title": "Proof all circles are similar Mathematics II High School Math Khan Academy.mp3", "Sentence": "We're asked to translate and dilate the unit circle to map it onto each circle. So this is the unit circle right over here. It's centered at zero comma zero. It has a radius of one. That's why we call it a unit circle. And when they say translate, they say move it around. So that would be a translation of it."}, {"video_title": "Proof all circles are similar Mathematics II High School Math Khan Academy.mp3", "Sentence": "It has a radius of one. That's why we call it a unit circle. And when they say translate, they say move it around. So that would be a translation of it. And then dilating it means making it larger. So dilating that unit circle would be doing, whoops, I just translated it, would be doing something like that. So we're gonna translate and dilate this unit circle to map it onto each circle."}, {"video_title": "Proof all circles are similar Mathematics II High School Math Khan Academy.mp3", "Sentence": "So that would be a translation of it. And then dilating it means making it larger. So dilating that unit circle would be doing, whoops, I just translated it, would be doing something like that. So we're gonna translate and dilate this unit circle to map it onto each circle. So for example, I can translate it so that the center is translated to the center of that magenta circle. And then I can dilate it so that it has been mapped onto that larger magenta circle. I can do that, let me do it for a few more."}, {"video_title": "Proof all circles are similar Mathematics II High School Math Khan Academy.mp3", "Sentence": "So we're gonna translate and dilate this unit circle to map it onto each circle. So for example, I can translate it so that the center is translated to the center of that magenta circle. And then I can dilate it so that it has been mapped onto that larger magenta circle. I can do that, let me do it for a few more. I'm not gonna do it for all of them. Just to give you the idea of what they're talking about. So now I'm translating the center of my, it's no longer a unit circle."}, {"video_title": "Proof all circles are similar Mathematics II High School Math Khan Academy.mp3", "Sentence": "I can do that, let me do it for a few more. I'm not gonna do it for all of them. Just to give you the idea of what they're talking about. So now I'm translating the center of my, it's no longer a unit circle. I'm translating the center of my circle to the center of the purple circle. And now I'm gonna dilate it so it has the same radius. And notice, I can map it."}, {"video_title": "Proof all circles are similar Mathematics II High School Math Khan Academy.mp3", "Sentence": "So now I'm translating the center of my, it's no longer a unit circle. I'm translating the center of my circle to the center of the purple circle. And now I'm gonna dilate it so it has the same radius. And notice, I can map it. And so if you can map one shape to another through translation and dilation, then the things are by definition, they are going to be similar. So this is really just an exercise in seeing that all circles are similar. If you just take any circle and you make it have the same center as another circle, then you can just scale it, you can just scale it up or down to match the circle that you moved it to the center of."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "What I'd like to do in this video is use some geometric arguments to prove that the slopes of perpendicular lines are negative reciprocals of each other. And so just to start off, we have lines L and M, and we're going to assume that they are perpendicular, so they intersect at a right angle. We see that depicted right over here. And so I'm gonna now construct some other lines here to help us make our geometric argument. So let me draw a horizontal line that intersects at this point right over here. Let's call that point A. And so let me see if I can do that."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "And so I'm gonna now construct some other lines here to help us make our geometric argument. So let me draw a horizontal line that intersects at this point right over here. Let's call that point A. And so let me see if I can do that. There you go. So that's a horizontal line that intersects at A. And now I'm gonna drop some verticals from that."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "And so let me see if I can do that. There you go. So that's a horizontal line that intersects at A. And now I'm gonna drop some verticals from that. So I'm gonna drop a vertical line right over here, and I'm gonna drop a vertical line right over here. And so that is 90 degrees, and that is 90 degrees, and I've constructed it that way. This top line is perfectly horizontal, and then I've dropped two vertical things, so they're at 90 degree angles."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "And now I'm gonna drop some verticals from that. So I'm gonna drop a vertical line right over here, and I'm gonna drop a vertical line right over here. And so that is 90 degrees, and that is 90 degrees, and I've constructed it that way. This top line is perfectly horizontal, and then I've dropped two vertical things, so they're at 90 degree angles. And let me now set up some points. So that already said, that's point A. Let's call this point B."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "This top line is perfectly horizontal, and then I've dropped two vertical things, so they're at 90 degree angles. And let me now set up some points. So that already said, that's point A. Let's call this point B. Let's call this point C. Let's call this point D. And let's call this point E right over here. Now, let's think about what the slope of line L is. So slope of, let me move this over a little bit."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "Let's call this point B. Let's call this point C. Let's call this point D. And let's call this point E right over here. Now, let's think about what the slope of line L is. So slope of, let me move this over a little bit. So slope, slope of L is going to be what? Well, that's, you could view line L as line, the line that connects point CA. So it's the slope of CA, you could say."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "So slope of, let me move this over a little bit. So slope, slope of L is going to be what? Well, that's, you could view line L as line, the line that connects point CA. So it's the slope of CA, you could say. This is the same thing as slope of line CA. L is line CA. And so to find the slope, that's change in Y over change in X."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "So it's the slope of CA, you could say. This is the same thing as slope of line CA. L is line CA. And so to find the slope, that's change in Y over change in X. So our change in Y is going to be CB. So it's gonna be the length of segment CB. That is our change in Y."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "And so to find the slope, that's change in Y over change in X. So our change in Y is going to be CB. So it's gonna be the length of segment CB. That is our change in Y. So it is CB over our change in X, which is the length of segment BA, which is the length of segment BA right over here. So that is BA. Now, what is the slope of line M?"}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "That is our change in Y. So it is CB over our change in X, which is the length of segment BA, which is the length of segment BA right over here. So that is BA. Now, what is the slope of line M? So slope, slope of M, and we could also say slope of, we could call line M line AE, line AE, like that. Well, if we're going to go between point A and point E, once again, it's just change in Y over change in X. Well, what's our change in Y going to be?"}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "Now, what is the slope of line M? So slope, slope of M, and we could also say slope of, we could call line M line AE, line AE, like that. Well, if we're going to go between point A and point E, once again, it's just change in Y over change in X. Well, what's our change in Y going to be? Well, our change in Y, well, we're gonna go from this level down to this level as we go from A to E. We could have done it over here as well. We're gonna go from A to E. That is our change in Y. So we might be tempted to say, well, that's just going to be the length of segment DE."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "Well, what's our change in Y going to be? Well, our change in Y, well, we're gonna go from this level down to this level as we go from A to E. We could have done it over here as well. We're gonna go from A to E. That is our change in Y. So we might be tempted to say, well, that's just going to be the length of segment DE. But remember, our Y is decreasing. So we're gonna subtract that length as we go from this Y level to that Y level over there. And what is our change in X?"}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "So we might be tempted to say, well, that's just going to be the length of segment DE. But remember, our Y is decreasing. So we're gonna subtract that length as we go from this Y level to that Y level over there. And what is our change in X? So our change in X, we're going to go, as we go from A to E, our change in X is going to be the length of segment AD. So AD. So our slope of M is going to be negative DE."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "And what is our change in X? So our change in X, we're going to go, as we go from A to E, our change in X is going to be the length of segment AD. So AD. So our slope of M is going to be negative DE. It's going to be the negative of this length because we're dropping by that much. That's our change in Y over segment A, over segment AD. So some of you might already be quite inspired by what we've already written because now we just have to establish that these two are, these two triangles, triangles CBA and triangle ADE, are similar."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "So our slope of M is going to be negative DE. It's going to be the negative of this length because we're dropping by that much. That's our change in Y over segment A, over segment AD. So some of you might already be quite inspired by what we've already written because now we just have to establish that these two are, these two triangles, triangles CBA and triangle ADE, are similar. And then we're going to be able to show that these are the negative reciprocal of each other. So let's show that these two triangles are similar. So let's say that we have this angle right over here."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "So some of you might already be quite inspired by what we've already written because now we just have to establish that these two are, these two triangles, triangles CBA and triangle ADE, are similar. And then we're going to be able to show that these are the negative reciprocal of each other. So let's show that these two triangles are similar. So let's say that we have this angle right over here. And let's say that angle has measure X, just for kicks. And let's say that we have, let me do another color for, let's say we have this angle right over here. And let's say that the measure, that that has measure Y."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "So let's say that we have this angle right over here. And let's say that angle has measure X, just for kicks. And let's say that we have, let me do another color for, let's say we have this angle right over here. And let's say that the measure, that that has measure Y. Well we know X plus Y plus 90 is equal to 180 because together they are supplementary. So I could write, I could write that X plus 90, plus 90, plus Y, plus Y, is going to be equal to, is going to be equal to 180 degrees. If you want you could subtract 90 from both sides of that and you could say, look, X plus Y is going to be equal to 90 degrees."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "And let's say that the measure, that that has measure Y. Well we know X plus Y plus 90 is equal to 180 because together they are supplementary. So I could write, I could write that X plus 90, plus 90, plus Y, plus Y, is going to be equal to, is going to be equal to 180 degrees. If you want you could subtract 90 from both sides of that and you could say, look, X plus Y is going to be equal to 90 degrees. Is going to be equal to 90 degrees. These are algebraically equivalent statements. So is equal to 90 degrees."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "If you want you could subtract 90 from both sides of that and you could say, look, X plus Y is going to be equal to 90 degrees. Is going to be equal to 90 degrees. These are algebraically equivalent statements. So is equal to 90 degrees. And how can we use this to fill out some of the other angles in these triangles? Well, let's see. X plus this angle down here has to be equal to 90 degrees."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "So is equal to 90 degrees. And how can we use this to fill out some of the other angles in these triangles? Well, let's see. X plus this angle down here has to be equal to 90 degrees. Or you could say X plus 90 plus what is going to be equal to 180. I'm looking at triangle CBA right over here. The interior angles of a triangle add up to 180."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "X plus this angle down here has to be equal to 90 degrees. Or you could say X plus 90 plus what is going to be equal to 180. I'm looking at triangle CBA right over here. The interior angles of a triangle add up to 180. So X plus 90 plus what is equal to 180? Well X plus 90 plus Y is equal to 180. We already established that."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "The interior angles of a triangle add up to 180. So X plus 90 plus what is equal to 180? Well X plus 90 plus Y is equal to 180. We already established that. Similarly over here. Y plus 90 plus what is going to be equal to 180? Well, same argument."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "We already established that. Similarly over here. Y plus 90 plus what is going to be equal to 180? Well, same argument. We already know Y plus 90 plus X is equal to 180. So Y plus 90 plus X is equal to, is equal to 180. And so notice, we have now established that triangle ABC and triangle EDA, that all of their interior angles, their corresponding interior angles are the same or that their three different angle measures all correspond to each other."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "Well, same argument. We already know Y plus 90 plus X is equal to 180. So Y plus 90 plus X is equal to, is equal to 180. And so notice, we have now established that triangle ABC and triangle EDA, that all of their interior angles, their corresponding interior angles are the same or that their three different angle measures all correspond to each other. They both have an angle of X, they both have a measure X, they both have an angle of measure Y, and they're both right triangles. So just by angle, angle, angle, so we could say by angle, angle, angle, one of our similarity postulates, we know that triangle E, triangle EDA, EDA is similar to triangle, to triangle ABC, to triangle ABC. And so that tells us that the ratio of corresponding sides are going to be the same."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "And so notice, we have now established that triangle ABC and triangle EDA, that all of their interior angles, their corresponding interior angles are the same or that their three different angle measures all correspond to each other. They both have an angle of X, they both have a measure X, they both have an angle of measure Y, and they're both right triangles. So just by angle, angle, angle, so we could say by angle, angle, angle, one of our similarity postulates, we know that triangle E, triangle EDA, EDA is similar to triangle, to triangle ABC, to triangle ABC. And so that tells us that the ratio of corresponding sides are going to be the same. And so for example, we know, let's find the ratio of corresponding sides. We know that the ratio of, let's say CB to BA, so let's write this down. We know that the ratio, so this tells us that the ratio of corresponding sides are going to be the same."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "And so that tells us that the ratio of corresponding sides are going to be the same. And so for example, we know, let's find the ratio of corresponding sides. We know that the ratio of, let's say CB to BA, so let's write this down. We know that the ratio, so this tells us that the ratio of corresponding sides are going to be the same. So the ratio of CB over BA, over BA is going to be equal to, is going to be equal to, well the corresponding side to CB, it's the side opposite the X degree angle right over here. So the corresponding side to CB is side AD, so that's going to be equal to AD over, what's the corresponding side to BA? Well BA is opposite the Y degree angle, so over here the corresponding side is DE, AD over DE, let me do that same color, over DE."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "We know that the ratio, so this tells us that the ratio of corresponding sides are going to be the same. So the ratio of CB over BA, over BA is going to be equal to, is going to be equal to, well the corresponding side to CB, it's the side opposite the X degree angle right over here. So the corresponding side to CB is side AD, so that's going to be equal to AD over, what's the corresponding side to BA? Well BA is opposite the Y degree angle, so over here the corresponding side is DE, AD over DE, let me do that same color, over DE. And so this right over here, this right over here we saw from the beginning, this is the slope, this is the slope of L. So slope, slope of L. And how does this relate to the slope of M? Notice the slope of M is the negative reciprocal of this. You take the reciprocal, you're going to get DE over AD, and then you have to take this negative right over here."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy.mp3", "Sentence": "Well BA is opposite the Y degree angle, so over here the corresponding side is DE, AD over DE, let me do that same color, over DE. And so this right over here, this right over here we saw from the beginning, this is the slope, this is the slope of L. So slope, slope of L. And how does this relate to the slope of M? Notice the slope of M is the negative reciprocal of this. You take the reciprocal, you're going to get DE over AD, and then you have to take this negative right over here. So we could write this as the negative reciprocal of slope of M. Negative reciprocal, reciprocal of M's, of M's slope. And there you have it. We've just shown that if we start with, if we assume these L and M are perpendicular, and we set up these similar triangles, and we were able to show that the slope of L is the negative reciprocal of the slope of M."}, {"video_title": "Finding arc measures Mathematics II High School Math Khan Academy.mp3", "Sentence": "So this first question says, what is the arc measure in degrees of arc AC on circle P below? So this is point A, that is point C. And when they're talking about arc AC, since they only have two letters here, we can assume that it's going to be the minor arc. And when we talk about the minor arc, there's two potential arcs that connect point A and point C. There's the one here on the left, and then there's the one on, there is the one on the right. And since C isn't exactly, isn't exactly straight down from A, it's a little bit to the right, the shorter arc, the arc with a smaller length, or the minor arc is going to be this one that I'm depicting here right on the right. So what is this arc measure going to be? Well, the measure of this arc is going to be exactly the same thing as in degrees as the measure of the central angle that intercepts the arc. So that central angle, let me do it in a different color, I'll do it in this blue color."}, {"video_title": "Finding arc measures Mathematics II High School Math Khan Academy.mp3", "Sentence": "And since C isn't exactly, isn't exactly straight down from A, it's a little bit to the right, the shorter arc, the arc with a smaller length, or the minor arc is going to be this one that I'm depicting here right on the right. So what is this arc measure going to be? Well, the measure of this arc is going to be exactly the same thing as in degrees as the measure of the central angle that intercepts the arc. So that central angle, let me do it in a different color, I'll do it in this blue color. That central angle is angle CPA, angle CPA, and the measure of that central angle is going to be 70 degrees plus 104 degrees. It's going to be this whole thing right over there. So it's going to be 174 degrees."}, {"video_title": "Finding arc measures Mathematics II High School Math Khan Academy.mp3", "Sentence": "So that central angle, let me do it in a different color, I'll do it in this blue color. That central angle is angle CPA, angle CPA, and the measure of that central angle is going to be 70 degrees plus 104 degrees. It's going to be this whole thing right over there. So it's going to be 174 degrees. 174 degrees. That's the arc measure in degrees of arc AC. Let's keep doing these."}, {"video_title": "Finding arc measures Mathematics II High School Math Khan Academy.mp3", "Sentence": "So it's going to be 174 degrees. 174 degrees. That's the arc measure in degrees of arc AC. Let's keep doing these. So let me do another one. So this next one asks us, in the figure below, in the figure below, segment AD, so this is point A, this is point D, so segment AD is this one right over here. Let me see if I can draw that."}, {"video_title": "Finding arc measures Mathematics II High School Math Khan Academy.mp3", "Sentence": "Let's keep doing these. So let me do another one. So this next one asks us, in the figure below, in the figure below, segment AD, so this is point A, this is point D, so segment AD is this one right over here. Let me see if I can draw that. That's AD right over there. AD and CE are diameters of the circle. So let me draw CE."}, {"video_title": "Finding arc measures Mathematics II High School Math Khan Academy.mp3", "Sentence": "Let me see if I can draw that. That's AD right over there. AD and CE are diameters of the circle. So let me draw CE. So CE is, we're going to connect points C and E. These are diameters. So let me, so they go straight. Whoops, I'm using the wrong tool."}, {"video_title": "Finding arc measures Mathematics II High School Math Khan Academy.mp3", "Sentence": "So let me draw CE. So CE is, we're going to connect points C and E. These are diameters. So let me, so they go straight. Whoops, I'm using the wrong tool. So those are, somehow I should, all right. So those are, whoops, how did that happen? So let me, somehow my pen got really big."}, {"video_title": "Finding arc measures Mathematics II High School Math Khan Academy.mp3", "Sentence": "Whoops, I'm using the wrong tool. So those are, somehow I should, all right. So those are, whoops, how did that happen? So let me, somehow my pen got really big. All right. That'll be almost there. Okay."}, {"video_title": "Finding arc measures Mathematics II High School Math Khan Academy.mp3", "Sentence": "So let me, somehow my pen got really big. All right. That'll be almost there. Okay. So CE, there you go. So those are both diameters of the circle. P, what is the arc measure of AB, of arc AB in degrees?"}, {"video_title": "Finding arc measures Mathematics II High School Math Khan Academy.mp3", "Sentence": "Okay. So CE, there you go. So those are both diameters of the circle. P, what is the arc measure of AB, of arc AB in degrees? So arc AB, once again, there's two potential arcs that connect point A and B. There's the minor arc, and since this only has two letters, we'll assume it's the minor arc. It's going to be this one over here."}, {"video_title": "Finding arc measures Mathematics II High School Math Khan Academy.mp3", "Sentence": "P, what is the arc measure of AB, of arc AB in degrees? So arc AB, once again, there's two potential arcs that connect point A and B. There's the minor arc, and since this only has two letters, we'll assume it's the minor arc. It's going to be this one over here. There's a major arc, but to denote the major arc, they would have said something like AEB or ADB or arc ACB to make us go this kind of the, this long way around. But this is arc AB. So we, in order to find the arc measure, we just really have to find the measure of the central angle."}, {"video_title": "Finding arc measures Mathematics II High School Math Khan Academy.mp3", "Sentence": "It's going to be this one over here. There's a major arc, but to denote the major arc, they would have said something like AEB or ADB or arc ACB to make us go this kind of the, this long way around. But this is arc AB. So we, in order to find the arc measure, we just really have to find the measure of the central angle. This is a central angle that intercepts that arc. We could even say it defines that arc in some way. So how can we figure out this angle?"}, {"video_title": "Finding arc measures Mathematics II High School Math Khan Academy.mp3", "Sentence": "So we, in order to find the arc measure, we just really have to find the measure of the central angle. This is a central angle that intercepts that arc. We could even say it defines that arc in some way. So how can we figure out this angle? This one's a little bit trickier. Well, the key here is to realize that this 93-degree angle, it is vertical to this whole angle right over here. And we know from geometry, which we're still learning as we do this example problem, that vertical angles are going to have the same measure."}, {"video_title": "Finding arc measures Mathematics II High School Math Khan Academy.mp3", "Sentence": "So how can we figure out this angle? This one's a little bit trickier. Well, the key here is to realize that this 93-degree angle, it is vertical to this whole angle right over here. And we know from geometry, which we're still learning as we do this example problem, that vertical angles are going to have the same measure. So if this one is 93 degrees, then this entire blue one right over here is also going to be, let me write it, this is also going to be 93 degrees. So 93 degrees, that's going to be made up of this red angle that we care about and the 38 degrees. So this red one, which is the measure of the central angle, it's also the arc measure of arc AB, is going to be 93 minus, 93 degrees minus 38 degrees."}, {"video_title": "Finding arc measures Mathematics II High School Math Khan Academy.mp3", "Sentence": "And we know from geometry, which we're still learning as we do this example problem, that vertical angles are going to have the same measure. So if this one is 93 degrees, then this entire blue one right over here is also going to be, let me write it, this is also going to be 93 degrees. So 93 degrees, that's going to be made up of this red angle that we care about and the 38 degrees. So this red one, which is the measure of the central angle, it's also the arc measure of arc AB, is going to be 93 minus, 93 degrees minus 38 degrees. So what is that going to be? Let's see, 93, I can write degrees there, minus 38 degrees. That is going to be equal to, let's see, if it was 93 minus 40, it would be 53, and it's going to be two more, it's going to be 55 degrees."}, {"video_title": "Finding arc measures Mathematics II High School Math Khan Academy.mp3", "Sentence": "So this red one, which is the measure of the central angle, it's also the arc measure of arc AB, is going to be 93 minus, 93 degrees minus 38 degrees. So what is that going to be? Let's see, 93, I can write degrees there, minus 38 degrees. That is going to be equal to, let's see, if it was 93 minus 40, it would be 53, and it's going to be two more, it's going to be 55 degrees. 55 degrees. And we are done. This angle right over here is 55 degrees."}, {"video_title": "Finding arc measures Mathematics II High School Math Khan Academy.mp3", "Sentence": "That is going to be equal to, let's see, if it was 93 minus 40, it would be 53, and it's going to be two more, it's going to be 55 degrees. 55 degrees. And we are done. This angle right over here is 55 degrees. If you were to add this angle measure, plus 38 degrees, you'd get 93 degrees. And that has the same measure, because it's vertical, with this angle right over here, with angle DPE. All right, let's do one more of these."}, {"video_title": "Finding arc measures Mathematics II High School Math Khan Academy.mp3", "Sentence": "This angle right over here is 55 degrees. If you were to add this angle measure, plus 38 degrees, you'd get 93 degrees. And that has the same measure, because it's vertical, with this angle right over here, with angle DPE. All right, let's do one more of these. So we have, in the figure below, and it doesn't quite fit on the page, but we'll scroll down in a second, AB is the diameter of circle P, is the diameter of circle P. All right, so AB is the diameter, let me label that. So AB is the diameter, so it's going straight across, straight across the circle. What is the arc measure of ABC in degrees?"}, {"video_title": "Finding arc measures Mathematics II High School Math Khan Academy.mp3", "Sentence": "All right, let's do one more of these. So we have, in the figure below, and it doesn't quite fit on the page, but we'll scroll down in a second, AB is the diameter of circle P, is the diameter of circle P. All right, so AB is the diameter, let me label that. So AB is the diameter, so it's going straight across, straight across the circle. What is the arc measure of ABC in degrees? So ABC, so they're making us go the long way around. This is a major arc they're talking about. Let me draw it."}, {"video_title": "Finding arc measures Mathematics II High School Math Khan Academy.mp3", "Sentence": "What is the arc measure of ABC in degrees? So ABC, so they're making us go the long way around. This is a major arc they're talking about. Let me draw it. Arc A, what is the arc measure of arc ABC? So we're going the long way around, so it's a major arc. So what is that going to be?"}, {"video_title": "Finding arc measures Mathematics II High School Math Khan Academy.mp3", "Sentence": "Let me draw it. Arc A, what is the arc measure of arc ABC? So we're going the long way around, so it's a major arc. So what is that going to be? Well, it's going to be, in degrees, the same measure as the angle, as the central angle that intercepts it. So it's going to be the same thing as this central angle right over here. Well, what is that central angle going to be?"}, {"video_title": "Finding arc measures Mathematics II High School Math Khan Academy.mp3", "Sentence": "So what is that going to be? Well, it's going to be, in degrees, the same measure as the angle, as the central angle that intercepts it. So it's going to be the same thing as this central angle right over here. Well, what is that central angle going to be? Well, since we know that this is a diameter, since AB is a diameter, we know that this part of it, this part of it is going to be 180 degrees. We're going halfway around the circle, 180 degrees. And so if we wanted to look at this whole angle, the angle that intercepts the major arc ABC, it's going to be the 180 degrees plus 69 degrees."}, {"video_title": "Finding arc measures Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, what is that central angle going to be? Well, since we know that this is a diameter, since AB is a diameter, we know that this part of it, this part of it is going to be 180 degrees. We're going halfway around the circle, 180 degrees. And so if we wanted to look at this whole angle, the angle that intercepts the major arc ABC, it's going to be the 180 degrees plus 69 degrees. So we're going to have 180 degrees plus 69 degrees, which is equal to, what is that? 249, 249 degrees. That's the arc measure of this major arc ABC."}, {"video_title": "Congruent shapes and transformations.mp3", "Sentence": "So let's see, this is triangle ABC, and it looks like at first he rotates triangle ABC about point C to get it right over here. So that's what they're depicting in this diagram. And then they say Cassan concluded it is not possible to map triangle ABC onto triangle GFE using a sequence of rigid transformations. So the triangles are not congruent. So what I want you to do is pause this video and think about is Cassan correct that they are not congruent because you cannot map ABC, triangle ABC onto triangle GFE with rigid transformations? All right, so the way I think about it, he was able to do the rotation that got us right over here. So it was rotation about point C. And so this point right over here, let me make sure I get this right, this would have become B prime, and then this is A prime, and then C is mapped to itself."}, {"video_title": "Congruent shapes and transformations.mp3", "Sentence": "So the triangles are not congruent. So what I want you to do is pause this video and think about is Cassan correct that they are not congruent because you cannot map ABC, triangle ABC onto triangle GFE with rigid transformations? All right, so the way I think about it, he was able to do the rotation that got us right over here. So it was rotation about point C. And so this point right over here, let me make sure I get this right, this would have become B prime, and then this is A prime, and then C is mapped to itself. So C is equal to C prime. But he's not done. There's another rigid transformation he could do, and that would be a reflection about the line FG."}, {"video_title": "Congruent shapes and transformations.mp3", "Sentence": "So it was rotation about point C. And so this point right over here, let me make sure I get this right, this would have become B prime, and then this is A prime, and then C is mapped to itself. So C is equal to C prime. But he's not done. There's another rigid transformation he could do, and that would be a reflection about the line FG. So if he reflects about the line FG, then this point is going to be mapped to point E, just like that. And then if you did that, you would see that there is a series of rigid transformations that maps triangle ABC onto triangle GFE. So Cassan is not correct."}, {"video_title": "Thinking about dilations Transformations Geometry Khan Academy.mp3", "Sentence": "Draw the image of ABCP under a dilation whose center is at P and a scale factor is 1 2 3rds. What are the lengths of the side AB and its image? So we're going to do a dilation centered at P. So if we're centering a dilation at P, that means that every other point is going, and its scale factor is 1 2 3rds. That means once we perform the dilation, every point is going to be 1 2 3rds times as far away from P. Well, P is zero away from P, so its image is still going to be at P. So let's put that point right over there. Now, point C is going to be 1 2 3rds times as far as it is right now. So let's see, right now it is six away. It's at negative three, and P is, or its X coordinate is the same, but in the Y direction, P is at three, C is at negative three, so it's six less."}, {"video_title": "Thinking about dilations Transformations Geometry Khan Academy.mp3", "Sentence": "That means once we perform the dilation, every point is going to be 1 2 3rds times as far away from P. Well, P is zero away from P, so its image is still going to be at P. So let's put that point right over there. Now, point C is going to be 1 2 3rds times as far as it is right now. So let's see, right now it is six away. It's at negative three, and P is, or its X coordinate is the same, but in the Y direction, P is at three, C is at negative three, so it's six less. We want to be 1 2 3rds times as far away. So what's 1 2 3rds of six? Well, 2 3rds of six is four, so it's going to be, six plus four is going to be, you're going to be ten away."}, {"video_title": "Thinking about dilations Transformations Geometry Khan Academy.mp3", "Sentence": "It's at negative three, and P is, or its X coordinate is the same, but in the Y direction, P is at three, C is at negative three, so it's six less. We want to be 1 2 3rds times as far away. So what's 1 2 3rds of six? Well, 2 3rds of six is four, so it's going to be, six plus four is going to be, you're going to be ten away. So three minus ten, that gets us to negative seven, so that gets us right over there. Now, point A, right now it is three more in the horizontal direction than point P's X coordinate, so we want to go 1 2 3rds as far. So what is 1 2 3rds times three?"}, {"video_title": "Thinking about dilations Transformations Geometry Khan Academy.mp3", "Sentence": "Well, 2 3rds of six is four, so it's going to be, six plus four is going to be, you're going to be ten away. So three minus ten, that gets us to negative seven, so that gets us right over there. Now, point A, right now it is three more in the horizontal direction than point P's X coordinate, so we want to go 1 2 3rds as far. So what is 1 2 3rds times three? Well, that's going to be three plus 2 3rds of three, which is another two, so that's going to be five. So we're going to get right over there, then we can complete the rectangle. And notice, point B is now 1 2 3rds times as far in the horizontal direction."}, {"video_title": "Thinking about dilations Transformations Geometry Khan Academy.mp3", "Sentence": "So what is 1 2 3rds times three? Well, that's going to be three plus 2 3rds of three, which is another two, so that's going to be five. So we're going to get right over there, then we can complete the rectangle. And notice, point B is now 1 2 3rds times as far in the horizontal direction. It was three away in the horizontal direction, now it is five away from P's X coordinate. And in the vertical direction, in the vertical direction, in the Y direction, it was six below P's Y coordinate. Now it is 1 2 3rds times as far."}, {"video_title": "Thinking about dilations Transformations Geometry Khan Academy.mp3", "Sentence": "And notice, point B is now 1 2 3rds times as far in the horizontal direction. It was three away in the horizontal direction, now it is five away from P's X coordinate. And in the vertical direction, in the vertical direction, in the Y direction, it was six below P's Y coordinate. Now it is 1 2 3rds times as far. It is ten below P's Y coordinate. So let's answer these questions. The length of segment AB, well, we already saw that, that is, we're going from three to negative three, that is six units long."}, {"video_title": "Thinking about dilations Transformations Geometry Khan Academy.mp3", "Sentence": "Now it is 1 2 3rds times as far. It is ten below P's Y coordinate. So let's answer these questions. The length of segment AB, well, we already saw that, that is, we're going from three to negative three, that is six units long. And its image, well, it's 1 2 3rds as long. We see it over here, we're going from three to negative seven. Three minus negative seven is ten."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "Let's say given this diagram right over here, we know that the length of segment AB is equal to the length of AC. So AB, which is this whole side right over here, the length of this entire side as a given is equal to the length of this entire side right over here. So that's the entire side right over there. And then we also know that angle ABF is equal to angle AC. You can see their measures are equal, or this implies that they're congruent or they have the same measures. It's equal to angle ACE. So this angle right over here is congruent to that angle right over there, or you could say that they have the same measure."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "And then we also know that angle ABF is equal to angle AC. You can see their measures are equal, or this implies that they're congruent or they have the same measures. It's equal to angle ACE. So this angle right over here is congruent to that angle right over there, or you could say that they have the same measure. Now the first thing that I want to attempt to prove in this video is whether BF has the same length as CE. So let's try to do that. So we already know a few things."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "So this angle right over here is congruent to that angle right over there, or you could say that they have the same measure. Now the first thing that I want to attempt to prove in this video is whether BF has the same length as CE. So let's try to do that. So we already know a few things. I could do a two column proof. Actually, let me just do it just so that in case you have to do two column proofs in your class, you can kind of see how to do it more formally. So let's write our statements."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "So we already know a few things. I could do a two column proof. Actually, let me just do it just so that in case you have to do two column proofs in your class, you can kind of see how to do it more formally. So let's write our statements. And then over here I'm going to write my reason for the statement. So let me just rewrite this. Just have a formal two column proof."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "So let's write our statements. And then over here I'm going to write my reason for the statement. So let me just rewrite this. Just have a formal two column proof. So we know AB is equal to AC. So this is statement one, and this is given. We know statement two, that angle ABF is equal to angle ACE."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "Just have a formal two column proof. So we know AB is equal to AC. So this is statement one, and this is given. We know statement two, that angle ABF is equal to angle ACE. Once again, that was given. Now the other interesting thing, we have an angle and we have a side on each of these triangles. And then what you can see is both of the triangles, and when I say both of the triangles, I'm talking about triangle ABF and triangle ACE."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "We know statement two, that angle ABF is equal to angle ACE. Once again, that was given. Now the other interesting thing, we have an angle and we have a side on each of these triangles. And then what you can see is both of the triangles, and when I say both of the triangles, I'm talking about triangle ABF and triangle ACE. And they both share this vertex at A. At point A is a vertex for both of these. So we could say angle AF, we could say angle BAF, we could say is equal to angle CAE."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "And then what you can see is both of the triangles, and when I say both of the triangles, I'm talking about triangle ABF and triangle ACE. And they both share this vertex at A. At point A is a vertex for both of these. So we could say angle AF, we could say angle BAF, we could say is equal to angle CAE. That makes it a little bit clearer that we're dealing with two different triangles right here, but it really is the exact same angle. It's equal to itself right there. That's our third statement."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "So we could say angle AF, we could say angle BAF, we could say is equal to angle CAE. That makes it a little bit clearer that we're dealing with two different triangles right here, but it really is the exact same angle. It's equal to itself right there. That's our third statement. And we could say it's obvious, some people would call this the reflexive property, that it's obvious that an angle is equal to itself. And so we could say it's obvious, or we could call it maybe the reflexive property, that an angle is clearly reflexive. Obviously equal to itself, even if we label it different ways, this angle is going to be the same measure."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "That's our third statement. And we could say it's obvious, some people would call this the reflexive property, that it's obvious that an angle is equal to itself. And so we could say it's obvious, or we could call it maybe the reflexive property, that an angle is clearly reflexive. Obviously equal to itself, even if we label it different ways, this angle is going to be the same measure. And now we have something interesting going on. We have an angle, a side, and an angle. So what we end up having is that triangle, so by angle side angle, we have the triangle BAF."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "Obviously equal to itself, even if we label it different ways, this angle is going to be the same measure. And now we have something interesting going on. We have an angle, a side, and an angle. So what we end up having is that triangle, so by angle side angle, we have the triangle BAF. So our statement number four, I'm running out of space right here, statement number, I'll go down here, statement number here is triangle BAF. Let me highlight it in a little blue right here. BAF."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "So what we end up having is that triangle, so by angle side angle, we have the triangle BAF. So our statement number four, I'm running out of space right here, statement number, I'll go down here, statement number here is triangle BAF. Let me highlight it in a little blue right here. BAF. So that's this entire triangle right over here. Half of the trick of some of these problems is just seeing the right triangle. So we started with this wide angle, we went through the side that we knew, and we went to this orange angle right over here."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "BAF. So that's this entire triangle right over here. Half of the trick of some of these problems is just seeing the right triangle. So we started with this wide angle, we went through the side that we knew, and we went to this orange angle right over here. We started at this angle, then we went to this orange angle across the side E that we know is congruent to that side over there, and then we went to the side, the angle, the vertex that's not labeled. So triangle BAF we now know is going to be congruent to triangle. We start at the wide angle, go to the orange angle, and then go to the unlabeled angle."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "So we started with this wide angle, we went through the side that we knew, and we went to this orange angle right over here. We started at this angle, then we went to this orange angle across the side E that we know is congruent to that side over there, and then we went to the side, the angle, the vertex that's not labeled. So triangle BAF we now know is going to be congruent to triangle. We start at the wide angle, go to the orange angle, and then go to the unlabeled angle. It's going to be congruent to angle to triangle CAF. So this is kind of a messily drawn version, but I think you get the idea. These two triangles are going to be congruent."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "We start at the wide angle, go to the orange angle, and then go to the unlabeled angle. It's going to be congruent to angle to triangle CAF. So this is kind of a messily drawn version, but I think you get the idea. These two triangles are going to be congruent. CAE, I should say, is congruent to triangle CAE. Wide angle, orange angle, and then the unlabeled angle in that triangle right over there. This comes straight out of angle, side, angle."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "These two triangles are going to be congruent. CAE, I should say, is congruent to triangle CAE. Wide angle, orange angle, and then the unlabeled angle in that triangle right over there. This comes straight out of angle, side, angle. This comes straight out of ASA. This is one angle, this is the side in between, and these are the two angles. It comes out of statements 1, 2, and 3."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "This comes straight out of angle, side, angle. This comes straight out of ASA. This is one angle, this is the side in between, and these are the two angles. It comes out of statements 1, 2, and 3. So they are congruent. We know that corresponding sides are going to be congruent. So we know our statement 5, we now know that BF is equal to CAE."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "It comes out of statements 1, 2, and 3. So they are congruent. We know that corresponding sides are going to be congruent. So we know our statement 5, we now know that BF is equal to CAE. This comes straight out of statement 4. We could say corresponding sides are congruent. Now let's take it up another notch."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "So we know our statement 5, we now know that BF is equal to CAE. This comes straight out of statement 4. We could say corresponding sides are congruent. Now let's take it up another notch. Let's see if we can prove whether ED is equal to EF. Let's just keep going down this and see if we can prove whether ED is equal to EF. I put a question mark there because we haven't necessarily proven it yet."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "Now let's take it up another notch. Let's see if we can prove whether ED is equal to EF. Let's just keep going down this and see if we can prove whether ED is equal to EF. I put a question mark there because we haven't necessarily proven it yet. I want to prove that this little short line segment ED is equal to DF. The interesting thing that we might, at first it might not be so obvious, how do we figure out some type of congruency over that, but we do already have some information here. We know that BAF is congruent to CAE."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "I put a question mark there because we haven't necessarily proven it yet. I want to prove that this little short line segment ED is equal to DF. The interesting thing that we might, at first it might not be so obvious, how do we figure out some type of congruency over that, but we do already have some information here. We know that BAF is congruent to CAE. We also know that this side, right over here, let me do it in a color that I haven't used yet. Let me see, I have not, I've been using a lot of the colors in my palettes. So we know that from these two congruent triangles, that side AE, which is part of CAE, we know that AE is going to be equal to AF, that these two sides are congruent."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "We know that BAF is congruent to CAE. We also know that this side, right over here, let me do it in a color that I haven't used yet. Let me see, I have not, I've been using a lot of the colors in my palettes. So we know that from these two congruent triangles, that side AE, which is part of CAE, we know that AE is going to be equal to AF, that these two sides are congruent. The reason why is because they're corresponding sides of congruent triangles. AF is the side opposite the white angle on BAF, triangle BAF, and AE is the side opposite the white angle on triangle CAE, which we know are congruent. So we know that AE is equal to AF."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "So we know that from these two congruent triangles, that side AE, which is part of CAE, we know that AE is going to be equal to AF, that these two sides are congruent. The reason why is because they're corresponding sides of congruent triangles. AF is the side opposite the white angle on BAF, triangle BAF, and AE is the side opposite the white angle on triangle CAE, which we know are congruent. So we know that AE is equal to AF. Once again, this comes from statement four, and we can even say corresponding sides congruent. Same reason as we gave right up here. Now what's interesting here is, this isn't even a triangle that we're seeing up here, but this information that these two characters are congruent, help us with this part over here."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "So we know that AE is equal to AF. Once again, this comes from statement four, and we can even say corresponding sides congruent. Same reason as we gave right up here. Now what's interesting here is, this isn't even a triangle that we're seeing up here, but this information that these two characters are congruent, help us with this part over here. Because we know that AB is equal to AC, that was given. And so we know that EB, let me write it over here, and I'll make it a little bit messy right over here. Statement seven will give us some space."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "Now what's interesting here is, this isn't even a triangle that we're seeing up here, but this information that these two characters are congruent, help us with this part over here. Because we know that AB is equal to AC, that was given. And so we know that EB, let me write it over here, and I'll make it a little bit messy right over here. Statement seven will give us some space. We know that BE is going to be equal to CF. Let me write that down. We know that BE is equal to CF."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "Statement seven will give us some space. We know that BE is going to be equal to CF. Let me write that down. We know that BE is equal to CF. And why do we know that? Let me put the reason right over here. Let me try to clean up my work a little bit."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "We know that BE is equal to CF. And why do we know that? Let me put the reason right over here. Let me try to clean up my work a little bit. This column has been slowly drifting to the left. But how do we know that BE is equal to CF? We know that the length of BE is equal to the length of BA minus AE."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "Let me try to clean up my work a little bit. This column has been slowly drifting to the left. But how do we know that BE is equal to CF? We know that the length of BE is equal to the length of BA minus AE. I should say AB, that's how I call it up here. So it's equal to AB minus AE, which is the same thing based on these last few things that we saw, as saying AC minus AF. Because AB is equal to AC, so that's equal to AC."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "We know that the length of BE is equal to the length of BA minus AE. I should say AB, that's how I call it up here. So it's equal to AB minus AE, which is the same thing based on these last few things that we saw, as saying AC minus AF. Because AB is equal to AC, so that's equal to AC. And AE, we already showed, is the same thing as AF. So AC minus AF, and AC minus AF is the same thing as CF right over here. It's equal to CF right over there."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "Because AB is equal to AC, so that's equal to AC. And AE, we already showed, is the same thing as AF. So AC minus AF, and AC minus AF is the same thing as CF right over here. It's equal to CF right over there. And we know that because we know this from statement one, we know it from statement five, and we know it from statement six. Actually, we didn't need statement five there. We just need one and six."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "It's equal to CF right over there. And we know that because we know this from statement one, we know it from statement five, and we know it from statement six. Actually, we didn't need statement five there. We just need one and six. Let's say we need this from one and six is what we had to do there. So we just know that, look, this side is equal to that side. This little part is equal to that part."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "We just need one and six. Let's say we need this from one and six is what we had to do there. So we just know that, look, this side is equal to that side. This little part is equal to that part. So if you subtract the big part minus the little part, this right over here is going to be equal to this right over here. That's all we're showing. So this yellow side is equal to this yellow side right over here."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "This little part is equal to that part. So if you subtract the big part minus the little part, this right over here is going to be equal to this right over here. That's all we're showing. So this yellow side is equal to this yellow side right over here. Now, the other thing that we know, and this is straight out of vertical angles, is that this angle, EDB, is going to be congruent to angle FDC. So let me write that down again. Eight, we know that angle EDB is going to be equal to angle FDC."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "So this yellow side is equal to this yellow side right over here. Now, the other thing that we know, and this is straight out of vertical angles, is that this angle, EDB, is going to be congruent to angle FDC. So let me write that down again. Eight, we know that angle EDB is going to be equal to angle FDC. That comes straight out of vertical angles. Vertical angles are congruent or their measures are equal. And now all of a sudden we have something interesting again."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "Eight, we know that angle EDB is going to be equal to angle FDC. That comes straight out of vertical angles. Vertical angles are congruent or their measures are equal. And now all of a sudden we have something interesting again. We have orange angle, white angle, side. So we know that these two smaller triangles are congruent. So now we know, and I don't want to lose my diagram, we know that triangle BED, so statement number nine, we know that triangle BED is congruent to triangle, same sides, white angle, yellow side, then orange angle."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "And now all of a sudden we have something interesting again. We have orange angle, white angle, side. So we know that these two smaller triangles are congruent. So now we know, and I don't want to lose my diagram, we know that triangle BED, so statement number nine, we know that triangle BED is congruent to triangle, same sides, white angle, yellow side, then orange angle. White angle, white angle, let me be careful here. White angle, so B is white angle, E is the unlabeled angle, and then D is the orange labeled angle. So we want to start at C, unlabeled angle, orange angle."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "So now we know, and I don't want to lose my diagram, we know that triangle BED, so statement number nine, we know that triangle BED is congruent to triangle, same sides, white angle, yellow side, then orange angle. White angle, white angle, let me be careful here. White angle, so B is white angle, E is the unlabeled angle, and then D is the orange labeled angle. So we want to start at C, unlabeled angle, orange angle. So CFD. So triangle CFD. And this comes straight from, once again, orange angle, white angle, side."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "So we want to start at C, unlabeled angle, orange angle. So CFD. So triangle CFD. And this comes straight from, once again, orange angle, white angle, side. So angle, angle, side. Orange angle, white angle, side. So this comes straight out of angle, angle, side congruency."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "And this comes straight from, once again, orange angle, white angle, side. So angle, angle, side. Orange angle, white angle, side. So this comes straight out of angle, angle, side congruency. And since we've now shown that this triangle is equal to that triangle, we know that their corresponding sides are equal. And then this is our home stretch. We now know, since these two triangles are congruent, we now know that ED is equal to DF, because their corresponding sides."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "So this comes straight out of angle, angle, side congruency. And since we've now shown that this triangle is equal to that triangle, we know that their corresponding sides are equal. And then this is our home stretch. We now know, since these two triangles are congruent, we now know that ED is equal to DF, because their corresponding sides. And I could write that right over here. ED is equal to DF. And once again, the reason here is the same thing up here."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "We now know, since these two triangles are congruent, we now know that ED is equal to DF, because their corresponding sides. And I could write that right over here. ED is equal to DF. And once again, the reason here is the same thing up here. Corresponding, so we know our statement 9, which means they're congruent. And corresponding sides, congruent. And we are done."}, {"video_title": "Congruent triangle example 2 Congruence Geometry Khan Academy.mp3", "Sentence": "And once again, the reason here is the same thing up here. Corresponding, so we know our statement 9, which means they're congruent. And corresponding sides, congruent. And we are done. So that was a pretty involved problem. But you see, once again, you go step by step. Just try to figure out whatever you can about each triangle, and you eventually get it."}, {"video_title": "Example with tangent and radius Circles Geometry Khan Academy.mp3", "Sentence": "What is this distance right over here between point A and point C? And I encourage you now to pause this video and try this out on your own. So I'm assuming you've given a go at it. So the key thing to realize here, if since AC is tangent to the circle at point C, that means it's going to be perpendicular to the radius between the center of the circle and point C. So this right over here is a right angle. And the reason why that is useful is now we know that triangle AOC is a right triangle. So we could use, if we know two of its sides, we could use the Pythagorean theorem to figure out the third. Now we clearly know OC."}, {"video_title": "Example with tangent and radius Circles Geometry Khan Academy.mp3", "Sentence": "So the key thing to realize here, if since AC is tangent to the circle at point C, that means it's going to be perpendicular to the radius between the center of the circle and point C. So this right over here is a right angle. And the reason why that is useful is now we know that triangle AOC is a right triangle. So we could use, if we know two of its sides, we could use the Pythagorean theorem to figure out the third. Now we clearly know OC. Now OA, we don't know the entire side. They only give us that AB is equal to two. But the thing that might jump out in your mind is OB is a radius."}, {"video_title": "Example with tangent and radius Circles Geometry Khan Academy.mp3", "Sentence": "Now we clearly know OC. Now OA, we don't know the entire side. They only give us that AB is equal to two. But the thing that might jump out in your mind is OB is a radius. OB is a radius. It's going to be the same length as any radius. So this is going to be three as well."}, {"video_title": "Example with tangent and radius Circles Geometry Khan Academy.mp3", "Sentence": "But the thing that might jump out in your mind is OB is a radius. OB is a radius. It's going to be the same length as any radius. So this is going to be three as well. It's the distance between the center of the circle and a point on the circle, just like the distance between O and C. So this is going to be three as well. And so now we are able to figure out that the hypotenuse of this triangle has length five. And so we need to figure out what side, the length of segment AC is."}, {"video_title": "Example with tangent and radius Circles Geometry Khan Academy.mp3", "Sentence": "So this is going to be three as well. It's the distance between the center of the circle and a point on the circle, just like the distance between O and C. So this is going to be three as well. And so now we are able to figure out that the hypotenuse of this triangle has length five. And so we need to figure out what side, the length of segment AC is. So let's just call that, I don't know, I'll call that, I will call that X. X. And so we know that X squared plus three squared, I'm just applying the Pythagorean theorem here, so plus three squared is going to be equal to the length of the hypotenuse squared, is going to be equal to five squared. And I know this is the hypotenuse."}, {"video_title": "Example with tangent and radius Circles Geometry Khan Academy.mp3", "Sentence": "And so we need to figure out what side, the length of segment AC is. So let's just call that, I don't know, I'll call that, I will call that X. X. And so we know that X squared plus three squared, I'm just applying the Pythagorean theorem here, so plus three squared is going to be equal to the length of the hypotenuse squared, is going to be equal to five squared. And I know this is the hypotenuse. It's the side opposite the 90 degree angle. It's the longest side of the right triangle. So X squared, X squared plus nine, plus nine is equal to 25, is equal to 25."}, {"video_title": "Example with tangent and radius Circles Geometry Khan Academy.mp3", "Sentence": "And I know this is the hypotenuse. It's the side opposite the 90 degree angle. It's the longest side of the right triangle. So X squared, X squared plus nine, plus nine is equal to 25, is equal to 25. Subtract nine from both sides, and you get, you get X squared is equal to 16. And so it should jump out at you that X is going to be equal to, X is equal to four. So X is equal to four."}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "So we're told circle P is below. This is circle P right over here. What is the arc measure of arc BC in degrees? So this is point B, this is point C. Let me pick a different color so you can see the arc. And since they only gave us two letters, we really want to find the minor arc. So we want to find the shorter arc between B and C. So the major arc would be the long way around. And if they wanted to specify the major arc, they would have had to give us three letters to force us to go the long way around."}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "So this is point B, this is point C. Let me pick a different color so you can see the arc. And since they only gave us two letters, we really want to find the minor arc. So we want to find the shorter arc between B and C. So the major arc would be the long way around. And if they wanted to specify the major arc, they would have had to give us three letters to force us to go the long way around. So if they said arc BAC or BDC, that would go the long way around. But since they just gave us just B and C, we assume it's going to be the minor arc. So we want to find that arc measure right over there."}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "And if they wanted to specify the major arc, they would have had to give us three letters to force us to go the long way around. So if they said arc BAC or BDC, that would go the long way around. But since they just gave us just B and C, we assume it's going to be the minor arc. So we want to find that arc measure right over there. Now the arc measure is going to be the exact same measure in degrees as the measure of the central angle that intercepts that arc. So it's going to be the same thing as the measure of this central angle, which is 4k plus 159 degrees. So if we can figure out what k is, we're going to know what this central angle measure is."}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "So we want to find that arc measure right over there. Now the arc measure is going to be the exact same measure in degrees as the measure of the central angle that intercepts that arc. So it's going to be the same thing as the measure of this central angle, which is 4k plus 159 degrees. So if we can figure out what k is, we're going to know what this central angle measure is. And then that's going to be the same thing as this arc measure. So how do we figure that out? Well, what might jump out at you is that this angle, angle BPC that we care about, is vertical to angle APD."}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "So if we can figure out what k is, we're going to know what this central angle measure is. And then that's going to be the same thing as this arc measure. So how do we figure that out? Well, what might jump out at you is that this angle, angle BPC that we care about, is vertical to angle APD. These are vertical angles. And so vertical angles are going to have the same measure. They're going to be congruent."}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, what might jump out at you is that this angle, angle BPC that we care about, is vertical to angle APD. These are vertical angles. And so vertical angles are going to have the same measure. They're going to be congruent. So let's set these two measures equal to each other. So we know that 4k plus 159 is going to be equal to 2k plus 153. So let's get all of our k terms on the left-hand side and all of the non-k terms on the right-hand side."}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "They're going to be congruent. So let's set these two measures equal to each other. So we know that 4k plus 159 is going to be equal to 2k plus 153. So let's get all of our k terms on the left-hand side and all of the non-k terms on the right-hand side. So let's subtract 2k from both sides. So we can subtract 2k from both sides. And let's subtract."}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "So let's get all of our k terms on the left-hand side and all of the non-k terms on the right-hand side. So let's subtract 2k from both sides. So we can subtract 2k from both sides. And let's subtract. Well, let me just do that first. I don't want to skip steps. And so I got rid of the k's on the right-hand side."}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "And let's subtract. Well, let me just do that first. I don't want to skip steps. And so I got rid of the k's on the right-hand side. So it's just going to be left with a 153. And on the left-hand side, 4k minus 2k is 2k. And I still have plus 159."}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "And so I got rid of the k's on the right-hand side. So it's just going to be left with a 153. And on the left-hand side, 4k minus 2k is 2k. And I still have plus 159. Let's get rid of this 159 on the left-hand side. So let's subtract it. But if I do it on the left-hand side, I need to do it on the right-hand side as well."}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "And I still have plus 159. Let's get rid of this 159 on the left-hand side. So let's subtract it. But if I do it on the left-hand side, I need to do it on the right-hand side as well. So subtract 159 from both sides. And I'm left with 2k is equal to 153 minus 159 is negative 6. So k is equal to, just divide both sides by 2, k is going to be equal to negative 3."}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "But if I do it on the left-hand side, I need to do it on the right-hand side as well. So subtract 159 from both sides. And I'm left with 2k is equal to 153 minus 159 is negative 6. So k is equal to, just divide both sides by 2, k is going to be equal to negative 3. Now, you might be tempted to say, oh, negative 3. But that's not what we're trying. We're not just trying to solve for k. We're trying to figure out this angle measure, which is going to be the same as the arc measure that we care about."}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "So k is equal to, just divide both sides by 2, k is going to be equal to negative 3. Now, you might be tempted to say, oh, negative 3. But that's not what we're trying. We're not just trying to solve for k. We're trying to figure out this angle measure, which is going to be the same as the arc measure that we care about. And that's just expressed in terms of k. So it's 4 times k plus 159. So that's going to be 4 times negative 3 plus 159. Well, what's that going to be?"}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "We're not just trying to solve for k. We're trying to figure out this angle measure, which is going to be the same as the arc measure that we care about. And that's just expressed in terms of k. So it's 4 times k plus 159. So that's going to be 4 times negative 3 plus 159. Well, what's that going to be? 4 times negative 3 is negative 12 plus 159 is going to be 147. So this angle right here is a measure of 147 degrees. And you can calculate."}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, what's that going to be? 4 times negative 3 is negative 12 plus 159 is going to be 147. So this angle right here is a measure of 147 degrees. And you can calculate. That's the same thing as over here. 2 times negative 3 is negative 6 plus 153 is 147 degrees. These two are the same."}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "And you can calculate. That's the same thing as over here. 2 times negative 3 is negative 6 plus 153 is 147 degrees. These two are the same. And so 147 degrees. This angle measure is the same as the measure of arc BC. Let's do one more of these."}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "These two are the same. And so 147 degrees. This angle measure is the same as the measure of arc BC. Let's do one more of these. Circle P is below. What is the arc measure of BC in degrees? Now, since once again, they only gave us two letters, we can assume it is the minor arc."}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "Let's do one more of these. Circle P is below. What is the arc measure of BC in degrees? Now, since once again, they only gave us two letters, we can assume it is the minor arc. So we care about BC. We care about this right over here. And so the measure of this arc is going to be the same thing as the measure of the central angle that intercepts that arc."}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "Now, since once again, they only gave us two letters, we can assume it is the minor arc. So we care about BC. We care about this right over here. And so the measure of this arc is going to be the same thing as the measure of the central angle that intercepts that arc. And that measure is going to be the sum of these two angles. So it's going to be 4y plus 6 plus 7y minus 7. So what's that?"}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "And so the measure of this arc is going to be the same thing as the measure of the central angle that intercepts that arc. And that measure is going to be the sum of these two angles. So it's going to be 4y plus 6 plus 7y minus 7. So what's that? 4y plus 7y. We can combine the y terms. It's going to be 11y."}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "So what's that? 4y plus 7y. We can combine the y terms. It's going to be 11y. And then 6 minus 7 is going to be negative 1. So it's going to be 11y minus 1. And how do we figure that out?"}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "It's going to be 11y. And then 6 minus 7 is going to be negative 1. So it's going to be 11y minus 1. And how do we figure that out? Well, how do we figure out what y is? We need to figure out what y is in order to figure out what 11y minus 1 is. Well, we know, let me write this down."}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "And how do we figure that out? Well, how do we figure out what y is? We need to figure out what y is in order to figure out what 11y minus 1 is. Well, we know, let me write this down. So the angle that we care about is 11y minus 1. We know that that angle plus this big angle that I'm going to show in blue, that if we add them together, that it's going to be 360 degrees. Because we would have gone all the way around the circle."}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, we know, let me write this down. So the angle that we care about is 11y minus 1. We know that that angle plus this big angle that I'm going to show in blue, that if we add them together, that it's going to be 360 degrees. Because we would have gone all the way around the circle. So we know that 11y minus 1 plus 20y minus 11 is going to be equal to 360 degrees. And so now we can just solve for y. What is, let me get some new colors involved, what is 11y plus 20y?"}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "Because we would have gone all the way around the circle. So we know that 11y minus 1 plus 20y minus 11 is going to be equal to 360 degrees. And so now we can just solve for y. What is, let me get some new colors involved, what is 11y plus 20y? Well, that's going to be 31y. And then if I have negative 1 and negative 11, that's going to be negative, let me do this in a different color. So that's going to be negative 1, negative 11."}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "What is, let me get some new colors involved, what is 11y plus 20y? Well, that's going to be 31y. And then if I have negative 1 and negative 11, that's going to be negative, let me do this in a different color. So that's going to be negative 1, negative 11. That's negative 12. And that's going to be equal to 360 degrees. So let's see, we could add 12 to both sides to get rid of that negative 12 right over there."}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "So that's going to be negative 1, negative 11. That's negative 12. And that's going to be equal to 360 degrees. So let's see, we could add 12 to both sides to get rid of that negative 12 right over there. And that's going to leave us with 31y is equal to 372. And so if we divide both sides by 31, it looks like 12. Yep, it'll go exactly 12 times."}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "So let's see, we could add 12 to both sides to get rid of that negative 12 right over there. And that's going to leave us with 31y is equal to 372. And so if we divide both sides by 31, it looks like 12. Yep, it'll go exactly 12 times. So y is equal to 12. And remember, we weren't trying to solve for y. We were trying to solve for 11y minus 1."}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "Yep, it'll go exactly 12 times. So y is equal to 12. And remember, we weren't trying to solve for y. We were trying to solve for 11y minus 1. So what is 11 times 12? We know that y is 12. 11 times 12 minus 1."}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "We were trying to solve for 11y minus 1. So what is 11 times 12? We know that y is 12. 11 times 12 minus 1. Let's see, 11 times 12 is 121. And then 121 minus 1 is going to be, oh, sorry, no. It's going to be, my multiplication tables are off."}, {"video_title": "Finding arc measures with equations Mathematics II High School Math Khan Academy.mp3", "Sentence": "11 times 12 minus 1. Let's see, 11 times 12 is 121. And then 121 minus 1 is going to be, oh, sorry, no. It's going to be, my multiplication tables are off. It's been a long day. 11 times 12 is going to be 132 minus 1 is going to be 131. And it's going to be in degrees."}, {"video_title": "Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3", "Sentence": "I have a circle here whose circumference is 18 pi. So if we were to measure all the way around the circle, we would get 18 pi. And we also have a central angle here. So this is the center of the circle. And this central angle that I'm about to draw has a measure of 10 degrees. So this angle right over here is 10 degrees. And what I'm curious about is the length of the arc that subtends that central angle."}, {"video_title": "Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3", "Sentence": "So this is the center of the circle. And this central angle that I'm about to draw has a measure of 10 degrees. So this angle right over here is 10 degrees. And what I'm curious about is the length of the arc that subtends that central angle. So what is the length of what I just did in magenta? And one way to think about it, or actually maybe the way to think about it, is that the ratio of this arc length to the entire circumference, let me write this down, arc length to the entire circumference should be the same as the ratio of the central angle to the total number of angles if you were to go all the way around the circle, so 2, 360 degrees. So let's just think about that."}, {"video_title": "Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3", "Sentence": "And what I'm curious about is the length of the arc that subtends that central angle. So what is the length of what I just did in magenta? And one way to think about it, or actually maybe the way to think about it, is that the ratio of this arc length to the entire circumference, let me write this down, arc length to the entire circumference should be the same as the ratio of the central angle to the total number of angles if you were to go all the way around the circle, so 2, 360 degrees. So let's just think about that. We know the circumference is 18 pi. So we know the circumference is 18 pi. We're looking for the arc length."}, {"video_title": "Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3", "Sentence": "So let's just think about that. We know the circumference is 18 pi. So we know the circumference is 18 pi. We're looking for the arc length. I'm just going to call that a, a for arc length. That's what we're going to try to solve for. We know that the central angle is 10 degrees."}, {"video_title": "Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3", "Sentence": "We're looking for the arc length. I'm just going to call that a, a for arc length. That's what we're going to try to solve for. We know that the central angle is 10 degrees. So you have 10 degrees over 360 degrees. Over 360. So we could simplify this by multiplying both sides by 18 pi."}, {"video_title": "Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3", "Sentence": "We know that the central angle is 10 degrees. So you have 10 degrees over 360 degrees. Over 360. So we could simplify this by multiplying both sides by 18 pi. Multiply both sides by 18 pi. And we get that our arc length is equal to, well, 10 over 360 is the same thing as 1 over 36. So it's equal to 1 over 36 times 18 pi, which is just so."}, {"video_title": "Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3", "Sentence": "So we could simplify this by multiplying both sides by 18 pi. Multiply both sides by 18 pi. And we get that our arc length is equal to, well, 10 over 360 is the same thing as 1 over 36. So it's equal to 1 over 36 times 18 pi, which is just so. It's 18 pi over 36, which is the same thing as pi over 2. So this arc right over here is going to be pi over 2, whatever units we're talking about, long. Now let's think about another scenario."}, {"video_title": "Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3", "Sentence": "So it's equal to 1 over 36 times 18 pi, which is just so. It's 18 pi over 36, which is the same thing as pi over 2. So this arc right over here is going to be pi over 2, whatever units we're talking about, long. Now let's think about another scenario. Let's imagine another, or the same circle. So it's the same circle here. Our circumference is still 18 pi."}, {"video_title": "Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3", "Sentence": "Now let's think about another scenario. Let's imagine another, or the same circle. So it's the same circle here. Our circumference is still 18 pi. Circumference is still 18 pi. There are people having a conference behind me or something. That's why you might hear those mumbling voices."}, {"video_title": "Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3", "Sentence": "Our circumference is still 18 pi. Circumference is still 18 pi. There are people having a conference behind me or something. That's why you might hear those mumbling voices. But this circumference is also 18 pi. So this is also 18 pi. But now I'm going to make the central angle, I'm going to make it an obtuse angle."}, {"video_title": "Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3", "Sentence": "That's why you might hear those mumbling voices. But this circumference is also 18 pi. So this is also 18 pi. But now I'm going to make the central angle, I'm going to make it an obtuse angle. So the central angle now, so let's say we were to start right over here. This is one side of the angle. I'm going to go and make a 350 degree angle."}, {"video_title": "Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3", "Sentence": "But now I'm going to make the central angle, I'm going to make it an obtuse angle. So the central angle now, so let's say we were to start right over here. This is one side of the angle. I'm going to go and make a 350 degree angle. So I'm going to go all the way around like that. So this right over here is a 350 degree angle. And now I'm curious about this arc that subtends this really huge angle."}, {"video_title": "Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3", "Sentence": "I'm going to go and make a 350 degree angle. So I'm going to go all the way around like that. So this right over here is a 350 degree angle. And now I'm curious about this arc that subtends this really huge angle. So now I want to figure out this arc length. So all of this, I want to figure out this arc length, the arc that subtends this really obtuse angle right over here. Well, same exact logic."}, {"video_title": "Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3", "Sentence": "And now I'm curious about this arc that subtends this really huge angle. So now I want to figure out this arc length. So all of this, I want to figure out this arc length, the arc that subtends this really obtuse angle right over here. Well, same exact logic. The ratio between our arc length, the ratio between our arc length, a, and the circumference of the entire circle, 18 pi, should be the same as the ratio between our central angle, our central angle that the arc subtends, so 350, over the total number of degrees in a circle, over 360. So multiply both sides by 18 pi, we get a is equal to, let's see, this is 35 times 18 over 36 pi. So 350 divided by 360 is 35 over 36."}, {"video_title": "Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3", "Sentence": "Well, same exact logic. The ratio between our arc length, the ratio between our arc length, a, and the circumference of the entire circle, 18 pi, should be the same as the ratio between our central angle, our central angle that the arc subtends, so 350, over the total number of degrees in a circle, over 360. So multiply both sides by 18 pi, we get a is equal to, let's see, this is 35 times 18 over 36 pi. So 350 divided by 360 is 35 over 36. So this is 35 times 18 times pi over 36. Well, both 36 and 18 are divisible by 18, so let's divide them both by 18. And so we are left with 35 over 2 pi, or we could even say, let me just write it that way, 35 pi over 2, or if you wanted to write it as a decimal, this would be 17.5 pi."}, {"video_title": "Length of an arc that subtends a central angle Circles Geometry Khan Academy.mp3", "Sentence": "So 350 divided by 360 is 35 over 36. So this is 35 times 18 times pi over 36. Well, both 36 and 18 are divisible by 18, so let's divide them both by 18. And so we are left with 35 over 2 pi, or we could even say, let me just write it that way, 35 pi over 2, or if you wanted to write it as a decimal, this would be 17.5 pi. Now, does this make sense? This right over here, this other arc length, when our central angle was 10 degrees, this had an arc length of 0.5 pi. So when you add these two together, this arc length and this arc length, 0.5 plus 17.5, you get to 18 pi, which was the circumference, which makes complete sense, because if you add these angles, 10 degrees and 350 degrees, you get 360 degrees in a circle."}, {"video_title": "Example reflecting quadrilateral over x axis.mp3", "Sentence": "And we have our little tool here on Khan Academy where we can construct a quadrilateral. And we need to construct the reflection of triangle ABCD. And so what we can do is, let me scroll down a little bit so we can see the entire coordinate axis. We wanna find the reflection across the x-axis. So I'm gonna reflect point by point. And actually, let me just move this whole thing down here so that we can see what is going on a little bit clearer. So let's just first reflect point, let me move this a little bit out of the way."}, {"video_title": "Example reflecting quadrilateral over x axis.mp3", "Sentence": "We wanna find the reflection across the x-axis. So I'm gonna reflect point by point. And actually, let me just move this whole thing down here so that we can see what is going on a little bit clearer. So let's just first reflect point, let me move this a little bit out of the way. So let's first reflect point A. So we're gonna reflect across the x-axis. A is four units above the x-axis."}, {"video_title": "Example reflecting quadrilateral over x axis.mp3", "Sentence": "So let's just first reflect point, let me move this a little bit out of the way. So let's first reflect point A. So we're gonna reflect across the x-axis. A is four units above the x-axis. One, two, three, four. So its image, A prime we could say, would be four units below the x-axis. So one, two, three, four."}, {"video_title": "Example reflecting quadrilateral over x axis.mp3", "Sentence": "A is four units above the x-axis. One, two, three, four. So its image, A prime we could say, would be four units below the x-axis. So one, two, three, four. So let's make this right over here A, A prime. I'm having trouble putting the, see if I move these other characters around. Okay, there you go."}, {"video_title": "Example reflecting quadrilateral over x axis.mp3", "Sentence": "So one, two, three, four. So let's make this right over here A, A prime. I'm having trouble putting the, see if I move these other characters around. Okay, there you go. So this is gonna be my A prime. Now let me try B. B is two units above the x-axis."}, {"video_title": "Example reflecting quadrilateral over x axis.mp3", "Sentence": "Okay, there you go. So this is gonna be my A prime. Now let me try B. B is two units above the x-axis. So B prime is gonna have the same x-coordinate, but it's gonna be two units below the x-axis. So let's make this our B. So this is our B right over here."}, {"video_title": "Example reflecting quadrilateral over x axis.mp3", "Sentence": "B is two units above the x-axis. So B prime is gonna have the same x-coordinate, but it's gonna be two units below the x-axis. So let's make this our B. So this is our B right over here. Now let's make this our C. C right here has the x-coordinate of negative five and a y-coordinate of negative four. Now C prime would have the same x-coordinate, but instead of being four units below the x-axis, it'll be four units above the x-axis. So we would have the coordinates negative five comma positive four."}, {"video_title": "Example reflecting quadrilateral over x axis.mp3", "Sentence": "So this is our B right over here. Now let's make this our C. C right here has the x-coordinate of negative five and a y-coordinate of negative four. Now C prime would have the same x-coordinate, but instead of being four units below the x-axis, it'll be four units above the x-axis. So we would have the coordinates negative five comma positive four. So this is gonna be our C here. So this goes to negative five. One, two, three, positive four."}, {"video_title": "Example reflecting quadrilateral over x axis.mp3", "Sentence": "So we would have the coordinates negative five comma positive four. So this is gonna be our C here. So this goes to negative five. One, two, three, positive four. And then last but not least, D. And so let's see, D right now is at negative two comma negative one. If we were to reflect across the x-axis, instead of being one unit below the x-axis, we'll be one unit above the x-axis, and we'll keep our x-coordinate of negative two. And so there you have it."}, {"video_title": "Example reflecting quadrilateral over x axis.mp3", "Sentence": "One, two, three, positive four. And then last but not least, D. And so let's see, D right now is at negative two comma negative one. If we were to reflect across the x-axis, instead of being one unit below the x-axis, we'll be one unit above the x-axis, and we'll keep our x-coordinate of negative two. And so there you have it. We have constructed the reflection of ABCD across the x-axis. And what's interesting about this example is that the original quadrilateral is on top of the x-axis. So you can kinda see this top part of the quadrilateral gets reflected below it, and this bottom part of the quadrilateral gets reflected above it."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "And I also touched a little bit on the idea of an axiom or postulate, but I want to be clear. Sometimes you will hear this referred to as a side, side, side theorem, and sometimes you'll hear it as a side, side, side postulate or axiom. Postulate or axiom. And I think it's worth differentiating what these mean. A postulate or an axiom is something that you just assume. You assume from the get-go, while a theorem is something you prove using more basic or using some postulates or axioms. So, in really, in all of mathematics, you make some core assumptions."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "And I think it's worth differentiating what these mean. A postulate or an axiom is something that you just assume. You assume from the get-go, while a theorem is something you prove using more basic or using some postulates or axioms. So, in really, in all of mathematics, you make some core assumptions. You make some core assumptions. You call these the axioms or the postulates. And then using those, you try to prove theorems."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "So, in really, in all of mathematics, you make some core assumptions. You make some core assumptions. You call these the axioms or the postulates. And then using those, you try to prove theorems. So maybe using that one, I can prove some theorem over here. And maybe using that theorem and then this axiom, I can prove another theorem over here. And then using both of those theorems, I can prove another theorem over here."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "And then using those, you try to prove theorems. So maybe using that one, I can prove some theorem over here. And maybe using that theorem and then this axiom, I can prove another theorem over here. And then using both of those theorems, I can prove another theorem over here. I think you get the picture. So this axiom might lead us to this theorem, and these two might lead us to this theorem right over here. And we essentially try to build our knowledge or we build a mathematics around these core assumptions."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "And then using both of those theorems, I can prove another theorem over here. I think you get the picture. So this axiom might lead us to this theorem, and these two might lead us to this theorem right over here. And we essentially try to build our knowledge or we build a mathematics around these core assumptions. In an introductory geometry class, we kind of, we don't rigorously prove side, side, side. We don't rigorously prove the side, side, side theorem. And that's why in a lot of geometry classes, you kind of just take it as a given, as a postulate or an axiom."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "And we essentially try to build our knowledge or we build a mathematics around these core assumptions. In an introductory geometry class, we kind of, we don't rigorously prove side, side, side. We don't rigorously prove the side, side, side theorem. And that's why in a lot of geometry classes, you kind of just take it as a given, as a postulate or an axiom. And the whole reason why I'm doing this is one, just so you know the difference between the words theorem and postulate or axiom, and also so that you don't get confused. It is just a given, but in a lot of books, and I've looked at several books, they do refer to it as a side, side, side theorem, even though they never prove it rigorously. They do just assume it."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "And that's why in a lot of geometry classes, you kind of just take it as a given, as a postulate or an axiom. And the whole reason why I'm doing this is one, just so you know the difference between the words theorem and postulate or axiom, and also so that you don't get confused. It is just a given, but in a lot of books, and I've looked at several books, they do refer to it as a side, side, side theorem, even though they never prove it rigorously. They do just assume it. So it really is more of a postulate or an axiom. Now with that out of the way, we're just going to assume going forward that we just know that this is true. We're going to take it as a given."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "They do just assume it. So it really is more of a postulate or an axiom. Now with that out of the way, we're just going to assume going forward that we just know that this is true. We're going to take it as a given. I want to show you that we can already do something pretty useful with it. So let's say that we have a circle. There's many useful things that we can already do with it."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "We're going to take it as a given. I want to show you that we can already do something pretty useful with it. So let's say that we have a circle. There's many useful things that we can already do with it. And this circle has a center right here at A. And let's say that we have a chord in this circle that is not a diameter. So let me draw a chord here."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "There's many useful things that we can already do with it. And this circle has a center right here at A. And let's say that we have a chord in this circle that is not a diameter. So let me draw a chord here. So let me draw a chord in this circle. So it's a kind of a segment of a secant line. And let's say that I have a line that bisects this chord from the center."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "So let me draw a chord here. So let me draw a chord in this circle. So it's a kind of a segment of a secant line. And let's say that I have a line that bisects this chord from the center. I guess I could call it a radius because I'm going to go from the center to the edge of the circle right over there. So I'm going to the center to the circle itself. And when I say bisects it, so these are all, I'm just setting up the problem right now."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "And let's say that I have a line that bisects this chord from the center. I guess I could call it a radius because I'm going to go from the center to the edge of the circle right over there. So I'm going to the center to the circle itself. And when I say bisects it, so these are all, I'm just setting up the problem right now. When I say bisecting it, it means it splits that line segment in half. So what it tells us is that the length of this segment right over here is going to be equivalent to the length of this segment right over there. And what I want to do is, so this is, I've set it up, I have a circle."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "And when I say bisects it, so these are all, I'm just setting up the problem right now. When I say bisecting it, it means it splits that line segment in half. So what it tells us is that the length of this segment right over here is going to be equivalent to the length of this segment right over there. And what I want to do is, so this is, I've set it up, I have a circle. This radius bisects this chord right over here. And what I want to do is prove, the goal here is to prove that it bisects this chord at a right angle. Or another way to say it, let me add some points here."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "And what I want to do is, so this is, I've set it up, I have a circle. This radius bisects this chord right over here. And what I want to do is prove, the goal here is to prove that it bisects this chord at a right angle. Or another way to say it, let me add some points here. Let's call this B, let's call this C, and let's call this D. I want to prove that segment AB is perpendicular, it intersects it at a right angle, is perpendicular to segment CD. And as you can imagine, I'm going to prove it pretty much using the side, side, side, whatever you want to call it, side, side, side theorem, postulate, or axiom. So let's do it, let's think about it this way."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "Or another way to say it, let me add some points here. Let's call this B, let's call this C, and let's call this D. I want to prove that segment AB is perpendicular, it intersects it at a right angle, is perpendicular to segment CD. And as you can imagine, I'm going to prove it pretty much using the side, side, side, whatever you want to call it, side, side, side theorem, postulate, or axiom. So let's do it, let's think about it this way. So you can imagine, if I'm going to use this, I need to have some triangles. There's no triangles here right now. But I can construct triangles, and I can construct triangles based on things I know."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "So let's do it, let's think about it this way. So you can imagine, if I'm going to use this, I need to have some triangles. There's no triangles here right now. But I can construct triangles, and I can construct triangles based on things I know. For example, I can construct, this has some radius, so let's call this, that's a radius right over here. The length of that is just going to be the radius of the circle. But I can also do it right over here."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "But I can construct triangles, and I can construct triangles based on things I know. For example, I can construct, this has some radius, so let's call this, that's a radius right over here. The length of that is just going to be the radius of the circle. But I can also do it right over here. The length of AC is also going to be the radius of the circle. So we know that these two lines have the same length, which is the radius of the circle. We could say that AD is congruent to AC, or they have the exact same lengths."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "But I can also do it right over here. The length of AC is also going to be the radius of the circle. So we know that these two lines have the same length, which is the radius of the circle. We could say that AD is congruent to AC, or they have the exact same lengths. We know from the setup in the problem, that this segment is equal in length to this segment over here. We could even, let me add a point here so I can refer to it. So if I call that point E, we know from the setup in the problem that CE is congruent to ED, or they have the same length."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "We could say that AD is congruent to AC, or they have the exact same lengths. We know from the setup in the problem, that this segment is equal in length to this segment over here. We could even, let me add a point here so I can refer to it. So if I call that point E, we know from the setup in the problem that CE is congruent to ED, or they have the same length. CE has the same length as ED. We also know that both of these triangles, the one here on the left and the one here on the right, they both share the side EA. So EA is clearly equal to EA."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "So if I call that point E, we know from the setup in the problem that CE is congruent to ED, or they have the same length. CE has the same length as ED. We also know that both of these triangles, the one here on the left and the one here on the right, they both share the side EA. So EA is clearly equal to EA. So this is clearly equal to itself. It's the same side. The same side is being used for both triangles."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "So EA is clearly equal to EA. So this is clearly equal to itself. It's the same side. The same side is being used for both triangles. The triangles are adjacent to each other. So we see a situation where we have two different triangles that have corresponding sides being equal. This side is equivalent to this side right over here."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "The same side is being used for both triangles. The triangles are adjacent to each other. So we see a situation where we have two different triangles that have corresponding sides being equal. This side is equivalent to this side right over here. This side is equal in length to that side over there. And then we have, obviously AE is equivalent to itself. It's a side on both of them."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "This side is equivalent to this side right over here. This side is equal in length to that side over there. And then we have, obviously AE is equivalent to itself. It's a side on both of them. It's the corresponding side on both of these triangles. So by side, side, side, we know that triangle ABC is congruent to triangle AE. Sorry, it's not ABC."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "It's a side on both of them. It's the corresponding side on both of these triangles. So by side, side, side, we know that triangle ABC is congruent to triangle AE. Sorry, it's not ABC. It's AEC. We know, actually let me write it over here. By side, side, side, we know that triangle AEC is congruent to triangle AED."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "Sorry, it's not ABC. It's AEC. We know, actually let me write it over here. By side, side, side, we know that triangle AEC is congruent to triangle AED. But how does that help us? How does that help us knowing that, you know, we used our little theorem, but how does that actually help us here? Well, what's cool is once we know that two triangles are congruent, so because they are congruent, that tells us, so from that we can deduce that all the angles are the same."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "By side, side, side, we know that triangle AEC is congruent to triangle AED. But how does that help us? How does that help us knowing that, you know, we used our little theorem, but how does that actually help us here? Well, what's cool is once we know that two triangles are congruent, so because they are congruent, that tells us, so from that we can deduce that all the angles are the same. And in particular, we can deduce that this angle right over here, that the measure of angle CEA is equivalent to the measure of angle DEA. And the reason why that's useful is that we also see just by looking at this that they are supplementary to each other. Their adjacent angles, their outer sides form a straight angle."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "Well, what's cool is once we know that two triangles are congruent, so because they are congruent, that tells us, so from that we can deduce that all the angles are the same. And in particular, we can deduce that this angle right over here, that the measure of angle CEA is equivalent to the measure of angle DEA. And the reason why that's useful is that we also see just by looking at this that they are supplementary to each other. Their adjacent angles, their outer sides form a straight angle. So CEA is supplementary and equivalent to DEA. So they're also supplementary. So we also have the measure of angle CEA plus the measure of angle DEA is equal to 180 degrees."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "Their adjacent angles, their outer sides form a straight angle. So CEA is supplementary and equivalent to DEA. So they're also supplementary. So we also have the measure of angle CEA plus the measure of angle DEA is equal to 180 degrees. But they're equivalent to each other. So I could replace the measure of DEA with the measure of CEA. Measure of angle CEA."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "So we also have the measure of angle CEA plus the measure of angle DEA is equal to 180 degrees. But they're equivalent to each other. So I could replace the measure of DEA with the measure of CEA. Measure of angle CEA. Or I could rewrite this as 2 times the measure of angle CEA is equal to 180 degrees. Or I could divide both sides by 2, and I say the measure of angle CEA is equal to 90 degrees, which is going to be the same as the measure of angle DEA because they're equivalent. So we know that this angle right over here is 90 degrees, so I can do it with that little box."}, {"video_title": "SSS to show a radius is perpendicular to a chord that it bisects Geometry Khan Academy.mp3", "Sentence": "Measure of angle CEA. Or I could rewrite this as 2 times the measure of angle CEA is equal to 180 degrees. Or I could divide both sides by 2, and I say the measure of angle CEA is equal to 90 degrees, which is going to be the same as the measure of angle DEA because they're equivalent. So we know that this angle right over here is 90 degrees, so I can do it with that little box. And this angle right over here is 90 degrees. And because AB intersects where it intersects CD, we have a 90-degree angle here and there, and we could also prove that it's over there as well. They are perpendicular to each other."}, {"video_title": "Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So let's just start by drawing ourselves a pyramid, and I'll draw one with a rectangular base, but depending on how we look at the formula, we could have a more general version. But a pyramid looks something like this. And you might get a sense of what the formula for the volume of a pyramid might be. If we say this dimension right over here is x, this dimension right over here, the length right over here is y, and then you have a height of this pyramid. If you were to go from the center straight to the top, or if you were to measure this distance right over here, which is the height of the pyramid, you just call that, let's call that z. And so you might say, well, I'm dealing with three dimensions, so maybe I multiply the three dimensions together, and that would give you volume in terms of units, but if you just multiplied xy times z, that would give you the volume of the entire rectangular prism that contains the pyramid. So that would give you the volume of this thing, which is clearly bigger, has a larger volume than the pyramid itself."}, {"video_title": "Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "If we say this dimension right over here is x, this dimension right over here, the length right over here is y, and then you have a height of this pyramid. If you were to go from the center straight to the top, or if you were to measure this distance right over here, which is the height of the pyramid, you just call that, let's call that z. And so you might say, well, I'm dealing with three dimensions, so maybe I multiply the three dimensions together, and that would give you volume in terms of units, but if you just multiplied xy times z, that would give you the volume of the entire rectangular prism that contains the pyramid. So that would give you the volume of this thing, which is clearly bigger, has a larger volume than the pyramid itself. The pyramid is fully contained inside of it, so this would be the tip of the pyramid on the surface, just like that. And so you might get a sense that, all right, maybe the volume of the pyramid is equal to x times y times z times some constant. And what we're going to do in this video is have an argument as to what that constant should be, assuming that this, the volume of the pyramid is roughly of this structure."}, {"video_title": "Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So that would give you the volume of this thing, which is clearly bigger, has a larger volume than the pyramid itself. The pyramid is fully contained inside of it, so this would be the tip of the pyramid on the surface, just like that. And so you might get a sense that, all right, maybe the volume of the pyramid is equal to x times y times z times some constant. And what we're going to do in this video is have an argument as to what that constant should be, assuming that this, the volume of the pyramid is roughly of this structure. And to help us with that, let's draw a larger rectangular prism and break it up into six pyramids that completely make up the volume of the rectangular prism. So first, let's imagine a pyramid that looks something like this, where its width is x, its depth is y, so that could be its base, and its height is halfway up the rectangular prism. So if the rectangular prism has height z, the pyramid's height is going to be z over two."}, {"video_title": "Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "And what we're going to do in this video is have an argument as to what that constant should be, assuming that this, the volume of the pyramid is roughly of this structure. And to help us with that, let's draw a larger rectangular prism and break it up into six pyramids that completely make up the volume of the rectangular prism. So first, let's imagine a pyramid that looks something like this, where its width is x, its depth is y, so that could be its base, and its height is halfway up the rectangular prism. So if the rectangular prism has height z, the pyramid's height is going to be z over two. Now, what would be the volume of that pyramid based on what we just saw over here? Well, that volume would be equal to some constant k times x times y, not times z, times the height of the pyramid, times z over two. So it'd be x times y times z over two."}, {"video_title": "Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So if the rectangular prism has height z, the pyramid's height is going to be z over two. Now, what would be the volume of that pyramid based on what we just saw over here? Well, that volume would be equal to some constant k times x times y, not times z, times the height of the pyramid, times z over two. So it'd be x times y times z over two. I'll just write times z over two. Or actually, we could even write it this way, x, y, z over two. Now, I can construct another pyramid that has the exact same dimensions."}, {"video_title": "Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So it'd be x times y times z over two. I'll just write times z over two. Or actually, we could even write it this way, x, y, z over two. Now, I can construct another pyramid that has the exact same dimensions. If I were to just flip that existing pyramid on its head and look something like this, this pyramid also has dimensions of an x-width, a y-depth, and a z over two height. So its volume would be this as well. Now, what is the combined volume of these two pyramids?"}, {"video_title": "Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "Now, I can construct another pyramid that has the exact same dimensions. If I were to just flip that existing pyramid on its head and look something like this, this pyramid also has dimensions of an x-width, a y-depth, and a z over two height. So its volume would be this as well. Now, what is the combined volume of these two pyramids? Well, it's just going to be this times two. So the combined volume of these pyramids, let me just draw it that way. So these two pyramids that look something like this, I'm gonna try to color-code it."}, {"video_title": "Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "Now, what is the combined volume of these two pyramids? Well, it's just going to be this times two. So the combined volume of these pyramids, let me just draw it that way. So these two pyramids that look something like this, I'm gonna try to color-code it. We have two of them, so two times their volume is going to be equal to, well, two times this is just going to be k times x, y, z. K, x, y, and z. And we have more pyramids to deal with. For example, I have this pyramid right over here where this face is its base."}, {"video_title": "Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So these two pyramids that look something like this, I'm gonna try to color-code it. We have two of them, so two times their volume is going to be equal to, well, two times this is just going to be k times x, y, z. K, x, y, and z. And we have more pyramids to deal with. For example, I have this pyramid right over here where this face is its base. And then if I try to draw the pyramid, it looks something like this, this one right over there. Now, what is its volume going to be? Its volume is going to be equal to k times its base is y times z."}, {"video_title": "Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "For example, I have this pyramid right over here where this face is its base. And then if I try to draw the pyramid, it looks something like this, this one right over there. Now, what is its volume going to be? Its volume is going to be equal to k times its base is y times z. So k, y, z. And what's its height? Well, its height is going to be half of x."}, {"video_title": "Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "Its volume is going to be equal to k times its base is y times z. So k, y, z. And what's its height? Well, its height is going to be half of x. So this height right over here is half of x. So it's k times y times z times x over two, or I could say times x, and then divide everything by two. Now, I have another pyramid that has the exact same dimensions."}, {"video_title": "Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "Well, its height is going to be half of x. So this height right over here is half of x. So it's k times y times z times x over two, or I could say times x, and then divide everything by two. Now, I have another pyramid that has the exact same dimensions. This one over here, if I try to draw it on the other face opposite the one we just saw, essentially if we just flip this one over, has the exact same dimensions. So one way to think about it, we have two pyramids that look like that with those types of dimensions. This is for an arbitrary rectangular prism that we are dealing with."}, {"video_title": "Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "Now, I have another pyramid that has the exact same dimensions. This one over here, if I try to draw it on the other face opposite the one we just saw, essentially if we just flip this one over, has the exact same dimensions. So one way to think about it, we have two pyramids that look like that with those types of dimensions. This is for an arbitrary rectangular prism that we are dealing with. So I have two of these. And so if you have two of their volumes, what's it going to be? It's going to be two times this expression."}, {"video_title": "Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "This is for an arbitrary rectangular prism that we are dealing with. So I have two of these. And so if you have two of their volumes, what's it going to be? It's going to be two times this expression. So it's going to be k times x, y, z. X, y, z. Interesting. And then last but not least, we have two more pyramids."}, {"video_title": "Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "It's going to be two times this expression. So it's going to be k times x, y, z. X, y, z. Interesting. And then last but not least, we have two more pyramids. We have this one that has a face that has the base right over here. That's its base, and if it was transparent, you'd be able to see where I'm drawing right here. And then you have one on the opposite side, right over there on the other side."}, {"video_title": "Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "And then last but not least, we have two more pyramids. We have this one that has a face that has the base right over here. That's its base, and if it was transparent, you'd be able to see where I'm drawing right here. And then you have one on the opposite side, right over there on the other side. And I could just say if you were to flip this around. And so by the exact same argument, so let me just draw it. So we have two of these, two of these pyramids."}, {"video_title": "Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "And then you have one on the opposite side, right over there on the other side. And I could just say if you were to flip this around. And so by the exact same argument, so let me just draw it. So we have two of these, two of these pyramids. My best to draw it. So times two. So each of them would have a volume of what?"}, {"video_title": "Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So we have two of these, two of these pyramids. My best to draw it. So times two. So each of them would have a volume of what? Each of them, their base is x times z. So it's going to be k times x times z. That's the area of their base."}, {"video_title": "Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So each of them would have a volume of what? Each of them, their base is x times z. So it's going to be k times x times z. That's the area of their base. And then what is their height? Well, each of them has a height of y over two. So times y over two."}, {"video_title": "Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "That's the area of their base. And then what is their height? Well, each of them has a height of y over two. So times y over two. And I have two of those pyramids, so I'm gonna multiply those by two. The twos cancel out, so I'm just left with k times x, y, z. So k times x, y, z."}, {"video_title": "Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So times y over two. And I have two of those pyramids, so I'm gonna multiply those by two. The twos cancel out, so I'm just left with k times x, y, z. So k times x, y, z. Now, one of the interesting things that we've just stumbled on in this is seeing that even though these pyramids have different dimensions and look different, they all have actually the same volume, which is interesting in and of themselves. And so if we were to add up the volumes of all of the pyramids here and use this formula to express them, so if I were to add all of them together, that should be equal to the volume of the entire rectangular prism. And then maybe we can figure out k. So the volume of the entire rectangular prism is x, y, z, x times y times z."}, {"video_title": "Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So k times x, y, z. Now, one of the interesting things that we've just stumbled on in this is seeing that even though these pyramids have different dimensions and look different, they all have actually the same volume, which is interesting in and of themselves. And so if we were to add up the volumes of all of the pyramids here and use this formula to express them, so if I were to add all of them together, that should be equal to the volume of the entire rectangular prism. And then maybe we can figure out k. So the volume of the entire rectangular prism is x, y, z, x times y times z. And then that's gotta be equal to the sum of these. So that's going to be equal to k x, y, z plus k x, y, z plus k x, y, z. Or you could say that's going to be equal to three k x, y, z."}, {"video_title": "Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "And then maybe we can figure out k. So the volume of the entire rectangular prism is x, y, z, x times y times z. And then that's gotta be equal to the sum of these. So that's going to be equal to k x, y, z plus k x, y, z plus k x, y, z. Or you could say that's going to be equal to three k x, y, z. All I did is I said, let me just add up the volume from all of these pyramids. And so what do we get for k? Well, we could divide both sides by three x, y, z to solve for k. Three x, y, z, three x, y, z."}, {"video_title": "Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "Or you could say that's going to be equal to three k x, y, z. All I did is I said, let me just add up the volume from all of these pyramids. And so what do we get for k? Well, we could divide both sides by three x, y, z to solve for k. Three x, y, z, three x, y, z. And we are left with, on the right-hand side, everything cancels out, we're just left with a k. And on the left-hand side, we're left with a 1 3rd. And so we get k is equal to 1 3rd. K is equal to 1 3rd."}, {"video_title": "Volume of pyramids intuition Solid geometry High school geometry Khan Academy.mp3", "Sentence": "Well, we could divide both sides by three x, y, z to solve for k. Three x, y, z, three x, y, z. And we are left with, on the right-hand side, everything cancels out, we're just left with a k. And on the left-hand side, we're left with a 1 3rd. And so we get k is equal to 1 3rd. K is equal to 1 3rd. And there you have it, that's our argument for why the volume of a pyramid is 1 3rd times the dimensions of the base times the height. So you might see it written that way, or you might see it written as 1 3rd times base. And so if x times y is the base, so the area of the base, so the base area, times the height, which in this case is z."}, {"video_title": "Geometric constructions congruent angles Congruence High school geometry Khan Academy.mp3", "Sentence": "What we're going to do in this video is learn to construct congruent angles. And we're going to do it with, of course, a pen or a pencil here. I'm going to use a ruler as a straight edge. And then I'm going to use a tool known as a compass, which looks a little bit fancy. But what it allows us to do, and we'll apply using it in a little bit, is it allows us to draw perfect circles or arcs of a given radius. You pivot on one point here, and then you use your pen or your pencil to trace out the arc or the circle. So let's just start with this angle right over here."}, {"video_title": "Geometric constructions congruent angles Congruence High school geometry Khan Academy.mp3", "Sentence": "And then I'm going to use a tool known as a compass, which looks a little bit fancy. But what it allows us to do, and we'll apply using it in a little bit, is it allows us to draw perfect circles or arcs of a given radius. You pivot on one point here, and then you use your pen or your pencil to trace out the arc or the circle. So let's just start with this angle right over here. And I'm going to construct an angle that is congruent to it. So let me make the vertex of my second angle right over there. And then let me draw one of the rays that originates at that vertex."}, {"video_title": "Geometric constructions congruent angles Congruence High school geometry Khan Academy.mp3", "Sentence": "So let's just start with this angle right over here. And I'm going to construct an angle that is congruent to it. So let me make the vertex of my second angle right over there. And then let me draw one of the rays that originates at that vertex. And I'm gonna put this angle in a different orientation just to show that they don't even have to have the same orientation. So it's going to look something like that. That's one of the rays."}, {"video_title": "Geometric constructions congruent angles Congruence High school geometry Khan Academy.mp3", "Sentence": "And then let me draw one of the rays that originates at that vertex. And I'm gonna put this angle in a different orientation just to show that they don't even have to have the same orientation. So it's going to look something like that. That's one of the rays. But then we have to figure out where do we put, where do we put the other ray so that the two angles are congruent? And this is where our compass is going to be really useful. So what I'm going to do is put the pivot point of a compass, of the compass, right at the vertex of the first angle."}, {"video_title": "Geometric constructions congruent angles Congruence High school geometry Khan Academy.mp3", "Sentence": "That's one of the rays. But then we have to figure out where do we put, where do we put the other ray so that the two angles are congruent? And this is where our compass is going to be really useful. So what I'm going to do is put the pivot point of a compass, of the compass, right at the vertex of the first angle. And I'm going to draw out an arc like this. And what's useful about the compass is you can keep the radius constant. And you can see it intersects our first two rays at points, let's just call this B and C. And I could call this point A right over here."}, {"video_title": "Geometric constructions congruent angles Congruence High school geometry Khan Academy.mp3", "Sentence": "So what I'm going to do is put the pivot point of a compass, of the compass, right at the vertex of the first angle. And I'm going to draw out an arc like this. And what's useful about the compass is you can keep the radius constant. And you can see it intersects our first two rays at points, let's just call this B and C. And I could call this point A right over here. And so let me, now that I have my compass with the exact right radius right now, let me draw that right over here. But this alone won't allow us to draw the angle just yet. But let me draw it like this."}, {"video_title": "Geometric constructions congruent angles Congruence High school geometry Khan Academy.mp3", "Sentence": "And you can see it intersects our first two rays at points, let's just call this B and C. And I could call this point A right over here. And so let me, now that I have my compass with the exact right radius right now, let me draw that right over here. But this alone won't allow us to draw the angle just yet. But let me draw it like this. And that is pretty good. And let's call this point right over here D, I don't know, I'll call this one E. And I want to figure out where to put my third point F so I can define ray EF so that these two angles are congruent. And what I can do is take my compass again and get a clear sense of the distance between C and B by adjusting my compass."}, {"video_title": "Geometric constructions congruent angles Congruence High school geometry Khan Academy.mp3", "Sentence": "But let me draw it like this. And that is pretty good. And let's call this point right over here D, I don't know, I'll call this one E. And I want to figure out where to put my third point F so I can define ray EF so that these two angles are congruent. And what I can do is take my compass again and get a clear sense of the distance between C and B by adjusting my compass. So one point is on C and my pencil is on B. So I have, let me get this right, so I have this distance right over here. I know this distance."}, {"video_title": "Geometric constructions congruent angles Congruence High school geometry Khan Academy.mp3", "Sentence": "And what I can do is take my compass again and get a clear sense of the distance between C and B by adjusting my compass. So one point is on C and my pencil is on B. So I have, let me get this right, so I have this distance right over here. I know this distance. And I have adjusted my compass accordingly. So I can get that same distance right over there. And so you can now imagine where I'm going to draw that second ray."}, {"video_title": "Geometric constructions congruent angles Congruence High school geometry Khan Academy.mp3", "Sentence": "I know this distance. And I have adjusted my compass accordingly. So I can get that same distance right over there. And so you can now imagine where I'm going to draw that second ray. That second ray, if I put point F right over here, my second ray I can just draw between, starting at point E right over here, going through point F. I could draw that a little bit neater. So it would look like that, my second ray. Ignore that first little line I drew, I'm using a pen, which I don't recommend for you to do it."}, {"video_title": "Geometric constructions congruent angles Congruence High school geometry Khan Academy.mp3", "Sentence": "And so you can now imagine where I'm going to draw that second ray. That second ray, if I put point F right over here, my second ray I can just draw between, starting at point E right over here, going through point F. I could draw that a little bit neater. So it would look like that, my second ray. Ignore that first little line I drew, I'm using a pen, which I don't recommend for you to do it. I'm doing it so that you can see it on this video. Now how do we know that this angle is now congruent to this angle right over here? Well, one way to do it is to think about triangle BAC, triangle BAC, and triangle, let's just call it DFE."}, {"video_title": "Geometric constructions congruent angles Congruence High school geometry Khan Academy.mp3", "Sentence": "Ignore that first little line I drew, I'm using a pen, which I don't recommend for you to do it. I'm doing it so that you can see it on this video. Now how do we know that this angle is now congruent to this angle right over here? Well, one way to do it is to think about triangle BAC, triangle BAC, and triangle, let's just call it DFE. So this triangle right over here. When we drew that first arc, we know that the distance between AC is equivalent to the distance between AB, and we kept the compass radius the same. So we know that's also the distance between EF and the distance between ED."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "So given this diagram, we need to figure out what the length of CF right over here is. And you might already guess that this will have to do something with similar triangles. Because at least it looks that triangle CFE is similar to ABE. And the intuition there is it's kind of embedded inside of it. And we're going to prove that to ourselves. And it also looks like triangle CFB is going to be similar to triangle DEB. But once again, we're going to have to prove that to ourselves."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "And the intuition there is it's kind of embedded inside of it. And we're going to prove that to ourselves. And it also looks like triangle CFB is going to be similar to triangle DEB. But once again, we're going to have to prove that to ourselves. And then maybe we can deal with all the ratios of the different sides to CF right over here. And then actually figure out what CF is going to be. So first, let's prove to ourselves that these definitely are similar triangles."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "But once again, we're going to have to prove that to ourselves. And then maybe we can deal with all the ratios of the different sides to CF right over here. And then actually figure out what CF is going to be. So first, let's prove to ourselves that these definitely are similar triangles. So you have this 90 degree angle in ABE. And you have this 90 degree angle in CFE. If we can prove just one other angle or one other set of corresponding angles is congruent in both, then we've proved that they're similar."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "So first, let's prove to ourselves that these definitely are similar triangles. So you have this 90 degree angle in ABE. And you have this 90 degree angle in CFE. If we can prove just one other angle or one other set of corresponding angles is congruent in both, then we've proved that they're similar. And we can either show that, look, they both share this angle right over here. Angle CEF is the same as angle AEB. So we've shown two angles, two corresponding angles in these triangles."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "If we can prove just one other angle or one other set of corresponding angles is congruent in both, then we've proved that they're similar. And we can either show that, look, they both share this angle right over here. Angle CEF is the same as angle AEB. So we've shown two angles, two corresponding angles in these triangles. This is an angle in both triangles. They are congruent, so the triangles are going to be similar. You could also show that this line is parallel to this line, because obviously these two angles are the same."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "So we've shown two angles, two corresponding angles in these triangles. This is an angle in both triangles. They are congruent, so the triangles are going to be similar. You could also show that this line is parallel to this line, because obviously these two angles are the same. And so these angles will also be the same. So they're definitely similar triangles. So let's just write that down, get that out of the way."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "You could also show that this line is parallel to this line, because obviously these two angles are the same. And so these angles will also be the same. So they're definitely similar triangles. So let's just write that down, get that out of the way. We know that triangle ABE is similar to triangle CFE. And you want to make sure you get in the right order. F is where the 90 degree angle is."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "So let's just write that down, get that out of the way. We know that triangle ABE is similar to triangle CFE. And you want to make sure you get in the right order. F is where the 90 degree angle is. B is where the 90 degree angle is. And then E is where this orange angle is. So CFE, it's similar to triangle CFE."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "F is where the 90 degree angle is. B is where the 90 degree angle is. And then E is where this orange angle is. So CFE, it's similar to triangle CFE. Now let's see if we can figure out that same statement going the other way, looking at triangle DEB. So once again, you have a 90 degree angle here. If this is 90, then this is definitely going to be 90 as well."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "So CFE, it's similar to triangle CFE. Now let's see if we can figure out that same statement going the other way, looking at triangle DEB. So once again, you have a 90 degree angle here. If this is 90, then this is definitely going to be 90 as well. You have a 90 degree angle here at CFB. You have a 90 degree angle at DEF or DEB, however you want to call it. So they have one set of corresponding angles that are congruent."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "If this is 90, then this is definitely going to be 90 as well. You have a 90 degree angle here at CFB. You have a 90 degree angle at DEF or DEB, however you want to call it. So they have one set of corresponding angles that are congruent. And then you'll also see that they both share this angle right over here on the smaller triangle. So I'm now looking at this triangle right over here as opposed to the one on the right. So they both share this angle right over here, DBE."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "So they have one set of corresponding angles that are congruent. And then you'll also see that they both share this angle right over here on the smaller triangle. So I'm now looking at this triangle right over here as opposed to the one on the right. So they both share this angle right over here, DBE. Angle DBE is the same as angle CBF. So I've shown you already that we have this angle is congruent to this angle. And we have this angle as a part of both."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "So they both share this angle right over here, DBE. Angle DBE is the same as angle CBF. So I've shown you already that we have this angle is congruent to this angle. And we have this angle as a part of both. So it's obviously congruent to itself. So we have two corresponding angles that are congruent to each other. So we know that this larger triangle over here is similar to this smaller triangle over there."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "And we have this angle as a part of both. So it's obviously congruent to itself. So we have two corresponding angles that are congruent to each other. So we know that this larger triangle over here is similar to this smaller triangle over there. So let me write this down. So we also know that triangle DEB is similar to triangle CFB. Now what can we do from here?"}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "So we know that this larger triangle over here is similar to this smaller triangle over there. So let me write this down. So we also know that triangle DEB is similar to triangle CFB. Now what can we do from here? Well, we know that the ratios of corresponding sides for each of those similar triangles are going to have to be the same. But we only have one side of one of the triangles. So in the case of AB and CFB, we've only been given one side."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "Now what can we do from here? Well, we know that the ratios of corresponding sides for each of those similar triangles are going to have to be the same. But we only have one side of one of the triangles. So in the case of AB and CFB, we've only been given one side. In the case of DEB and CFB, we've only been given one side right over here. So there doesn't seem to be a lot to work with. And this is why this is a slightly more challenging problem here."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "So in the case of AB and CFB, we've only been given one side. In the case of DEB and CFB, we've only been given one side right over here. So there doesn't seem to be a lot to work with. And this is why this is a slightly more challenging problem here. Let's just go ahead and see if we can assume one of the sides, or maybe a side that's shared by both of these larger triangles, and then maybe things will work out from there. So let's just assume that this length right over here, let's just assume that BE is equal to Y. So let me just write this down."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "And this is why this is a slightly more challenging problem here. Let's just go ahead and see if we can assume one of the sides, or maybe a side that's shared by both of these larger triangles, and then maybe things will work out from there. So let's just assume that this length right over here, let's just assume that BE is equal to Y. So let me just write this down. This whole length is going to be equal to Y, because this at least gives us something to work with. And Y is shared by both ABE and DEB, so that seems useful. And then we're going to have to think about the smaller triangles right over there."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "So let me just write this down. This whole length is going to be equal to Y, because this at least gives us something to work with. And Y is shared by both ABE and DEB, so that seems useful. And then we're going to have to think about the smaller triangles right over there. So maybe we'll call this length BFx. And then let's call FE. Well, if this is X, then this is Y minus X."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "And then we're going to have to think about the smaller triangles right over there. So maybe we'll call this length BFx. And then let's call FE. Well, if this is X, then this is Y minus X. So we've introduced a bunch of variables here, but maybe with all the proportionalities and things, just maybe things will work out, or at least we'll have a little bit more sense of where we can go with this actual problem. But now we can start dealing with the similar triangles. For example, so we want to figure out what CF is."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "Well, if this is X, then this is Y minus X. So we've introduced a bunch of variables here, but maybe with all the proportionalities and things, just maybe things will work out, or at least we'll have a little bit more sense of where we can go with this actual problem. But now we can start dealing with the similar triangles. For example, so we want to figure out what CF is. We now know that for these two triangles right here, the ratio of the corresponding sides are going to be constant. So for example, the ratio between CF and 9, their corresponding sides, the ratio between CF and 9 has got to be equal to the ratio between Y minus X, that's that side right there, Y minus X and the corresponding side of the larger triangle. Well, the corresponding side of the larger triangle is this entire length."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "For example, so we want to figure out what CF is. We now know that for these two triangles right here, the ratio of the corresponding sides are going to be constant. So for example, the ratio between CF and 9, their corresponding sides, the ratio between CF and 9 has got to be equal to the ratio between Y minus X, that's that side right there, Y minus X and the corresponding side of the larger triangle. Well, the corresponding side of the larger triangle is this entire length. And that entire length right over there is Y. So it's equal to Y minus X over Y. So we could simplify this a little bit."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "Well, the corresponding side of the larger triangle is this entire length. And that entire length right over there is Y. So it's equal to Y minus X over Y. So we could simplify this a little bit. Well, I'll hold off for a second. Let's see if we can do something similar with this thing on the right. So once again, we have CF, its corresponding side on DEB."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "So we could simplify this a little bit. Well, I'll hold off for a second. Let's see if we can do something similar with this thing on the right. So once again, we have CF, its corresponding side on DEB. So now we're looking at the triangle CFB, not looking at triangle CFE anymore. So now when we're looking at this triangle, CF corresponds to DE. So we have CF over DE is going to be equal to, so CF over DE is going to be equal to X is going to be equal to, let me do that in a different color, it's going to be equal to, I'm using all my colors, it's going to be equal to X over this entire base right over here, so this entire B, which once again we know is Y, so over Y."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "So once again, we have CF, its corresponding side on DEB. So now we're looking at the triangle CFB, not looking at triangle CFE anymore. So now when we're looking at this triangle, CF corresponds to DE. So we have CF over DE is going to be equal to, so CF over DE is going to be equal to X is going to be equal to, let me do that in a different color, it's going to be equal to, I'm using all my colors, it's going to be equal to X over this entire base right over here, so this entire B, which once again we know is Y, so over Y. And now this looks interesting because it looks like we have three unknowns. We have CF, sorry, we know what DE is already. This is 12."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "So we have CF over DE is going to be equal to, so CF over DE is going to be equal to X is going to be equal to, let me do that in a different color, it's going to be equal to, I'm using all my colors, it's going to be equal to X over this entire base right over here, so this entire B, which once again we know is Y, so over Y. And now this looks interesting because it looks like we have three unknowns. We have CF, sorry, we know what DE is already. This is 12. I could have written CF over 12. The ratio between CF and 12 is going to be the ratio between X and Y. So we have three unknowns and only two equations, so it seems hard to solve it first because there's one unknown, another unknown, another unknown, another unknown, another unknown, and another unknown."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "This is 12. I could have written CF over 12. The ratio between CF and 12 is going to be the ratio between X and Y. So we have three unknowns and only two equations, so it seems hard to solve it first because there's one unknown, another unknown, another unknown, another unknown, another unknown, and another unknown. But it looks like I can write this right here, this expression, in terms of X over Y, and then we could do a substitution. So that's why this was a little tricky. So this one right here we can rewrite as CF, let me do it in that same green color, we can rewrite it as CF over 9 is equal to Y minus X over Y, the same thing as Y over Y minus X over Y, over, or 1 minus X over Y."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "So we have three unknowns and only two equations, so it seems hard to solve it first because there's one unknown, another unknown, another unknown, another unknown, another unknown, and another unknown. But it looks like I can write this right here, this expression, in terms of X over Y, and then we could do a substitution. So that's why this was a little tricky. So this one right here we can rewrite as CF, let me do it in that same green color, we can rewrite it as CF over 9 is equal to Y minus X over Y, the same thing as Y over Y minus X over Y, over, or 1 minus X over Y. All I did is I essentially, I guess you could say, distributed the 1 over Y times both of these terms. Y over Y minus X over Y, or 1 minus X minus Y. And this is useful because we already know what X minus Y is equal to."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "So this one right here we can rewrite as CF, let me do it in that same green color, we can rewrite it as CF over 9 is equal to Y minus X over Y, the same thing as Y over Y minus X over Y, over, or 1 minus X over Y. All I did is I essentially, I guess you could say, distributed the 1 over Y times both of these terms. Y over Y minus X over Y, or 1 minus X minus Y. And this is useful because we already know what X minus Y is equal to. Sorry, X over Y is equal to. We already know that X over Y is equal to CF over 12. So this right over here I can replace with this, CF over 12."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "And this is useful because we already know what X minus Y is equal to. Sorry, X over Y is equal to. We already know that X over Y is equal to CF over 12. So this right over here I can replace with this, CF over 12. So then we get, this is a home stretch here, CF, which is what we care about, CF over 9 is equal to 1 minus CF over 12. And now we have one equation with one unknown, and we should be able to solve this right over here. So we could add CF over 12 to both sides."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "So this right over here I can replace with this, CF over 12. So then we get, this is a home stretch here, CF, which is what we care about, CF over 9 is equal to 1 minus CF over 12. And now we have one equation with one unknown, and we should be able to solve this right over here. So we could add CF over 12 to both sides. So you have CF over 9 plus CF over 12 is equal to 1. We just have to find a common denominator here, and I think 36 will do the trick. So 9 times 4 is 36, so if you had to multiply 9 times 4, you have to multiply CF times 4."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "So we could add CF over 12 to both sides. So you have CF over 9 plus CF over 12 is equal to 1. We just have to find a common denominator here, and I think 36 will do the trick. So 9 times 4 is 36, so if you had to multiply 9 times 4, you have to multiply CF times 4. So you have 4 CF, 4 CF over 36 is the same thing as CF over 9, and then plus CF over 12 is the same thing as 3 CF over 36. And this is going to be equal to 1. And then we are left with 4 CF plus 3 CF is 7."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "So 9 times 4 is 36, so if you had to multiply 9 times 4, you have to multiply CF times 4. So you have 4 CF, 4 CF over 36 is the same thing as CF over 9, and then plus CF over 12 is the same thing as 3 CF over 36. And this is going to be equal to 1. And then we are left with 4 CF plus 3 CF is 7. CF over 36 is equal to 1. And then to solve for CF, we can multiply both sides times the reciprocal of 7 over 36. So 36 over 7, multiply both sides times that, 36 over 7."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "And then we are left with 4 CF plus 3 CF is 7. CF over 36 is equal to 1. And then to solve for CF, we can multiply both sides times the reciprocal of 7 over 36. So 36 over 7, multiply both sides times that, 36 over 7. This side things cancel out. And we are left with our final, we get our drum roll now. CF is equal to CF, so all of this stuff cancels out."}, {"video_title": "Challenging similarity problem Similarity Geometry Khan Academy.mp3", "Sentence": "So 36 over 7, multiply both sides times that, 36 over 7. This side things cancel out. And we are left with our final, we get our drum roll now. CF is equal to CF, so all of this stuff cancels out. CF is equal to 1 times 36 over 7, or it's just 36 over 7. And this was a pretty cool problem, because what it shows you is if you have two things, let's see, this thing is some type of a pole or a stick or maybe the wall of a building or who knows what it is. If this is 9 feet tall or 9 yards tall or 9 meters tall, and this over here, this other one is 12 meters tall or 12 yards or whatever units you want to use it, if you were to drape a string from either of them to the base of the other, from the top of one of them to the base of the other, regardless of how far apart these two things are going to be, we just said they're y apart, regardless of how far apart they are, the place where those two strings would intersect are going to be 36 sevenths high, or I guess 5 and 1 sevenths high, regardless of how far they are."}, {"video_title": "Inscribed shapes find inscribed angle High School Math Khan Academy.mp3", "Sentence": "So try to figure out the measure of this angle. I encourage you to pause the video now and try it on your own. All right, now let's work through this together. And the key realization here is to think about this angle. Well, it is an inscribed angle. We see its vertex is sitting on the circle itself. And then think about the arc that it intercepts."}, {"video_title": "Inscribed shapes find inscribed angle High School Math Khan Academy.mp3", "Sentence": "And the key realization here is to think about this angle. Well, it is an inscribed angle. We see its vertex is sitting on the circle itself. And then think about the arc that it intercepts. And we see that it intercepts. So let me draw these two sides of the angle. We see that it intercepts arc CD."}, {"video_title": "Inscribed shapes find inscribed angle High School Math Khan Academy.mp3", "Sentence": "And then think about the arc that it intercepts. And we see that it intercepts. So let me draw these two sides of the angle. We see that it intercepts arc CD. It intercepts arc CD. And so the measure of this angle, since it's an inscribed angle, is going to be half the measure of arc CD. So if we could figure out the measure of arc CD, then we're going to be in good shape."}, {"video_title": "Inscribed shapes find inscribed angle High School Math Khan Academy.mp3", "Sentence": "We see that it intercepts arc CD. It intercepts arc CD. And so the measure of this angle, since it's an inscribed angle, is going to be half the measure of arc CD. So if we could figure out the measure of arc CD, then we're going to be in good shape. So if we figure out the measure of arc CD, then we take half of that and we'll figure out what we care about. Well, what you might notice is that there is another inscribed angle that also intercepts arc CD. We have this angle right over here."}, {"video_title": "Inscribed shapes find inscribed angle High School Math Khan Academy.mp3", "Sentence": "So if we could figure out the measure of arc CD, then we're going to be in good shape. So if we figure out the measure of arc CD, then we take half of that and we'll figure out what we care about. Well, what you might notice is that there is another inscribed angle that also intercepts arc CD. We have this angle right over here. It also intercepts arc CD. So you could call this angle C, whoops, you could call this angle CFD. This also intercepts the same arc."}, {"video_title": "Inscribed shapes find inscribed angle High School Math Khan Academy.mp3", "Sentence": "We have this angle right over here. It also intercepts arc CD. So you could call this angle C, whoops, you could call this angle CFD. This also intercepts the same arc. So there's two ways you could think about it. Two inscribed angles that intercept the same arc are going to have the same angle measure. So just off of that, you could say that this is going to be, that these two angles are going to have the same measure."}, {"video_title": "Inscribed shapes find inscribed angle High School Math Khan Academy.mp3", "Sentence": "This also intercepts the same arc. So there's two ways you could think about it. Two inscribed angles that intercept the same arc are going to have the same angle measure. So just off of that, you could say that this is going to be, that these two angles are going to have the same measure. So you could say this is going to be 50 degrees. Or you could actually solve what the measure of arc CD is. It's going to be twice the measure of the inscribed angle that intercepts it."}, {"video_title": "Inscribed shapes find inscribed angle High School Math Khan Academy.mp3", "Sentence": "So just off of that, you could say that this is going to be, that these two angles are going to have the same measure. So you could say this is going to be 50 degrees. Or you could actually solve what the measure of arc CD is. It's going to be twice the measure of the inscribed angle that intercepts it. So the measure of arc CD is going to be 100 degrees, twice the 50 degrees. And then you use that and you say, well, if the measure of that arc is 100 degrees, then an inscribed angle that intercepts it is going to have half its measure. It's going to be 50 degrees."}, {"video_title": "Cavalieri's principle in 3D Solid geometry High school geometry Khan Academy.mp3", "Sentence": "Now what I'm going to do is start cutting up this left cylinder here and shifting things around. So if I just cut it in two and take that bottom cylinder, that bottom half, and shift it a bit, have I changed its volume? Well, clearly I have not changed its volume. I still have the same volume, the combined volume of both of these half cylinders, I could say, are equal to the original cylinder. Now what if I were to cut it up even more? So let me cut it up now into three. Well, once again, I still haven't changed my original volume."}, {"video_title": "Cavalieri's principle in 3D Solid geometry High school geometry Khan Academy.mp3", "Sentence": "I still have the same volume, the combined volume of both of these half cylinders, I could say, are equal to the original cylinder. Now what if I were to cut it up even more? So let me cut it up now into three. Well, once again, I still haven't changed my original volume. It's still the same volume as the original, and I'll just cut it up in 2 3rds, and if I shift them around a little bit, I'm not changing the volume, and I could keep doing that. I could cut it up into a bunch of them. Notice, this still has the same original volume."}, {"video_title": "Cavalieri's principle in 3D Solid geometry High school geometry Khan Academy.mp3", "Sentence": "Well, once again, I still haven't changed my original volume. It's still the same volume as the original, and I'll just cut it up in 2 3rds, and if I shift them around a little bit, I'm not changing the volume, and I could keep doing that. I could cut it up into a bunch of them. Notice, this still has the same original volume. I've just cut it up into a bunch of sections, a bunch, I've cut it horizontally, and now I'm just shifting things around, but that doesn't change the volume. And I can do it a bunch of times. This looks like some type of poker chips or gambling chips, where I could have my original cylinder, and now I've cut it horizontally into a bunch of these, I guess you could say chips, but clearly it has the same combined volume."}, {"video_title": "Cavalieri's principle in 3D Solid geometry High school geometry Khan Academy.mp3", "Sentence": "Notice, this still has the same original volume. I've just cut it up into a bunch of sections, a bunch, I've cut it horizontally, and now I'm just shifting things around, but that doesn't change the volume. And I can do it a bunch of times. This looks like some type of poker chips or gambling chips, where I could have my original cylinder, and now I've cut it horizontally into a bunch of these, I guess you could say chips, but clearly it has the same combined volume. I can shift it around a bit, but it has the same volume. And this leads us to an interesting question, and it's actually a principle known as Cavalieri's Principle, which is if I have two figures that have the same height, and at any point along that height, the cross-sectional area is the same, then the two figures have the same volume. Now, how does what I just say apply to what's going on here?"}, {"video_title": "Cavalieri's principle in 3D Solid geometry High school geometry Khan Academy.mp3", "Sentence": "This looks like some type of poker chips or gambling chips, where I could have my original cylinder, and now I've cut it horizontally into a bunch of these, I guess you could say chips, but clearly it has the same combined volume. I can shift it around a bit, but it has the same volume. And this leads us to an interesting question, and it's actually a principle known as Cavalieri's Principle, which is if I have two figures that have the same height, and at any point along that height, the cross-sectional area is the same, then the two figures have the same volume. Now, how does what I just say apply to what's going on here? Well, clearly both of these figures have the same height, and then at any point here, wherever I did the cuts, at this point, at the same point on this original cylinder, well, my cross-sectional area is going to be the same, because it's going to be the same area as the base in the case of this cylinder, and so it meets Cavalieri's Principle. But Cavalieri's Principle's nothing exotic. It comes straight out of common sense."}, {"video_title": "Cavalieri's principle in 3D Solid geometry High school geometry Khan Academy.mp3", "Sentence": "Now, how does what I just say apply to what's going on here? Well, clearly both of these figures have the same height, and then at any point here, wherever I did the cuts, at this point, at the same point on this original cylinder, well, my cross-sectional area is going to be the same, because it's going to be the same area as the base in the case of this cylinder, and so it meets Cavalieri's Principle. But Cavalieri's Principle's nothing exotic. It comes straight out of common sense. I can just do more cuts like this, and you can see that I have, you could say a more continuous-looking skewed cylinder, but this will have the same volume as our original cylinder. When I shift it around like this, it's not changing the volume, and that's not just true for cylinders. I could do the exact same argument with some form of a prism."}, {"video_title": "Cavalieri's principle in 3D Solid geometry High school geometry Khan Academy.mp3", "Sentence": "It comes straight out of common sense. I can just do more cuts like this, and you can see that I have, you could say a more continuous-looking skewed cylinder, but this will have the same volume as our original cylinder. When I shift it around like this, it's not changing the volume, and that's not just true for cylinders. I could do the exact same argument with some form of a prism. Once again, they have the same volume. I could cut the left one in half and shift it around. It doesn't change its volume."}, {"video_title": "Cavalieri's principle in 3D Solid geometry High school geometry Khan Academy.mp3", "Sentence": "I could do the exact same argument with some form of a prism. Once again, they have the same volume. I could cut the left one in half and shift it around. It doesn't change its volume. I could cut it more and shift those around. It still doesn't change the volume. So Cavalieri's Principle seems to make a lot of intuitive sense here."}, {"video_title": "Cavalieri's principle in 3D Solid geometry High school geometry Khan Academy.mp3", "Sentence": "It doesn't change its volume. I could cut it more and shift those around. It still doesn't change the volume. So Cavalieri's Principle seems to make a lot of intuitive sense here. If I have two figures that have the same height, and at any point along that height, the cross-sectional area is the same, then the figures have the same volume. So these figures also have the same volume. And I could do it with interesting things like, say, a pyramid."}, {"video_title": "Cavalieri's principle in 3D Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So Cavalieri's Principle seems to make a lot of intuitive sense here. If I have two figures that have the same height, and at any point along that height, the cross-sectional area is the same, then the figures have the same volume. So these figures also have the same volume. And I could do it with interesting things like, say, a pyramid. These two pyramids have the same volume, and if I were to cut the left pyramid halfway along its height and shift the bottom like this, that doesn't change its volume. And I can keep doing that with more and more cuts. And, because at any point here, these figures have the same height, and at any point on that height, the cross-sectional area is the same, and so they have the same volumes."}, {"video_title": "Cavalieri's principle in 3D Solid geometry High school geometry Khan Academy.mp3", "Sentence": "And I could do it with interesting things like, say, a pyramid. These two pyramids have the same volume, and if I were to cut the left pyramid halfway along its height and shift the bottom like this, that doesn't change its volume. And I can keep doing that with more and more cuts. And, because at any point here, these figures have the same height, and at any point on that height, the cross-sectional area is the same, and so they have the same volumes. But once again, it is intuitive. And it goes all the way to the case where you have, you could view it as a continuous pyramid right over here that has been skewed. So no matter how much you skew it, it's going to have the same volume as our original pyramid because they have the same height, and the cross-sectional area at any point in the height is going to be the same."}, {"video_title": "Cavalieri's principle in 3D Solid geometry High school geometry Khan Academy.mp3", "Sentence": "And, because at any point here, these figures have the same height, and at any point on that height, the cross-sectional area is the same, and so they have the same volumes. But once again, it is intuitive. And it goes all the way to the case where you have, you could view it as a continuous pyramid right over here that has been skewed. So no matter how much you skew it, it's going to have the same volume as our original pyramid because they have the same height, and the cross-sectional area at any point in the height is going to be the same. We could actually do this with any figure. So these spheres have the same volume. I could cut the left one in half, halfway along its height, and shift it like this."}, {"video_title": "Cavalieri's principle in 3D Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So no matter how much you skew it, it's going to have the same volume as our original pyramid because they have the same height, and the cross-sectional area at any point in the height is going to be the same. We could actually do this with any figure. So these spheres have the same volume. I could cut the left one in half, halfway along its height, and shift it like this. Clearly, I'm not changing the volume. And I could make more cuts like that. And clearly, it has still the same volume."}, {"video_title": "Cavalieri's principle in 3D Solid geometry High school geometry Khan Academy.mp3", "Sentence": "I could cut the left one in half, halfway along its height, and shift it like this. Clearly, I'm not changing the volume. And I could make more cuts like that. And clearly, it has still the same volume. And this meets Cavalieri's principle because they have the same height, and the cross-section at any point along that height is going to be the same. So even though I can cut that one up and I can shift it, it looks like a different type of object, a different type of thing, but they have the same height, and the cross-sections at any point are the same area. So we have the same volume, which is a useful thing to know, not just to know the principle, but hopefully this video helps you gain some of the intuition for why it makes intuitive sense."}, {"video_title": "Constructing equilateral triangle inscribed in circle Geometry Khan Academy.mp3", "Sentence": "Looks pretty good. And now let me move this center so it sits on our original circle. So they now sit on each other, or their centers now sit on each other. So I can make it. That looks pretty good. And now let's think about something. If I were to draw this segment right over here, this of course has the length of the radius."}, {"video_title": "Constructing equilateral triangle inscribed in circle Geometry Khan Academy.mp3", "Sentence": "So I can make it. That looks pretty good. And now let's think about something. If I were to draw this segment right over here, this of course has the length of the radius. Now let's do another one. And that's either of their radii, because they have the same length. Now let's center this at our new circle."}, {"video_title": "Constructing equilateral triangle inscribed in circle Geometry Khan Academy.mp3", "Sentence": "If I were to draw this segment right over here, this of course has the length of the radius. Now let's do another one. And that's either of their radii, because they have the same length. Now let's center this at our new circle. And take it out here. Now this is equal to the radius of the new circle, which is the same as the radius of the old circle. It's going to be the same as this length here."}, {"video_title": "Constructing equilateral triangle inscribed in circle Geometry Khan Academy.mp3", "Sentence": "Now let's center this at our new circle. And take it out here. Now this is equal to the radius of the new circle, which is the same as the radius of the old circle. It's going to be the same as this length here. So these two segments have the same length. Now if I were to connect that point to that point, this is a radius of our original circle. And so it's going to have the same length as these two."}, {"video_title": "Constructing equilateral triangle inscribed in circle Geometry Khan Academy.mp3", "Sentence": "It's going to be the same as this length here. So these two segments have the same length. Now if I were to connect that point to that point, this is a radius of our original circle. And so it's going to have the same length as these two. So this right over here, I have constructed an equilateral triangle. Now why is this at all useful? Well, we know that the angles in an equilateral triangle are 60 degrees."}, {"video_title": "Constructing equilateral triangle inscribed in circle Geometry Khan Academy.mp3", "Sentence": "And so it's going to have the same length as these two. So this right over here, I have constructed an equilateral triangle. Now why is this at all useful? Well, we know that the angles in an equilateral triangle are 60 degrees. So we know that this angle right over here is 60 degrees. Now why is this being 60 degrees interesting? Well, imagine if we constructed another triangle out here, just symmetrically, but kind of flipped down, just like this."}, {"video_title": "Constructing equilateral triangle inscribed in circle Geometry Khan Academy.mp3", "Sentence": "Well, we know that the angles in an equilateral triangle are 60 degrees. So we know that this angle right over here is 60 degrees. Now why is this being 60 degrees interesting? Well, imagine if we constructed another triangle out here, just symmetrically, but kind of flipped down, just like this. Well, the same argument, this angle right over here between these two edges, this is also going to be 60 degrees. So this entire interior angle, if we add those two up, are going to be 120 degrees. Now why is that interesting?"}, {"video_title": "Constructing equilateral triangle inscribed in circle Geometry Khan Academy.mp3", "Sentence": "Well, imagine if we constructed another triangle out here, just symmetrically, but kind of flipped down, just like this. Well, the same argument, this angle right over here between these two edges, this is also going to be 60 degrees. So this entire interior angle, if we add those two up, are going to be 120 degrees. Now why is that interesting? Well, if this interior angle is 120 degrees, then that means that this arc right over here is 120 degrees, or it's a third of the way around the triangle. This right over here is a third of the way around the triangle. Since that's a third of the way around the triangle, if I were to connect these two dots, that is going to be, this right over here, is going to be a side of our equilateral triangle."}, {"video_title": "Constructing equilateral triangle inscribed in circle Geometry Khan Academy.mp3", "Sentence": "Now why is that interesting? Well, if this interior angle is 120 degrees, then that means that this arc right over here is 120 degrees, or it's a third of the way around the triangle. This right over here is a third of the way around the triangle. Since that's a third of the way around the triangle, if I were to connect these two dots, that is going to be, this right over here, is going to be a side of our equilateral triangle. This right over here, it's secant to an arc that is 1 third of the entire circle. And now I can keep doing this. Let's move, I'll reuse these, let's move our circle around."}, {"video_title": "Constructing equilateral triangle inscribed in circle Geometry Khan Academy.mp3", "Sentence": "Since that's a third of the way around the triangle, if I were to connect these two dots, that is going to be, this right over here, is going to be a side of our equilateral triangle. This right over here, it's secant to an arc that is 1 third of the entire circle. And now I can keep doing this. Let's move, I'll reuse these, let's move our circle around. And so now I'm going to move my circle along the circle. And what I want, again, I just want to intersect these two points. And so now, let's see, I could take one of these, take it there, take it there, same exact argument."}, {"video_title": "Constructing equilateral triangle inscribed in circle Geometry Khan Academy.mp3", "Sentence": "Let's move, I'll reuse these, let's move our circle around. And so now I'm going to move my circle along the circle. And what I want, again, I just want to intersect these two points. And so now, let's see, I could take one of these, take it there, take it there, same exact argument. This angle that I haven't fully drawn, or this arc, you could say, is 120 degrees. So this is going to be one side of our equilateral triangle. It's secant to an arc of 120 degrees."}, {"video_title": "Constructing equilateral triangle inscribed in circle Geometry Khan Academy.mp3", "Sentence": "And so now, let's see, I could take one of these, take it there, take it there, same exact argument. This angle that I haven't fully drawn, or this arc, you could say, is 120 degrees. So this is going to be one side of our equilateral triangle. It's secant to an arc of 120 degrees. So let's move this around again. Actually, we don't even have to move this around anymore. We could just connect those last two dots."}, {"video_title": "Constructing equilateral triangle inscribed in circle Geometry Khan Academy.mp3", "Sentence": "It's secant to an arc of 120 degrees. So let's move this around again. Actually, we don't even have to move this around anymore. We could just connect those last two dots. So we could just connect this one. Actually, I just want to connect that one to that one. And just like that, and we're done."}, {"video_title": "Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3", "Sentence": "And I want to think about which of these figures are going to be unchanged if I were to rotate it 180 degrees. So let's do two examples of that. So I have two copies of this square. If I were to take one of these copies and rotate it 180 degrees, so let me show you what that looks like. And we're going to rotate around its center 180 degrees. We're going to rotate around the center. So this is it."}, {"video_title": "Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3", "Sentence": "If I were to take one of these copies and rotate it 180 degrees, so let me show you what that looks like. And we're going to rotate around its center 180 degrees. We're going to rotate around the center. So this is it. So we're rotating it. That's rotated 90 degrees. And then we've rotated 180 degrees."}, {"video_title": "Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3", "Sentence": "So this is it. So we're rotating it. That's rotated 90 degrees. And then we've rotated 180 degrees. And notice, the figure looks exactly the same. This one, the square, is unchanged by 180 degree rotation. Now what about this trapezoid right over here?"}, {"video_title": "Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3", "Sentence": "And then we've rotated 180 degrees. And notice, the figure looks exactly the same. This one, the square, is unchanged by 180 degree rotation. Now what about this trapezoid right over here? Let's think about what happens when it's rotated by 180 degrees. So that is 90 degrees and 180 degrees. So this has now been changed."}, {"video_title": "Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3", "Sentence": "Now what about this trapezoid right over here? Let's think about what happens when it's rotated by 180 degrees. So that is 90 degrees and 180 degrees. So this has now been changed. Now I have the short side, or I have my base is short and my top is long before my base was long and my top was short. So when I rotated 180 degrees, I didn't get to the exact same figure. I have essentially an upside down version of it."}, {"video_title": "Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3", "Sentence": "So this has now been changed. Now I have the short side, or I have my base is short and my top is long before my base was long and my top was short. So when I rotated 180 degrees, I didn't get to the exact same figure. I have essentially an upside down version of it. So what I want you to do for the rest of these is pause the video and think about which of these will be unchanged and which of them will be changed when you rotate by 180 degrees. So let's look at this star thing. And one way that my brain visualizes is imagine the center."}, {"video_title": "Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3", "Sentence": "I have essentially an upside down version of it. So what I want you to do for the rest of these is pause the video and think about which of these will be unchanged and which of them will be changed when you rotate by 180 degrees. So let's look at this star thing. And one way that my brain visualizes is imagine the center. That's what we're rotating around. And then if you rotate 180 degrees, imagine any point, say this point, relative to the center. If you were to rotate it 90 degrees, you would get over here."}, {"video_title": "Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3", "Sentence": "And one way that my brain visualizes is imagine the center. That's what we're rotating around. And then if you rotate 180 degrees, imagine any point, say this point, relative to the center. If you were to rotate it 90 degrees, you would get over here. And then if you rotate 180 degrees, you go over here. You go on the opposite side of the center from where it is. So from that point to the center, you keep going that same distance, you'll end up right over there."}, {"video_title": "Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3", "Sentence": "If you were to rotate it 90 degrees, you would get over here. And then if you rotate 180 degrees, you go over here. You go on the opposite side of the center from where it is. So from that point to the center, you keep going that same distance, you'll end up right over there. So this one looks like it won't be changed, but let's verify it. So we're gonna rotate 90 degrees, and then we have 180 degrees. It is unchanged."}, {"video_title": "Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3", "Sentence": "So from that point to the center, you keep going that same distance, you'll end up right over there. So this one looks like it won't be changed, but let's verify it. So we're gonna rotate 90 degrees, and then we have 180 degrees. It is unchanged. Now let's look at this parallelogram right over here. So its center, if we think about its center where my cursor is right now, think about this point. The distance between that point and the center, if we were to keep going that same distance again, you would get to that point."}, {"video_title": "Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3", "Sentence": "It is unchanged. Now let's look at this parallelogram right over here. So its center, if we think about its center where my cursor is right now, think about this point. The distance between that point and the center, if we were to keep going that same distance again, you would get to that point. Likewise, the distance between this point and the center, if we were to go that same distance again, you would get to that point. So it seems like that point would end up there, that point would end up there, and vice versa. So I think this one will be unchanged by rotation."}, {"video_title": "Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3", "Sentence": "The distance between that point and the center, if we were to keep going that same distance again, you would get to that point. Likewise, the distance between this point and the center, if we were to go that same distance again, you would get to that point. So it seems like that point would end up there, that point would end up there, and vice versa. So I think this one will be unchanged by rotation. So let's verify it. So you go 90 degrees, and then you go 180 degrees. Or I should say, it will be unchanged by rotation of 180 degrees around its center."}, {"video_title": "Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3", "Sentence": "So I think this one will be unchanged by rotation. So let's verify it. So you go 90 degrees, and then you go 180 degrees. Or I should say, it will be unchanged by rotation of 180 degrees around its center. We got the same figure. Now let's think about this triangle. So if you think about the center of the figure, let's say the center of the figure is right around here."}, {"video_title": "Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3", "Sentence": "Or I should say, it will be unchanged by rotation of 180 degrees around its center. We got the same figure. Now let's think about this triangle. So if you think about the center of the figure, let's say the center of the figure is right around here. If you take this point, go to the center of the figure, and then go that distance again, you end up in a place where there's no point right now. So that point is going to end up there, this point is going to end up there, this point is gonna end up here. So you're not going to have the same figure anymore."}, {"video_title": "Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3", "Sentence": "So if you think about the center of the figure, let's say the center of the figure is right around here. If you take this point, go to the center of the figure, and then go that distance again, you end up in a place where there's no point right now. So that point is going to end up there, this point is going to end up there, this point is gonna end up here. So you're not going to have the same figure anymore. And so we can rotate it to verify. So that's rotated 90 degrees, and then that's rotated 180 degrees. So we've kind of turned this thing on its side."}, {"video_title": "Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3", "Sentence": "So you're not going to have the same figure anymore. And so we can rotate it to verify. So that's rotated 90 degrees, and then that's rotated 180 degrees. So we've kind of turned this thing on its side. It is not the same thing. Now let's think about this figure right over here. Well, this figure, if you rotate it 180 degrees, this point is now going to be down here, and this point is going to be up here."}, {"video_title": "Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3", "Sentence": "So we've kind of turned this thing on its side. It is not the same thing. Now let's think about this figure right over here. Well, this figure, if you rotate it 180 degrees, this point is now going to be down here, and this point is going to be up here. So you're gonna make, essentially it's going to be an upside down version of the same kite. And we can view that, we can visualize that now. So it's going to be different, but let's just show it."}, {"video_title": "Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3", "Sentence": "Well, this figure, if you rotate it 180 degrees, this point is now going to be down here, and this point is going to be up here. So you're gonna make, essentially it's going to be an upside down version of the same kite. And we can view that, we can visualize that now. So it's going to be different, but let's just show it. So that is 90 degrees, and now this is 100 and 180 degrees. If it was actually symmetric, if it actually was symmetric about the horizontal axis, then we would have a different scenario. We would have a different scenario with this thing right over here."}, {"video_title": "Positive and negative rotaion of points example.mp3", "Sentence": "We're told that point P was rotated about the origin zero comma zero by 60 degrees. Which point is the image of P? Pause this video and see if you can figure that out. Alright, now let's think about it. It's being, this is point P. It's being rotated around the origin, zero comma zero by 60 degrees. So if originally point P is right over here, and we're rotating by positive 60 degrees, that means we go counterclockwise by 60 degrees. So this looks like about 60 degrees right over here."}, {"video_title": "Positive and negative rotaion of points example.mp3", "Sentence": "Alright, now let's think about it. It's being, this is point P. It's being rotated around the origin, zero comma zero by 60 degrees. So if originally point P is right over here, and we're rotating by positive 60 degrees, that means we go counterclockwise by 60 degrees. So this looks like about 60 degrees right over here. One way to think about 60 degrees is that that's 1 3rd of 180 degrees. So does this look like 1 3rd of 180 degrees? Remember, 180 degrees would be almost a full line."}, {"video_title": "Positive and negative rotaion of points example.mp3", "Sentence": "So this looks like about 60 degrees right over here. One way to think about 60 degrees is that that's 1 3rd of 180 degrees. So does this look like 1 3rd of 180 degrees? Remember, 180 degrees would be almost a full line. So that indeed does look like 1 3rd of 180 degrees, 60 degrees. It gets us to point C. Point C, and it looks like it's the same distance from the origin we have just rotated by 60 degrees. Point D looks like it's more than 60 degree rotation, so I won't go with that one."}, {"video_title": "Positive and negative rotaion of points example.mp3", "Sentence": "Remember, 180 degrees would be almost a full line. So that indeed does look like 1 3rd of 180 degrees, 60 degrees. It gets us to point C. Point C, and it looks like it's the same distance from the origin we have just rotated by 60 degrees. Point D looks like it's more than 60 degree rotation, so I won't go with that one. Alright, let's do one more of these. So we're told point P was rotated by negative 90 degrees. The center of rotation is indicated."}, {"video_title": "Positive and negative rotaion of points example.mp3", "Sentence": "Point D looks like it's more than 60 degree rotation, so I won't go with that one. Alright, let's do one more of these. So we're told point P was rotated by negative 90 degrees. The center of rotation is indicated. Which point is the image of P? So once again, pause this video and try to think about it. Alright, so we have our center of rotation, this is our point P, and we're rotating by negative 90 degrees."}, {"video_title": "Positive and negative rotaion of points example.mp3", "Sentence": "The center of rotation is indicated. Which point is the image of P? So once again, pause this video and try to think about it. Alright, so we have our center of rotation, this is our point P, and we're rotating by negative 90 degrees. So this means we are going clockwise. So we're going in that direction, and 90 degrees is easy to spot, it's a right angle. And so it would look like that, and it looks like it is getting us right to point A."}, {"video_title": "Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "And I want to imagine what type of a shape I would get if I were to make a vertical cut. And just to refresh ourselves or give us a sense of what a vertical cut is, imagine if this was made out of Jell-O or something kind of fairly soft, but it's still a three-dimensional solid, and I were to make a cut. Let's say I were to make a cut right over here. So let's say I have this big, sharp metal thing. Let me draw it like that. So you have this big, sharp metal thing. Let me draw it a little bit neater."}, {"video_title": "Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "So let's say I have this big, sharp metal thing. Let me draw it like that. So you have this big, sharp metal thing. Let me draw it a little bit neater. So you have this big, sharp metal thing that I'm going to cut right over here. So this is the thing that I'm going to make the cut. And I'm going to go straight down."}, {"video_title": "Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "Let me draw it a little bit neater. So you have this big, sharp metal thing that I'm going to cut right over here. So this is the thing that I'm going to make the cut. And I'm going to go straight down. This is a vertical cut that we're talking about. So this is the thing that I'm going to cut with. Let me make it big enough so that it can capture the shape that will result."}, {"video_title": "Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "And I'm going to go straight down. This is a vertical cut that we're talking about. So this is the thing that I'm going to cut with. Let me make it big enough so that it can capture the shape that will result. So this thing right over here, it's right in front. And I'm going to make it go straight down and cut through this Jell-O or whatever you want to call it, this rectangular pyramid of Jell-O. And what would be the resulting shape of the intersection between the Jell-O and this thing that I'm using to cut it?"}, {"video_title": "Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "Let me make it big enough so that it can capture the shape that will result. So this thing right over here, it's right in front. And I'm going to make it go straight down and cut through this Jell-O or whatever you want to call it, this rectangular pyramid of Jell-O. And what would be the resulting shape of the intersection between the Jell-O and this thing that I'm using to cut it? And now I encourage you to pause your video and think about what the resulting shape would be. And the shape would be in two dimensions, right? This purple surface is a two-dimensional."}, {"video_title": "Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "And what would be the resulting shape of the intersection between the Jell-O and this thing that I'm using to cut it? And now I encourage you to pause your video and think about what the resulting shape would be. And the shape would be in two dimensions, right? This purple surface is a two-dimensional. You could view it as part of a plane. And so where this intersects when you cut down this rectangular pyramid is the shape we're looking for. So I encourage you to pause the video and think about it or try to come up with it on your own."}, {"video_title": "Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "This purple surface is a two-dimensional. You could view it as part of a plane. And so where this intersects when you cut down this rectangular pyramid is the shape we're looking for. So I encourage you to pause the video and think about it or try to come up with it on your own. So let's think about it. And let me draw the rectangular pyramid again. So that's the same one."}, {"video_title": "Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "So I encourage you to pause the video and think about it or try to come up with it on your own. So let's think about it. And let me draw the rectangular pyramid again. So that's the same one. And now let me see what it would look like once I've done my cut, once I've brought this thing down. So then this is where I cut. So I cut it right over here."}, {"video_title": "Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "So that's the same one. And now let me see what it would look like once I've done my cut, once I've brought this thing down. So then this is where I cut. So I cut it right over here. And then it'll exit the bottom. It'll cut along this side like that, cut along that side like that. And then it'll exit the bottom right over there."}, {"video_title": "Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "So I cut it right over here. And then it'll exit the bottom. It'll cut along this side like that, cut along that side like that. And then it'll exit the bottom right over there. And so let me draw my whole thing. And so once I slice it down, it will look like this. My best shot at drawing it."}, {"video_title": "Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "And then it'll exit the bottom right over there. And so let me draw my whole thing. And so once I slice it down, it will look like this. My best shot at drawing it. It will look like this. This is a rectangular. This is a vertical cut."}, {"video_title": "Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "My best shot at drawing it. It will look like this. This is a rectangular. This is a vertical cut. So I brought this thing down. And now the intersection between the thing that I'm cutting with and this pyramid is going to be this shape right over here. It cut into the top right over there."}, {"video_title": "Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "This is a vertical cut. So I brought this thing down. And now the intersection between the thing that I'm cutting with and this pyramid is going to be this shape right over here. It cut into the top right over there. It would get all the way to the bottom right over there. And along this side, it would cut right there. And along that side, it would cut right over there."}, {"video_title": "Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "It cut into the top right over there. It would get all the way to the bottom right over there. And along this side, it would cut right there. And along that side, it would cut right over there. So what's the resulting shape in two dimensions of essentially the intersection between the slicer and the jello? Well, it would be this thing. It would look like a trapezoid."}, {"video_title": "Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "And along that side, it would cut right over there. So what's the resulting shape in two dimensions of essentially the intersection between the slicer and the jello? Well, it would be this thing. It would look like a trapezoid. Let me do that in a new color. I'm overusing that one color. It would look like a trapezoid."}, {"video_title": "Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "It would look like a trapezoid. Let me do that in a new color. I'm overusing that one color. It would look like a trapezoid. So this would be the resulting shape. So the resulting shape would look like this. It would look like this, just like that if you made the cut right over there."}, {"video_title": "Dilations and shape properties.mp3", "Sentence": "I have this little dilation tool. So the first question is, are the coordinates of the vertices going to be preserved? Well, pause the video and try to think about that. Well, let's just try it out experimentally. We can see under an arbitrary dilation here, the coordinates are not preserved. The point that corresponds to D now has a different coordinate. The vertices, the vertex that corresponds to A now has different coordinates."}, {"video_title": "Dilations and shape properties.mp3", "Sentence": "Well, let's just try it out experimentally. We can see under an arbitrary dilation here, the coordinates are not preserved. The point that corresponds to D now has a different coordinate. The vertices, the vertex that corresponds to A now has different coordinates. Same thing for B and C. The corresponding points after the dilation now sit on a different part of the coordinate plane. So in this case, the coordinates of the vertices are not preserved. Now the next question, let me go back to where we were."}, {"video_title": "Dilations and shape properties.mp3", "Sentence": "The vertices, the vertex that corresponds to A now has different coordinates. Same thing for B and C. The corresponding points after the dilation now sit on a different part of the coordinate plane. So in this case, the coordinates of the vertices are not preserved. Now the next question, let me go back to where we were. So the next question, the corresponding line segments after dilation, are they sitting on the same line? And so let me dilate again. And so you can see, if you consider this point B prime, because it corresponds to point B, the segment B prime, C prime, this does not sit on the same line as BC."}, {"video_title": "Dilations and shape properties.mp3", "Sentence": "Now the next question, let me go back to where we were. So the next question, the corresponding line segments after dilation, are they sitting on the same line? And so let me dilate again. And so you can see, if you consider this point B prime, because it corresponds to point B, the segment B prime, C prime, this does not sit on the same line as BC. But the segment A prime, D prime, the corresponding line segment to line segment AD, that does sit on the same line. And if you think about why that is, well, if we originally draw a line that if we look at the line that contains segment AD, it also goes through point P. And so as we expand out, this segment right over here is going to expand and shift outward along the same lines. But that's not going to be true of these other segments because they don't, because the point P does not sit on the line that those segments sit on."}, {"video_title": "Dilations and shape properties.mp3", "Sentence": "And so you can see, if you consider this point B prime, because it corresponds to point B, the segment B prime, C prime, this does not sit on the same line as BC. But the segment A prime, D prime, the corresponding line segment to line segment AD, that does sit on the same line. And if you think about why that is, well, if we originally draw a line that if we look at the line that contains segment AD, it also goes through point P. And so as we expand out, this segment right over here is going to expand and shift outward along the same lines. But that's not going to be true of these other segments because they don't, because the point P does not sit on the line that those segments sit on. And so let's just expand it again. So you see that right over there. Now the next question, are angle measures preserved?"}, {"video_title": "Dilations and shape properties.mp3", "Sentence": "But that's not going to be true of these other segments because they don't, because the point P does not sit on the line that those segments sit on. And so let's just expand it again. So you see that right over there. Now the next question, are angle measures preserved? Well, it looks like they are. And this is one of the things that is true about a dilation, is that you're going to preserve angle measures. This angle is still a right angle."}, {"video_title": "Dilations and shape properties.mp3", "Sentence": "Now the next question, are angle measures preserved? Well, it looks like they are. And this is one of the things that is true about a dilation, is that you're going to preserve angle measures. This angle is still a right angle. This angle here, I guess you could call it angle, the measure of angle B, is the same as the measure of angle B prime. And you can see it with all of these points right over there. And then the last question, are side lengths, perimeter, and area preserved?"}, {"video_title": "Dilations and shape properties.mp3", "Sentence": "This angle is still a right angle. This angle here, I guess you could call it angle, the measure of angle B, is the same as the measure of angle B prime. And you can see it with all of these points right over there. And then the last question, are side lengths, perimeter, and area preserved? Well, we can immediately see as we dilate outwards, for example, the segment corresponding to AD has gotten longer. In fact, if we dilate outwards, all of the segments, the corresponding segments are getting larger. And if they're all getting larger, then the perimeter is getting larger and the area is getting larger."}, {"video_title": "Dilations and shape properties.mp3", "Sentence": "And then the last question, are side lengths, perimeter, and area preserved? Well, we can immediately see as we dilate outwards, for example, the segment corresponding to AD has gotten longer. In fact, if we dilate outwards, all of the segments, the corresponding segments are getting larger. And if they're all getting larger, then the perimeter is getting larger and the area is getting larger. Likewise, if we dilate in like this, they're all getting smaller. So side lengths, perimeter, and area are not preserved. Now let's ask the same questions with another dilation."}, {"video_title": "Dilations and shape properties.mp3", "Sentence": "And if they're all getting larger, then the perimeter is getting larger and the area is getting larger. Likewise, if we dilate in like this, they're all getting smaller. So side lengths, perimeter, and area are not preserved. Now let's ask the same questions with another dilation. And this is going to be interesting because we're going to look at a dilation that is centered at one of the vertices of our shape. So let me scroll down here. And so I have the same tool again."}, {"video_title": "Dilations and shape properties.mp3", "Sentence": "Now let's ask the same questions with another dilation. And this is going to be interesting because we're going to look at a dilation that is centered at one of the vertices of our shape. So let me scroll down here. And so I have the same tool again. And now here we have a triangle, triangle ABC, and we're gonna dilate about point C. So first of all, do we think the vertices, the coordinates of the vertices are going to be preserved? Let's dilate out. Well, you can see point C is preserved."}, {"video_title": "Dilations and shape properties.mp3", "Sentence": "And so I have the same tool again. And now here we have a triangle, triangle ABC, and we're gonna dilate about point C. So first of all, do we think the vertices, the coordinates of the vertices are going to be preserved? Let's dilate out. Well, you can see point C is preserved. When it gets mapped after the dilation, it sits in the exact same place, but the things that correspond to A and B are not preserved. You could call this A prime, and this definitely has different coordinates than A, and B prime definitely has different coordinates than B. Now what about corresponding line segments?"}, {"video_title": "Dilations and shape properties.mp3", "Sentence": "Well, you can see point C is preserved. When it gets mapped after the dilation, it sits in the exact same place, but the things that correspond to A and B are not preserved. You could call this A prime, and this definitely has different coordinates than A, and B prime definitely has different coordinates than B. Now what about corresponding line segments? Are they on the same line? Well, some of them are and some of them aren't. So for example, when we dilate, so let's look at the segment AC and the segment BC."}, {"video_title": "Dilations and shape properties.mp3", "Sentence": "Now what about corresponding line segments? Are they on the same line? Well, some of them are and some of them aren't. So for example, when we dilate, so let's look at the segment AC and the segment BC. When we dilate, we can see, whoops, when we dilate, we can see the corresponding segments, you could call this A prime, C prime, or B prime, C prime, do still sit on that same line. And that's because the point that we are dilating about, point C, sat on those original segments. So we're essentially just lengthening out on the point that is not the center of dilation."}, {"video_title": "Dilations and shape properties.mp3", "Sentence": "So for example, when we dilate, so let's look at the segment AC and the segment BC. When we dilate, we can see, whoops, when we dilate, we can see the corresponding segments, you could call this A prime, C prime, or B prime, C prime, do still sit on that same line. And that's because the point that we are dilating about, point C, sat on those original segments. So we're essentially just lengthening out on the point that is not the center of dilation. We're lengthening out away from it, or if the dilation is going in, we would be shortening along that same line. But some of the segments are not overlapping on the same line. So for example, A prime, B prime does not sit along the same line as A, B."}, {"video_title": "Dilations and shape properties.mp3", "Sentence": "So we're essentially just lengthening out on the point that is not the center of dilation. We're lengthening out away from it, or if the dilation is going in, we would be shortening along that same line. But some of the segments are not overlapping on the same line. So for example, A prime, B prime does not sit along the same line as A, B. Now what about the angle measures? Well, we already talked about it. Angle measures are preserved under dilations."}, {"video_title": "Dilations and shape properties.mp3", "Sentence": "So for example, A prime, B prime does not sit along the same line as A, B. Now what about the angle measures? Well, we already talked about it. Angle measures are preserved under dilations. The measure of angle C here, this is the exact same angle, and so is the measure of angle, you could call this A prime and B prime right over here. And then finally, what about side lengths? Well, you can clearly see that when I dilate out, my side lengths increase, or if I dilate in, my side lengths decrease."}, {"video_title": "Dilations and shape properties.mp3", "Sentence": "Angle measures are preserved under dilations. The measure of angle C here, this is the exact same angle, and so is the measure of angle, you could call this A prime and B prime right over here. And then finally, what about side lengths? Well, you can clearly see that when I dilate out, my side lengths increase, or if I dilate in, my side lengths decrease. And so side lengths are not preserved. And if side lengths are not preserved, then the perimeter is not preserved, and also the area is not preserved. You could view area as a function of the side lengths."}, {"video_title": "Proving the SAS triangle congruence criterion using transformations Geometry Khan Academy.mp3", "Sentence": "What we're going to do in this video is see that if we have two different triangles and we have two sets of corresponding sides that have the same length, for example, this blue side has the same length as this blue side here, and this orange side has the same length as this orange side here, and the angle that is formed between those sides, so we have two corresponding angles right over here, that they also have the equal measure. So we could think about, we have a side, an angle, a side, a side, an angle, and a side. If those have the same lengths or measures, then we can deduce that these two triangles must be congruent by the rigid motion definition of congruency, or the shorthand is, if we have side, angle, side in common and the angle is between the two sides, then the two triangles will be congruent. So to be able to prove this, in order to make this deduction, we just have to say that there's always a rigid transformation if we have a side, angle, side in common that will allow us to map one triangle onto the other, because if there is a series of rigid transformations that allow us to do it, then by the rigid transformation definition, the true triangles are congruent. So the first thing that we could do is we could reference back to where we saw that if we have two segments that have the same length, like segment AB and segment DE, if we have two segments with the same length, that they are congruent. You can always map one segment onto the other with a series of rigid transformations. The way that we could do that in this case is we could map point B onto point E, so this would be now, I'll put B prime right over here, and if we just did a transformation to do that, so if we just translated like that, then side, whoops, then side BA would, that orange side would be something like that, but then we could do another rigid transformation that rotates about point E, or B prime, that rotates that orange side and the whole triangle with it onto DE, in which case, once we do that second rigid transformation, point A will now coincide with D, or we could say A prime is equal to D. But the question is, where would C now sit?"}, {"video_title": "Proving the SAS triangle congruence criterion using transformations Geometry Khan Academy.mp3", "Sentence": "So to be able to prove this, in order to make this deduction, we just have to say that there's always a rigid transformation if we have a side, angle, side in common that will allow us to map one triangle onto the other, because if there is a series of rigid transformations that allow us to do it, then by the rigid transformation definition, the true triangles are congruent. So the first thing that we could do is we could reference back to where we saw that if we have two segments that have the same length, like segment AB and segment DE, if we have two segments with the same length, that they are congruent. You can always map one segment onto the other with a series of rigid transformations. The way that we could do that in this case is we could map point B onto point E, so this would be now, I'll put B prime right over here, and if we just did a transformation to do that, so if we just translated like that, then side, whoops, then side BA would, that orange side would be something like that, but then we could do another rigid transformation that rotates about point E, or B prime, that rotates that orange side and the whole triangle with it onto DE, in which case, once we do that second rigid transformation, point A will now coincide with D, or we could say A prime is equal to D. But the question is, where would C now sit? Well, we can see the distance between A and C, in fact, we can use our compass for it. The distance between A and C is, right, is just like that. And so, since all of these rigid transformations preserve distance, we know that C prime, the point that C gets mapped to after those first two transformations, C prime, its distance is going to stay the same from A prime, so C prime is going to be someplace, someplace along this curve right over here."}, {"video_title": "Proving the SAS triangle congruence criterion using transformations Geometry Khan Academy.mp3", "Sentence": "The way that we could do that in this case is we could map point B onto point E, so this would be now, I'll put B prime right over here, and if we just did a transformation to do that, so if we just translated like that, then side, whoops, then side BA would, that orange side would be something like that, but then we could do another rigid transformation that rotates about point E, or B prime, that rotates that orange side and the whole triangle with it onto DE, in which case, once we do that second rigid transformation, point A will now coincide with D, or we could say A prime is equal to D. But the question is, where would C now sit? Well, we can see the distance between A and C, in fact, we can use our compass for it. The distance between A and C is, right, is just like that. And so, since all of these rigid transformations preserve distance, we know that C prime, the point that C gets mapped to after those first two transformations, C prime, its distance is going to stay the same from A prime, so C prime is going to be someplace, someplace along this curve right over here. We also know that the rigid transformations preserve angle measures, and so we also know that as we do the mapping, the angle will be preserved. So either side AC will be mapped to this side right over here, and if that's the case, then F would be equal to C prime, and we would have found our rigid transformation based on SAS, and so therefore, the two triangles would be congruent. But there's another possibility that the angle gets conserved, but side AC is mapped down here."}, {"video_title": "Proving the SAS triangle congruence criterion using transformations Geometry Khan Academy.mp3", "Sentence": "And so, since all of these rigid transformations preserve distance, we know that C prime, the point that C gets mapped to after those first two transformations, C prime, its distance is going to stay the same from A prime, so C prime is going to be someplace, someplace along this curve right over here. We also know that the rigid transformations preserve angle measures, and so we also know that as we do the mapping, the angle will be preserved. So either side AC will be mapped to this side right over here, and if that's the case, then F would be equal to C prime, and we would have found our rigid transformation based on SAS, and so therefore, the two triangles would be congruent. But there's another possibility that the angle gets conserved, but side AC is mapped down here. So there's another possibility that side AC, due to our rigid transformations, or after our first set of rigid transformations, looks something like this. So it looks, it looks something like that, in which case C prime would be mapped right over there. And in that case, we can just do one more rigid transformation."}, {"video_title": "Proving the SAS triangle congruence criterion using transformations Geometry Khan Academy.mp3", "Sentence": "But there's another possibility that the angle gets conserved, but side AC is mapped down here. So there's another possibility that side AC, due to our rigid transformations, or after our first set of rigid transformations, looks something like this. So it looks, it looks something like that, in which case C prime would be mapped right over there. And in that case, we can just do one more rigid transformation. We can just do a reflection about DE, or A prime, B prime, to reflect point C prime over that to get right over there. How do we know that C prime would then be mapped to F? Well, this angle would be preserved due to the rigid transformation."}, {"video_title": "Proving the SAS triangle congruence criterion using transformations Geometry Khan Academy.mp3", "Sentence": "And in that case, we can just do one more rigid transformation. We can just do a reflection about DE, or A prime, B prime, to reflect point C prime over that to get right over there. How do we know that C prime would then be mapped to F? Well, this angle would be preserved due to the rigid transformation. So as we flip it over, as we do the reflection over DE, the angle would be preserved, and A prime, C prime will then map to DF. And then we'd be done. We have just shown that there's always a series of rigid transformations, as long as you meet this SAS criteria, that can map one triangle onto the other, and therefore, they are congruent."}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "It's transforming from one thing to another. So what would transformation mean in a mathematical context? Well, it could mean that you're taking something mathematical and you're changing it into something else mathematical, and that's exactly what it is. It's talking about taking a set of coordinates or a set of points and then changing them into a different set of coordinates or a different set of points. For example, this right over here, this is a quadrilateral, we've plotted it on the coordinate plane. This is a set of points, not just the four points that represent the vertices of the quadrilateral, but all the points along the sides too. There's a bunch of points along this."}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "It's talking about taking a set of coordinates or a set of points and then changing them into a different set of coordinates or a different set of points. For example, this right over here, this is a quadrilateral, we've plotted it on the coordinate plane. This is a set of points, not just the four points that represent the vertices of the quadrilateral, but all the points along the sides too. There's a bunch of points along this. You could argue there's an infinite, or there are an infinite number of points along this quadrilateral. This right over here, the point x equals zero, y equals negative four, this is a point on the quadrilateral. Now, we can apply a transformation to this."}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "There's a bunch of points along this. You could argue there's an infinite, or there are an infinite number of points along this quadrilateral. This right over here, the point x equals zero, y equals negative four, this is a point on the quadrilateral. Now, we can apply a transformation to this. The first one I'm going to show you is a translation, which just means moving all the points in the same direction and the same amount in that same direction. I'm using the Khan Academy translation widget to do it. Let's translate this, and I can do it by grabbing onto one of the vertices."}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "Now, we can apply a transformation to this. The first one I'm going to show you is a translation, which just means moving all the points in the same direction and the same amount in that same direction. I'm using the Khan Academy translation widget to do it. Let's translate this, and I can do it by grabbing onto one of the vertices. Notice, I've now shifted it to the right by two. Every point here, not just the orange points, has shifted to the right by two. This one has shifted to the right by two."}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "Let's translate this, and I can do it by grabbing onto one of the vertices. Notice, I've now shifted it to the right by two. Every point here, not just the orange points, has shifted to the right by two. This one has shifted to the right by two. This point right over here has shifted to the right by two. Every point has shifted in the same direction by the same amount. That's what a translation is."}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "This one has shifted to the right by two. This point right over here has shifted to the right by two. Every point has shifted in the same direction by the same amount. That's what a translation is. Now, I've shifted, let's see, if I put it here, everything, every point has shifted to the right one and up one. They've all shifted by the same amount in the same direction. That is a translation, but you can imagine a translation is not the only kind of transformation."}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "That's what a translation is. Now, I've shifted, let's see, if I put it here, everything, every point has shifted to the right one and up one. They've all shifted by the same amount in the same direction. That is a translation, but you can imagine a translation is not the only kind of transformation. In fact, there's an unlimited variation, there's an unlimited number of different transformations. For example, I could do a rotation. I have another set of points here that's represented by quadrilateral, I guess we could call it CD or BCDE, and I could rotate it."}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "That is a translation, but you can imagine a translation is not the only kind of transformation. In fact, there's an unlimited variation, there's an unlimited number of different transformations. For example, I could do a rotation. I have another set of points here that's represented by quadrilateral, I guess we could call it CD or BCDE, and I could rotate it. I would rotate it around a point. For example, I could rotate it around the point D. This is what I start with. If I, let me see if I can do this."}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "I have another set of points here that's represented by quadrilateral, I guess we could call it CD or BCDE, and I could rotate it. I would rotate it around a point. For example, I could rotate it around the point D. This is what I start with. If I, let me see if I can do this. I could rotate it like, actually let me see. If I start like this, I could rotate it 90 degrees. I could rotate 90 degrees."}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "If I, let me see if I can do this. I could rotate it like, actually let me see. If I start like this, I could rotate it 90 degrees. I could rotate 90 degrees. I could rotate it like, that looks pretty close to a 90 degree rotation. Every point that was on the original or in the original set of points, I've now shifted it relative to that point that I'm rotating around, I've now rotated it 90 degrees. This point has now mapped to this point over here."}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "I could rotate 90 degrees. I could rotate it like, that looks pretty close to a 90 degree rotation. Every point that was on the original or in the original set of points, I've now shifted it relative to that point that I'm rotating around, I've now rotated it 90 degrees. This point has now mapped to this point over here. This point has now mapped to this point over here. I'm just picking the vertices because those are a little bit easier to think about. This point has mapped to this point."}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "This point has now mapped to this point over here. This point has now mapped to this point over here. I'm just picking the vertices because those are a little bit easier to think about. This point has mapped to this point. The point of rotation, actually, since D is actually the point of rotation, that one actually has not shifted. Just so you get some terminology, the set of points after you apply the transformation, this is called the image of the transformation. I had quadrilateral BCDE."}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "This point has mapped to this point. The point of rotation, actually, since D is actually the point of rotation, that one actually has not shifted. Just so you get some terminology, the set of points after you apply the transformation, this is called the image of the transformation. I had quadrilateral BCDE. I applied a 90 degree counterclockwise rotation around the point D. This new set of points, this is the image of our original quadrilateral after the transformation. I don't have to just, let me undo this, I don't have to rotate around just one of the points that are on the original set, that are on our quadrilateral. I could rotate around the origin."}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "I had quadrilateral BCDE. I applied a 90 degree counterclockwise rotation around the point D. This new set of points, this is the image of our original quadrilateral after the transformation. I don't have to just, let me undo this, I don't have to rotate around just one of the points that are on the original set, that are on our quadrilateral. I could rotate around the origin. I could do something like that. Notice it's a different rotation now. It's a different rotation."}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "I could rotate around the origin. I could do something like that. Notice it's a different rotation now. It's a different rotation. I could rotate around any point. Let's look at another transformation. That would be the notion of a reflection."}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "It's a different rotation. I could rotate around any point. Let's look at another transformation. That would be the notion of a reflection. You know what reflection means in everyday life. You kind of imagine the reflection of an image in a mirror or on the water. That's exactly what we're going to do over here."}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "That would be the notion of a reflection. You know what reflection means in everyday life. You kind of imagine the reflection of an image in a mirror or on the water. That's exactly what we're going to do over here. If we reflect, we reflect across a line. Let me do that. This, what is this, one, two, three, four, five, this not irregular pentagon, let's reflect it."}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "That's exactly what we're going to do over here. If we reflect, we reflect across a line. Let me do that. This, what is this, one, two, three, four, five, this not irregular pentagon, let's reflect it. To reflect it, let me actually make a line like this. I could reflect it across a whole series of lines. Whoops, let me see if I can."}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "This, what is this, one, two, three, four, five, this not irregular pentagon, let's reflect it. To reflect it, let me actually make a line like this. I could reflect it across a whole series of lines. Whoops, let me see if I can. Let's reflect it across this. What does it mean to reflect across something? One way I imagine is if this was, we're going to get its mirror image and you kind of imagine this as kind of the line of symmetry, that the image and the original set, the original shape, they should be mirror images across this line."}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "Whoops, let me see if I can. Let's reflect it across this. What does it mean to reflect across something? One way I imagine is if this was, we're going to get its mirror image and you kind of imagine this as kind of the line of symmetry, that the image and the original set, the original shape, they should be mirror images across this line. We could see that. Let's do the reflection. There you go."}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "One way I imagine is if this was, we're going to get its mirror image and you kind of imagine this as kind of the line of symmetry, that the image and the original set, the original shape, they should be mirror images across this line. We could see that. Let's do the reflection. There you go. You see we have a mirror image. This is this far away from the line. This corresponding point in the image is on the other side of the line, but the same distance."}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "There you go. You see we have a mirror image. This is this far away from the line. This corresponding point in the image is on the other side of the line, but the same distance. This point over here is this distance from the line and this point over here is the same distance, but on the other side. Now all of the transformations that I've just showed you, the translation, the reflection, the rotation, these are called rigid transformations. Once again, you can just think about what does rigid mean in everyday life?"}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "This corresponding point in the image is on the other side of the line, but the same distance. This point over here is this distance from the line and this point over here is the same distance, but on the other side. Now all of the transformations that I've just showed you, the translation, the reflection, the rotation, these are called rigid transformations. Once again, you can just think about what does rigid mean in everyday life? It means something that's not flexible. It means something that you can't kind of stretch or scale up or scale down. It kind of maintains its shape."}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "Once again, you can just think about what does rigid mean in everyday life? It means something that's not flexible. It means something that you can't kind of stretch or scale up or scale down. It kind of maintains its shape. That's what rigid transformations are fundamentally about. If you want to think a little bit more mathematically, a rigid transformation is one in which lengths and angles are preserved. You can see in this transformation right over here, the distance between this point and this point, between points T and R, and the distance between their corresponding image points, that distance is the same."}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "It kind of maintains its shape. That's what rigid transformations are fundamentally about. If you want to think a little bit more mathematically, a rigid transformation is one in which lengths and angles are preserved. You can see in this transformation right over here, the distance between this point and this point, between points T and R, and the distance between their corresponding image points, that distance is the same. The angle here, angle RTY, the measure of this angle over here, if you look at the corresponding angle in the image, it's going to be the same angle. So you might, and the same thing is true if you're doing a translation. You can imagine these are kind of acting like rigid objects."}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "You can see in this transformation right over here, the distance between this point and this point, between points T and R, and the distance between their corresponding image points, that distance is the same. The angle here, angle RTY, the measure of this angle over here, if you look at the corresponding angle in the image, it's going to be the same angle. So you might, and the same thing is true if you're doing a translation. You can imagine these are kind of acting like rigid objects. You can't stretch them. They're not flexible. They're maintaining their shape."}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "You can imagine these are kind of acting like rigid objects. You can't stretch them. They're not flexible. They're maintaining their shape. Now what would be examples of transformations that are not rigid transformations? Well, you can imagine scaling things up and down. If I were to zoom, if I were to scale this out where it has maybe the angles are preserved, but the lengths aren't preserved, that would not be a rigid transformation."}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "They're maintaining their shape. Now what would be examples of transformations that are not rigid transformations? Well, you can imagine scaling things up and down. If I were to zoom, if I were to scale this out where it has maybe the angles are preserved, but the lengths aren't preserved, that would not be a rigid transformation. If I were to just stretch one side of it, or if I were to just pull this point while the other point stayed where they are, I'd be kind of distorting it or stretching it. That would not be a rigid transformation. So hopefully this gets you, it's actually very, very interesting."}, {"video_title": "Introduction to transformations Transformations Geometry Khan Academy.mp3", "Sentence": "If I were to zoom, if I were to scale this out where it has maybe the angles are preserved, but the lengths aren't preserved, that would not be a rigid transformation. If I were to just stretch one side of it, or if I were to just pull this point while the other point stayed where they are, I'd be kind of distorting it or stretching it. That would not be a rigid transformation. So hopefully this gets you, it's actually very, very interesting. I mean, when you use an art program or actually use a lot of computer graphics or you play a video game, most of what the video game is doing is actually doing transformations, sometimes in two dimensions, sometimes in three dimensions. And once you get into more advanced math, especially things like linear algebra, there's a whole field that's really focused around transformations. In fact, some of the computers with really good graphics processors, a graphics processor is just a piece of hardware that is really good at performing mathematical transformations so that you can immerse yourself in a 3D reality or whatever else."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "And what I want to do is look at the midpoints of each of the sides of ABC. So this is the midpoint of one of the sides of side BC, let's call that point D. Let's call this midpoint E, and let's call this midpoint right over here F. And since it's the midpoint, we know that the distance between BD is equal to the distance from D to C. So this distance is equal to this distance. We know that AE is equal to EC, so this distance is equal to that distance. And we know that AF is equal to FB, so this distance is equal to this distance. Instead of drawing medians going from these midpoints to the vertices, what I want to do is I want to connect these midpoints and see what happens. So if I connect them, I clearly have three points. So if you connect three nonlinear points like this, you will get another triangle."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "And we know that AF is equal to FB, so this distance is equal to this distance. Instead of drawing medians going from these midpoints to the vertices, what I want to do is I want to connect these midpoints and see what happens. So if I connect them, I clearly have three points. So if you connect three nonlinear points like this, you will get another triangle. And this triangle that's formed from the midpoints of the sides of this larger triangle, we call this a medial triangle. And that's all nice and cute by itself. But what we're going to see in this video is that the medial triangle actually has some very neat properties."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "So if you connect three nonlinear points like this, you will get another triangle. And this triangle that's formed from the midpoints of the sides of this larger triangle, we call this a medial triangle. And that's all nice and cute by itself. But what we're going to see in this video is that the medial triangle actually has some very neat properties. What we're actually going to show is that it divides any triangle into four smaller triangles that are congruent to each other, that all four of these triangles are identical to each other and they're all similar to the larger triangle. And you could think of them each as having 1 4th of the area of the larger triangle. So let's go about proving it."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "But what we're going to see in this video is that the medial triangle actually has some very neat properties. What we're actually going to show is that it divides any triangle into four smaller triangles that are congruent to each other, that all four of these triangles are identical to each other and they're all similar to the larger triangle. And you could think of them each as having 1 4th of the area of the larger triangle. So let's go about proving it. So first, let's focus on this triangle down here, triangle CDE. And it looks similar to the larger triangle, to triangle CBA. But let's prove it to ourselves."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "So let's go about proving it. So first, let's focus on this triangle down here, triangle CDE. And it looks similar to the larger triangle, to triangle CBA. But let's prove it to ourselves. So one thing we can say is, well, look, both of them share this angle right over here. Both the larger triangle, triangle CBA, shares this angle. And the smaller triangle, CDE, has this angle."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "But let's prove it to ourselves. So one thing we can say is, well, look, both of them share this angle right over here. Both the larger triangle, triangle CBA, shares this angle. And the smaller triangle, CDE, has this angle. So they definitely share that angle. And then let's think about the ratios of the sides. We know that the ratio of CD to CB is equal to 1 over 2."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "And the smaller triangle, CDE, has this angle. So they definitely share that angle. And then let's think about the ratios of the sides. We know that the ratio of CD to CB is equal to 1 over 2. This is half of this entire side, is equal to 1 over 2. And that's the same thing as the ratio of CE to CA. CE is exactly half of CA, because E is the midpoint."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "We know that the ratio of CD to CB is equal to 1 over 2. This is half of this entire side, is equal to 1 over 2. And that's the same thing as the ratio of CE to CA. CE is exactly half of CA, because E is the midpoint. It's equal to CE over CA. So we have an angle, and corresponding angles that are congruent. And then the ratios of two corresponding sides on either side of that angle are the same."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "CE is exactly half of CA, because E is the midpoint. It's equal to CE over CA. So we have an angle, and corresponding angles that are congruent. And then the ratios of two corresponding sides on either side of that angle are the same. CD over CB is 1 half. CE over CA is 1 half. And the angle in between is congruent."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "And then the ratios of two corresponding sides on either side of that angle are the same. CD over CB is 1 half. CE over CA is 1 half. And the angle in between is congruent. So by SAS similarity, we know that triangle CDE is similar to triangle CBA. And just from that, you can get some interesting results. Because then we know that the ratio of this side of the smaller triangle to the longer triangle is also going to be 1 half."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "And the angle in between is congruent. So by SAS similarity, we know that triangle CDE is similar to triangle CBA. And just from that, you can get some interesting results. Because then we know that the ratio of this side of the smaller triangle to the longer triangle is also going to be 1 half. Because the other two sides have a ratio of 1 half, and we're dealing with similar triangles. So this is going to be 1 half of that. And we know 1 half of AB is just going to be the length of FA."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "Because then we know that the ratio of this side of the smaller triangle to the longer triangle is also going to be 1 half. Because the other two sides have a ratio of 1 half, and we're dealing with similar triangles. So this is going to be 1 half of that. And we know 1 half of AB is just going to be the length of FA. So we know that this length right over here is going to be the same as FA or FB. And we get that straight from similar triangles. Because these are similar, we know that DE over BA has got to be equal to these ratios, the other corresponding sides, which is equal to 1 half."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "And we know 1 half of AB is just going to be the length of FA. So we know that this length right over here is going to be the same as FA or FB. And we get that straight from similar triangles. Because these are similar, we know that DE over BA has got to be equal to these ratios, the other corresponding sides, which is equal to 1 half. And so that's how we got that right over there. Now let's think about this triangle up here. Triangle, we could call it BDF."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "Because these are similar, we know that DE over BA has got to be equal to these ratios, the other corresponding sides, which is equal to 1 half. And so that's how we got that right over there. Now let's think about this triangle up here. Triangle, we could call it BDF. So first of all, if you compare triangle BDF to the larger triangle, they both share this angle right over here, angle ABC. They both have that angle in common. And we're going to have the exact same argument."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "Triangle, we could call it BDF. So first of all, if you compare triangle BDF to the larger triangle, they both share this angle right over here, angle ABC. They both have that angle in common. And we're going to have the exact same argument. You can just look at this diagram, and you know that the ratio of BA, the ratio of BF to BA is equal to 1 half, which is also the ratio of BD to BC. The ratio of this to that is the same as the ratio of this to that, which is 1 half. Because BD is half of this whole length, BF is half of that whole length."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "And we're going to have the exact same argument. You can just look at this diagram, and you know that the ratio of BA, the ratio of BF to BA is equal to 1 half, which is also the ratio of BD to BC. The ratio of this to that is the same as the ratio of this to that, which is 1 half. Because BD is half of this whole length, BF is half of that whole length. And so you have corresponding sides have the same ratio on the two triangles. And they share an angle in between. So once again, by SAS similarity, we know that triangle, I'll write it this way, DBF is similar to triangle CBA."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "Because BD is half of this whole length, BF is half of that whole length. And so you have corresponding sides have the same ratio on the two triangles. And they share an angle in between. So once again, by SAS similarity, we know that triangle, I'll write it this way, DBF is similar to triangle CBA. And once again, we use this exact same kind of argument that we did with this triangle. Well, if it's similar, the ratio of all the corresponding sides have to be the same. And that ratio is 1 half."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "So once again, by SAS similarity, we know that triangle, I'll write it this way, DBF is similar to triangle CBA. And once again, we use this exact same kind of argument that we did with this triangle. Well, if it's similar, the ratio of all the corresponding sides have to be the same. And that ratio is 1 half. So the ratio of this side to this side, the ratio of FD to AC has to be 1 half. Or FD has to be 1 half of AC. And 1 half of AC is just the length of AE."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "And that ratio is 1 half. So the ratio of this side to this side, the ratio of FD to AC has to be 1 half. Or FD has to be 1 half of AC. And 1 half of AC is just the length of AE. So that is just going to be that length right over there. I think you see where this is going. And also, because it's similar, all of the corresponding angles have to be the same."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "And 1 half of AC is just the length of AE. So that is just going to be that length right over there. I think you see where this is going. And also, because it's similar, all of the corresponding angles have to be the same. And we know that the larger triangle has a yellow angle right over there. So we'd have that yellow angle right over here. And this triangle right over here was also similar to the larger triangle."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "And also, because it's similar, all of the corresponding angles have to be the same. And we know that the larger triangle has a yellow angle right over there. So we'd have that yellow angle right over here. And this triangle right over here was also similar to the larger triangle. So it will have that same angle measure up here. We already showed that in this first part. So now let's go to this third triangle."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "And this triangle right over here was also similar to the larger triangle. So it will have that same angle measure up here. We already showed that in this first part. So now let's go to this third triangle. I think you see the pattern. I'm sure you might be able to just pause this video and prove it for yourself. But we see that the ratio of AF over AB is going to be the same as the ratio of AE over AC, which is equal to 1 half."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "So now let's go to this third triangle. I think you see the pattern. I'm sure you might be able to just pause this video and prove it for yourself. But we see that the ratio of AF over AB is going to be the same as the ratio of AE over AC, which is equal to 1 half. So we have two corresponding sides where the ratio is 1 half from the smaller to the larger triangle. And they share a common angle. They share this angle in between the two sides."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "But we see that the ratio of AF over AB is going to be the same as the ratio of AE over AC, which is equal to 1 half. So we have two corresponding sides where the ratio is 1 half from the smaller to the larger triangle. And they share a common angle. They share this angle in between the two sides. So by SAS similarity, this is getting repetitive now, we know that triangle EFA is similar to triangle CBA. And so the ratio of all of the corresponding sides need to be 1 half. So the ratio of FE to BC needs to be 1 half, or FE needs to be half of that, which is just the length of BD."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "They share this angle in between the two sides. So by SAS similarity, this is getting repetitive now, we know that triangle EFA is similar to triangle CBA. And so the ratio of all of the corresponding sides need to be 1 half. So the ratio of FE to BC needs to be 1 half, or FE needs to be half of that, which is just the length of BD. So this is just going to be that length right over there. And you can also say that this triangle, this triangle, and this triangle, we haven't talked about this middle one yet, they're all similar to the larger triangle so they're also all going to be similar to each other. So they're all going to have the same corresponding angles."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "So the ratio of FE to BC needs to be 1 half, or FE needs to be half of that, which is just the length of BD. So this is just going to be that length right over there. And you can also say that this triangle, this triangle, and this triangle, we haven't talked about this middle one yet, they're all similar to the larger triangle so they're also all going to be similar to each other. So they're all going to have the same corresponding angles. So if the larger triangle had this yellow angle here, then all of the triangles are going to have this yellow angle right over there. As the larger triangle had this blue angle, right over here, then in the corresponding vertex, all of the triangles are going to have that blue angle. All of the ones that we've shown are similar."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "So they're all going to have the same corresponding angles. So if the larger triangle had this yellow angle here, then all of the triangles are going to have this yellow angle right over there. As the larger triangle had this blue angle, right over here, then in the corresponding vertex, all of the triangles are going to have that blue angle. All of the ones that we've shown are similar. We haven't thought about this middle triangle just yet. And of course, if this is similar to the whole, it'll also have this angle at this vertex right over here, because this corresponds to that vertex based on the similarity. So that's interesting."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "All of the ones that we've shown are similar. We haven't thought about this middle triangle just yet. And of course, if this is similar to the whole, it'll also have this angle at this vertex right over here, because this corresponds to that vertex based on the similarity. So that's interesting. Now let's compare the triangles to each other. We've now shown that all of these triangles have the exact same three sides. It has this blue side, or actually, I don't know, this one marked side, this two marked side, this three marked side, one marked, two marked, three marked, one marked, two marked, three marked."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "So that's interesting. Now let's compare the triangles to each other. We've now shown that all of these triangles have the exact same three sides. It has this blue side, or actually, I don't know, this one marked side, this two marked side, this three marked side, one marked, two marked, three marked, one marked, two marked, three marked. And that even applies to this middle triangle right over here. So by side, side, side congruency, we now know, and we want to be careful to get our corresponding sides right, we now know that triangle CDE is congruent to triangle DBF, I want to get the corresponding sides. I'm looking at the colors."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "It has this blue side, or actually, I don't know, this one marked side, this two marked side, this three marked side, one marked, two marked, three marked, one marked, two marked, three marked. And that even applies to this middle triangle right over here. So by side, side, side congruency, we now know, and we want to be careful to get our corresponding sides right, we now know that triangle CDE is congruent to triangle DBF, I want to get the corresponding sides. I'm looking at the colors. I went from yellow to magenta to blue, yellow, magenta to blue, which is going to be congruent to triangle EFA, which is going to be congruent to this triangle in here, but we want to make sure that we're getting the right corresponding sides here. So to make sure we do that, we just have to think about the angles. So we know, and this is interesting, because the interior angles of a triangle add up to 180 degrees, we know this magenta angle plus this blue angle plus this yellow angle equal 180."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "I'm looking at the colors. I went from yellow to magenta to blue, yellow, magenta to blue, which is going to be congruent to triangle EFA, which is going to be congruent to this triangle in here, but we want to make sure that we're getting the right corresponding sides here. So to make sure we do that, we just have to think about the angles. So we know, and this is interesting, because the interior angles of a triangle add up to 180 degrees, we know this magenta angle plus this blue angle plus this yellow angle equal 180. Here, we have the blue angle and the magenta angle, and clearly they will all add up to 180, so you must have the blue angle must be right over here. Same argument, yellow angle and blue angle, we must have the magenta angle right over here. They add up to 180, so this must be the magenta angle."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "So we know, and this is interesting, because the interior angles of a triangle add up to 180 degrees, we know this magenta angle plus this blue angle plus this yellow angle equal 180. Here, we have the blue angle and the magenta angle, and clearly they will all add up to 180, so you must have the blue angle must be right over here. Same argument, yellow angle and blue angle, we must have the magenta angle right over here. They add up to 180, so this must be the magenta angle. And then finally, magenta and blue, this must be the yellow angle right over there. And so when we wrote the congruency here, we started at CDE. We went yellow, magenta, blue."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "They add up to 180, so this must be the magenta angle. And then finally, magenta and blue, this must be the yellow angle right over there. And so when we wrote the congruency here, we started at CDE. We went yellow, magenta, blue. So over here, we're going to go yellow, magenta, blue. So it's going to be congruent to triangle FED. And so that's pretty cool."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "We went yellow, magenta, blue. So over here, we're going to go yellow, magenta, blue. So it's going to be congruent to triangle FED. And so that's pretty cool. We just showed that all three, that this triangle, this triangle, this triangle, and that triangle are congruent. And also, we can look at the corresponding, and that they all have ratios relative to the large. They're all similar to the larger triangle, the triangle ABC, and that the ratio between the sides is 1 to 2."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "And so that's pretty cool. We just showed that all three, that this triangle, this triangle, this triangle, and that triangle are congruent. And also, we can look at the corresponding, and that they all have ratios relative to the large. They're all similar to the larger triangle, the triangle ABC, and that the ratio between the sides is 1 to 2. And also, because we've looked at corresponding angles, we see, for example, that this angle is the same as that angle. So if you viewed DC or if you viewed BC as a transversal, all of a sudden, it becomes pretty clear that FD is going to be parallel to AC because the corresponding angles are congruent. So this is going to be parallel to that right over there."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "They're all similar to the larger triangle, the triangle ABC, and that the ratio between the sides is 1 to 2. And also, because we've looked at corresponding angles, we see, for example, that this angle is the same as that angle. So if you viewed DC or if you viewed BC as a transversal, all of a sudden, it becomes pretty clear that FD is going to be parallel to AC because the corresponding angles are congruent. So this is going to be parallel to that right over there. And then you could use that same exact argument to say, well, then this side, because once again, corresponding angles here and here, you could say that this is going to be parallel to that right over there. And then finally, you make the same argument over here. You have, I want to make sure I get the right corresponding angles, you have this line and this line and this angle corresponds to that angle."}, {"video_title": "Exploring medial triangles Special properties and parts of triangles Geometry Khan Academy.mp3", "Sentence": "So this is going to be parallel to that right over there. And then you could use that same exact argument to say, well, then this side, because once again, corresponding angles here and here, you could say that this is going to be parallel to that right over there. And then finally, you make the same argument over here. You have, I want to make sure I get the right corresponding angles, you have this line and this line and this angle corresponds to that angle. They're the same. So this DE must be parallel to BA. So that's another neat property of this medial triangle."}, {"video_title": "Volume of a sphere Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "So if I have a sphere, so this isn't just a circle, this is a sphere, you could view it as a globe of some kind, so I'm going to shade it a little bit so you can tell that it's three dimensional. They're giving us the diameter, so if we go from one side of the sphere straight through the center of it, so we're imagining that we can see through the sphere, and we go straight through the centimeter, that distance right over there is 14 centimeters. Now, to find the volume of a sphere, we prove this, or you will see a proof for this later when you learn calculus, but the formula for the volume of a sphere is volume is equal to 4 thirds pi r cubed, where r is the radius of the sphere. So they've given us the diameter, and just like for circles, the radius of the sphere is half of the diameter. So in this example, our radius is going to be 7 centimeters. In fact, the sphere itself is a set of all points in three dimensions that is exactly the radius away from the center. But with that out of the way, let's just apply this radius being 7 centimeters to this formula right over here."}, {"video_title": "Volume of a sphere Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "So they've given us the diameter, and just like for circles, the radius of the sphere is half of the diameter. So in this example, our radius is going to be 7 centimeters. In fact, the sphere itself is a set of all points in three dimensions that is exactly the radius away from the center. But with that out of the way, let's just apply this radius being 7 centimeters to this formula right over here. So we're going to have a volume is equal to 4 thirds pi times 7 centimeters to the third power. So I'll do that in that pink color. So times 7 centimeters to the third power."}, {"video_title": "Volume of a sphere Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "But with that out of the way, let's just apply this radius being 7 centimeters to this formula right over here. So we're going to have a volume is equal to 4 thirds pi times 7 centimeters to the third power. So I'll do that in that pink color. So times 7 centimeters to the third power. And since it already involves pi, and you can approximate pi with 3.14, some people even approximate it with 22 over 7, but we'll actually just get the calculator out to get the exact value for this volume. So this is going to be, so my volume is going to be 4 divided by 3, and then I don't want to just put a pi there, because that might interpret it as 4 divided by 3 pi. So 4 divided by 3 times pi times 7 to the third power."}, {"video_title": "Volume of a sphere Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "So times 7 centimeters to the third power. And since it already involves pi, and you can approximate pi with 3.14, some people even approximate it with 22 over 7, but we'll actually just get the calculator out to get the exact value for this volume. So this is going to be, so my volume is going to be 4 divided by 3, and then I don't want to just put a pi there, because that might interpret it as 4 divided by 3 pi. So 4 divided by 3 times pi times 7 to the third power. In order of operations, it'll do the exponent before it does the multiplication, so this should work out. And the units are going to be in centimeters cubed, or cubic centimeters. So we get 1436, they don't tell us what to round it to, so I'll just round it to the nearest tenth."}, {"video_title": "Congruent triangle proof example Congruence Geometry Khan Academy.mp3", "Sentence": "So we have these two parallel lines, line segment A, B, and line segment C, D. They are parallel, I should say, they're parallel line segments. And then we have these transversals that go across them. So you have this transversal B, C right over here, and you have this transversal A, D. And what this diagram tells us is that the distance between A and E, this little hash mark, says that this line segment is the same distance as the distance between E and D. Or another way to think about it is that point E is at the midpoint, or is the midpoint of line segment A, D. And what I want to think about in this video is, is point E also the midpoint of line segment B, C? So this is the question right over here. So is E the midpoint of line segment B, C? And you can imagine, based on a lot of the videos we've been seeing lately, maybe it has something to do with congruent triangles. So let's see if we can set up some congruency relationship between the two obvious triangles in this diagram."}, {"video_title": "Congruent triangle proof example Congruence Geometry Khan Academy.mp3", "Sentence": "So this is the question right over here. So is E the midpoint of line segment B, C? And you can imagine, based on a lot of the videos we've been seeing lately, maybe it has something to do with congruent triangles. So let's see if we can set up some congruency relationship between the two obvious triangles in this diagram. We have this triangle up here on the left, and we have this diagram down here. This one kind of looks like it's pointing up, this one looks like it's pointing down. So there's a bunch of things we know about vertical angles and angles of transversals."}, {"video_title": "Congruent triangle proof example Congruence Geometry Khan Academy.mp3", "Sentence": "So let's see if we can set up some congruency relationship between the two obvious triangles in this diagram. We have this triangle up here on the left, and we have this diagram down here. This one kind of looks like it's pointing up, this one looks like it's pointing down. So there's a bunch of things we know about vertical angles and angles of transversals. The most obvious one is that we have this vertical. We know that angle AEB is going to be congruent, or its measure is going to be equal to the measure of angle CED. So we know that angle AEB is going to be congruent to angle DEC, which really just means they have the exact same measure."}, {"video_title": "Congruent triangle proof example Congruence Geometry Khan Academy.mp3", "Sentence": "So there's a bunch of things we know about vertical angles and angles of transversals. The most obvious one is that we have this vertical. We know that angle AEB is going to be congruent, or its measure is going to be equal to the measure of angle CED. So we know that angle AEB is going to be congruent to angle DEC, which really just means they have the exact same measure. And we know that because they are vertical angles. Now we also know that AB and CD are parallel, so this line right over here is a transversal. So we know, for example, and there's actually several ways that we can do this problem, but we know that this is a transversal, and there's a couple of ways to think about it right over here."}, {"video_title": "Congruent triangle proof example Congruence Geometry Khan Academy.mp3", "Sentence": "So we know that angle AEB is going to be congruent to angle DEC, which really just means they have the exact same measure. And we know that because they are vertical angles. Now we also know that AB and CD are parallel, so this line right over here is a transversal. So we know, for example, and there's actually several ways that we can do this problem, but we know that this is a transversal, and there's a couple of ways to think about it right over here. So let me just continue the transversal so we get to see all of the different angles. You could say that this angle right here, angle ABE, so this is its measure right over here, you could say that it is the alternate interior angle to angle ECD, to this angle right over there. And if that didn't jump out of view, you would say that the corresponding angle to this one right over here is this angle right up here."}, {"video_title": "Congruent triangle proof example Congruence Geometry Khan Academy.mp3", "Sentence": "So we know, for example, and there's actually several ways that we can do this problem, but we know that this is a transversal, and there's a couple of ways to think about it right over here. So let me just continue the transversal so we get to see all of the different angles. You could say that this angle right here, angle ABE, so this is its measure right over here, you could say that it is the alternate interior angle to angle ECD, to this angle right over there. And if that didn't jump out of view, you would say that the corresponding angle to this one right over here is this angle right up here. If you were to continue this line off a little bit, these are the corresponding angles, and then this one is vertical. But either way, angle ABE is going to be congruent to angle DCE. And we could say because it's alternate interior angles."}, {"video_title": "Congruent triangle proof example Congruence Geometry Khan Academy.mp3", "Sentence": "And if that didn't jump out of view, you would say that the corresponding angle to this one right over here is this angle right up here. If you were to continue this line off a little bit, these are the corresponding angles, and then this one is vertical. But either way, angle ABE is going to be congruent to angle DCE. And we could say because it's alternate interior angles. And then we have an interesting relationship. We have an angle congruent to an angle, another angle congruent to an angle, and then the next side is congruent to the next side over here. So pink-green side."}, {"video_title": "Congruent triangle proof example Congruence Geometry Khan Academy.mp3", "Sentence": "And we could say because it's alternate interior angles. And then we have an interesting relationship. We have an angle congruent to an angle, another angle congruent to an angle, and then the next side is congruent to the next side over here. So pink-green side. So we can employ AAS, angle-angle side. And it's in the right order. So now we know that triangle, we have to make sure that we get the letters right here, that we have the right corresponding vertices."}, {"video_title": "Congruent triangle proof example Congruence Geometry Khan Academy.mp3", "Sentence": "So pink-green side. So we can employ AAS, angle-angle side. And it's in the right order. So now we know that triangle, we have to make sure that we get the letters right here, that we have the right corresponding vertices. We can say that triangle AEB, triangle AE, actually let me start with the angle just to make it interesting. Angle BEA, so we're starting with the magenta angle going to the green angle and then going to the one that we haven't labeled. So angle BEA we can say is congruent to angle, we start with the magenta vertices, C, go to the center, E, and then go to the unlabeled one, D. And we know this because of angle-angle side and they correspond to each other."}, {"video_title": "Congruent triangle proof example Congruence Geometry Khan Academy.mp3", "Sentence": "So now we know that triangle, we have to make sure that we get the letters right here, that we have the right corresponding vertices. We can say that triangle AEB, triangle AE, actually let me start with the angle just to make it interesting. Angle BEA, so we're starting with the magenta angle going to the green angle and then going to the one that we haven't labeled. So angle BEA we can say is congruent to angle, we start with the magenta vertices, C, go to the center, E, and then go to the unlabeled one, D. And we know this because of angle-angle side and they correspond to each other. Magenta-green side, magenta-green side, they're all congruent. So this is from AAS. And then if we know that they are congruent, that means corresponding sides are congruent."}, {"video_title": "Congruent triangle proof example Congruence Geometry Khan Academy.mp3", "Sentence": "So angle BEA we can say is congruent to angle, we start with the magenta vertices, C, go to the center, E, and then go to the unlabeled one, D. And we know this because of angle-angle side and they correspond to each other. Magenta-green side, magenta-green side, they're all congruent. So this is from AAS. And then if we know that they are congruent, that means corresponding sides are congruent. So then we know that this side, so we know these two triangles are congruent, so that means that the corresponding sides are congruent. So then we know that length of BE, that we know that BE, the length of that segment BE is going to be equal. And that's the segment that's between the magenta and the green angles."}, {"video_title": "Congruent triangle proof example Congruence Geometry Khan Academy.mp3", "Sentence": "And then if we know that they are congruent, that means corresponding sides are congruent. So then we know that this side, so we know these two triangles are congruent, so that means that the corresponding sides are congruent. So then we know that length of BE, that we know that BE, the length of that segment BE is going to be equal. And that's the segment that's between the magenta and the green angles. The corresponding side is side CE, between the magenta and the green angles, is equal to CE. And this just comes out of the previous statement. If we number them, that's 1, that's 2, and that's 3."}, {"video_title": "Congruent triangle proof example Congruence Geometry Khan Academy.mp3", "Sentence": "And that's the segment that's between the magenta and the green angles. The corresponding side is side CE, between the magenta and the green angles, is equal to CE. And this just comes out of the previous statement. If we number them, that's 1, that's 2, and that's 3. And so that comes out of statement 3. And so we have proven this. E is the midpoint of BC."}, {"video_title": "Congruent triangle proof example Congruence Geometry Khan Academy.mp3", "Sentence": "If we number them, that's 1, that's 2, and that's 3. And so that comes out of statement 3. And so we have proven this. E is the midpoint of BC. It comes straight out of the fact that BE is equal to CE. So I can mark this off with a hash. This line segment right over here is congruent to this line segment right over here because we know that those two triangles are congruent."}, {"video_title": "Congruent triangle proof example Congruence Geometry Khan Academy.mp3", "Sentence": "E is the midpoint of BC. It comes straight out of the fact that BE is equal to CE. So I can mark this off with a hash. This line segment right over here is congruent to this line segment right over here because we know that those two triangles are congruent. And I've inadvertently right here done a little two-column proof. This over here on the left-hand side is my statement. And then on the right-hand side, I gave my reason."}, {"video_title": "Applying volume of solids Solid geometry High school geometry Khan Academy.mp3", "Sentence": "It's something that would store grain and then it can kind of fall out of the bottom, has a radius of 10 meters at the top and is eight meters tall. So let's draw that. So it's cone-shaped and it has a radius at the top. So the top must be where the base is. I guess one way to think about it, it must be the wider part of the cone. So it looks like this, something like that. That's what this first sentence tells us."}, {"video_title": "Applying volume of solids Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So the top must be where the base is. I guess one way to think about it, it must be the wider part of the cone. So it looks like this, something like that. That's what this first sentence tells us. It has a radius of 10 meters. So this distance right over here is 10 meters. And the height is eight meters."}, {"video_title": "Applying volume of solids Solid geometry High school geometry Khan Academy.mp3", "Sentence": "That's what this first sentence tells us. It has a radius of 10 meters. So this distance right over here is 10 meters. And the height is eight meters. They say it's eight meters tall. So this right over here is eight meters. Then they tell us it is filled up to two meters from the top with grain."}, {"video_title": "Applying volume of solids Solid geometry High school geometry Khan Academy.mp3", "Sentence": "And the height is eight meters. They say it's eight meters tall. So this right over here is eight meters. Then they tell us it is filled up to two meters from the top with grain. So one way to think about it, it's filled about this high with grain. So it's filled about that high with grain. So this distance is going to be eight minus these two."}, {"video_title": "Applying volume of solids Solid geometry High school geometry Khan Academy.mp3", "Sentence": "Then they tell us it is filled up to two meters from the top with grain. So one way to think about it, it's filled about this high with grain. So it's filled about that high with grain. So this distance is going to be eight minus these two. So this is going to be six meters high. That's what that second sentence tells us. The hopper will pour the grain into boxes with dimensions of 0.5 meters by 0.5 meters by 0.4 meters."}, {"video_title": "Applying volume of solids Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So this distance is going to be eight minus these two. So this is going to be six meters high. That's what that second sentence tells us. The hopper will pour the grain into boxes with dimensions of 0.5 meters by 0.5 meters by 0.4 meters. The hopper pours grain at a rate of eight cubic meters per minute. So the first, a lot of information there. The first question is what is the volume of grain in the hopper?"}, {"video_title": "Applying volume of solids Solid geometry High school geometry Khan Academy.mp3", "Sentence": "The hopper will pour the grain into boxes with dimensions of 0.5 meters by 0.5 meters by 0.4 meters. The hopper pours grain at a rate of eight cubic meters per minute. So the first, a lot of information there. The first question is what is the volume of grain in the hopper? So before we even get to these other questions, let's see if we can answer that. So that's going to be this volume right over here of the red part, the cone made up of the grain. Pause this video and try to figure it out."}, {"video_title": "Applying volume of solids Solid geometry High school geometry Khan Academy.mp3", "Sentence": "The first question is what is the volume of grain in the hopper? So before we even get to these other questions, let's see if we can answer that. So that's going to be this volume right over here of the red part, the cone made up of the grain. Pause this video and try to figure it out. Well, from previous videos, we know that volume of a cone is going to be 1 3rd times the area of the base times the height. Now we know the height is six meters, but what we need to do is figure out the area of the base. Well, how do we do that?"}, {"video_title": "Applying volume of solids Solid geometry High school geometry Khan Academy.mp3", "Sentence": "Pause this video and try to figure it out. Well, from previous videos, we know that volume of a cone is going to be 1 3rd times the area of the base times the height. Now we know the height is six meters, but what we need to do is figure out the area of the base. Well, how do we do that? Well, we'd have to figure out the radius of the base. Let's call that R right over here. And how do we figure that out?"}, {"video_title": "Applying volume of solids Solid geometry High school geometry Khan Academy.mp3", "Sentence": "Well, how do we do that? Well, we'd have to figure out the radius of the base. Let's call that R right over here. And how do we figure that out? Well, we can look at these two triangles that you can see on my screen and realize that they are similar triangles. This line is parallel to that line. This is a right angle."}, {"video_title": "Applying volume of solids Solid geometry High school geometry Khan Academy.mp3", "Sentence": "And how do we figure that out? Well, we can look at these two triangles that you can see on my screen and realize that they are similar triangles. This line is parallel to that line. This is a right angle. This is a right angle because both of these cuts of these surfaces are going to be parallel to the ground. And then this angle is going to be congruent to this angle because you could view this line as a transversal between parallel lines and these are corresponding angles. And then both triangles share this."}, {"video_title": "Applying volume of solids Solid geometry High school geometry Khan Academy.mp3", "Sentence": "This is a right angle. This is a right angle because both of these cuts of these surfaces are going to be parallel to the ground. And then this angle is going to be congruent to this angle because you could view this line as a transversal between parallel lines and these are corresponding angles. And then both triangles share this. So you have angle, angle, angle. These are similar triangles. And so we can set up a proportion here."}, {"video_title": "Applying volume of solids Solid geometry High school geometry Khan Academy.mp3", "Sentence": "And then both triangles share this. So you have angle, angle, angle. These are similar triangles. And so we can set up a proportion here. We can say the ratio between R and 10 meters, the ratio of R to 10 is equal to the ratio of six to eight, is equal to the ratio of six to eight. And then we could try to solve for R. R is going to be equal to, R is equal to multiply both sides by 10, multiply both sides by 10, and you're going to get 60 over eight. 60 over eight, eight goes into 60 seven times with four left over."}, {"video_title": "Applying volume of solids Solid geometry High school geometry Khan Academy.mp3", "Sentence": "And so we can set up a proportion here. We can say the ratio between R and 10 meters, the ratio of R to 10 is equal to the ratio of six to eight, is equal to the ratio of six to eight. And then we could try to solve for R. R is going to be equal to, R is equal to multiply both sides by 10, multiply both sides by 10, and you're going to get 60 over eight. 60 over eight, eight goes into 60 seven times with four left over. So it's seven and four eighths, or it's also 7.5. And so if you wanna know the area of the base right over here, if you wanted to know this B, it would be pi times the radius squared. So B in this case is going to be pi times 7.5."}, {"video_title": "Applying volume of solids Solid geometry High school geometry Khan Academy.mp3", "Sentence": "60 over eight, eight goes into 60 seven times with four left over. So it's seven and four eighths, or it's also 7.5. And so if you wanna know the area of the base right over here, if you wanted to know this B, it would be pi times the radius squared. So B in this case is going to be pi times 7.5. We're dealing with meters squared. And so the volume, to answer the first question, the volume is going to be one third times the area of the base, this area up here, which is pi times 7.5 meters squared times the height. So times six meters."}, {"video_title": "Applying volume of solids Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So B in this case is going to be pi times 7.5. We're dealing with meters squared. And so the volume, to answer the first question, the volume is going to be one third times the area of the base, this area up here, which is pi times 7.5 meters squared times the height. So times six meters. And let's see, we could simplify this a little bit. Six divided by three, or six times one third, is just going to be equal to two. And so let me get my calculator."}, {"video_title": "Applying volume of solids Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So times six meters. And let's see, we could simplify this a little bit. Six divided by three, or six times one third, is just going to be equal to two. And so let me get my calculator. They say round to the nearest tenth of a cubic meter. So we have 7.5 squared times two times pi is equal to, if we round to the nearest tenth, it's gonna be 353.4 cubic meters. So the volume is approximately 353.4 cubic meters."}, {"video_title": "Applying volume of solids Solid geometry High school geometry Khan Academy.mp3", "Sentence": "And so let me get my calculator. They say round to the nearest tenth of a cubic meter. So we have 7.5 squared times two times pi is equal to, if we round to the nearest tenth, it's gonna be 353.4 cubic meters. So the volume is approximately 353.4 cubic meters. So that's the answer to the first part right over there. And then they say how many complete boxes will the grain fill? Well, they talk about the boxes right over here."}, {"video_title": "Applying volume of solids Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So the volume is approximately 353.4 cubic meters. So that's the answer to the first part right over there. And then they say how many complete boxes will the grain fill? Well, they talk about the boxes right over here. The hopper will pour the grain into boxes with dimensions of 0.5 meters by 0.5 meters by 0.4 meters. So we can imagine these boxes. They look like this."}, {"video_title": "Applying volume of solids Solid geometry High school geometry Khan Academy.mp3", "Sentence": "Well, they talk about the boxes right over here. The hopper will pour the grain into boxes with dimensions of 0.5 meters by 0.5 meters by 0.4 meters. So we can imagine these boxes. They look like this. And they are 0.5 meters by 0.5 meters by 0.4 meters. So the volume of each box is just going to be the product of these three numbers. So the volume of each box is going to be the width times the depth times the height."}, {"video_title": "Applying volume of solids Solid geometry High school geometry Khan Academy.mp3", "Sentence": "They look like this. And they are 0.5 meters by 0.5 meters by 0.4 meters. So the volume of each box is just going to be the product of these three numbers. So the volume of each box is going to be the width times the depth times the height. So 0.5 meters times 0.5 meters times 0.4 meters. And we should be able to do this in our head because five times five is 25. 25 times four is equal to 100."}, {"video_title": "Applying volume of solids Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So the volume of each box is going to be the width times the depth times the height. So 0.5 meters times 0.5 meters times 0.4 meters. And we should be able to do this in our head because five times five is 25. 25 times four is equal to 100. But then we have to think we have one, two, three digits to the right of the decimal point. So one, two, three. So this is going to be 1 0.100 cubic meters."}, {"video_title": "Applying volume of solids Solid geometry High school geometry Khan Academy.mp3", "Sentence": "25 times four is equal to 100. But then we have to think we have one, two, three digits to the right of the decimal point. So one, two, three. So this is going to be 1 0.100 cubic meters. So a tenth of a cubic meter. So how many tenths of cubic meters can I fill up with this much grain? Well, it's just going to be this number divided by a tenth."}, {"video_title": "Applying volume of solids Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So this is going to be 1 0.100 cubic meters. So a tenth of a cubic meter. So how many tenths of cubic meters can I fill up with this much grain? Well, it's just going to be this number divided by a tenth. Well, if you divide by a tenth, that's the same thing as multiplying by 10. And so if you multiply this by 10, you're going to get 3,534 boxes. Now, once again, let's just appreciate."}, {"video_title": "Applying volume of solids Solid geometry High school geometry Khan Academy.mp3", "Sentence": "Well, it's just going to be this number divided by a tenth. Well, if you divide by a tenth, that's the same thing as multiplying by 10. And so if you multiply this by 10, you're going to get 3,534 boxes. Now, once again, let's just appreciate. Every cubic meter, you can fill 10 of these boxes. And this is how many cubic meters we have. So if you multiply this by 10, it tells you how many boxes you fill up."}, {"video_title": "Applying volume of solids Solid geometry High school geometry Khan Academy.mp3", "Sentence": "Now, once again, let's just appreciate. Every cubic meter, you can fill 10 of these boxes. And this is how many cubic meters we have. So if you multiply this by 10, it tells you how many boxes you fill up. And one way to think about it, we've seen this in other videos, we're shifting the decimal one place over to the right to get this many boxes. And it's important to realize complete boxes because when we got to 353.4, we did round down. So we do have that amount, but we're not going to fill up another box with whatever this rounding error that we rounded down from."}, {"video_title": "Applying volume of solids Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So if you multiply this by 10, it tells you how many boxes you fill up. And one way to think about it, we've seen this in other videos, we're shifting the decimal one place over to the right to get this many boxes. And it's important to realize complete boxes because when we got to 353.4, we did round down. So we do have that amount, but we're not going to fill up another box with whatever this rounding error that we rounded down from. So the last question is, to the nearest minute, how long does it take to fill the boxes? Well, this is the total volume, and we're going to fill eight cubic meters per minute. So the answer over here is going to be our total volume, it's going to be 353.4 cubic meters."}, {"video_title": "Applying volume of solids Solid geometry High school geometry Khan Academy.mp3", "Sentence": "So we do have that amount, but we're not going to fill up another box with whatever this rounding error that we rounded down from. So the last question is, to the nearest minute, how long does it take to fill the boxes? Well, this is the total volume, and we're going to fill eight cubic meters per minute. So the answer over here is going to be our total volume, it's going to be 353.4 cubic meters. And we're going to divide that by our rate, eight cubic meters per minute. And that is going to give us 353.4 divided by eight is equal to, and if we want to round to the nearest minute, 44 minutes is equal to approximately 44 minutes to fill all the boxes. And we're done."}, {"video_title": "Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "You can't be just perpendicular by yourself. And perpendicular lines, just so you have a visualization for what perpendicular lines look like. Two lines are perpendicular if they intersect at right angles. So if this is one line right there, a perpendicular line will look like this. A perpendicular line will intersect it, but it won't just be any intersection. It will intersect at right angles. It will intersect at right angles."}, {"video_title": "Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So if this is one line right there, a perpendicular line will look like this. A perpendicular line will intersect it, but it won't just be any intersection. It will intersect at right angles. It will intersect at right angles. So these two lines are perpendicular. Now, if two lines are perpendicular, if the slope of this orange line is m, so let's say its equation is y is equal to mx plus, let's say it's b1, so it's some y-intercept, then the equation of this yellow line, its slope is going to be the negative inverse of this guy. This guy right here is going to be y is equal to negative 1 over mx plus some other y-intercept."}, {"video_title": "Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "It will intersect at right angles. So these two lines are perpendicular. Now, if two lines are perpendicular, if the slope of this orange line is m, so let's say its equation is y is equal to mx plus, let's say it's b1, so it's some y-intercept, then the equation of this yellow line, its slope is going to be the negative inverse of this guy. This guy right here is going to be y is equal to negative 1 over mx plus some other y-intercept. Or another way to think about it is if two lines are perpendicular, the product of their slopes is going to be negative 1. And so you could write that there. m times negative 1 over m. That's going to be, these two guys are going to cancel out, that's going to be equal to negative 1."}, {"video_title": "Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "This guy right here is going to be y is equal to negative 1 over mx plus some other y-intercept. Or another way to think about it is if two lines are perpendicular, the product of their slopes is going to be negative 1. And so you could write that there. m times negative 1 over m. That's going to be, these two guys are going to cancel out, that's going to be equal to negative 1. So let's figure out the slopes of each of these lines and figure out if any of them are the negative inverse of any of the other ones. So line A, the slope is pretty easy to figure out. It's already in slope-intercept form."}, {"video_title": "Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "m times negative 1 over m. That's going to be, these two guys are going to cancel out, that's going to be equal to negative 1. So let's figure out the slopes of each of these lines and figure out if any of them are the negative inverse of any of the other ones. So line A, the slope is pretty easy to figure out. It's already in slope-intercept form. Its slope is 3. So line A has a slope of 3. Line B, it's in standard form, not too hard to put it in slope-intercept form, so let's try to do it."}, {"video_title": "Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "It's already in slope-intercept form. Its slope is 3. So line A has a slope of 3. Line B, it's in standard form, not too hard to put it in slope-intercept form, so let's try to do it. So let's do line B over here. Line B, we have x plus 3y is equal to negative 21. Let's subtract x from both sides so that it ends up on the right-hand side."}, {"video_title": "Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "Line B, it's in standard form, not too hard to put it in slope-intercept form, so let's try to do it. So let's do line B over here. Line B, we have x plus 3y is equal to negative 21. Let's subtract x from both sides so that it ends up on the right-hand side. So this, we end up with 3y is equal to negative x minus 21. And now let's divide both sides of this equation by 3. And we get y is equal to negative 1 third x minus 7."}, {"video_title": "Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "Let's subtract x from both sides so that it ends up on the right-hand side. So this, we end up with 3y is equal to negative x minus 21. And now let's divide both sides of this equation by 3. And we get y is equal to negative 1 third x minus 7. So this character's slope is negative 1 third. So here, m is equal to negative 1 third. So we already see they are the negative inverse of each other."}, {"video_title": "Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "And we get y is equal to negative 1 third x minus 7. So this character's slope is negative 1 third. So here, m is equal to negative 1 third. So we already see they are the negative inverse of each other. You take the inverse of 3, it's 1 third, and then it's the negative of that. Or you take the inverse of negative 1 third, it's negative 3, and then this is the negative of that. So these two lines are definitely perpendicular."}, {"video_title": "Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So we already see they are the negative inverse of each other. You take the inverse of 3, it's 1 third, and then it's the negative of that. Or you take the inverse of negative 1 third, it's negative 3, and then this is the negative of that. So these two lines are definitely perpendicular. Let's see this third line over here. So line C is 3x plus y is equal to 10. If we subtract 3x from both sides, we get y is equal to negative 3x plus 10."}, {"video_title": "Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "So these two lines are definitely perpendicular. Let's see this third line over here. So line C is 3x plus y is equal to 10. If we subtract 3x from both sides, we get y is equal to negative 3x plus 10. So our slope in this case is negative 3. So our slope here is equal to negative 3. Now this guy is the negative of that guy."}, {"video_title": "Perpendicular lines from equation Mathematics I High School Math Khan Academy.mp3", "Sentence": "If we subtract 3x from both sides, we get y is equal to negative 3x plus 10. So our slope in this case is negative 3. So our slope here is equal to negative 3. Now this guy is the negative of that guy. This guy's slope is the negative of that, but not the negative inverse. So it's not perpendicular. And this guy is the inverse of that guy, but not the negative inverse."}, {"video_title": "Examples recognizing transformations.mp3", "Sentence": "And the transformations we're going to look at are things like rotations, where you are spinning something around a point. We're gonna look at translations, where you're shifting all the points of a figure. We're gonna look at reflection, where you flip a figure over some type of a line. And we'll look at dilations, where you're essentially going to either shrink or expand some type of a figure. So with that out of the way, let's think about this question. What single transformation was applied to triangle A to get triangle B? So it looks like triangle A and triangle B, they're the same size."}, {"video_title": "Examples recognizing transformations.mp3", "Sentence": "And we'll look at dilations, where you're essentially going to either shrink or expand some type of a figure. So with that out of the way, let's think about this question. What single transformation was applied to triangle A to get triangle B? So it looks like triangle A and triangle B, they're the same size. And what's really happened is that every one of these points has been shifted, or another way I could say it, they have all been translated a little bit to the right and up. And so, right like this, they have all been translated. So this right over here is clearly a translation."}, {"video_title": "Examples recognizing transformations.mp3", "Sentence": "So it looks like triangle A and triangle B, they're the same size. And what's really happened is that every one of these points has been shifted, or another way I could say it, they have all been translated a little bit to the right and up. And so, right like this, they have all been translated. So this right over here is clearly a translation. Let's do another example. What single transformation was applied to get, what is applied to triangle A to get to triangle B? So if I look at this, these diagrams, this point seems to correspond with that one."}, {"video_title": "Examples recognizing transformations.mp3", "Sentence": "So this right over here is clearly a translation. Let's do another example. What single transformation was applied to get, what is applied to triangle A to get to triangle B? So if I look at this, these diagrams, this point seems to correspond with that one. This one corresponds with that one. So it doesn't look like straight translation, because they would have been translated in different ways. So it's definitely not a straight translation."}, {"video_title": "Examples recognizing transformations.mp3", "Sentence": "So if I look at this, these diagrams, this point seems to correspond with that one. This one corresponds with that one. So it doesn't look like straight translation, because they would have been translated in different ways. So it's definitely not a straight translation. Let's think about it. It looks like there might be a rotation here. So maybe it looks like that point went over there, that point went over there, this point went over here."}, {"video_title": "Examples recognizing transformations.mp3", "Sentence": "So it's definitely not a straight translation. Let's think about it. It looks like there might be a rotation here. So maybe it looks like that point went over there, that point went over there, this point went over here. And so we could be rotating around some point right about here. And if you rotate around that point, you could get to a situation that looks like triangle B. And I don't know the exact point that we're rotating around, but this looks pretty clear, like a rotation."}, {"video_title": "Examples recognizing transformations.mp3", "Sentence": "So maybe it looks like that point went over there, that point went over there, this point went over here. And so we could be rotating around some point right about here. And if you rotate around that point, you could get to a situation that looks like triangle B. And I don't know the exact point that we're rotating around, but this looks pretty clear, like a rotation. Let's do another example. What single transformation was applied to quadrilateral A to get to quadrilateral B? So let's see, it looks like this point corresponds to that point."}, {"video_title": "Examples recognizing transformations.mp3", "Sentence": "And I don't know the exact point that we're rotating around, but this looks pretty clear, like a rotation. Let's do another example. What single transformation was applied to quadrilateral A to get to quadrilateral B? So let's see, it looks like this point corresponds to that point. And so, and then this point corresponds to that point, and that point corresponds to that point. So they actually look like reflections of each other. So we're to imagine some type of a mirror right over here."}, {"video_title": "Examples recognizing transformations.mp3", "Sentence": "So let's see, it looks like this point corresponds to that point. And so, and then this point corresponds to that point, and that point corresponds to that point. So they actually look like reflections of each other. So we're to imagine some type of a mirror right over here. They're actually mirror images. This got flipped over the line, that got flipped over the line, and that got flipped over the line. So it's pretty clear that this right over here is a reflection."}, {"video_title": "Examples recognizing transformations.mp3", "Sentence": "So we're to imagine some type of a mirror right over here. They're actually mirror images. This got flipped over the line, that got flipped over the line, and that got flipped over the line. So it's pretty clear that this right over here is a reflection. All right, let's do one more of these. What single transformation was applied to quadrilateral A to get to quadrilateral B? All right."}, {"video_title": "Examples recognizing transformations.mp3", "Sentence": "So it's pretty clear that this right over here is a reflection. All right, let's do one more of these. What single transformation was applied to quadrilateral A to get to quadrilateral B? All right. So this looks like, so quadrilateral B is clearly bigger. So this is a non-rigid transformation. The distance between corresponding points looks like it has increased."}, {"video_title": "Examples recognizing transformations.mp3", "Sentence": "All right. So this looks like, so quadrilateral B is clearly bigger. So this is a non-rigid transformation. The distance between corresponding points looks like it has increased. Now you might be saying, well, wouldn't that be, it looks like if you're making something bigger or smaller, that looks like a dilation. But it looks like this has been moved as well, as it has been translated. And the key here to realize is around, what is your center of dilation?"}, {"video_title": "Examples recognizing transformations.mp3", "Sentence": "The distance between corresponding points looks like it has increased. Now you might be saying, well, wouldn't that be, it looks like if you're making something bigger or smaller, that looks like a dilation. But it looks like this has been moved as well, as it has been translated. And the key here to realize is around, what is your center of dilation? So for example, if your center of dilation is, let's say right over here, then all of these things are going to be stretched that way. And so this point might go to there, that point might go over there, this point might go, this point might go over here, and then that point might go over here. So this is definitely a dilation, where you are, your center, where everything is expanding from, is just outside of our trapezoid A."}, {"video_title": "Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3", "Sentence": "And the reason why I wrote angle side angle here and angle angle side is to realize that these are equivalent, because if you have two angles, then you know what the third angle is going to be. So for example, in this case right over here, if we know that we have two pairs of angles that have the same measure, then that means that the third pair must have the same measure as well. So we'll know this as well. So if you really think about it, if you have the side between the two angles, that's equivalent to having an angle, an angle, and a side, because as long as you have two angles, the third angle is also going to have the same measure as the corresponding third angle on the other triangle. So let's just show a series of rigid transformations that can get us from ABC to DEF. So the first step you might imagine, we've already shown that if you have two segments of equal length, that they are congruent, you can have a series of rigid transformations that maps one onto the other. So what I want to do is map segment AC onto DF."}, {"video_title": "Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3", "Sentence": "So if you really think about it, if you have the side between the two angles, that's equivalent to having an angle, an angle, and a side, because as long as you have two angles, the third angle is also going to have the same measure as the corresponding third angle on the other triangle. So let's just show a series of rigid transformations that can get us from ABC to DEF. So the first step you might imagine, we've already shown that if you have two segments of equal length, that they are congruent, you can have a series of rigid transformations that maps one onto the other. So what I want to do is map segment AC onto DF. And the way that I could do that is I could translate point A to be on top of point D. So then I'll call this A prime. And then when I do that, this segment AC is going to look something like this. I'm just sketching it right now."}, {"video_title": "Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3", "Sentence": "So what I want to do is map segment AC onto DF. And the way that I could do that is I could translate point A to be on top of point D. So then I'll call this A prime. And then when I do that, this segment AC is going to look something like this. I'm just sketching it right now. It's going to be in that direction. But then, and the whole, the rest of the triangle is going to come with it. So let's see, the rest of that orange side, side AB is going to look something like that."}, {"video_title": "Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3", "Sentence": "I'm just sketching it right now. It's going to be in that direction. But then, and the whole, the rest of the triangle is going to come with it. So let's see, the rest of that orange side, side AB is going to look something like that. But then we could do another rigid transformation, which is rotate about point D or point A prime. They're the same point now. So that point C coincides with point F. And so just like that, you would have two rigid transformations that get us, that map AC onto DF."}, {"video_title": "Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3", "Sentence": "So let's see, the rest of that orange side, side AB is going to look something like that. But then we could do another rigid transformation, which is rotate about point D or point A prime. They're the same point now. So that point C coincides with point F. And so just like that, you would have two rigid transformations that get us, that map AC onto DF. And so A prime, where A is mapped, is now equal to D, and F is now equal to C prime. But the question is, where does point B now sit? And the realization here is that angle measures are preserved."}, {"video_title": "Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3", "Sentence": "So that point C coincides with point F. And so just like that, you would have two rigid transformations that get us, that map AC onto DF. And so A prime, where A is mapped, is now equal to D, and F is now equal to C prime. But the question is, where does point B now sit? And the realization here is that angle measures are preserved. And since angle measures are preserved, we are either going to have a situation where this angle, let's see, this angle is angle CAB gets preserved. So then it would be C prime, A prime, and then B prime would have to sit someplace on this ray for if we're gonna preserve the measure of angle CAB, B prime is going to sit someplace along that ray because an angle is defined by two rays that intersect at the vertex or start at the vertex. And because this angle is preserved, that's the angle that is formed by these two rays."}, {"video_title": "Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3", "Sentence": "And the realization here is that angle measures are preserved. And since angle measures are preserved, we are either going to have a situation where this angle, let's see, this angle is angle CAB gets preserved. So then it would be C prime, A prime, and then B prime would have to sit someplace on this ray for if we're gonna preserve the measure of angle CAB, B prime is going to sit someplace along that ray because an angle is defined by two rays that intersect at the vertex or start at the vertex. And because this angle is preserved, that's the angle that is formed by these two rays. You could say ray CA and ray CB. We know that B prime also has to sit someplace on this ray as well. So B prime also has to sit someplace on this ray."}, {"video_title": "Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3", "Sentence": "And because this angle is preserved, that's the angle that is formed by these two rays. You could say ray CA and ray CB. We know that B prime also has to sit someplace on this ray as well. So B prime also has to sit someplace on this ray. And I think you see where this is going. If B prime, because these two angles are preserved, because this angle and this angle are preserved, have to sit someplace on both of these rays, they intersect at one point, this point right over here that coincides with point E. So this is where B prime would be. So that's one scenario in which case we've shown that you can get a series of rigid transformations from this triangle to this triangle."}, {"video_title": "Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3", "Sentence": "So B prime also has to sit someplace on this ray. And I think you see where this is going. If B prime, because these two angles are preserved, because this angle and this angle are preserved, have to sit someplace on both of these rays, they intersect at one point, this point right over here that coincides with point E. So this is where B prime would be. So that's one scenario in which case we've shown that you can get a series of rigid transformations from this triangle to this triangle. But there's another one. There is a circumstance where the angles get preserved, but instead of being on, instead of the angles being on the, I guess you could say the bottom right side of this blue line, you could imagine the angles get preserved such that they are on the other side. So the angles get preserved so that they are on the other side of that blue line."}, {"video_title": "Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3", "Sentence": "So that's one scenario in which case we've shown that you can get a series of rigid transformations from this triangle to this triangle. But there's another one. There is a circumstance where the angles get preserved, but instead of being on, instead of the angles being on the, I guess you could say the bottom right side of this blue line, you could imagine the angles get preserved such that they are on the other side. So the angles get preserved so that they are on the other side of that blue line. And then the question is, in that situation, where would B prime end up? Well, actually, let me draw this a little bit, let me do this a little bit more exact. Let me replicate these angles."}, {"video_title": "Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3", "Sentence": "So the angles get preserved so that they are on the other side of that blue line. And then the question is, in that situation, where would B prime end up? Well, actually, let me draw this a little bit, let me do this a little bit more exact. Let me replicate these angles. So I'm going to draw an arc like this, an arc like this, and then I'll measure this distance. It's just like this. We've done this in other videos when we're trying to replicate angles."}, {"video_title": "Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3", "Sentence": "Let me replicate these angles. So I'm going to draw an arc like this, an arc like this, and then I'll measure this distance. It's just like this. We've done this in other videos when we're trying to replicate angles. So it's like that far. And so let me draw that on this point right over here, this far. So if the angles are on that side of line, I guess we could say DF or A prime, C prime, we know that B prime would have to sit someplace on this ray."}, {"video_title": "Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3", "Sentence": "We've done this in other videos when we're trying to replicate angles. So it's like that far. And so let me draw that on this point right over here, this far. So if the angles are on that side of line, I guess we could say DF or A prime, C prime, we know that B prime would have to sit someplace on this ray. So let me draw that as neatly as I can. Someplace on this ray. And it would have to sit someplace on the ray formed by the other angle."}, {"video_title": "Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3", "Sentence": "So if the angles are on that side of line, I guess we could say DF or A prime, C prime, we know that B prime would have to sit someplace on this ray. So let me draw that as neatly as I can. Someplace on this ray. And it would have to sit someplace on the ray formed by the other angle. So let me see if I can draw that as neatly as possible. So let me make a, make a arc like this. I probably did that a little bit bigger than I need to, but hopefully it serves our purposes."}, {"video_title": "Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3", "Sentence": "And it would have to sit someplace on the ray formed by the other angle. So let me see if I can draw that as neatly as possible. So let me make a, make a arc like this. I probably did that a little bit bigger than I need to, but hopefully it serves our purposes. I measured this distance right over here. If I measure that distance over here, it would get us right over there. So B prime either sits on this ray, or it could sit, or and it has to sit, I should really say, on this ray."}, {"video_title": "Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3", "Sentence": "I probably did that a little bit bigger than I need to, but hopefully it serves our purposes. I measured this distance right over here. If I measure that distance over here, it would get us right over there. So B prime either sits on this ray, or it could sit, or and it has to sit, I should really say, on this ray. It goes through this point and this point. And it has to sit on this ray. And you can see where these two rays intersect is right over there."}, {"video_title": "Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3", "Sentence": "So B prime either sits on this ray, or it could sit, or and it has to sit, I should really say, on this ray. It goes through this point and this point. And it has to sit on this ray. And you can see where these two rays intersect is right over there. So the other scenario is if the angles get preserved in a way that they're on the other side of that blue line, well then B prime is there. And then we could just add one more rigid transformation to our series of rigid transformations, which is essentially, or is, a reflection across line DF, or A prime, C prime. Why will that work to map B prime onto E?"}, {"video_title": "Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3", "Sentence": "And you can see where these two rays intersect is right over there. So the other scenario is if the angles get preserved in a way that they're on the other side of that blue line, well then B prime is there. And then we could just add one more rigid transformation to our series of rigid transformations, which is essentially, or is, a reflection across line DF, or A prime, C prime. Why will that work to map B prime onto E? Well, because reflection is also a rigid transformation, so angles are preserved. And so as this angle gets flipped over, it's preserved. As this angle gets flipped over, the measure of it, I should say, is preserved."}, {"video_title": "Proving the ASA and AAS triangle congruence criteria using transformations Geometry Khan Academy.mp3", "Sentence": "Why will that work to map B prime onto E? Well, because reflection is also a rigid transformation, so angles are preserved. And so as this angle gets flipped over, it's preserved. As this angle gets flipped over, the measure of it, I should say, is preserved. And so that means we'll go to that first case where then these rays would be flipped onto these rays, and B prime would have to sit on that intersection. And there you have it. If you have two angles, and if you have two angles, you're gonna know the third."}, {"video_title": "Parallel & perpendicular lines from graph.mp3", "Sentence": "So you have parallel, you have perpendicular, perpendicular, and of course you have lines that are neither parallel nor perpendicular. And just as a bit of a review, if you've never seen this before, parallel lines, they never intersect. So let me draw some axes. So if those are my coordinate axes right there, that's my x-axis, that is my y-axis. If this is a line that I'm drawing in magenta, a parallel line might look something like this. It's not the exact same line, but they have the exact same slope. If this moves a certain amount, if this change in y over change in x is a certain amount, this change in y over change in x is the same amount, and that's why they never intersect."}, {"video_title": "Parallel & perpendicular lines from graph.mp3", "Sentence": "So if those are my coordinate axes right there, that's my x-axis, that is my y-axis. If this is a line that I'm drawing in magenta, a parallel line might look something like this. It's not the exact same line, but they have the exact same slope. If this moves a certain amount, if this change in y over change in x is a certain amount, this change in y over change in x is the same amount, and that's why they never intersect. So they have the same slope. Parallel lines have the same slope. Perpendicular lines, depending on how you want to view it, they're kind of the opposite."}, {"video_title": "Parallel & perpendicular lines from graph.mp3", "Sentence": "If this moves a certain amount, if this change in y over change in x is a certain amount, this change in y over change in x is the same amount, and that's why they never intersect. So they have the same slope. Parallel lines have the same slope. Perpendicular lines, depending on how you want to view it, they're kind of the opposite. Let's say that this is some line. A line that is perpendicular to that will not only intersect the line, it will intersect it at a right angle, at a 90 degree angle. I'm not going to prove it for you here, I actually prove it in the linear algebra playlist."}, {"video_title": "Parallel & perpendicular lines from graph.mp3", "Sentence": "Perpendicular lines, depending on how you want to view it, they're kind of the opposite. Let's say that this is some line. A line that is perpendicular to that will not only intersect the line, it will intersect it at a right angle, at a 90 degree angle. I'm not going to prove it for you here, I actually prove it in the linear algebra playlist. But a perpendicular line's slope, so let's say that this one right here, let's say that yellow line has a slope of m, then this orange line that's perpendicular to the yellow line is going to have a slope of negative 1 over m. Their slopes are going to be the negative inverse of each other. Now, given this information, let's look at a bunch of lines and figure out if they're parallel, if they're perpendicular, or if they are neither. To do that, we just have to keep looking at the slopes."}, {"video_title": "Parallel & perpendicular lines from graph.mp3", "Sentence": "I'm not going to prove it for you here, I actually prove it in the linear algebra playlist. But a perpendicular line's slope, so let's say that this one right here, let's say that yellow line has a slope of m, then this orange line that's perpendicular to the yellow line is going to have a slope of negative 1 over m. Their slopes are going to be the negative inverse of each other. Now, given this information, let's look at a bunch of lines and figure out if they're parallel, if they're perpendicular, or if they are neither. To do that, we just have to keep looking at the slopes. So let's see. They say one line passes through the points 4, negative 3, and negative 8, 0. Another line passes through the points negative 1, negative 1, and negative 2, 6."}, {"video_title": "Parallel & perpendicular lines from graph.mp3", "Sentence": "To do that, we just have to keep looking at the slopes. So let's see. They say one line passes through the points 4, negative 3, and negative 8, 0. Another line passes through the points negative 1, negative 1, and negative 2, 6. Let's figure out the slopes of each of these lines. I'll first do this one in pink. This slope right here, so line 1, I'll call it slope 1."}, {"video_title": "Parallel & perpendicular lines from graph.mp3", "Sentence": "Another line passes through the points negative 1, negative 1, and negative 2, 6. Let's figure out the slopes of each of these lines. I'll first do this one in pink. This slope right here, so line 1, I'll call it slope 1. Slope 1 is, let's just say it is, well, let's take this as the finishing point. Negative 3 minus 0, remember, change in y, over 4 minus negative 8. So this is equal to negative 3 over, this is the same thing as 4 plus 8, negative 3 over 12, which is equal to negative 1 fourth, divide the numerator and denominator by 3."}, {"video_title": "Parallel & perpendicular lines from graph.mp3", "Sentence": "This slope right here, so line 1, I'll call it slope 1. Slope 1 is, let's just say it is, well, let's take this as the finishing point. Negative 3 minus 0, remember, change in y, over 4 minus negative 8. So this is equal to negative 3 over, this is the same thing as 4 plus 8, negative 3 over 12, which is equal to negative 1 fourth, divide the numerator and denominator by 3. That's this line, that's the first line. Now what about the second line? The second line, the slope for that second line is, well, let's take here, negative 1 minus 6 over negative 1 minus negative 2, is equal to negative 1 minus 6 is negative 7, over negative 1 minus negative 2, that's the same thing as negative 1 plus 2."}, {"video_title": "Parallel & perpendicular lines from graph.mp3", "Sentence": "So this is equal to negative 3 over, this is the same thing as 4 plus 8, negative 3 over 12, which is equal to negative 1 fourth, divide the numerator and denominator by 3. That's this line, that's the first line. Now what about the second line? The second line, the slope for that second line is, well, let's take here, negative 1 minus 6 over negative 1 minus negative 2, is equal to negative 1 minus 6 is negative 7, over negative 1 minus negative 2, that's the same thing as negative 1 plus 2. Well, that's just 1. So the slope here is negative 7. So here their slopes are neither equal, so they're not parallel, nor are they the negative inverse of each other."}, {"video_title": "Parallel & perpendicular lines from graph.mp3", "Sentence": "The second line, the slope for that second line is, well, let's take here, negative 1 minus 6 over negative 1 minus negative 2, is equal to negative 1 minus 6 is negative 7, over negative 1 minus negative 2, that's the same thing as negative 1 plus 2. Well, that's just 1. So the slope here is negative 7. So here their slopes are neither equal, so they're not parallel, nor are they the negative inverse of each other. So this is neither. This is neither parallel nor perpendicular. So these two lines, they intersect, but they're not going to intersect at a 90 degree angle."}, {"video_title": "Parallel & perpendicular lines from graph.mp3", "Sentence": "So here their slopes are neither equal, so they're not parallel, nor are they the negative inverse of each other. So this is neither. This is neither parallel nor perpendicular. So these two lines, they intersect, but they're not going to intersect at a 90 degree angle. Let's do a couple more of these. So I have here, once again, one line passing through these points, and then another line passing through these points. So let's just look at their slopes."}, {"video_title": "Parallel & perpendicular lines from graph.mp3", "Sentence": "So these two lines, they intersect, but they're not going to intersect at a 90 degree angle. Let's do a couple more of these. So I have here, once again, one line passing through these points, and then another line passing through these points. So let's just look at their slopes. So this one in green, what's the slope? The slope of the green one, I'll call that the first line. We could say, let's see, change in y, so we could do negative 2 minus 14 over, I did negative 2 first, so I'll do 1 first, over 1 minus negative 3."}, {"video_title": "Parallel & perpendicular lines from graph.mp3", "Sentence": "So let's just look at their slopes. So this one in green, what's the slope? The slope of the green one, I'll call that the first line. We could say, let's see, change in y, so we could do negative 2 minus 14 over, I did negative 2 first, so I'll do 1 first, over 1 minus negative 3. So negative 2 minus 14 is negative 16. 1 minus negative 3, the same thing as 1 plus 3, that's over 4. So this is negative 4."}, {"video_title": "Parallel & perpendicular lines from graph.mp3", "Sentence": "We could say, let's see, change in y, so we could do negative 2 minus 14 over, I did negative 2 first, so I'll do 1 first, over 1 minus negative 3. So negative 2 minus 14 is negative 16. 1 minus negative 3, the same thing as 1 plus 3, that's over 4. So this is negative 4. Now what's the slope of that second line right there? So we have the slope of that second line, let's say 5 minus negative 3, that's our change in y, over negative 2 minus 0. So this is equal to 5 minus negative 3, that's the same thing as 5 plus 3, that's 8."}, {"video_title": "Parallel & perpendicular lines from graph.mp3", "Sentence": "So this is negative 4. Now what's the slope of that second line right there? So we have the slope of that second line, let's say 5 minus negative 3, that's our change in y, over negative 2 minus 0. So this is equal to 5 minus negative 3, that's the same thing as 5 plus 3, that's 8. And then negative 2 minus 0 is negative 2. So this is also equal to negative 4. So these two lines are parallel."}, {"video_title": "Parallel & perpendicular lines from graph.mp3", "Sentence": "So this is equal to 5 minus negative 3, that's the same thing as 5 plus 3, that's 8. And then negative 2 minus 0 is negative 2. So this is also equal to negative 4. So these two lines are parallel. These two lines are parallel. They have the exact same slope. And I encourage you to find the equations of both of these lines and graph both of these lines and verify for yourself that they are indeed parallel."}, {"video_title": "Parallel & perpendicular lines from graph.mp3", "Sentence": "So these two lines are parallel. These two lines are parallel. They have the exact same slope. And I encourage you to find the equations of both of these lines and graph both of these lines and verify for yourself that they are indeed parallel. Let's do this one. Once again, this is just an exercise in finding slopes. So this first line has those points."}, {"video_title": "Parallel & perpendicular lines from graph.mp3", "Sentence": "And I encourage you to find the equations of both of these lines and graph both of these lines and verify for yourself that they are indeed parallel. Let's do this one. Once again, this is just an exercise in finding slopes. So this first line has those points. Let's figure out its slope. The slope of this first line, one line passes through these points. So let's see, 3 minus negative 3, that's our change in y, over 3 minus negative 6."}, {"video_title": "Parallel & perpendicular lines from graph.mp3", "Sentence": "So this first line has those points. Let's figure out its slope. The slope of this first line, one line passes through these points. So let's see, 3 minus negative 3, that's our change in y, over 3 minus negative 6. So this is the same thing as 3 plus 3, which is 6, over 3 plus 6, which is 9. So this first line has a slope of 2 thirds. What is the second line's slope?"}, {"video_title": "Parallel & perpendicular lines from graph.mp3", "Sentence": "So let's see, 3 minus negative 3, that's our change in y, over 3 minus negative 6. So this is the same thing as 3 plus 3, which is 6, over 3 plus 6, which is 9. So this first line has a slope of 2 thirds. What is the second line's slope? This is the second line there. That's the other line passing through these points. So the other line's slope, let's see, we could say negative 8 minus 4, over 2 minus negative 6."}, {"video_title": "Parallel & perpendicular lines from graph.mp3", "Sentence": "What is the second line's slope? This is the second line there. That's the other line passing through these points. So the other line's slope, let's see, we could say negative 8 minus 4, over 2 minus negative 6. So what is this equal to? Negative 8 minus 4 is negative 12. 2 minus negative 6, that's the same thing as 2 plus 6."}, {"video_title": "Parallel & perpendicular lines from graph.mp3", "Sentence": "So the other line's slope, let's see, we could say negative 8 minus 4, over 2 minus negative 6. So what is this equal to? Negative 8 minus 4 is negative 12. 2 minus negative 6, that's the same thing as 2 plus 6. The negatives cancel out. So it's negative 12 over 8, which is the same thing if we divide the numerator and denominator by 4. That's negative 3 halves."}, {"video_title": "Parallel & perpendicular lines from graph.mp3", "Sentence": "2 minus negative 6, that's the same thing as 2 plus 6. The negatives cancel out. So it's negative 12 over 8, which is the same thing if we divide the numerator and denominator by 4. That's negative 3 halves. Notice these guys are the negative inverse of each other. If I take negative 1 over 2 thirds, that is equal to negative 1 times 3 halves, which is equal to negative 3 halves. These guys are the negative inverses of each other."}, {"video_title": "Parallel & perpendicular lines from graph.mp3", "Sentence": "That's negative 3 halves. Notice these guys are the negative inverse of each other. If I take negative 1 over 2 thirds, that is equal to negative 1 times 3 halves, which is equal to negative 3 halves. These guys are the negative inverses of each other. You swap the numerator and denominator, make them negative, and they become equal to each other. So these two lines are perpendicular. I encourage you to find the equations."}, {"video_title": "Volume of a cone Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "Let's think a little bit about the volume of a cone. So a cone would have a circular base, or I guess depends on how you want to draw it. Think of like a conical hat of some kind. It would have a circle as a base, and it would come to some point. So it would look something like that. This you could consider this to be a cone, just like that. Or you can make it upside down if you're thinking of an ice cream cone."}, {"video_title": "Volume of a cone Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "It would have a circle as a base, and it would come to some point. So it would look something like that. This you could consider this to be a cone, just like that. Or you can make it upside down if you're thinking of an ice cream cone. So it might look something like that. That's the top of it. And then it comes down like this."}, {"video_title": "Volume of a cone Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "Or you can make it upside down if you're thinking of an ice cream cone. So it might look something like that. That's the top of it. And then it comes down like this. This also is kind of those disposable cups of water. You might see at some water coolers. And the important things that we need to think about when we want to know about the volume of a cone is we definitely want to know the radius of the base."}, {"video_title": "Volume of a cone Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "And then it comes down like this. This also is kind of those disposable cups of water. You might see at some water coolers. And the important things that we need to think about when we want to know about the volume of a cone is we definitely want to know the radius of the base. So that's the radius of the base. Or here's the radius of the top part. You definitely want to know that radius."}, {"video_title": "Volume of a cone Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "And the important things that we need to think about when we want to know about the volume of a cone is we definitely want to know the radius of the base. So that's the radius of the base. Or here's the radius of the top part. You definitely want to know that radius. And you want to know the height of the cone. So let's call that h. Or right over here, you could call this distance right over here h. And the formula for the volume of a cone, and it's interesting because it's close to the formula for the volume of a cylinder in a very clean way, which is somewhat surprising. And that's what's neat about a lot of this three dimensional geometry is that it's not as messy as you would think it would be."}, {"video_title": "Volume of a cone Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "You definitely want to know that radius. And you want to know the height of the cone. So let's call that h. Or right over here, you could call this distance right over here h. And the formula for the volume of a cone, and it's interesting because it's close to the formula for the volume of a cylinder in a very clean way, which is somewhat surprising. And that's what's neat about a lot of this three dimensional geometry is that it's not as messy as you would think it would be. It is the area of the base. Well, what's the area of the base? Well, that's going to be the area of the base is going to be pi r squared times the height."}, {"video_title": "Volume of a cone Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "And that's what's neat about a lot of this three dimensional geometry is that it's not as messy as you would think it would be. It is the area of the base. Well, what's the area of the base? Well, that's going to be the area of the base is going to be pi r squared times the height. And if you just multiply the height times pi r squared, that would give you the volume of an entire cylinder that looks something like that. So this would give you this entire volume that looks of the figure that looks like this, where its center of the top is the tip right over here. So if I just left it as pi r squared h, or h times pi r squared, it's the volume of this entire can, this entire cylinder."}, {"video_title": "Volume of a cone Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "Well, that's going to be the area of the base is going to be pi r squared times the height. And if you just multiply the height times pi r squared, that would give you the volume of an entire cylinder that looks something like that. So this would give you this entire volume that looks of the figure that looks like this, where its center of the top is the tip right over here. So if I just left it as pi r squared h, or h times pi r squared, it's the volume of this entire can, this entire cylinder. But if you just want the cone, it's 1 third of that. It is 1 third of that. And that's what I mean when I say it's surprisingly clean, that this cone right over here is 1 third the volume of this cylinder that is essentially, you can kind of think of the cylinder as bounding around it."}, {"video_title": "Volume of a cone Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "So if I just left it as pi r squared h, or h times pi r squared, it's the volume of this entire can, this entire cylinder. But if you just want the cone, it's 1 third of that. It is 1 third of that. And that's what I mean when I say it's surprisingly clean, that this cone right over here is 1 third the volume of this cylinder that is essentially, you can kind of think of the cylinder as bounding around it. Or if you wanted to rewrite this, you could write this as 1 third times pi, or pi over 3 times h r squared, however you want to view it. The easy way I remember it, for me, the volume of a cylinder is very intuitive. You take the area of the base, and then you multiply that times the height."}, {"video_title": "Volume of a cone Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "And that's what I mean when I say it's surprisingly clean, that this cone right over here is 1 third the volume of this cylinder that is essentially, you can kind of think of the cylinder as bounding around it. Or if you wanted to rewrite this, you could write this as 1 third times pi, or pi over 3 times h r squared, however you want to view it. The easy way I remember it, for me, the volume of a cylinder is very intuitive. You take the area of the base, and then you multiply that times the height. And so a volume of a cone is just 1 third of that. It's just 1 third the volume of the bounding cylinder is one way to think about it. But let's just apply these numbers just to make sure that it makes sense to us."}, {"video_title": "Volume of a cone Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "You take the area of the base, and then you multiply that times the height. And so a volume of a cone is just 1 third of that. It's just 1 third the volume of the bounding cylinder is one way to think about it. But let's just apply these numbers just to make sure that it makes sense to us. So let's say that this is some type of a conical glass, the type that you might see at the water cooler. And let's say that we're told that it holds, so this thing holds 131 cubic centimeters of water. And let's say that we're also told that its height right over here, actually I want to do that in a different color."}, {"video_title": "Volume of a cone Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "But let's just apply these numbers just to make sure that it makes sense to us. So let's say that this is some type of a conical glass, the type that you might see at the water cooler. And let's say that we're told that it holds, so this thing holds 131 cubic centimeters of water. And let's say that we're also told that its height right over here, actually I want to do that in a different color. We're told that the height of this cone is 5 centimeters. And so given that, what is roughly the radius of the top of this glass? Let's just say to the nearest tenth of a centimeter."}, {"video_title": "Volume of a cone Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "And let's say that we're also told that its height right over here, actually I want to do that in a different color. We're told that the height of this cone is 5 centimeters. And so given that, what is roughly the radius of the top of this glass? Let's just say to the nearest tenth of a centimeter. Well, we just once again have to apply the formula. The volume, which is 131 cubic centimeters, is going to be equal to 1 third times pi times the height, which is 5 centimeters, times the radius squared. Or if we wanted to solve for the radius squared, we could just divide both sides by all of this business."}, {"video_title": "Volume of a cone Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "Let's just say to the nearest tenth of a centimeter. Well, we just once again have to apply the formula. The volume, which is 131 cubic centimeters, is going to be equal to 1 third times pi times the height, which is 5 centimeters, times the radius squared. Or if we wanted to solve for the radius squared, we could just divide both sides by all of this business. And we would get radius squared is equal to 131 cubic centimeters. You divide by 1 third. That's the same thing as multiplying by 3."}, {"video_title": "Volume of a cone Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "Or if we wanted to solve for the radius squared, we could just divide both sides by all of this business. And we would get radius squared is equal to 131 cubic centimeters. You divide by 1 third. That's the same thing as multiplying by 3. And then of course, you're going to divide by pi. And you're going to divide by 5 centimeters. Now let's see if we can clean this up."}, {"video_title": "Volume of a cone Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "That's the same thing as multiplying by 3. And then of course, you're going to divide by pi. And you're going to divide by 5 centimeters. Now let's see if we can clean this up. Centimeters will cancel out with one of these centimeters, so you'll just be left with square centimeters, only in the numerator. And so this is going to be, and then to solve for r, we could take the square root of both sides. So we could say that r is going to be equal to the square root of 3 times 131 is 393 over 5 pi."}, {"video_title": "Volume of a cone Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "Now let's see if we can clean this up. Centimeters will cancel out with one of these centimeters, so you'll just be left with square centimeters, only in the numerator. And so this is going to be, and then to solve for r, we could take the square root of both sides. So we could say that r is going to be equal to the square root of 3 times 131 is 393 over 5 pi. So that's this part right over here. And the square root, once again, remember, we can treat units just like algebraic quantities. The square root of centimeters squared, well, that's just going to be centimeters, which is nice because we want our units in centimeters."}, {"video_title": "Volume of a cone Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "So we could say that r is going to be equal to the square root of 3 times 131 is 393 over 5 pi. So that's this part right over here. And the square root, once again, remember, we can treat units just like algebraic quantities. The square root of centimeters squared, well, that's just going to be centimeters, which is nice because we want our units in centimeters. So let's get our calculator out to actually calculate this kind of messy expression. Turn it on. Let's see."}, {"video_title": "Volume of a cone Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "The square root of centimeters squared, well, that's just going to be centimeters, which is nice because we want our units in centimeters. So let's get our calculator out to actually calculate this kind of messy expression. Turn it on. Let's see. Square root of 393 divided by 5 times pi is equal to 5. Well, that's pretty close. To the nearest, it's pretty much 5 centimeters."}, {"video_title": "Volume density.mp3", "Sentence": "And one simple way to think about density is it's a quantity of something, and we're going to think about examples of it, per unit volume, so per volume. So for example, let's say that I have a cubic meter right over here. Let me have two different cubic meters, just to give you an example. So these are both cubic meters. And let's say in the one on the left, I have a quantity of, let's say, six of these dots per cubic meter. And over here, I only have three of these dots per cubic meter. Well, here I have a higher density."}, {"video_title": "Volume density.mp3", "Sentence": "So these are both cubic meters. And let's say in the one on the left, I have a quantity of, let's say, six of these dots per cubic meter. And over here, I only have three of these dots per cubic meter. Well, here I have a higher density. And in general, we're gonna take the quantity and divide it by the volume. And the units are going to be some quantity per unit volume. Now, you're typically going to see mass per unit volume, but density, especially in the volume context, can refer to any quantity per unit volume."}, {"video_title": "Volume density.mp3", "Sentence": "Well, here I have a higher density. And in general, we're gonna take the quantity and divide it by the volume. And the units are going to be some quantity per unit volume. Now, you're typically going to see mass per unit volume, but density, especially in the volume context, can refer to any quantity per unit volume. Now, with that out of the way, let's give ourselves a little bit of an example. So here we're told that stone spheres thought to be carved by the Deke people, I'm not sure if I'm pronouncing that correctly, more than 1,000 years ago, are a national symbol of Costa Rica. One such sphere has a diameter of about 1.8 meters and masses about 8,300 kilograms."}, {"video_title": "Volume density.mp3", "Sentence": "Now, you're typically going to see mass per unit volume, but density, especially in the volume context, can refer to any quantity per unit volume. Now, with that out of the way, let's give ourselves a little bit of an example. So here we're told that stone spheres thought to be carved by the Deke people, I'm not sure if I'm pronouncing that correctly, more than 1,000 years ago, are a national symbol of Costa Rica. One such sphere has a diameter of about 1.8 meters and masses about 8,300 kilograms. Based on these measurements, what is the density of this sphere in kilograms per cubic meter? Round to the nearest 100 kilograms per cubic meter. So pause this video and see if you can figure that out."}, {"video_title": "Volume density.mp3", "Sentence": "One such sphere has a diameter of about 1.8 meters and masses about 8,300 kilograms. Based on these measurements, what is the density of this sphere in kilograms per cubic meter? Round to the nearest 100 kilograms per cubic meter. So pause this video and see if you can figure that out. All right, so we're going to wanna get kilograms per cubic meter. So we know the total number of kilograms in one point, in a sphere that has a diameter of 1.8 meters, so that's the total number of kilograms, but we don't know the volume just yet. So we have a sphere like this, this would be a cross section of it, its diameter is 1.8 meters."}, {"video_title": "Volume density.mp3", "Sentence": "So pause this video and see if you can figure that out. All right, so we're going to wanna get kilograms per cubic meter. So we know the total number of kilograms in one point, in a sphere that has a diameter of 1.8 meters, so that's the total number of kilograms, but we don't know the volume just yet. So we have a sphere like this, this would be a cross section of it, its diameter is 1.8 meters. Now, you may or may not already know that the volume of a sphere is given by 4 3rds pi r cubed. And so the radius here is 0.9 meters, and so that would be the r right over here. So the volume of one of these spheres is going to be, let me write it over here, the volume is going to be 4 3rds pi times 0.9 to the 3rd power."}, {"video_title": "Volume density.mp3", "Sentence": "So we have a sphere like this, this would be a cross section of it, its diameter is 1.8 meters. Now, you may or may not already know that the volume of a sphere is given by 4 3rds pi r cubed. And so the radius here is 0.9 meters, and so that would be the r right over here. So the volume of one of these spheres is going to be, let me write it over here, the volume is going to be 4 3rds pi times 0.9 to the 3rd power. And we know what the mass is. The mass in that volume is 8,300 kilograms. So we would know that the density, the density in this situation is going to be 8,300 kilograms, 8,300 kilograms per this many cubic meters, 4 3rds pi times 0.9 to the 3rd power cubic meters."}, {"video_title": "Volume density.mp3", "Sentence": "So the volume of one of these spheres is going to be, let me write it over here, the volume is going to be 4 3rds pi times 0.9 to the 3rd power. And we know what the mass is. The mass in that volume is 8,300 kilograms. So we would know that the density, the density in this situation is going to be 8,300 kilograms, 8,300 kilograms per this many cubic meters, 4 3rds pi times 0.9 to the 3rd power cubic meters. And we're gonna need a calculator for this, and we're gonna round to the nearest 100 kilograms. So we have 8,300 divided by, and let me just open parentheses here, 4 divided by 3 times pi times 0.9 to the 3rd power. And then I'm going to close my parentheses, is equal to this right over here."}, {"video_title": "Another isosceles example problem Congruence Geometry Khan Academy.mp3", "Sentence": "The measure of two angles of an isosceles triangle are three x plus five degrees, we'll say, and x plus 16 degrees. Find all possible values of x. So let's think about this. Let's draw ourselves an isosceles triangle or two. So it's an isosceles triangle, like that and like that. And actually, let me draw a couple of them, just because we want to think about all of the different possibilities here. All of the different possibilities."}, {"video_title": "Another isosceles example problem Congruence Geometry Khan Academy.mp3", "Sentence": "Let's draw ourselves an isosceles triangle or two. So it's an isosceles triangle, like that and like that. And actually, let me draw a couple of them, just because we want to think about all of the different possibilities here. All of the different possibilities. So we know from what we know about isosceles triangles that the base angles are going to be congruent. So that angle's going to be equal to that angle, that angle's going to be equal to that angle. And so what could the three x plus five degrees and the x plus 16, what could they be measures of?"}, {"video_title": "Another isosceles example problem Congruence Geometry Khan Academy.mp3", "Sentence": "All of the different possibilities. So we know from what we know about isosceles triangles that the base angles are going to be congruent. So that angle's going to be equal to that angle, that angle's going to be equal to that angle. And so what could the three x plus five degrees and the x plus 16, what could they be measures of? Well, maybe this one right over here has a measure of three x plus five degrees, and the vertex is the other one. So maybe this one up here is the x plus 16, x plus 16 degrees. The other possibility, the other possibilities is that this is describing both base angles, in which case they would be equal."}, {"video_title": "Another isosceles example problem Congruence Geometry Khan Academy.mp3", "Sentence": "And so what could the three x plus five degrees and the x plus 16, what could they be measures of? Well, maybe this one right over here has a measure of three x plus five degrees, and the vertex is the other one. So maybe this one up here is the x plus 16, x plus 16 degrees. The other possibility, the other possibilities is that this is describing both base angles, in which case they would be equal. So maybe this one is three x plus five, and maybe this one over here is x plus 16. X plus 16. And that is, and then the final possibility, actually we haven't exhausted all of them, is if we swap these two."}, {"video_title": "Another isosceles example problem Congruence Geometry Khan Academy.mp3", "Sentence": "The other possibility, the other possibilities is that this is describing both base angles, in which case they would be equal. So maybe this one is three x plus five, and maybe this one over here is x plus 16. X plus 16. And that is, and then the final possibility, actually we haven't exhausted all of them, is if we swap these two. If this one is x plus 16 and that one is three x plus five. So let me draw ourselves another triangle. Let me draw ourselves another triangle."}, {"video_title": "Another isosceles example problem Congruence Geometry Khan Academy.mp3", "Sentence": "And that is, and then the final possibility, actually we haven't exhausted all of them, is if we swap these two. If this one is x plus 16 and that one is three x plus five. So let me draw ourselves another triangle. Let me draw ourselves another triangle. And obviously swapping these two aren't going to make a difference because they are equal to each other. And then we can make that one equal to three x plus five, but that's not going to change anything either because they're equal to each other. So the last situation is where this angle down here is x plus 16, and this angle up here is x plus, is three x plus five."}, {"video_title": "Another isosceles example problem Congruence Geometry Khan Academy.mp3", "Sentence": "Let me draw ourselves another triangle. And obviously swapping these two aren't going to make a difference because they are equal to each other. And then we can make that one equal to three x plus five, but that's not going to change anything either because they're equal to each other. So the last situation is where this angle down here is x plus 16, and this angle up here is x plus, is three x plus five. This is three x plus five. So let's just work through each of these. So in this situation, if this base angle is three x plus five, so is this base angle."}, {"video_title": "Another isosceles example problem Congruence Geometry Khan Academy.mp3", "Sentence": "So the last situation is where this angle down here is x plus 16, and this angle up here is x plus, is three x plus five. This is three x plus five. So let's just work through each of these. So in this situation, if this base angle is three x plus five, so is this base angle. And then we know that all three of these are going to have to add up to 180 degrees. So we get three x plus five, plus three x plus five, plus x plus 16, plus x plus 16 is going to be equal to 180 degrees. We have three x, let's just add up the, you have three x plus three x, which gives you six x, plus another x gives you seven x."}, {"video_title": "Another isosceles example problem Congruence Geometry Khan Academy.mp3", "Sentence": "So in this situation, if this base angle is three x plus five, so is this base angle. And then we know that all three of these are going to have to add up to 180 degrees. So we get three x plus five, plus three x plus five, plus x plus 16, plus x plus 16 is going to be equal to 180 degrees. We have three x, let's just add up the, you have three x plus three x, which gives you six x, plus another x gives you seven x. Seven x. And then you have five plus five, which is 10, plus 16 is equal to 26. Is equal to 26, and that is going to be equal to 180."}, {"video_title": "Another isosceles example problem Congruence Geometry Khan Academy.mp3", "Sentence": "We have three x, let's just add up the, you have three x plus three x, which gives you six x, plus another x gives you seven x. Seven x. And then you have five plus five, which is 10, plus 16 is equal to 26. Is equal to 26, and that is going to be equal to 180. And then we have, let's see, 180 minus 26. If we subtract 26 from both sides, we get 180 minus, 180 minus 20 is 160, minus another six is 154. 150, 154."}, {"video_title": "Another isosceles example problem Congruence Geometry Khan Academy.mp3", "Sentence": "Is equal to 26, and that is going to be equal to 180. And then we have, let's see, 180 minus 26. If we subtract 26 from both sides, we get 180 minus, 180 minus 20 is 160, minus another six is 154. 150, 154. You have seven x is equal to 154. Seven x is equal to 154. And let's see how many times this, if we divide both sides by seven, seven will go into 140, 20 times."}, {"video_title": "Another isosceles example problem Congruence Geometry Khan Academy.mp3", "Sentence": "150, 154. You have seven x is equal to 154. Seven x is equal to 154. And let's see how many times this, if we divide both sides by seven, seven will go into 140, 20 times. And then you have another 14, so it looks like it's 22 times. So x is equal to 22. Is that right?"}, {"video_title": "Another isosceles example problem Congruence Geometry Khan Academy.mp3", "Sentence": "And let's see how many times this, if we divide both sides by seven, seven will go into 140, 20 times. And then you have another 14, so it looks like it's 22 times. So x is equal to 22. Is that right? 20 times seven is 140. 140 plus 14 is 154. So we have x is equal to 22 in this first, 22 degrees in this first scenario."}, {"video_title": "Another isosceles example problem Congruence Geometry Khan Academy.mp3", "Sentence": "Is that right? 20 times seven is 140. 140 plus 14 is 154. So we have x is equal to 22 in this first, 22 degrees in this first scenario. Now let's think about the second scenario over here. Now we have, now we have these two characters are going to be equal to each other, because they're both the base angles. So you have three x plus five is equal to x plus 16."}, {"video_title": "Another isosceles example problem Congruence Geometry Khan Academy.mp3", "Sentence": "So we have x is equal to 22 in this first, 22 degrees in this first scenario. Now let's think about the second scenario over here. Now we have, now we have these two characters are going to be equal to each other, because they're both the base angles. So you have three x plus five is equal to x plus 16. Well, you can subtract x from both sides, and so this becomes two x plus five is equal to 16. We can subtract five from both sides, and you get two x is equal to 11. And then you can divide both sides, you can divide both sides by two, and you get x is equal to 11, is equal to 11 halves."}, {"video_title": "Another isosceles example problem Congruence Geometry Khan Academy.mp3", "Sentence": "So you have three x plus five is equal to x plus 16. Well, you can subtract x from both sides, and so this becomes two x plus five is equal to 16. We can subtract five from both sides, and you get two x is equal to 11. And then you can divide both sides, you can divide both sides by two, and you get x is equal to 11, is equal to 11 halves. So that is our second scenario. And then we do our third scenario right over here. If this base angle is x plus 16, then this base angle right over here is also going to be x plus 16."}, {"video_title": "Another isosceles example problem Congruence Geometry Khan Academy.mp3", "Sentence": "And then you can divide both sides, you can divide both sides by two, and you get x is equal to 11, is equal to 11 halves. So that is our second scenario. And then we do our third scenario right over here. If this base angle is x plus 16, then this base angle right over here is also going to be x plus 16. They are congruent. And then we can do the same thing that we did for the first scenario. All of these angles are going to have to add up to 180 degrees."}, {"video_title": "Another isosceles example problem Congruence Geometry Khan Academy.mp3", "Sentence": "If this base angle is x plus 16, then this base angle right over here is also going to be x plus 16. They are congruent. And then we can do the same thing that we did for the first scenario. All of these angles are going to have to add up to 180 degrees. So we have x plus 16 plus x plus 16 plus three x plus five, plus three x plus five. When you add them all together, you're going to get 180 degrees. Now let's add up all the x terms."}, {"video_title": "Another isosceles example problem Congruence Geometry Khan Academy.mp3", "Sentence": "All of these angles are going to have to add up to 180 degrees. So we have x plus 16 plus x plus 16 plus three x plus five, plus three x plus five. When you add them all together, you're going to get 180 degrees. Now let's add up all the x terms. X plus x is two x plus three x is five x. So we get five x, five x, and then you have 16 plus 16 is 32. 32 plus five is 37."}, {"video_title": "Another isosceles example problem Congruence Geometry Khan Academy.mp3", "Sentence": "Now let's add up all the x terms. X plus x is two x plus three x is five x. So we get five x, five x, and then you have 16 plus 16 is 32. 32 plus five is 37. Plus 37 is equal to 180 degrees. Is equal to 180 degrees. Subtract 37 from both sides, and we get five x."}, {"video_title": "Another isosceles example problem Congruence Geometry Khan Academy.mp3", "Sentence": "32 plus five is 37. Plus 37 is equal to 180 degrees. Is equal to 180 degrees. Subtract 37 from both sides, and we get five x. Five x is equal to, 180 minus 30 is 150, so that gets us to 143. So it's not going to divide nicely. Divide both sides by five, and you get x is equal to 143 over five, which we could just leave as an improper fraction."}, {"video_title": "Another isosceles example problem Congruence Geometry Khan Academy.mp3", "Sentence": "Subtract 37 from both sides, and we get five x. Five x is equal to, 180 minus 30 is 150, so that gets us to 143. So it's not going to divide nicely. Divide both sides by five, and you get x is equal to 143 over five, which we could just leave as an improper fraction. You could write it as a mixed number, or however else you might want to write it. And we're done. These are the three possible values, the three possible values of x, given the information that they gave us right up there."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So let's start with an example. Let's say I have the point. I'll do it in a darker color so we can see it on the graph paper, let's say I have the point 3, negative 4. So if I were to graph it, I'd go 1, 2, 3, and then I'd go down 4. 1, 2, 3, 4 right there is 3, negative 4. And let's say I also have the point 6, 0. So 1, 2, 3, 4, 5, 6."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So if I were to graph it, I'd go 1, 2, 3, and then I'd go down 4. 1, 2, 3, 4 right there is 3, negative 4. And let's say I also have the point 6, 0. So 1, 2, 3, 4, 5, 6. And then there's no movement in the y direction. We're just sitting on the x-axis. The y-coordinate is 0, so that's 6, 0."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So 1, 2, 3, 4, 5, 6. And then there's no movement in the y direction. We're just sitting on the x-axis. The y-coordinate is 0, so that's 6, 0. And what I want to figure out is the distance between these two points. How far is this blue point away from this orange point? And at first you're like, gee, Sal, I don't think I've ever seen anything about how to solve for a distance like this, and what are you even talking about, the Pythagorean theorem?"}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "The y-coordinate is 0, so that's 6, 0. And what I want to figure out is the distance between these two points. How far is this blue point away from this orange point? And at first you're like, gee, Sal, I don't think I've ever seen anything about how to solve for a distance like this, and what are you even talking about, the Pythagorean theorem? I don't see a triangle there. And if you don't see a triangle, let me draw it for you. Let me draw the triangle."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And at first you're like, gee, Sal, I don't think I've ever seen anything about how to solve for a distance like this, and what are you even talking about, the Pythagorean theorem? I don't see a triangle there. And if you don't see a triangle, let me draw it for you. Let me draw the triangle. Let me draw this triangle right there, just like that. Let me actually do several colors here just to really hit the point home. So there's our triangle."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Let me draw the triangle. Let me draw this triangle right there, just like that. Let me actually do several colors here just to really hit the point home. So there's our triangle. And you might immediately recognize this is a right triangle. This is a right angle right there. The base goes straight left to right."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So there's our triangle. And you might immediately recognize this is a right triangle. This is a right angle right there. The base goes straight left to right. The right side goes straight up and down. So we're dealing with a right triangle. So if we could just figure out what the base length is and what this height is, we could use the Pythagorean theorem to figure out this long side, the side that is opposite the right angle, the hypotenuse."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "The base goes straight left to right. The right side goes straight up and down. So we're dealing with a right triangle. So if we could just figure out what the base length is and what this height is, we could use the Pythagorean theorem to figure out this long side, the side that is opposite the right angle, the hypotenuse. This right here, the distance is the hypotenuse of this right triangle. Let me write that down. The distance is equal to the hypotenuse of this right triangle."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So if we could just figure out what the base length is and what this height is, we could use the Pythagorean theorem to figure out this long side, the side that is opposite the right angle, the hypotenuse. This right here, the distance is the hypotenuse of this right triangle. Let me write that down. The distance is equal to the hypotenuse of this right triangle. So let me draw it a little bit bigger. So this is the hypotenuse right there. And then we have the side on the right, the side that goes straight up and down."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "The distance is equal to the hypotenuse of this right triangle. So let me draw it a little bit bigger. So this is the hypotenuse right there. And then we have the side on the right, the side that goes straight up and down. And then we have our base. Now, how do we figure out? Let's call this d for distance."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And then we have the side on the right, the side that goes straight up and down. And then we have our base. Now, how do we figure out? Let's call this d for distance. That's the length of our hypotenuse. How do we figure out the lengths of this up and down side and the base side right here? So let's look at the base first."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Let's call this d for distance. That's the length of our hypotenuse. How do we figure out the lengths of this up and down side and the base side right here? So let's look at the base first. What is this distance? You could even count it on this graph paper. But here, we're at x is equal to 3."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So let's look at the base first. What is this distance? You could even count it on this graph paper. But here, we're at x is equal to 3. And here, we're at x is equal to 6. We're just moving straight right. This is the same distance as that distance right there."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "But here, we're at x is equal to 3. And here, we're at x is equal to 6. We're just moving straight right. This is the same distance as that distance right there. So to figure out that distance, it's literally the end x point. And you can actually go either way, because you're going to square everything. So it doesn't matter if you get negative numbers."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "This is the same distance as that distance right there. So to figure out that distance, it's literally the end x point. And you can actually go either way, because you're going to square everything. So it doesn't matter if you get negative numbers. So it's going to be 6. The distance here is going to be 6 minus 3. That's this distance right here, which is equal to 3."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So it doesn't matter if you get negative numbers. So it's going to be 6. The distance here is going to be 6 minus 3. That's this distance right here, which is equal to 3. So we figured out the base. And just to remind ourselves, that is equal to the change in x. That was equal to your finishing x minus your starting x."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "That's this distance right here, which is equal to 3. So we figured out the base. And just to remind ourselves, that is equal to the change in x. That was equal to your finishing x minus your starting x. 6 minus 3. This is our delta x. Now, by the same exact line of reasoning, this height right here is going to be your change in y."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "That was equal to your finishing x minus your starting x. 6 minus 3. This is our delta x. Now, by the same exact line of reasoning, this height right here is going to be your change in y. Up here, you're at y is equal to 0. That's kind of where you finish. That's your higher y point."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Now, by the same exact line of reasoning, this height right here is going to be your change in y. Up here, you're at y is equal to 0. That's kind of where you finish. That's your higher y point. And over here, you're at y is equal to negative 4. So change in y is equal to 0 minus negative 4. I'm just taking the larger y value minus the smaller y value, the larger x value minus the smaller x value."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "That's your higher y point. And over here, you're at y is equal to negative 4. So change in y is equal to 0 minus negative 4. I'm just taking the larger y value minus the smaller y value, the larger x value minus the smaller x value. But you're going to see, we're going to square it in a second. So even if you did it the other way around, you get a negative number, but you still get the same answer. So this is equal to 4."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "I'm just taking the larger y value minus the smaller y value, the larger x value minus the smaller x value. But you're going to see, we're going to square it in a second. So even if you did it the other way around, you get a negative number, but you still get the same answer. So this is equal to 4. So this side is equal to 4. You could even count it on the graph paper if you like. And this side is equal to 3."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So this is equal to 4. So this side is equal to 4. You could even count it on the graph paper if you like. And this side is equal to 3. And now we can do the Pythagorean theorem. This distance is the distance squared. Be careful."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And this side is equal to 3. And now we can do the Pythagorean theorem. This distance is the distance squared. Be careful. The distance squared is going to be equal to this delta x squared, the change in x squared plus the change in y squared. This is nothing fancy. Sometimes people will call this the distance formula."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Be careful. The distance squared is going to be equal to this delta x squared, the change in x squared plus the change in y squared. This is nothing fancy. Sometimes people will call this the distance formula. It's just the Pythagorean theorem. This side squared plus that side squared is equal to hypotenuse squared, because this is a right triangle. So let's apply it with these numbers, the numbers that we have at hand."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Sometimes people will call this the distance formula. It's just the Pythagorean theorem. This side squared plus that side squared is equal to hypotenuse squared, because this is a right triangle. So let's apply it with these numbers, the numbers that we have at hand. So the distance squared is going to be equal to delta x squared is 3 squared plus delta y squared plus 4 squared, which is equal to 9 plus 16, which is equal to 25. So the distance is equal to, let me write that, d squared is equal to 25. d, our distance, is equal to, you don't want to take the negative square root, because you can't have a negative distance, so only the principal root, the positive square root of 25, which is equal to 5. So this distance right here is 5."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So let's apply it with these numbers, the numbers that we have at hand. So the distance squared is going to be equal to delta x squared is 3 squared plus delta y squared plus 4 squared, which is equal to 9 plus 16, which is equal to 25. So the distance is equal to, let me write that, d squared is equal to 25. d, our distance, is equal to, you don't want to take the negative square root, because you can't have a negative distance, so only the principal root, the positive square root of 25, which is equal to 5. So this distance right here is 5. Or if we look at this distance right here, that was the original problem. How far is this point from that point? It is 5 units away."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So this distance right here is 5. Or if we look at this distance right here, that was the original problem. How far is this point from that point? It is 5 units away. So what you'll see here, they call it the distance formula, but it's just the Pythagorean theorem. And just so you're exposed to all of the ways that you'll see the distance formula, sometimes people will say, oh, if I have two points, if I have one point, let's call it x1 and y1. So that's just a particular point."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "It is 5 units away. So what you'll see here, they call it the distance formula, but it's just the Pythagorean theorem. And just so you're exposed to all of the ways that you'll see the distance formula, sometimes people will say, oh, if I have two points, if I have one point, let's call it x1 and y1. So that's just a particular point. And let's say I have another point, that is x2, y2. Sometimes you'll see this formula, that the distance, you'll see it in different ways, but you'll see that the distance is equal to, and it looks like this really complicated formula, but I want you to see that this is really just the Pythagorean theorem. You'll see that the distance is equal to x2 minus x1 squared plus y2 minus y1 squared."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So that's just a particular point. And let's say I have another point, that is x2, y2. Sometimes you'll see this formula, that the distance, you'll see it in different ways, but you'll see that the distance is equal to, and it looks like this really complicated formula, but I want you to see that this is really just the Pythagorean theorem. You'll see that the distance is equal to x2 minus x1 squared plus y2 minus y1 squared. You'll see this written in a lot of textbooks as the distance formula. It's a complete waste of your time to memorize it, because it's really just the Pythagorean theorem. This is your change in x."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "You'll see that the distance is equal to x2 minus x1 squared plus y2 minus y1 squared. You'll see this written in a lot of textbooks as the distance formula. It's a complete waste of your time to memorize it, because it's really just the Pythagorean theorem. This is your change in x. And it really doesn't matter which x you pick to be first or second, because even if you get the negative of this value, when you square it, the negative disappears. This right here is your change in y. So it's just saying that the distance squared, remember if you square both sides of this equation, the radical will disappear, and this will be the distance squared is equal to this expression squared, delta x squared, change in x. Delta means change in."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "This is your change in x. And it really doesn't matter which x you pick to be first or second, because even if you get the negative of this value, when you square it, the negative disappears. This right here is your change in y. So it's just saying that the distance squared, remember if you square both sides of this equation, the radical will disappear, and this will be the distance squared is equal to this expression squared, delta x squared, change in x. Delta means change in. Delta x squared plus delta y squared. Don't want to confuse you. Delta y just means change in y. I should have probably said that earlier in the video."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So it's just saying that the distance squared, remember if you square both sides of this equation, the radical will disappear, and this will be the distance squared is equal to this expression squared, delta x squared, change in x. Delta means change in. Delta x squared plus delta y squared. Don't want to confuse you. Delta y just means change in y. I should have probably said that earlier in the video. But let's apply it to a couple of more. And I'll just pick some points at random. Let's say I have the point 1, 2, 3, 4, 5, 6."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Delta y just means change in y. I should have probably said that earlier in the video. But let's apply it to a couple of more. And I'll just pick some points at random. Let's say I have the point 1, 2, 3, 4, 5, 6. Negative 6 comma negative 4. And let's say I want to find the distance between that and 1 comma 1, 2, 3, 4, 5, 6, 7. And the point 1 comma 7."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Let's say I have the point 1, 2, 3, 4, 5, 6. Negative 6 comma negative 4. And let's say I want to find the distance between that and 1 comma 1, 2, 3, 4, 5, 6, 7. And the point 1 comma 7. So I want to find this distance right here. So it's the exact same idea. We just use the Pythagorean theorem."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And the point 1 comma 7. So I want to find this distance right here. So it's the exact same idea. We just use the Pythagorean theorem. You figure out this distance, which is our change in x. This distance, which is our change in y. This distance squared plus this distance squared is going to equal that distance squared."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "We just use the Pythagorean theorem. You figure out this distance, which is our change in x. This distance, which is our change in y. This distance squared plus this distance squared is going to equal that distance squared. So let's do it. So our change in x, you just take, it doesn't matter. I mean, in general, you want to take the larger x value minus the smaller x value, but you could do it either way."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "This distance squared plus this distance squared is going to equal that distance squared. So let's do it. So our change in x, you just take, it doesn't matter. I mean, in general, you want to take the larger x value minus the smaller x value, but you could do it either way. So we could write the distance squared is equal to, what's our change in x? So let's take the larger x minus the smaller x. 1 minus negative 6 squared plus the change in y."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "I mean, in general, you want to take the larger x value minus the smaller x value, but you could do it either way. So we could write the distance squared is equal to, what's our change in x? So let's take the larger x minus the smaller x. 1 minus negative 6 squared plus the change in y. The larger y is here. It's 7 minus negative 4 squared. And I just picked these numbers at random, so they're probably not going to come out too cleanly."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "1 minus negative 6 squared plus the change in y. The larger y is here. It's 7 minus negative 4 squared. And I just picked these numbers at random, so they're probably not going to come out too cleanly. So we get that the distance squared is equal to 1 minus negative 6, that is 7 squared. And you'll even see it over here if you count it. You go 1, 2, 3, 4, 5, 6, 7."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And I just picked these numbers at random, so they're probably not going to come out too cleanly. So we get that the distance squared is equal to 1 minus negative 6, that is 7 squared. And you'll even see it over here if you count it. You go 1, 2, 3, 4, 5, 6, 7. That's that number right here. That's what your change in x is. Plus 7 minus negative 4."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "You go 1, 2, 3, 4, 5, 6, 7. That's that number right here. That's what your change in x is. Plus 7 minus negative 4. That's 11. This is this distance right here. And you can count it on the blocks."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Plus 7 minus negative 4. That's 11. This is this distance right here. And you can count it on the blocks. We're going up 11. We're just taking 7 minus negative 4 to get a distance of 11. So plus 11 squared is equal to d squared."}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And you can count it on the blocks. We're going up 11. We're just taking 7 minus negative 4 to get a distance of 11. So plus 11 squared is equal to d squared. So let me just take the calculator out. So the distance, let's just take, if we get 7 squared plus 11 squared is equal to 170. That distance is going to be the square root of that, right?"}, {"video_title": "Distance formula Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So plus 11 squared is equal to d squared. So let me just take the calculator out. So the distance, let's just take, if we get 7 squared plus 11 squared is equal to 170. That distance is going to be the square root of that, right? d squared is equal to 170. So let's take the square root of 170 and we get roughly 13.04. So this distance right here, we tried to figure out, is 13.04."}, {"video_title": "45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "Or another way to think about it is, if the shortest side is 1, and I'll do the shortest side, then the medium side, then the longest side. So if the side opposite the 30 degree side is 1, then the side opposite the 60 degree side is square root of 3 times that, so it's going to be square root of 3, and then the hypotenuse is going to be twice that. In the last video, we started with x and we said that the 30 degree side is x over 2, but if the 30 degree side is 1, then this is going to be twice that, so it's going to be 2. This right here is the side opposite the 30 degree side, opposite the 60 degree side, and then the hypotenuse opposite the 90 degree side. And so in general, if you see any triangle that has those ratios, you say, hey, that's a 30-60-90 triangle. Or if you see a triangle that you know is a 30-60-90 triangle, you could say, hey, I know how to figure out one of the sides based on this ratio right over here. And just as an example, if you see a triangle that looks like this, where the sides are 2, 2 square roots of 3, and 4."}, {"video_title": "45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "This right here is the side opposite the 30 degree side, opposite the 60 degree side, and then the hypotenuse opposite the 90 degree side. And so in general, if you see any triangle that has those ratios, you say, hey, that's a 30-60-90 triangle. Or if you see a triangle that you know is a 30-60-90 triangle, you could say, hey, I know how to figure out one of the sides based on this ratio right over here. And just as an example, if you see a triangle that looks like this, where the sides are 2, 2 square roots of 3, and 4. Once again, the ratio of 2 to 2 square roots of 3 is 1 to square root of 3. The ratio of 2 to 4 is the same thing as 1 to 2. This right here must be a 30-60-90 triangle."}, {"video_title": "45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And just as an example, if you see a triangle that looks like this, where the sides are 2, 2 square roots of 3, and 4. Once again, the ratio of 2 to 2 square roots of 3 is 1 to square root of 3. The ratio of 2 to 4 is the same thing as 1 to 2. This right here must be a 30-60-90 triangle. What I want to introduce you to in this video is another important type of triangle that shows up a lot in geometry and a lot in trigonometry, and this is a 45-45-90 triangle. Or another way to think about it is if I have a right triangle that is also isosceles. You obviously can't have a right triangle that is equilateral, because an equilateral triangle has all of their angles have to be 60 degrees, but you can have a right triangle that is isosceles."}, {"video_title": "45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "This right here must be a 30-60-90 triangle. What I want to introduce you to in this video is another important type of triangle that shows up a lot in geometry and a lot in trigonometry, and this is a 45-45-90 triangle. Or another way to think about it is if I have a right triangle that is also isosceles. You obviously can't have a right triangle that is equilateral, because an equilateral triangle has all of their angles have to be 60 degrees, but you can have a right triangle that is isosceles. And isosceles, let me write this, this is a right isosceles triangle. And if it's isosceles, that means two of the sides are equal. So these are the two sides that are equal."}, {"video_title": "45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "You obviously can't have a right triangle that is equilateral, because an equilateral triangle has all of their angles have to be 60 degrees, but you can have a right triangle that is isosceles. And isosceles, let me write this, this is a right isosceles triangle. And if it's isosceles, that means two of the sides are equal. So these are the two sides that are equal. And then if the two sides are equal, we have proved to ourselves that the base angles are equal. And if we call the measure of these base angles x, then we know that x plus x plus 90 have to be equal to 180. x plus x plus 90 need to be equal to 180. Or if we subtract 90 from both sides, you get x plus x is equal to 90, or 2x is equal to 90."}, {"video_title": "45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So these are the two sides that are equal. And then if the two sides are equal, we have proved to ourselves that the base angles are equal. And if we call the measure of these base angles x, then we know that x plus x plus 90 have to be equal to 180. x plus x plus 90 need to be equal to 180. Or if we subtract 90 from both sides, you get x plus x is equal to 90, or 2x is equal to 90. Or if you divide both sides by 2, you get x is equal to 45 degrees. So a right isosceles triangle can also be called, and this is the more typical name for it, it can also be called a 45-45-90 triangle. And what I want to do in this video is come up with the ratios for the sides of a 45-45-90 triangle, just like we did for a 30-60-90 triangle."}, {"video_title": "45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "Or if we subtract 90 from both sides, you get x plus x is equal to 90, or 2x is equal to 90. Or if you divide both sides by 2, you get x is equal to 45 degrees. So a right isosceles triangle can also be called, and this is the more typical name for it, it can also be called a 45-45-90 triangle. And what I want to do in this video is come up with the ratios for the sides of a 45-45-90 triangle, just like we did for a 30-60-90 triangle. This was actually more straightforward. Because in a 45-45-90 triangle, if we call one of the legs x, the other leg is also going to be x, and then we can use the Pythagorean theorem to figure out the length of the hypotenuse. So the length of the hypotenuse, let's call that c. So we get x squared plus x squared."}, {"video_title": "45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And what I want to do in this video is come up with the ratios for the sides of a 45-45-90 triangle, just like we did for a 30-60-90 triangle. This was actually more straightforward. Because in a 45-45-90 triangle, if we call one of the legs x, the other leg is also going to be x, and then we can use the Pythagorean theorem to figure out the length of the hypotenuse. So the length of the hypotenuse, let's call that c. So we get x squared plus x squared. That's the length of both of the legs. So when we sum those up, that's going to have to be equal to c squared. This is just straight out of the Pythagorean theorem."}, {"video_title": "45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So the length of the hypotenuse, let's call that c. So we get x squared plus x squared. That's the length of both of the legs. So when we sum those up, that's going to have to be equal to c squared. This is just straight out of the Pythagorean theorem. So we get 2x squared is equal to c squared. And I'll take the principal root of both sides of that. I want to just change it to yellow, and it keeps not letting me."}, {"video_title": "45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "This is just straight out of the Pythagorean theorem. So we get 2x squared is equal to c squared. And I'll take the principal root of both sides of that. I want to just change it to yellow, and it keeps not letting me. Okay, to c squared. And I'll take the principal root of both sides of that. The left-hand side, you get principal root of 2 is just square root of 2."}, {"video_title": "45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "I want to just change it to yellow, and it keeps not letting me. Okay, to c squared. And I'll take the principal root of both sides of that. The left-hand side, you get principal root of 2 is just square root of 2. And then the principal root of x squared is just going to be x. So you're going to have x times the square root of 2 is equal to c. So if you have a right isosceles triangle, whatever the two legs are, they're going to have the same length. That's why it's isosceles."}, {"video_title": "45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "The left-hand side, you get principal root of 2 is just square root of 2. And then the principal root of x squared is just going to be x. So you're going to have x times the square root of 2 is equal to c. So if you have a right isosceles triangle, whatever the two legs are, they're going to have the same length. That's why it's isosceles. The hypotenuse is going to be square root of 2 times that. So c is equal to x times the square root of 2. So, for example, if you have a triangle that looks like this."}, {"video_title": "45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "That's why it's isosceles. The hypotenuse is going to be square root of 2 times that. So c is equal to x times the square root of 2. So, for example, if you have a triangle that looks like this. Let me draw it a slightly different way. It's good to have to orient ourselves in different ways every time. So if we see a triangle that's 90 degrees, 45, and 45 like this, and you really just have to know two of these angles to know what the other one is going to be."}, {"video_title": "45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So, for example, if you have a triangle that looks like this. Let me draw it a slightly different way. It's good to have to orient ourselves in different ways every time. So if we see a triangle that's 90 degrees, 45, and 45 like this, and you really just have to know two of these angles to know what the other one is going to be. And if I tell you that this side right over here is 3, I actually don't even have to tell you that this other side is going to be 3. This is an isosceles triangle, so the two legs are going to be the same. And you won't even have to apply the Pythagorean theorem if you know this, and this is a good one to know, that the hypotenuse here, the side opposite the 90-degree side, is just going to be square root of 2 times the length of either of the legs."}, {"video_title": "45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So if we see a triangle that's 90 degrees, 45, and 45 like this, and you really just have to know two of these angles to know what the other one is going to be. And if I tell you that this side right over here is 3, I actually don't even have to tell you that this other side is going to be 3. This is an isosceles triangle, so the two legs are going to be the same. And you won't even have to apply the Pythagorean theorem if you know this, and this is a good one to know, that the hypotenuse here, the side opposite the 90-degree side, is just going to be square root of 2 times the length of either of the legs. So it's going to be 3 times the square root of 2. So the ratio of the sides in a 45-45-90 triangle, or a right isosceles triangle, the ratio of the sides are one of the legs can be 1, then the other leg's going to have the same measure, the same length, and then the hypotenuse is going to be square root of 2 times either of those. 1 to 1, 2 square root of 2."}, {"video_title": "45-45-90 triangle side ratios Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And you won't even have to apply the Pythagorean theorem if you know this, and this is a good one to know, that the hypotenuse here, the side opposite the 90-degree side, is just going to be square root of 2 times the length of either of the legs. So it's going to be 3 times the square root of 2. So the ratio of the sides in a 45-45-90 triangle, or a right isosceles triangle, the ratio of the sides are one of the legs can be 1, then the other leg's going to have the same measure, the same length, and then the hypotenuse is going to be square root of 2 times either of those. 1 to 1, 2 square root of 2. So this is 45-45-90. Let me write this as 45-45-90. That's the ratios."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3", "Sentence": "What I'd like to do in this video is use some geometric arguments to prove that the slopes of perpendicular lines are negative reciprocals of each other. And so just to start off, we have lines L and M, and we're going to assume that they are perpendicular, so they intersect at a right angle. We see that depicted right over here. And so I'm gonna now construct some other lines here to help us make our geometric argument. So let me draw a horizontal line that intersects at this point right over here. Let's call that point A. And so let me see if I can do that."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3", "Sentence": "And so I'm gonna now construct some other lines here to help us make our geometric argument. So let me draw a horizontal line that intersects at this point right over here. Let's call that point A. And so let me see if I can do that. There you go. So that's a horizontal line that intersects at A. And now I'm gonna drop some verticals from that."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3", "Sentence": "And so let me see if I can do that. There you go. So that's a horizontal line that intersects at A. And now I'm gonna drop some verticals from that. So I'm gonna drop a vertical line right over here, and I'm gonna drop a vertical line right over here. And so that is 90 degrees, and that is 90 degrees, and I've constructed it that way. This top line is perfectly horizontal, and then I've dropped two vertical things, so they're at 90 degree angles."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3", "Sentence": "And now I'm gonna drop some verticals from that. So I'm gonna drop a vertical line right over here, and I'm gonna drop a vertical line right over here. And so that is 90 degrees, and that is 90 degrees, and I've constructed it that way. This top line is perfectly horizontal, and then I've dropped two vertical things, so they're at 90 degree angles. And let me now set up some points. So that already said, that's point A. Let's call this point B."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3", "Sentence": "This top line is perfectly horizontal, and then I've dropped two vertical things, so they're at 90 degree angles. And let me now set up some points. So that already said, that's point A. Let's call this point B. Let's call this point C. Let's call this point D. And let's call this point E right over here. Now, let's think about what the slope of line L is. So slope of, let me move this over a little bit."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3", "Sentence": "Let's call this point B. Let's call this point C. Let's call this point D. And let's call this point E right over here. Now, let's think about what the slope of line L is. So slope of, let me move this over a little bit. So slope, slope of L is going to be what? Well, that's, you could view line L as line, the line that connects point CA. So it's the slope of CA, you could say."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3", "Sentence": "So slope of, let me move this over a little bit. So slope, slope of L is going to be what? Well, that's, you could view line L as line, the line that connects point CA. So it's the slope of CA, you could say. This is the same thing as slope of line CA. L is line CA. And so to find the slope, that's change in Y over change in X."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3", "Sentence": "So it's the slope of CA, you could say. This is the same thing as slope of line CA. L is line CA. And so to find the slope, that's change in Y over change in X. So our change in Y is going to be CB. So it's gonna be the length of segment CB. That is our change in Y."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3", "Sentence": "And so to find the slope, that's change in Y over change in X. So our change in Y is going to be CB. So it's gonna be the length of segment CB. That is our change in Y. So it is CB over our change in X, which is the length of segment BA, which is the length of segment BA right over here. So that is BA. Now, what is the slope of line M?"}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3", "Sentence": "That is our change in Y. So it is CB over our change in X, which is the length of segment BA, which is the length of segment BA right over here. So that is BA. Now, what is the slope of line M? So slope, slope of M, and we could also say slope of, we could call line M line AE, line AE, like that. Well, if we're going to go between point A and point E, once again, it's just change in Y over change in X. Well, what's our change in Y going to be?"}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3", "Sentence": "Now, what is the slope of line M? So slope, slope of M, and we could also say slope of, we could call line M line AE, line AE, like that. Well, if we're going to go between point A and point E, once again, it's just change in Y over change in X. Well, what's our change in Y going to be? Well, our change in Y, well, we're gonna go from this level down to this level as we go from A to E. We could have done it over here as well. We're gonna go from A to E. That is our change in Y. So we might be tempted to say, well, that's just going to be the length of segment DE."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3", "Sentence": "Well, what's our change in Y going to be? Well, our change in Y, well, we're gonna go from this level down to this level as we go from A to E. We could have done it over here as well. We're gonna go from A to E. That is our change in Y. So we might be tempted to say, well, that's just going to be the length of segment DE. But remember, our Y is decreasing. So we're gonna subtract that length as we go from this Y level to that Y level over there. And what is our change in X?"}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3", "Sentence": "So we might be tempted to say, well, that's just going to be the length of segment DE. But remember, our Y is decreasing. So we're gonna subtract that length as we go from this Y level to that Y level over there. And what is our change in X? So our change in X, we're going to go, as we go from A to E, our change in X is going to be the length of segment AD. So AD. So our slope of M is going to be negative DE."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3", "Sentence": "And what is our change in X? So our change in X, we're going to go, as we go from A to E, our change in X is going to be the length of segment AD. So AD. So our slope of M is going to be negative DE. It's going to be the negative of this length because we're dropping by that much. That's our change in Y over segment A, over segment AD. So some of you might already be quite inspired by what we've already written because now we just have to establish that these two are, these two triangles, triangles CBA and triangle ADE, are similar."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3", "Sentence": "So our slope of M is going to be negative DE. It's going to be the negative of this length because we're dropping by that much. That's our change in Y over segment A, over segment AD. So some of you might already be quite inspired by what we've already written because now we just have to establish that these two are, these two triangles, triangles CBA and triangle ADE, are similar. And then we're going to be able to show that these are the negative reciprocal of each other. So let's show that these two triangles are similar. So let's say that we have this angle right over here."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3", "Sentence": "So some of you might already be quite inspired by what we've already written because now we just have to establish that these two are, these two triangles, triangles CBA and triangle ADE, are similar. And then we're going to be able to show that these are the negative reciprocal of each other. So let's show that these two triangles are similar. So let's say that we have this angle right over here. And let's say that angle has measure X, just for kicks. And let's say that we have, let me do another color for, let's say we have this angle right over here. And let's say that the measure, that that has measure Y."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3", "Sentence": "So let's say that we have this angle right over here. And let's say that angle has measure X, just for kicks. And let's say that we have, let me do another color for, let's say we have this angle right over here. And let's say that the measure, that that has measure Y. Well we know X plus Y plus 90 is equal to 180 because together they are supplementary. So I could write, I could write that X plus 90, plus 90, plus Y, plus Y, is going to be equal to, is going to be equal to 180 degrees. If you want you could subtract 90 from both sides of that and you could say, look, X plus Y is going to be equal to 90 degrees."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3", "Sentence": "And let's say that the measure, that that has measure Y. Well we know X plus Y plus 90 is equal to 180 because together they are supplementary. So I could write, I could write that X plus 90, plus 90, plus Y, plus Y, is going to be equal to, is going to be equal to 180 degrees. If you want you could subtract 90 from both sides of that and you could say, look, X plus Y is going to be equal to 90 degrees. Is going to be equal to 90 degrees. These are algebraically equivalent statements. So is equal to 90 degrees."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3", "Sentence": "If you want you could subtract 90 from both sides of that and you could say, look, X plus Y is going to be equal to 90 degrees. Is going to be equal to 90 degrees. These are algebraically equivalent statements. So is equal to 90 degrees. And how can we use this to fill out some of the other angles in these triangles? Well, let's see. X plus this angle down here has to be equal to 90 degrees."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3", "Sentence": "So is equal to 90 degrees. And how can we use this to fill out some of the other angles in these triangles? Well, let's see. X plus this angle down here has to be equal to 90 degrees. Or you could say X plus 90 plus what is going to be equal to 180. I'm looking at triangle CBA right over here. The interior angles of a triangle add up to 180."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3", "Sentence": "X plus this angle down here has to be equal to 90 degrees. Or you could say X plus 90 plus what is going to be equal to 180. I'm looking at triangle CBA right over here. The interior angles of a triangle add up to 180. So X plus 90 plus what is equal to 180? Well X plus 90 plus Y is equal to 180. We already established that."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3", "Sentence": "The interior angles of a triangle add up to 180. So X plus 90 plus what is equal to 180? Well X plus 90 plus Y is equal to 180. We already established that. Similarly over here. Y plus 90 plus what is going to be equal to 180? Well, same argument."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3", "Sentence": "We already established that. Similarly over here. Y plus 90 plus what is going to be equal to 180? Well, same argument. We already know Y plus 90 plus X is equal to 180. So Y plus 90 plus X is equal to, is equal to 180. And so notice, we have now established that triangle ABC and triangle EDA, that all of their interior angles, their corresponding interior angles are the same or that their three different angle measures all correspond to each other."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3", "Sentence": "Well, same argument. We already know Y plus 90 plus X is equal to 180. So Y plus 90 plus X is equal to, is equal to 180. And so notice, we have now established that triangle ABC and triangle EDA, that all of their interior angles, their corresponding interior angles are the same or that their three different angle measures all correspond to each other. They both have an angle of X, they both have a measure X, they both have an angle of measure Y, and they're both right triangles. So just by angle, angle, angle, so we could say by angle, angle, angle, one of our similarity postulates, we know that triangle E, triangle EDA, EDA is similar to triangle, to triangle ABC, to triangle ABC. And so that tells us that the ratio of corresponding sides are going to be the same."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3", "Sentence": "And so notice, we have now established that triangle ABC and triangle EDA, that all of their interior angles, their corresponding interior angles are the same or that their three different angle measures all correspond to each other. They both have an angle of X, they both have a measure X, they both have an angle of measure Y, and they're both right triangles. So just by angle, angle, angle, so we could say by angle, angle, angle, one of our similarity postulates, we know that triangle E, triangle EDA, EDA is similar to triangle, to triangle ABC, to triangle ABC. And so that tells us that the ratio of corresponding sides are going to be the same. And so for example, we know, let's find the ratio of corresponding sides. We know that the ratio of, let's say CB to BA, so let's write this down. We know that the ratio, so this tells us that the ratio of corresponding sides are going to be the same."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3", "Sentence": "And so that tells us that the ratio of corresponding sides are going to be the same. And so for example, we know, let's find the ratio of corresponding sides. We know that the ratio of, let's say CB to BA, so let's write this down. We know that the ratio, so this tells us that the ratio of corresponding sides are going to be the same. So the ratio of CB over BA, over BA is going to be equal to, is going to be equal to, well the corresponding side to CB, it's the side opposite the X degree angle right over here. So the corresponding side to CB is side AD, so that's going to be equal to AD over, what's the corresponding side to BA? Well BA is opposite the Y degree angle, so over here the corresponding side is DE, AD over DE, let me do that same color, over DE."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3", "Sentence": "We know that the ratio, so this tells us that the ratio of corresponding sides are going to be the same. So the ratio of CB over BA, over BA is going to be equal to, is going to be equal to, well the corresponding side to CB, it's the side opposite the X degree angle right over here. So the corresponding side to CB is side AD, so that's going to be equal to AD over, what's the corresponding side to BA? Well BA is opposite the Y degree angle, so over here the corresponding side is DE, AD over DE, let me do that same color, over DE. And so this right over here, this right over here we saw from the beginning, this is the slope, this is the slope of L. So slope, slope of L. And how does this relate to the slope of M? Notice the slope of M is the negative reciprocal of this. You take the reciprocal, you're going to get DE over AD, and then you have to take this negative right over here."}, {"video_title": "Proof perpendicular lines have negative reciprocal slope High School Math Khan Academy (2).mp3", "Sentence": "Well BA is opposite the Y degree angle, so over here the corresponding side is DE, AD over DE, let me do that same color, over DE. And so this right over here, this right over here we saw from the beginning, this is the slope, this is the slope of L. So slope, slope of L. And how does this relate to the slope of M? Notice the slope of M is the negative reciprocal of this. You take the reciprocal, you're going to get DE over AD, and then you have to take this negative right over here. So we could write this as the negative reciprocal of slope of M. Negative reciprocal, reciprocal of M's, of M's slope. And there you have it. We've just shown that if we start with, if we assume these L and M are perpendicular, and we set up these similar triangles, and we were able to show that the slope of L is the negative reciprocal of the slope of M."}, {"video_title": "Area of a circle intuition High School Geometry High School Math Khan Academy.mp3", "Sentence": "We see that we're leaving out less We see when we have now. This is a one two three four five six seven sided polygon We're leaving we're leaving a little bit less We're underestimating still but we're underestimating by less this area that we're giving up isn't as large as this area right over here So in this approximation we have What I say seven triangles one two three four five six seven triangles and the area of each of those triangles is once again a b over two now a and b here are different than a and b over here and notice what's happening as As we increase as we increase the number of triangles Not only is it approximating the area of the circle better But a is getting longer And you can see as you could imagine is as we increase many many many more triangles a is going to approach our Now another thing to think about is what is seven? What is seven times B approaching? so we're saying that a is approaching R as we add more of these sides of the polygon as we add more triangles and what is the number of Triangles times the base of the triangle. What is that approaching? Well, this is going to approach is going to approach the perimeter of the pot Or this is going to be the perimeter of the polygon. So seven times B is that Plus let me actually let me draw this it's that plus that Plus that you get the point plus that plus that plus that plus that So once again Seven let me write this as seven times B."}, {"video_title": "Area of a circle intuition High School Geometry High School Math Khan Academy.mp3", "Sentence": "so we're saying that a is approaching R as we add more of these sides of the polygon as we add more triangles and what is the number of Triangles times the base of the triangle. What is that approaching? Well, this is going to approach is going to approach the perimeter of the pot Or this is going to be the perimeter of the polygon. So seven times B is that Plus let me actually let me draw this it's that plus that Plus that you get the point plus that plus that plus that plus that So once again Seven let me write this as seven times B. That is the perimeter of the polygon perimeter Perimeter of the polygon So think about what's happening as we as we have more and more sides of the polygon our a Our height of each of our triangles is going to approach Our radius is going to approach The radius is going to get longed the height of each of the triangles gonna get longer and longer It's going to approach the radius as we have as we approach an infinite number of triangles And then the number of polygons we have are again the number of polygons the number of sides we have Times the bases that's going to be the perimeter of the polygon and as we add more and more sides as we add more and the perimeter of the polygon is going to approach the circumference of the circle. I'll write it out, circumference. And you see that even more clearly right over here."}, {"video_title": "Area of a circle intuition High School Geometry High School Math Khan Academy.mp3", "Sentence": "So seven times B is that Plus let me actually let me draw this it's that plus that Plus that you get the point plus that plus that plus that plus that So once again Seven let me write this as seven times B. That is the perimeter of the polygon perimeter Perimeter of the polygon So think about what's happening as we as we have more and more sides of the polygon our a Our height of each of our triangles is going to approach Our radius is going to approach The radius is going to get longed the height of each of the triangles gonna get longer and longer It's going to approach the radius as we have as we approach an infinite number of triangles And then the number of polygons we have are again the number of polygons the number of sides we have Times the bases that's going to be the perimeter of the polygon and as we add more and more sides as we add more and the perimeter of the polygon is going to approach the circumference of the circle. I'll write it out, circumference. And you see that even more clearly right over here. So once again, how many sides do I have here? I have one, two, three, four, five, six, seven, eight, nine, ten sides. So this, I can write the perimeter of the polygon as ten times b, and then if I multiply that times a over two, if I multiply that times this another color, a over, let me just write it like this, times a over two, I'm once again approximating the area of the circle, because a times b over two, that's the area of each of these triangles, and then I have ten of these triangles."}, {"video_title": "Area of a circle intuition High School Geometry High School Math Khan Academy.mp3", "Sentence": "And you see that even more clearly right over here. So once again, how many sides do I have here? I have one, two, three, four, five, six, seven, eight, nine, ten sides. So this, I can write the perimeter of the polygon as ten times b, and then if I multiply that times a over two, if I multiply that times this another color, a over, let me just write it like this, times a over two, I'm once again approximating the area of the circle, because a times b over two, that's the area of each of these triangles, and then I have ten of these triangles. But now let's think about this more generally. Let's think about it if I were to have n, if I were to have an n-sided polygon. So I have an n-sided polygon, then I'd be approximating the area as n times b, n times b."}, {"video_title": "Area of a circle intuition High School Geometry High School Math Khan Academy.mp3", "Sentence": "So this, I can write the perimeter of the polygon as ten times b, and then if I multiply that times a over two, if I multiply that times this another color, a over, let me just write it like this, times a over two, I'm once again approximating the area of the circle, because a times b over two, that's the area of each of these triangles, and then I have ten of these triangles. But now let's think about this more generally. Let's think about it if I were to have n, if I were to have an n-sided polygon. So I have an n-sided polygon, then I'd be approximating the area as n times b, n times b. We see this right over here, when n is equal to ten, we have ten times b, so it's n times b times a over two. This isn't something mysterious. The base times the height divided by two, this right over here, that's the area of each triangle, and then I'm going to have n of these triangles."}, {"video_title": "Area of a circle intuition High School Geometry High School Math Khan Academy.mp3", "Sentence": "So I have an n-sided polygon, then I'd be approximating the area as n times b, n times b. We see this right over here, when n is equal to ten, we have ten times b, so it's n times b times a over two. This isn't something mysterious. The base times the height divided by two, this right over here, that's the area of each triangle, and then I'm going to have n of these triangles. So this is our approximation for the area, so let me write this, the area is going to be approximately that right over there. It's going to be n, the number of triangles I have, times the area of each triangle. Now what's going to happen as n approaches infinity, as I approach having an infinite-sided polygon, as I have an infinite number of triangles?"}, {"video_title": "Area of a circle intuition High School Geometry High School Math Khan Academy.mp3", "Sentence": "The base times the height divided by two, this right over here, that's the area of each triangle, and then I'm going to have n of these triangles. So this is our approximation for the area, so let me write this, the area is going to be approximately that right over there. It's going to be n, the number of triangles I have, times the area of each triangle. Now what's going to happen as n approaches infinity, as I approach having an infinite-sided polygon, as I have an infinite number of triangles? So let's just think this through a little bit, because this is where it gets interesting. This is the informal argument. To do this better, I'd have to dig out a little bit of calculus, but this gives you the essence."}, {"video_title": "Area of a circle intuition High School Geometry High School Math Khan Academy.mp3", "Sentence": "Now what's going to happen as n approaches infinity, as I approach having an infinite-sided polygon, as I have an infinite number of triangles? So let's just think this through a little bit, because this is where it gets interesting. This is the informal argument. To do this better, I'd have to dig out a little bit of calculus, but this gives you the essence. So let's just think about what happens as n approaches infinity. So as n approaches infinity, we've already said, as we have more and more sides, and we have more and more triangles, more and more triangles, a approaches r. So let's write that down. So a is going to approach r, the height of the triangles is going to approach the radius."}, {"video_title": "Area of a circle intuition High School Geometry High School Math Khan Academy.mp3", "Sentence": "To do this better, I'd have to dig out a little bit of calculus, but this gives you the essence. So let's just think about what happens as n approaches infinity. So as n approaches infinity, we've already said, as we have more and more sides, and we have more and more triangles, more and more triangles, a approaches r. So let's write that down. So a is going to approach r, the height of the triangles is going to approach the radius. And what else is going to happen? Well, n times b, the perimeter of the polygon, the perimeter of the polygon, is going to approach the circumference. So a is going to approach r, and n times b is going to approach the circumference."}, {"video_title": "Area of a circle intuition High School Geometry High School Math Khan Academy.mp3", "Sentence": "So a is going to approach r, the height of the triangles is going to approach the radius. And what else is going to happen? Well, n times b, the perimeter of the polygon, the perimeter of the polygon, is going to approach the circumference. So a is going to approach r, and n times b is going to approach the circumference. Or another way of thinking about it, if it's approaching the circumference, we could say that n times b is going to approach 2 pi times the radius, because that's what the circumference is going to be equal to. So if a is approaching the radius, and nb is approaching 2 pi r, well then, what is the area of our polygon? Or what is the area of our polygon going to, or the area of our circle going to be?"}, {"video_title": "Area of a circle intuition High School Geometry High School Math Khan Academy.mp3", "Sentence": "So a is going to approach r, and n times b is going to approach the circumference. Or another way of thinking about it, if it's approaching the circumference, we could say that n times b is going to approach 2 pi times the radius, because that's what the circumference is going to be equal to. So if a is approaching the radius, and nb is approaching 2 pi r, well then, what is the area of our polygon? Or what is the area of our polygon going to, or the area of our circle going to be? Well, it is going to approach, or I should say the area of our polygon is going to approach, nb is going to approach 2 pi r, so instead of nb, I'm writing 2 pi r there. a is going to approach r, and then I'm dividing it by 2. So as n approaches infinity, as we have an infinite number of sides of our polygon, an infinite number of triangles, the area of our polygon will approach this, which is equal to what?"}, {"video_title": "Area of a circle intuition High School Geometry High School Math Khan Academy.mp3", "Sentence": "Or what is the area of our polygon going to, or the area of our circle going to be? Well, it is going to approach, or I should say the area of our polygon is going to approach, nb is going to approach 2 pi r, so instead of nb, I'm writing 2 pi r there. a is going to approach r, and then I'm dividing it by 2. So as n approaches infinity, as we have an infinite number of sides of our polygon, an infinite number of triangles, the area of our polygon will approach this, which is equal to what? Well, you have 2 divided by 2, and then pi r times r is equal to pi r squared. So as we approach having an infinite number of triangles, an infinite number of sides, we see that we approach the area of the circle. And as we approach the area of the circle, we are approaching pi r squared."}, {"video_title": "Similar shapes & transformations.mp3", "Sentence": "We're told that Shui concluded the quadrilaterals, these two over here, have four pairs of congruent corresponding angles. We can see these right over there. And so based on that, she concludes that the figures are similar. What error, if any, did Shui make in her conclusion? Pause this video and try to figure this out on your own. All right, so let's just remind ourselves one definition of similarity that we often use in geometry class, and that's two figures are similar as if you can, through a series of rigid transformations and dilations, if you can map one figure on to the other. Now, when I look at these two figures, you could try to do something."}, {"video_title": "Similar shapes & transformations.mp3", "Sentence": "What error, if any, did Shui make in her conclusion? Pause this video and try to figure this out on your own. All right, so let's just remind ourselves one definition of similarity that we often use in geometry class, and that's two figures are similar as if you can, through a series of rigid transformations and dilations, if you can map one figure on to the other. Now, when I look at these two figures, you could try to do something. You could say, okay, let me shift it so that k gets mapped on to h, and if you did that, it looks like l would get mapped on to g, but these sides, kn and lm right over here, they seem a good bit longer. So, and then if you tried to dilate it down so that the length of kn is the same as the length of hi, well then, the lengths of kl and gh would be different. So it doesn't seem like you can do this."}, {"video_title": "Similar shapes & transformations.mp3", "Sentence": "Now, when I look at these two figures, you could try to do something. You could say, okay, let me shift it so that k gets mapped on to h, and if you did that, it looks like l would get mapped on to g, but these sides, kn and lm right over here, they seem a good bit longer. So, and then if you tried to dilate it down so that the length of kn is the same as the length of hi, well then, the lengths of kl and gh would be different. So it doesn't seem like you can do this. So it is strange that Shui concluded that they are similar. So let's find the mistake. I'm already, I'll already rule out c, that it's a correct conclusion, because I don't think they are similar."}, {"video_title": "Similar shapes & transformations.mp3", "Sentence": "So it doesn't seem like you can do this. So it is strange that Shui concluded that they are similar. So let's find the mistake. I'm already, I'll already rule out c, that it's a correct conclusion, because I don't think they are similar. So let's see. Is the error that a rigid transformation, a translation, would map hg on to kl? Yep, we just talked about that."}, {"video_title": "Similar shapes & transformations.mp3", "Sentence": "I'm already, I'll already rule out c, that it's a correct conclusion, because I don't think they are similar. So let's see. Is the error that a rigid transformation, a translation, would map hg on to kl? Yep, we just talked about that. Hg can be mapped on to kl. So the quadrilaterals are congruent, not similar. Oh, choice A is making an even stronger statement, because anything that is congruent is going to be similar."}, {"video_title": "Similar shapes & transformations.mp3", "Sentence": "Yep, we just talked about that. Hg can be mapped on to kl. So the quadrilaterals are congruent, not similar. Oh, choice A is making an even stronger statement, because anything that is congruent is going to be similar. You actually can't have something that's congruent and not similar, and so choice A does not make any sense. So our deductive reasoning tells us it's probably choice B, but let's just read it. It's impossible to map quadrilateral ghij onto quadrilateral lknm using only rigid transformations and dilations, so the figures are not similar."}, {"video_title": "Similar shapes & transformations.mp3", "Sentence": "Oh, choice A is making an even stronger statement, because anything that is congruent is going to be similar. You actually can't have something that's congruent and not similar, and so choice A does not make any sense. So our deductive reasoning tells us it's probably choice B, but let's just read it. It's impossible to map quadrilateral ghij onto quadrilateral lknm using only rigid transformations and dilations, so the figures are not similar. Yeah, that's right. You could try. You could map hg on to kl, but then segment ij would look something like this."}, {"video_title": "Similar shapes & transformations.mp3", "Sentence": "It's impossible to map quadrilateral ghij onto quadrilateral lknm using only rigid transformations and dilations, so the figures are not similar. Yeah, that's right. You could try. You could map hg on to kl, but then segment ij would look something like this. Ij would go right over here. And then if you tried to dilate it so that the length of hi and gj matched kn or lm, then you're going to make hg bigger as well, so you're never going to be able to map them on to each other, even if you can use dilations. So I like choice B."}, {"video_title": "Dilating triangles find the error Performing transformations Geometry Khan Academy.mp3", "Sentence": "So pause this video and see if you can figure this out on your own. All right, now before I even look at the choices, I like to just think about what would that dilation actually look like? So our center of dilation is P and it's a scale factor of 3 4th. So one way to think about it is, however far any point was from P before, it's not going to be 3 4ths as far, but along the same line. So I'm just going to estimate it. So if C was there, 3 1 2 would be this far. So 3 4ths would be right about there."}, {"video_title": "Dilating triangles find the error Performing transformations Geometry Khan Academy.mp3", "Sentence": "So one way to think about it is, however far any point was from P before, it's not going to be 3 4ths as far, but along the same line. So I'm just going to estimate it. So if C was there, 3 1 2 would be this far. So 3 4ths would be right about there. So C prime should be about there. If we have this line connecting B and P like this, 3, let's see, half of that is there. 3 4ths is going to be there."}, {"video_title": "Dilating triangles find the error Performing transformations Geometry Khan Academy.mp3", "Sentence": "So 3 4ths would be right about there. So C prime should be about there. If we have this line connecting B and P like this, 3, let's see, half of that is there. 3 4ths is going to be there. So B prime should be there. And then on this line, halfway is roughly there. I'm just eyeballing it."}, {"video_title": "Dilating triangles find the error Performing transformations Geometry Khan Academy.mp3", "Sentence": "3 4ths is going to be there. So B prime should be there. And then on this line, halfway is roughly there. I'm just eyeballing it. So 3 4ths is there. So A prime, A prime should be there. And so A prime, B prime, C prime should look something like this, which we can see is exactly what we see for choice C. So choice C, it looks correct."}, {"video_title": "Dilating triangles find the error Performing transformations Geometry Khan Academy.mp3", "Sentence": "I'm just eyeballing it. So 3 4ths is there. So A prime, A prime should be there. And so A prime, B prime, C prime should look something like this, which we can see is exactly what we see for choice C. So choice C, it looks correct. So I'm gonna just circle that or select it just like that. But let's just make sure we understand why these other two choices were not correct. So choice A, it looks like it is a dilation with a 3 4ths scale factor."}, {"video_title": "Dilating triangles find the error Performing transformations Geometry Khan Academy.mp3", "Sentence": "And so A prime, B prime, C prime should look something like this, which we can see is exactly what we see for choice C. So choice C, it looks correct. So I'm gonna just circle that or select it just like that. But let's just make sure we understand why these other two choices were not correct. So choice A, it looks like it is a dilation with a 3 4ths scale factor. Each of the dimensions, each of the sides of these triangles, of this triangle looks like it's about 3 4ths of what it originally was. But it doesn't look like the center of dilation is P. Here, the center of dilation looks like it is probably the midpoint between or the midpoint of segment AC, because now it looks like everything is 3 4ths of the distance it was to that point. So they have this other center of dilation in choice A."}, {"video_title": "Dilating triangles find the error Performing transformations Geometry Khan Academy.mp3", "Sentence": "So choice A, it looks like it is a dilation with a 3 4ths scale factor. Each of the dimensions, each of the sides of these triangles, of this triangle looks like it's about 3 4ths of what it originally was. But it doesn't look like the center of dilation is P. Here, the center of dilation looks like it is probably the midpoint between or the midpoint of segment AC, because now it looks like everything is 3 4ths of the distance it was to that point. So they have this other center of dilation in choice A. The center of dilation is not P, and that's why we can rule that one out. And then for choice B right over here, it looks like they just got the scale factor wrong. Actually, they got the center of dilation and the scale factor wrong."}, {"video_title": "Dilating triangles find the error Performing transformations Geometry Khan Academy.mp3", "Sentence": "So they have this other center of dilation in choice A. The center of dilation is not P, and that's why we can rule that one out. And then for choice B right over here, it looks like they just got the scale factor wrong. Actually, they got the center of dilation and the scale factor wrong. It still looks like they are using this as a center of dilation, but this scale factor looks like it's much closer to 1 4th or 1 3rd, not 3 4ths. So that's why we can rule that one out as well. We like our choice C."}, {"video_title": "Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "Well, let's say we wanted to construct a square. How could we slice a cube with a plane to get the intersection of this cube and that plane to be a square? Well, imagine if that plane were to cut just like this. A square is maybe the most obvious one. So it cuts the top right over there. It cuts the top right over there. It cuts this side right over here."}, {"video_title": "Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "A square is maybe the most obvious one. So it cuts the top right over there. It cuts the top right over there. It cuts this side right over here. It cuts this side right over here. It cuts this side, I guess, in the back. If it's a glass cube, you'd be able to see it."}, {"video_title": "Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "It cuts this side right over here. It cuts this side right over here. It cuts this side, I guess, in the back. If it's a glass cube, you'd be able to see it. Right over there, a dotted line. And then it cuts this right over here. So you could imagine a plane that did this."}, {"video_title": "Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "If it's a glass cube, you'd be able to see it. Right over there, a dotted line. And then it cuts this right over here. So you could imagine a plane that did this. And if I wanted to draw the broader plane, I could draw it like this. Let me see if I can do a decent and adequate job of drawing the actual, I guess you could say, part of the plane that is cutting this cube. It could look something like this."}, {"video_title": "Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "So you could imagine a plane that did this. And if I wanted to draw the broader plane, I could draw it like this. Let me see if I can do a decent and adequate job of drawing the actual, I guess you could say, part of the plane that is cutting this cube. It could look something like this. And I could even color in the part of the plane that you could actually see if the cube were opaque. If you couldn't see through it. But if you could see it through it, you'd see this dotted line and the plane would look like that."}, {"video_title": "Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "It could look something like this. And I could even color in the part of the plane that you could actually see if the cube were opaque. If you couldn't see through it. But if you could see it through it, you'd see this dotted line and the plane would look like that. So a square is a pretty straightforward thing to get if you're doing a planar slice of a cube. But what about a rectangle? How can you get that?"}, {"video_title": "Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "But if you could see it through it, you'd see this dotted line and the plane would look like that. So a square is a pretty straightforward thing to get if you're doing a planar slice of a cube. But what about a rectangle? How can you get that? And at any point, I encourage you, pause the video and try to think about it on your own. How can you get these shapes that I'm talking about? Well, a rectangle, you can actually cut like this."}, {"video_title": "Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "How can you get that? And at any point, I encourage you, pause the video and try to think about it on your own. How can you get these shapes that I'm talking about? Well, a rectangle, you can actually cut like this. So if you cut this side like this, and then cut that side like that, and then you cut this side like that, I think you see where this is going, this side like that, and then you cut the bottom right over there, then the intersection of the plane that you're cutting with, so the intersection, let's see, this could be the plane that I'm actually cutting with, so the intersection of the plane that I'm cutting with and my cube is going to be a rectangle. So it might look like this, and once again, let me shade in the stuff. If you kind of view this, if you imagine the plane is like one of those huge blades that magicians use to saw people in half, or pretend like they're, or give us the illusion of sawing people in half, it might look something like this."}, {"video_title": "Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "Well, a rectangle, you can actually cut like this. So if you cut this side like this, and then cut that side like that, and then you cut this side like that, I think you see where this is going, this side like that, and then you cut the bottom right over there, then the intersection of the plane that you're cutting with, so the intersection, let's see, this could be the plane that I'm actually cutting with, so the intersection of the plane that I'm cutting with and my cube is going to be a rectangle. So it might look like this, and once again, let me shade in the stuff. If you kind of view this, if you imagine the plane is like one of those huge blades that magicians use to saw people in half, or pretend like they're, or give us the illusion of sawing people in half, it might look something like this. Okay, so you're like, okay, that's not so hard to digest, that I can, if I intersect a plane with a cube, I can get a square, I can get a rectangle. But what about triangles? Well, once again, and pause the video if you think you can figure it out, triangles, not so bad."}, {"video_title": "Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "If you kind of view this, if you imagine the plane is like one of those huge blades that magicians use to saw people in half, or pretend like they're, or give us the illusion of sawing people in half, it might look something like this. Okay, so you're like, okay, that's not so hard to digest, that I can, if I intersect a plane with a cube, I can get a square, I can get a rectangle. But what about triangles? Well, once again, and pause the video if you think you can figure it out, triangles, not so bad. You could cut this side right over here, this side right over here, and this side right over here, and then this is, and of course I could keep drawing the plane, but I think you get the idea. This would be a triangle, and there's different types of triangles that you can construct. You could construct an equilateral triangle, so as long as this cut is the same length as this cut right over here, is the same length as, or the length that it intersects on this space of the cube, that's going to be an equilateral triangle."}, {"video_title": "Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "Well, once again, and pause the video if you think you can figure it out, triangles, not so bad. You could cut this side right over here, this side right over here, and this side right over here, and then this is, and of course I could keep drawing the plane, but I think you get the idea. This would be a triangle, and there's different types of triangles that you can construct. You could construct an equilateral triangle, so as long as this cut is the same length as this cut right over here, is the same length as, or the length that it intersects on this space of the cube, that's going to be an equilateral triangle. If you pushed this point out more, actually let me do that in a different color, if you pushed it out more, you're going to have an isosceles triangle. You're going to have an isosceles triangle. If you were to bring this point really, really, really close, like here, you would approach having a right angle, but it wouldn't be quite a right angle."}, {"video_title": "Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "You could construct an equilateral triangle, so as long as this cut is the same length as this cut right over here, is the same length as, or the length that it intersects on this space of the cube, that's going to be an equilateral triangle. If you pushed this point out more, actually let me do that in a different color, if you pushed it out more, you're going to have an isosceles triangle. You're going to have an isosceles triangle. If you were to bring this point really, really, really close, like here, you would approach having a right angle, but it wouldn't be quite a right angle. You'd still have, these angles would still be less than 90 degrees. You can approach 90 degrees, so you can't quite have an exactly a right angle. And so you can't get to 90 degrees, you're definitely not going to get to 91 degrees, so you're actually not going to be able to do an obtuse triangle either."}, {"video_title": "Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "If you were to bring this point really, really, really close, like here, you would approach having a right angle, but it wouldn't be quite a right angle. You'd still have, these angles would still be less than 90 degrees. You can approach 90 degrees, so you can't quite have an exactly a right angle. And so you can't get to 90 degrees, you're definitely not going to get to 91 degrees, so you're actually not going to be able to do an obtuse triangle either. But you can do an equilateral, you can do an isosceles, you can do scalene triangles. You could do the different types of acute triangles. But now let's do some really interesting things."}, {"video_title": "Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "And so you can't get to 90 degrees, you're definitely not going to get to 91 degrees, so you're actually not going to be able to do an obtuse triangle either. But you can do an equilateral, you can do an isosceles, you can do scalene triangles. You could do the different types of acute triangles. But now let's do some really interesting things. Can you get a pentagon by slicing a cube with a plane? And I really want you to pause the video and think about it here, because that's just a fun thing to think about. How can you get a pentagon by slicing a cube with a plane?"}, {"video_title": "Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "But now let's do some really interesting things. Can you get a pentagon by slicing a cube with a plane? And I really want you to pause the video and think about it here, because that's just a fun thing to think about. How can you get a pentagon by slicing a cube with a plane? All right, so here I go. This is how you can get a pentagon by slicing a cube with a plane. Imagine slicing the top, let me do it a little bit different."}, {"video_title": "Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "How can you get a pentagon by slicing a cube with a plane? All right, so here I go. This is how you can get a pentagon by slicing a cube with a plane. Imagine slicing the top, let me do it a little bit different. So imagine slicing the top right over there. So we imagine slicing the top like this. But imagine slicing this back side like that."}, {"video_title": "Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "Imagine slicing the top, let me do it a little bit different. So imagine slicing the top right over there. So we imagine slicing the top like this. But imagine slicing this back side like that. This back side that you can't see quite like that. Now you slice this side right over here like this. And then you slice this side right over here like this."}, {"video_title": "Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "But imagine slicing this back side like that. This back side that you can't see quite like that. Now you slice this side right over here like this. And then you slice this side right over here like this. This could be, if I wanted to draw the plane, it maybe won't be so obvious if I try to draw the plane, but you get the actual idea. If I slice the right angle, not at a right angle, at the right, at a right angle, actually I shouldn't use the word right angle, that'll confuse everything. If I slice it in the proper angle, then the intersection of my plane and my cube is going to be this pentagon right over here."}, {"video_title": "Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "And then you slice this side right over here like this. This could be, if I wanted to draw the plane, it maybe won't be so obvious if I try to draw the plane, but you get the actual idea. If I slice the right angle, not at a right angle, at the right, at a right angle, actually I shouldn't use the word right angle, that'll confuse everything. If I slice it in the proper angle, then the intersection of my plane and my cube is going to be this pentagon right over here. Now let's up the stakes something. Let's up the stakes even more. What about a hexagon?"}, {"video_title": "Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "If I slice it in the proper angle, then the intersection of my plane and my cube is going to be this pentagon right over here. Now let's up the stakes something. Let's up the stakes even more. What about a hexagon? Can I slice a cube in a way with a two-dimensional plane to get the intersection of the plane and the cube being a hexagon? Well, as you can imagine, I wouldn't have asked you the question unless I could. So let's see if we can do it."}, {"video_title": "Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "What about a hexagon? Can I slice a cube in a way with a two-dimensional plane to get the intersection of the plane and the cube being a hexagon? Well, as you can imagine, I wouldn't have asked you the question unless I could. So let's see if we can do it. If we slice this right over there, if we slice this bottom piece right over there, and then you slice this back side like that, you slice this back side like that, and then you slice this side that we can see right over there, and then this side that we can see right over there, and this side I could have written it much straighter. Hopefully you get the idea. I can slice this cube so that I can actually get a hexagon."}, {"video_title": "Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "So let's see if we can do it. If we slice this right over there, if we slice this bottom piece right over there, and then you slice this back side like that, you slice this back side like that, and then you slice this side that we can see right over there, and then this side that we can see right over there, and this side I could have written it much straighter. Hopefully you get the idea. I can slice this cube so that I can actually get a hexagon. Hopefully this gives you a better appreciation for what you can actually do with a cube, especially if you're busy slicing it with large planar planes or large planar blades in some way. There's actually more to a cube than maybe you might have imagined in the past. One way to think about it is there are six sides to a cube or six surfaces to a cube."}, {"video_title": "Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "I can slice this cube so that I can actually get a hexagon. Hopefully this gives you a better appreciation for what you can actually do with a cube, especially if you're busy slicing it with large planar planes or large planar blades in some way. There's actually more to a cube than maybe you might have imagined in the past. One way to think about it is there are six sides to a cube or six surfaces to a cube. You can cut as many as six of the surfaces when you intersect it with a plane, and every time you cut into one of those surfaces, it forms a side. Here we're cutting into four sides. Here we're cutting into four surfaces or four sides."}, {"video_title": "Ways to cut a cube Perimeter, area, and volume Geometry Khan Academy.mp3", "Sentence": "One way to think about it is there are six sides to a cube or six surfaces to a cube. You can cut as many as six of the surfaces when you intersect it with a plane, and every time you cut into one of those surfaces, it forms a side. Here we're cutting into four sides. Here we're cutting into four surfaces or four sides. Here we're cutting into three. Here we're cutting into five. We're not cutting into the bottom of the cube here."}, {"video_title": "Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3", "Sentence": "And then they say use one of the triangles, so use one of these three triangles, to approximate the ratio, the ratio is the length of segment PN divided by the length of segment MN. So they want us to figure out the ratio PN over MN. So pause this video and see if you can figure this out. All right, now let's work through this together. Now, given that they want us to figure out this ratio, and they want us to actually evaluate it or be able to approximate it, we are probably dealing with similarity. And so what I would wanna look for is, are one of these triangles similar to the triangle we have here? And we're dealing with similar triangles if we have two angles in common, because if we have two angles in common, then that means that we definitely have the third angle as well, because the third angle is completely determined by what the other two angles are."}, {"video_title": "Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3", "Sentence": "All right, now let's work through this together. Now, given that they want us to figure out this ratio, and they want us to actually evaluate it or be able to approximate it, we are probably dealing with similarity. And so what I would wanna look for is, are one of these triangles similar to the triangle we have here? And we're dealing with similar triangles if we have two angles in common, because if we have two angles in common, then that means that we definitely have the third angle as well, because the third angle is completely determined by what the other two angles are. So we have a 35 degree angle here, and we have a 90 degree angle here. And of all of these choices, this doesn't have a 35 degree angle, it has a 90. This doesn't have a 35, it has a 90."}, {"video_title": "Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3", "Sentence": "And we're dealing with similar triangles if we have two angles in common, because if we have two angles in common, then that means that we definitely have the third angle as well, because the third angle is completely determined by what the other two angles are. So we have a 35 degree angle here, and we have a 90 degree angle here. And of all of these choices, this doesn't have a 35 degree angle, it has a 90. This doesn't have a 35, it has a 90. But triangle two here has a 35 degree angle, has a 90 degree angle, and has a 55 degree angle. And if you did the math, knowing that 35 plus 90 plus this have to add up to 180 degrees, you would see that this too has a measure of 55 degrees. And so given that all of our angle measures are the same between triangle PNM and triangle number two right over here, we know that these two are similar triangles."}, {"video_title": "Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3", "Sentence": "This doesn't have a 35, it has a 90. But triangle two here has a 35 degree angle, has a 90 degree angle, and has a 55 degree angle. And if you did the math, knowing that 35 plus 90 plus this have to add up to 180 degrees, you would see that this too has a measure of 55 degrees. And so given that all of our angle measures are the same between triangle PNM and triangle number two right over here, we know that these two are similar triangles. And so the ratios between corresponding sides are going to be the same. We could either take the ratio across triangles, or we could say the ratio within when we just look at one triangle. And so if you look at PN over MN, let me try to color code it."}, {"video_title": "Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3", "Sentence": "And so given that all of our angle measures are the same between triangle PNM and triangle number two right over here, we know that these two are similar triangles. And so the ratios between corresponding sides are going to be the same. We could either take the ratio across triangles, or we could say the ratio within when we just look at one triangle. And so if you look at PN over MN, let me try to color code it. So PN right over here, that corresponds to the side that's opposite the 35 degree angle. So that would correspond to this side right over here on triangle two. And then MN, that's this that I'm coloring in this bluish color."}, {"video_title": "Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3", "Sentence": "And so if you look at PN over MN, let me try to color code it. So PN right over here, that corresponds to the side that's opposite the 35 degree angle. So that would correspond to this side right over here on triangle two. And then MN, that's this that I'm coloring in this bluish color. Not so well. Probably spent more time coloring. That's opposite the 55 degree angle."}, {"video_title": "Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3", "Sentence": "And then MN, that's this that I'm coloring in this bluish color. Not so well. Probably spent more time coloring. That's opposite the 55 degree angle. And so opposite the 55 degree angle would be right over there. Now, since these triangles are similar, the ratio of the length of the red side over the length of the blue side is going to be the same in either triangle. So PN, let me write it this way."}, {"video_title": "Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3", "Sentence": "That's opposite the 55 degree angle. And so opposite the 55 degree angle would be right over there. Now, since these triangles are similar, the ratio of the length of the red side over the length of the blue side is going to be the same in either triangle. So PN, let me write it this way. The length of segment PN over the length of segment MN is going to be equivalent to 5.7 over 8.2. Because this ratio is going to be the same for the corresponding sides, regardless of which triangle you look at. So if you take the side that's opposite 35 degrees, that's 5.7 over 8.2."}, {"video_title": "Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3", "Sentence": "So PN, let me write it this way. The length of segment PN over the length of segment MN is going to be equivalent to 5.7 over 8.2. Because this ratio is going to be the same for the corresponding sides, regardless of which triangle you look at. So if you take the side that's opposite 35 degrees, that's 5.7 over 8.2. Now, to be very clear, it doesn't mean that somehow the length of this side is 5.7 or that the length of this side is 8.2. We would only be able to make that conclusion if they were congruent. But with similarity, we know that the ratios, if we look at the ratio of the red side to the blue side on each of those triangles, that would be the same."}, {"video_title": "Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3", "Sentence": "So if you take the side that's opposite 35 degrees, that's 5.7 over 8.2. Now, to be very clear, it doesn't mean that somehow the length of this side is 5.7 or that the length of this side is 8.2. We would only be able to make that conclusion if they were congruent. But with similarity, we know that the ratios, if we look at the ratio of the red side to the blue side on each of those triangles, that would be the same. And so this gives us that ratio. And let's see, 5.7 over 8.2. Which of these choices get close to that?"}, {"video_title": "Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3", "Sentence": "But with similarity, we know that the ratios, if we look at the ratio of the red side to the blue side on each of those triangles, that would be the same. And so this gives us that ratio. And let's see, 5.7 over 8.2. Which of these choices get close to that? Well, we could say that this is roughly, if I am approximating it, let's see, it's going to be larger than 0.57 because 8.2 is less than 10. And so we are going to rule this choice out. And 5.7 is less than 8.2, so it can't be over one."}, {"video_title": "Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3", "Sentence": "Which of these choices get close to that? Well, we could say that this is roughly, if I am approximating it, let's see, it's going to be larger than 0.57 because 8.2 is less than 10. And so we are going to rule this choice out. And 5.7 is less than 8.2, so it can't be over one. And so we have to think between these two choices. Well, the simplest thing I can do is actually just try to start dividing it by hand. So 8.2 goes into 5.7 the same number of times as 82 goes into 57."}, {"video_title": "Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3", "Sentence": "And 5.7 is less than 8.2, so it can't be over one. And so we have to think between these two choices. Well, the simplest thing I can do is actually just try to start dividing it by hand. So 8.2 goes into 5.7 the same number of times as 82 goes into 57. And I'll add some decimals here. So it doesn't go into 57, but how many times does 82 go into 570? I would assume it's about six times, maybe seven times, looks like."}, {"video_title": "Using similarity to estimate ratio between side lengths High school geometry Khan Academy.mp3", "Sentence": "So 8.2 goes into 5.7 the same number of times as 82 goes into 57. And I'll add some decimals here. So it doesn't go into 57, but how many times does 82 go into 570? I would assume it's about six times, maybe seven times, looks like. So seven times two is 14, and then seven times eight is 56, I guess it's 57. So it's actually a little less than 0.7. This got, made me go a little bit too high."}, {"video_title": "Dilating shapes shrinking Performing transformations High school geometry Khan Academy.mp3", "Sentence": "We're told draw the image of triangle ABC under a dilation whose center is P and scale factor is 1 4th. And what we see here is the widget on Khan Academy where we can do that. So we have this figure, this triangle, ABC, A, B, C, right over here. And what we wanna do is dilate it. So that means scaling it up or down. And the center of that dilation is this point P. So one way to think about it is, let's think about the distance between point P and each of these points, and we wanna scale it by 1 4th. So the distance is gonna be 1 4th of what it was before."}, {"video_title": "Dilating shapes shrinking Performing transformations High school geometry Khan Academy.mp3", "Sentence": "And what we wanna do is dilate it. So that means scaling it up or down. And the center of that dilation is this point P. So one way to think about it is, let's think about the distance between point P and each of these points, and we wanna scale it by 1 4th. So the distance is gonna be 1 4th of what it was before. So for example, this point right over here, if we just even look diagonally from P to A, we can see that we are crossing one square, two squares, three squares, four squares. So if we have a scale factor of 1 4th, instead of crossing four squares diagonally, we would only cross one square diagonally. So I'll put the corresponding point to A right over there."}, {"video_title": "Dilating shapes shrinking Performing transformations High school geometry Khan Academy.mp3", "Sentence": "So the distance is gonna be 1 4th of what it was before. So for example, this point right over here, if we just even look diagonally from P to A, we can see that we are crossing one square, two squares, three squares, four squares. So if we have a scale factor of 1 4th, instead of crossing four squares diagonally, we would only cross one square diagonally. So I'll put the corresponding point to A right over there. Now what about for point C? It's not quite as obvious, but one way we could think about it is, we could think about how far are we going horizontally from P to C, and then how far do we go vertically? So horizontally, we're going one, two, three, four, five, six, seven, eight of these units, and then vertically, we're going one, two, three, four."}, {"video_title": "Dilating shapes shrinking Performing transformations High school geometry Khan Academy.mp3", "Sentence": "So I'll put the corresponding point to A right over there. Now what about for point C? It's not quite as obvious, but one way we could think about it is, we could think about how far are we going horizontally from P to C, and then how far do we go vertically? So horizontally, we're going one, two, three, four, five, six, seven, eight of these units, and then vertically, we're going one, two, three, four. So we're going to the left eight and up four. Now if we have a scale factor of 1 4th, we just multiply each of those by 1 4th. So instead of going to the left eight, we would go to the left two."}, {"video_title": "Dilating shapes shrinking Performing transformations High school geometry Khan Academy.mp3", "Sentence": "So horizontally, we're going one, two, three, four, five, six, seven, eight of these units, and then vertically, we're going one, two, three, four. So we're going to the left eight and up four. Now if we have a scale factor of 1 4th, we just multiply each of those by 1 4th. So instead of going to the left eight, we would go to the left two. Eight times 1 4th is two. Instead of going up four, we would go up one. So this would be the corresponding point to point C. And then we'll do the same thing for point B."}, {"video_title": "Dilating shapes shrinking Performing transformations High school geometry Khan Academy.mp3", "Sentence": "So instead of going to the left eight, we would go to the left two. Eight times 1 4th is two. Instead of going up four, we would go up one. So this would be the corresponding point to point C. And then we'll do the same thing for point B. When we go from P to B, we're going one, two, three, four, five, six, seven, eight up, and we're going four to the left. So if we have a scale factor of 1 4th, instead of going eight up, we'll go two up, and instead of going four to the left, we will go one to the left. So there you have it."}, {"video_title": "Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3", "Sentence": "We have a bunch of triangles here and some lengths of sides and a couple of right angles. Maybe we can establish similarity between some of the triangles. There's actually three different triangles that I can see here. This triangle, this triangle, and this larger triangle. If we can establish some similarity here, maybe we can use ratios between sides somehow to figure out what BC is. When you look at it, you have a right angle right over here. In triangle BDC, you have one right angle."}, {"video_title": "Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3", "Sentence": "This triangle, this triangle, and this larger triangle. If we can establish some similarity here, maybe we can use ratios between sides somehow to figure out what BC is. When you look at it, you have a right angle right over here. In triangle BDC, you have one right angle. In triangle ABC, you have another right angle. If we can show that they have another angle or another corresponding set of angles that are congruent to each other, then we can show that they're similar. Actually, both of those triangles, both BDC and ABC, both share this angle right over here."}, {"video_title": "Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3", "Sentence": "In triangle BDC, you have one right angle. In triangle ABC, you have another right angle. If we can show that they have another angle or another corresponding set of angles that are congruent to each other, then we can show that they're similar. Actually, both of those triangles, both BDC and ABC, both share this angle right over here. If they share that angle, then they definitely share two angles. They both share that angle right over there. Let me do that in a different color just to make it different than those right angles."}, {"video_title": "Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3", "Sentence": "Actually, both of those triangles, both BDC and ABC, both share this angle right over here. If they share that angle, then they definitely share two angles. They both share that angle right over there. Let me do that in a different color just to make it different than those right angles. They both share that angle there. We know that two triangles that have at least two of their angles, that have at least two congruent angles, they are going to be similar triangles. We know that triangle, I'll write this, triangle ABC, we went from the unlabeled angle to the yellow right angle to the orange angle."}, {"video_title": "Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3", "Sentence": "Let me do that in a different color just to make it different than those right angles. They both share that angle there. We know that two triangles that have at least two of their angles, that have at least two congruent angles, they are going to be similar triangles. We know that triangle, I'll write this, triangle ABC, we went from the unlabeled angle to the yellow right angle to the orange angle. Let me write it this way. We went from the unlabeled angle right over here to the orange angle, or sorry, to the yellow angle, I'm having trouble with colors, to the orange angle, ABC. We want to do this very carefully here because the same points or the same vertices might not play the same role in both triangles."}, {"video_title": "Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3", "Sentence": "We know that triangle, I'll write this, triangle ABC, we went from the unlabeled angle to the yellow right angle to the orange angle. Let me write it this way. We went from the unlabeled angle right over here to the orange angle, or sorry, to the yellow angle, I'm having trouble with colors, to the orange angle, ABC. We want to do this very carefully here because the same points or the same vertices might not play the same role in both triangles. We want to make sure we're getting the similarity right. White vertex to the 90 degree angle vertex to the orange vertex, that is going to be similar to triangle. Which is the one that is neither right angle, so we're looking at the smaller triangle right over here, which is the one that is neither a right angle or the orange angle?"}, {"video_title": "Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3", "Sentence": "We want to do this very carefully here because the same points or the same vertices might not play the same role in both triangles. We want to make sure we're getting the similarity right. White vertex to the 90 degree angle vertex to the orange vertex, that is going to be similar to triangle. Which is the one that is neither right angle, so we're looking at the smaller triangle right over here, which is the one that is neither a right angle or the orange angle? It's going to be vertex B. Vertex B had the right angle when you think about the larger triangle, but we haven't thought about just that little angle right over there. We're going to start at vertex B, then we're going to go to the right angle. The right angle is vertex D, and then we go to vertex C, which is in orange."}, {"video_title": "Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3", "Sentence": "Which is the one that is neither right angle, so we're looking at the smaller triangle right over here, which is the one that is neither a right angle or the orange angle? It's going to be vertex B. Vertex B had the right angle when you think about the larger triangle, but we haven't thought about just that little angle right over there. We're going to start at vertex B, then we're going to go to the right angle. The right angle is vertex D, and then we go to vertex C, which is in orange. We have shown that they are similar. Now that we know that they are similar, we can attempt to take ratios between the sides. Let's think about it."}, {"video_title": "Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3", "Sentence": "The right angle is vertex D, and then we go to vertex C, which is in orange. We have shown that they are similar. Now that we know that they are similar, we can attempt to take ratios between the sides. Let's think about it. We know what the length of AC is. AC is going to be equal to 8, 6 plus 2. We know that AC, what's the corresponding side on this triangle right over here?"}, {"video_title": "Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3", "Sentence": "Let's think about it. We know what the length of AC is. AC is going to be equal to 8, 6 plus 2. We know that AC, what's the corresponding side on this triangle right over here? You can literally look at the letters. A and C is going to correspond to BC. The first and the third, first and the third."}, {"video_title": "Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3", "Sentence": "We know that AC, what's the corresponding side on this triangle right over here? You can literally look at the letters. A and C is going to correspond to BC. The first and the third, first and the third. AC is going to correspond to BC. This is interesting because we're already involving BC. What is going to correspond to, and then if we look at BC on the larger triangle, so if we look at BC on the larger triangle, BC is going to correspond to what on the smaller triangle?"}, {"video_title": "Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3", "Sentence": "The first and the third, first and the third. AC is going to correspond to BC. This is interesting because we're already involving BC. What is going to correspond to, and then if we look at BC on the larger triangle, so if we look at BC on the larger triangle, BC is going to correspond to what on the smaller triangle? It's going to correspond to DC. It's good because we know what AC is, and we know what DC is, and so we can solve for BC. I want to take one more step to show you what we just did here because BC is playing two different roles."}, {"video_title": "Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3", "Sentence": "What is going to correspond to, and then if we look at BC on the larger triangle, so if we look at BC on the larger triangle, BC is going to correspond to what on the smaller triangle? It's going to correspond to DC. It's good because we know what AC is, and we know what DC is, and so we can solve for BC. I want to take one more step to show you what we just did here because BC is playing two different roles. On this first statement right over here, we're thinking of BC. BC on our smaller triangle corresponds to AC on our larger triangle. Then in the second statement, BC on our larger triangle corresponds to DC on our smaller triangle."}, {"video_title": "Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3", "Sentence": "I want to take one more step to show you what we just did here because BC is playing two different roles. On this first statement right over here, we're thinking of BC. BC on our smaller triangle corresponds to AC on our larger triangle. Then in the second statement, BC on our larger triangle corresponds to DC on our smaller triangle. In both of these cases, these are our larger triangles, and then this is from the smaller triangle right over here, corresponding sides. This is a cool problem because BC plays two different roles in both triangles. Now we have enough information to solve for BC."}, {"video_title": "Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3", "Sentence": "Then in the second statement, BC on our larger triangle corresponds to DC on our smaller triangle. In both of these cases, these are our larger triangles, and then this is from the smaller triangle right over here, corresponding sides. This is a cool problem because BC plays two different roles in both triangles. Now we have enough information to solve for BC. We know that AC is equal to 9. We know that AC is equal to 8. 6 plus 2 is 8."}, {"video_title": "Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3", "Sentence": "Now we have enough information to solve for BC. We know that AC is equal to 9. We know that AC is equal to 8. 6 plus 2 is 8. We know that DC is equal to 2. That's given. Now we can cross multiply."}, {"video_title": "Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3", "Sentence": "6 plus 2 is 8. We know that DC is equal to 2. That's given. Now we can cross multiply. 8 times 2 is 16, is equal to BC times BC, is equal to BC squared. BC is going to be equal to the principal root of 16, which is 4. BC is equal to 4."}, {"video_title": "Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3", "Sentence": "Now we can cross multiply. 8 times 2 is 16, is equal to BC times BC, is equal to BC squared. BC is going to be equal to the principal root of 16, which is 4. BC is equal to 4. BC is equal to 4, and we're done. The hardest part about this problem is just realizing that BC plays two different roles and just keeping your head straight on those two different roles. Just to make it clear, let me actually draw these two triangles separately."}, {"video_title": "Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3", "Sentence": "BC is equal to 4. BC is equal to 4, and we're done. The hardest part about this problem is just realizing that BC plays two different roles and just keeping your head straight on those two different roles. Just to make it clear, let me actually draw these two triangles separately. If I drew ABC separately, it would look like this. It would look like this. This is my triangle ABC."}, {"video_title": "Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3", "Sentence": "Just to make it clear, let me actually draw these two triangles separately. If I drew ABC separately, it would look like this. It would look like this. This is my triangle ABC. Then this is a right angle. This is our orange angle. We know that the length of this side right over here is 8, and we know that the length of this side when we figure it out through this problem is 4."}, {"video_title": "Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3", "Sentence": "This is my triangle ABC. Then this is a right angle. This is our orange angle. We know that the length of this side right over here is 8, and we know that the length of this side when we figure it out through this problem is 4. Then if we wanted to draw BDC, we would draw it like this. BDC looks like this. This is BDC."}, {"video_title": "Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3", "Sentence": "We know that the length of this side right over here is 8, and we know that the length of this side when we figure it out through this problem is 4. Then if we wanted to draw BDC, we would draw it like this. BDC looks like this. This is BDC. That's a little bit easier to visualize because this is our right angle. This is our orange angle. This is 4, and this right over here is 2."}, {"video_title": "Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3", "Sentence": "This is BDC. That's a little bit easier to visualize because this is our right angle. This is our orange angle. This is 4, and this right over here is 2. I did it this way to show you that you kind of have to flip this triangle over and rotate it just to have kind of a similar orientation. Then it might make it look a little bit clearer. If you found this part confusing, I encourage you to try to flip and rotate BDC in such a way that it seems to look a lot like ABC."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3", "Sentence": "Give the lengths to the nearest tenth. So when they say solve the right triangle, we can assume that they're saying, hey, figure out the lengths of all the sides, so whatever a is equal to, whatever b is equal to, and also what are all the angles of the right triangle? They've given two of them, we might have to figure out this third right over here. So there's multiple ways to tackle this, but we'll just go and we'll just try to tackle side xw first, try to figure out what a is, and I'll give you a hint. You can use a calculator, and using a calculator, you can use your trigonometric functions that we've looked at a good bit now. So I'll give you a few seconds to think about how to figure out what a is. Well, what do we know?"}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3", "Sentence": "So there's multiple ways to tackle this, but we'll just go and we'll just try to tackle side xw first, try to figure out what a is, and I'll give you a hint. You can use a calculator, and using a calculator, you can use your trigonometric functions that we've looked at a good bit now. So I'll give you a few seconds to think about how to figure out what a is. Well, what do we know? We know this angle y right over here, we know the side adjacent to angle y, and length a, this is the side, that's the length of the side that is opposite, that is opposite to angle y. So what trigonometric ratio, what trigonometric ratio deals with the opposite and the adjacent? So if we're looking at angle y, relative to angle y, this is the opposite, and this right over here is the adjacent."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3", "Sentence": "Well, what do we know? We know this angle y right over here, we know the side adjacent to angle y, and length a, this is the side, that's the length of the side that is opposite, that is opposite to angle y. So what trigonometric ratio, what trigonometric ratio deals with the opposite and the adjacent? So if we're looking at angle y, relative to angle y, this is the opposite, and this right over here is the adjacent. Well, if we don't remember, we can go back to SOHCAHTOA. SOHCAHTOA. Sine deals with opposite and hypotenuse, cosine deals with adjacent and hypotenuse, tangent deals with opposite over adjacent, opposite over adjacent."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3", "Sentence": "So if we're looking at angle y, relative to angle y, this is the opposite, and this right over here is the adjacent. Well, if we don't remember, we can go back to SOHCAHTOA. SOHCAHTOA. Sine deals with opposite and hypotenuse, cosine deals with adjacent and hypotenuse, tangent deals with opposite over adjacent, opposite over adjacent. So we can say that the tangent, the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a, over the length of the adjacent side, which they gave us in the diagram, which has length, which has length five. And you might say, well, how do I figure out a? Well, we can use our calculator to evaluate what the tangent of 65 degrees are, and then we can solve for a."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3", "Sentence": "Sine deals with opposite and hypotenuse, cosine deals with adjacent and hypotenuse, tangent deals with opposite over adjacent, opposite over adjacent. So we can say that the tangent, the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a, over the length of the adjacent side, which they gave us in the diagram, which has length, which has length five. And you might say, well, how do I figure out a? Well, we can use our calculator to evaluate what the tangent of 65 degrees are, and then we can solve for a. And actually, if we just wanna get the expression, explicitly solving for a, we can just multiply both sides of this equation times five. So let's do that. Five times, times five, these cancel out, and we are left with, if we flip the equal around, we're left with a is equal to five times the tangent of 65, of 65 degrees."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3", "Sentence": "Well, we can use our calculator to evaluate what the tangent of 65 degrees are, and then we can solve for a. And actually, if we just wanna get the expression, explicitly solving for a, we can just multiply both sides of this equation times five. So let's do that. Five times, times five, these cancel out, and we are left with, if we flip the equal around, we're left with a is equal to five times the tangent of 65, of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest 10th. My handy TI-85 out, and I have five times the, not the, the tangent, no, I didn't need to press that second right over there, just the regular tangent, of 65 degrees. And I am, I get, if I round to the nearest 10th like they asked me to, I get 10.7."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3", "Sentence": "Five times, times five, these cancel out, and we are left with, if we flip the equal around, we're left with a is equal to five times the tangent of 65, of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest 10th. My handy TI-85 out, and I have five times the, not the, the tangent, no, I didn't need to press that second right over there, just the regular tangent, of 65 degrees. And I am, I get, if I round to the nearest 10th like they asked me to, I get 10.7. So this is, so a is approximately equal to 10.7. I say approximately, because I rounded it, I rounded it down. I didn't, this is not the exact number, but a is equal to 10.7."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3", "Sentence": "And I am, I get, if I round to the nearest 10th like they asked me to, I get 10.7. So this is, so a is approximately equal to 10.7. I say approximately, because I rounded it, I rounded it down. I didn't, this is not the exact number, but a is equal to 10.7. So we now know that this has length 10.7 approximately. There's several ways that we can try to tackle b, and I'll let you pick the way you want to, but then I'll just do it the way I would like to. So my next question to you is, what is the length of the side yw, or what is the value of b?"}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3", "Sentence": "I didn't, this is not the exact number, but a is equal to 10.7. So we now know that this has length 10.7 approximately. There's several ways that we can try to tackle b, and I'll let you pick the way you want to, but then I'll just do it the way I would like to. So my next question to you is, what is the length of the side yw, or what is the value of b? Well, there's several ways to do it. This is a hypotenuse, so we could use trigonometric functions that deal with adjacent over hypotenuse or opposite over hypotenuse, or we could just use the Pythagorean theorem. We know two sides of a right triangle."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3", "Sentence": "So my next question to you is, what is the length of the side yw, or what is the value of b? Well, there's several ways to do it. This is a hypotenuse, so we could use trigonometric functions that deal with adjacent over hypotenuse or opposite over hypotenuse, or we could just use the Pythagorean theorem. We know two sides of a right triangle. We can come up with the third side. I will go with using trigonometric ratios, since that's what we've been working on a good bit. So this length b, that's the length of the hypotenuse."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3", "Sentence": "We know two sides of a right triangle. We can come up with the third side. I will go with using trigonometric ratios, since that's what we've been working on a good bit. So this length b, that's the length of the hypotenuse. So this side wy is the hypotenuse. And so what trigonometric ratios, and we can decide what we want to use. We could use opposite and hypotenuse."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3", "Sentence": "So this length b, that's the length of the hypotenuse. So this side wy is the hypotenuse. And so what trigonometric ratios, and we can decide what we want to use. We could use opposite and hypotenuse. We could use adjacent and hypotenuse. Since we know that xy is exactly five, we don't have to deal with this approximation. Let's use that side."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3", "Sentence": "We could use opposite and hypotenuse. We could use adjacent and hypotenuse. Since we know that xy is exactly five, we don't have to deal with this approximation. Let's use that side. So what trigonometric ratios deal with adjacent and hypotenuse? Well, we see from Sohcahtoa, cosine deals with adjacent over hypotenuse. So we could say that the cosine of 65 degrees, cosine of 65 degrees, is equal to the length of the adjacent side, which is five, over the length of the hypotenuse, which has length b."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3", "Sentence": "Let's use that side. So what trigonometric ratios deal with adjacent and hypotenuse? Well, we see from Sohcahtoa, cosine deals with adjacent over hypotenuse. So we could say that the cosine of 65 degrees, cosine of 65 degrees, is equal to the length of the adjacent side, which is five, over the length of the hypotenuse, which has length b. And then we can try to solve for b. You multiply both sides times b. You're left with b times cosine of 65 degrees is equal to five."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3", "Sentence": "So we could say that the cosine of 65 degrees, cosine of 65 degrees, is equal to the length of the adjacent side, which is five, over the length of the hypotenuse, which has length b. And then we can try to solve for b. You multiply both sides times b. You're left with b times cosine of 65 degrees is equal to five. And then to solve for b, you can divide both sides by cosine of 65 degrees. This is just a number here. So we're just dividing."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3", "Sentence": "You're left with b times cosine of 65 degrees is equal to five. And then to solve for b, you can divide both sides by cosine of 65 degrees. This is just a number here. So we're just dividing. We have to figure it out with our calculator, but this is just going to evaluate to some number. So we can divide both sides by that, by cosine of 65 degrees, cosine of 65 degrees. And we're left with b is equal to five over the cosine of 65 degrees."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3", "Sentence": "So we're just dividing. We have to figure it out with our calculator, but this is just going to evaluate to some number. So we can divide both sides by that, by cosine of 65 degrees, cosine of 65 degrees. And we're left with b is equal to five over the cosine of 65 degrees. So let us now use our calculator to figure out the length of b. Length of b is five divided by cosine of 65 degrees. And I get, if I round to the nearest tenth, 11.8."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3", "Sentence": "And we're left with b is equal to five over the cosine of 65 degrees. So let us now use our calculator to figure out the length of b. Length of b is five divided by cosine of 65 degrees. And I get, if I round to the nearest tenth, 11.8. So b is approximately equal to, rounded to the nearest tenth, 11.8. So b is equal to 11.8. And then we're almost done solving this right triangle."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3", "Sentence": "And I get, if I round to the nearest tenth, 11.8. So b is approximately equal to, rounded to the nearest tenth, 11.8. So b is equal to 11.8. And then we're almost done solving this right triangle. And you could have figured this out using the Pythagorean theorem as well, saying that five squared plus 10.7 squared should be equal to b squared. And hopefully you would get the exact same answer. And the last thing we have to figure out is the measure of angle w. Angle w right over here."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3", "Sentence": "And then we're almost done solving this right triangle. And you could have figured this out using the Pythagorean theorem as well, saying that five squared plus 10.7 squared should be equal to b squared. And hopefully you would get the exact same answer. And the last thing we have to figure out is the measure of angle w. Angle w right over here. So I'll give you a few seconds to think about what the measure of angle w is. Well, here we just have to remember that the sum of the angles of a triangle add up to 180 degrees. So angle w plus 65 degrees, that's this angle right up here, plus the right angle, this is a right triangle, they're going to add up to 180 degrees."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy (2).mp3", "Sentence": "And the last thing we have to figure out is the measure of angle w. Angle w right over here. So I'll give you a few seconds to think about what the measure of angle w is. Well, here we just have to remember that the sum of the angles of a triangle add up to 180 degrees. So angle w plus 65 degrees, that's this angle right up here, plus the right angle, this is a right triangle, they're going to add up to 180 degrees. So all we need to do is, well, we could simplify the left-hand side right over here. 65 plus 90 is 155. So angle w plus 155 degrees is equal to 180 degrees."}, {"video_title": "Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3", "Sentence": "What we've got over here is a triangle where all three sides have the same length, or all three sides are congruent to each other. And a triangle like this we call equilateral. This is an equilateral triangle. Now what I want to do is prove that if all three sides are the same, then we know that all three angles are going to have the same measure. So let's think how we can do this. Well, first of all, we could just look at, we know that AB is equal to AC. So let's just pretend that we don't even know that this also happens to be equal to BC."}, {"video_title": "Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3", "Sentence": "Now what I want to do is prove that if all three sides are the same, then we know that all three angles are going to have the same measure. So let's think how we can do this. Well, first of all, we could just look at, we know that AB is equal to AC. So let's just pretend that we don't even know that this also happens to be equal to BC. And we know for isosceles triangles, if two legs have the same length, then the base angles have the same length. So let's write this down. We know that angle ABC is going to be congruent to angle ACB."}, {"video_title": "Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3", "Sentence": "So let's just pretend that we don't even know that this also happens to be equal to BC. And we know for isosceles triangles, if two legs have the same length, then the base angles have the same length. So let's write this down. We know that angle ABC is going to be congruent to angle ACB. So let me write this down. We know angle ABC is congruent to angle ACB because, so maybe this is my statement right over here, statement, and then we have reason. And the reason here, and I'll write it in just kind of shorthand, is that they're base angles of, I guess you could say, an isosceles because we know that this side is equal to that side."}, {"video_title": "Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3", "Sentence": "We know that angle ABC is going to be congruent to angle ACB. So let me write this down. We know angle ABC is congruent to angle ACB because, so maybe this is my statement right over here, statement, and then we have reason. And the reason here, and I'll write it in just kind of shorthand, is that they're base angles of, I guess you could say, an isosceles because we know that this side is equal to that side. Now obviously this is an equilateral. All of the sides are equal, but the fact that these two legs are equal show that the base angles are equal. So we say two legs equal imply base angles are going to be equal."}, {"video_title": "Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3", "Sentence": "And the reason here, and I'll write it in just kind of shorthand, is that they're base angles of, I guess you could say, an isosceles because we know that this side is equal to that side. Now obviously this is an equilateral. All of the sides are equal, but the fact that these two legs are equal show that the base angles are equal. So we say two legs equal imply base angles are going to be equal. And that just comes from what we actually did in the last video with isosceles triangles. But we could also view this triangle the other way. We could also say that maybe this angle over here is the vertex angle, maybe these two are the base angles."}, {"video_title": "Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3", "Sentence": "So we say two legs equal imply base angles are going to be equal. And that just comes from what we actually did in the last video with isosceles triangles. But we could also view this triangle the other way. We could also say that maybe this angle over here is the vertex angle, maybe these two are the base angles. Because then you have a situation where this side and this side are congruent to each other, and then that angle and that angle are going to be base angles. So you could say angle CAB is going to be congruent to angle ABC really for the same reason. So now looking at different legs here and different base angles, this would now be the base in this example."}, {"video_title": "Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3", "Sentence": "We could also say that maybe this angle over here is the vertex angle, maybe these two are the base angles. Because then you have a situation where this side and this side are congruent to each other, and then that angle and that angle are going to be base angles. So you could say angle CAB is going to be congruent to angle ABC really for the same reason. So now looking at different legs here and different base angles, this would now be the base in this example. You could imagine turning an isosceles triangle on its side, but it's the exact same logic. So let's just review what I talked about. These two sides are equal, which imply these two base angles are equal."}, {"video_title": "Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3", "Sentence": "So now looking at different legs here and different base angles, this would now be the base in this example. You could imagine turning an isosceles triangle on its side, but it's the exact same logic. So let's just review what I talked about. These two sides are equal, which imply these two base angles are equal. These two sides being equal imply these two base angles are equal. Well, if ABC is congruent to ACB and it's congruent to CAB, all of these angles are congruent to each other. So then we get angle ABC is congruent to angle ACB, which is congruent to angle CAB."}, {"video_title": "Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3", "Sentence": "These two sides are equal, which imply these two base angles are equal. These two sides being equal imply these two base angles are equal. Well, if ABC is congruent to ACB and it's congruent to CAB, all of these angles are congruent to each other. So then we get angle ABC is congruent to angle ACB, which is congruent to angle CAB. And that pretty much gives us all of the angles. So if you have an equilateral triangle, it's actually an equiangular triangle as well. All of the angles are going to be the same."}, {"video_title": "Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3", "Sentence": "So then we get angle ABC is congruent to angle ACB, which is congruent to angle CAB. And that pretty much gives us all of the angles. So if you have an equilateral triangle, it's actually an equiangular triangle as well. All of the angles are going to be the same. And you actually know what that measure is if you have three things that are the same. So let's call that x, x, x, and they add up to 180. You get x plus x plus x is equal to 180, or 3x is equal to 180."}, {"video_title": "Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3", "Sentence": "All of the angles are going to be the same. And you actually know what that measure is if you have three things that are the same. So let's call that x, x, x, and they add up to 180. You get x plus x plus x is equal to 180, or 3x is equal to 180. Divide both sides by 3. You get x is equal to 60 degrees. So in an equilateral triangle, not only are they all the same angles, but they're all equal to exactly 60 degree angles."}, {"video_title": "Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3", "Sentence": "You get x plus x plus x is equal to 180, or 3x is equal to 180. Divide both sides by 3. You get x is equal to 60 degrees. So in an equilateral triangle, not only are they all the same angles, but they're all equal to exactly 60 degree angles. Now let's think about it the other way around. Let's say I have a triangle. Let's say we've got ourselves a triangle where all of the angles are the same."}, {"video_title": "Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3", "Sentence": "So in an equilateral triangle, not only are they all the same angles, but they're all equal to exactly 60 degree angles. Now let's think about it the other way around. Let's say I have a triangle. Let's say we've got ourselves a triangle where all of the angles are the same. So let's say that's point x, point y, and point z. And we know that all the angles are the same. So we know that this angle is congruent to this angle is congruent to that angle."}, {"video_title": "Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3", "Sentence": "Let's say we've got ourselves a triangle where all of the angles are the same. So let's say that's point x, point y, and point z. And we know that all the angles are the same. So we know that this angle is congruent to this angle is congruent to that angle. So what we showed in the last video on isosceles triangles is that if two base angles are the same, then the corresponding legs are also going to be the same. So we know, for example, that yx is congruent to yz. We know yx is congruent to yz."}, {"video_title": "Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3", "Sentence": "So we know that this angle is congruent to this angle is congruent to that angle. So what we showed in the last video on isosceles triangles is that if two base angles are the same, then the corresponding legs are also going to be the same. So we know, for example, that yx is congruent to yz. We know yx is congruent to yz. And we know that because the base angles are congruent. We also know that yz is congruent to xz by the same argument. But here we're dealing with different base angles."}, {"video_title": "Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3", "Sentence": "We know yx is congruent to yz. And we know that because the base angles are congruent. We also know that yz is congruent to xz by the same argument. But here we're dealing with different base angles. So now once again you can view this as almost an isosceles triangle turned on its side. This is the vertex angle right over here. These are the two base angles."}, {"video_title": "Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3", "Sentence": "But here we're dealing with different base angles. So now once again you can view this as almost an isosceles triangle turned on its side. This is the vertex angle right over here. These are the two base angles. This would be the base now. And we know that because these two base angles are congruent by the same logic. In this first case, the base angles were this angle and that angle."}, {"video_title": "Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3", "Sentence": "These are the two base angles. This would be the base now. And we know that because these two base angles are congruent by the same logic. In this first case, the base angles were this angle and that angle. In the second case, the base angles are that angle and that angle. And actually let me write it down. The base angles in this first case, let me do that same magenta, are y, x, z."}, {"video_title": "Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3", "Sentence": "In this first case, the base angles were this angle and that angle. In the second case, the base angles are that angle and that angle. And actually let me write it down. The base angles in this first case, let me do that same magenta, are y, x, z. So this angle y, x, z is congruent to angle y, z, x. That was in the first case. These were the base angles."}, {"video_title": "Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3", "Sentence": "The base angles in this first case, let me do that same magenta, are y, x, z. So this angle y, x, z is congruent to angle y, z, x. That was in the first case. These were the base angles. So based on the proof we saw in the last video, that implies these sides are congruent. Here we have these two base angles. So here we're saying angle, let me do that in green, angle x, y, z is congruent to angle y, x, z."}, {"video_title": "Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3", "Sentence": "These were the base angles. So based on the proof we saw in the last video, that implies these sides are congruent. Here we have these two base angles. So here we're saying angle, let me do that in green, angle x, y, z is congruent to angle y, x, z. And so that implies that these two guys right over here are congruent. Well there we've proved it. We said that this side, y, x, is congruent to y, z."}, {"video_title": "Equilateral triangle sides and angles congruent Congruence Geometry Khan Academy.mp3", "Sentence": "So here we're saying angle, let me do that in green, angle x, y, z is congruent to angle y, x, z. And so that implies that these two guys right over here are congruent. Well there we've proved it. We said that this side, y, x, is congruent to y, z. And we've shown that y, z is congruent to x, z. So all of the sides are congruent to each other. So once again, if you have all the angles equal, and they're going to have to be 60 degrees, then you know that all of the sides are going to be equal as well."}, {"video_title": "Dividing line segments according to ratio.mp3", "Sentence": "Find the coordinates of point B on segment, line segment AC, such that the ratio of AB to AC is three to five. So pause this video and see if you can figure that out. All right, now let's work through this together, and to help us visualize, let's plot these points. So first, let us plot point A, which is at negative one comma four. So negative one comma one, two, three, four. So that right over there is point A. And then let's think about point C, which is at four comma negative six."}, {"video_title": "Dividing line segments according to ratio.mp3", "Sentence": "So first, let us plot point A, which is at negative one comma four. So negative one comma one, two, three, four. So that right over there is point A. And then let's think about point C, which is at four comma negative six. So one, two, three, four, comma negative six. Negative one, negative two, negative three, negative four, negative five, negative six. Just like that."}, {"video_title": "Dividing line segments according to ratio.mp3", "Sentence": "And then let's think about point C, which is at four comma negative six. So one, two, three, four, comma negative six. Negative one, negative two, negative three, negative four, negative five, negative six. Just like that. And so the segment AC, I get my ruler tool out here, segment AC is going to look like that. And the ratio between the distance of A to B and A to C is three to five. Or another way to think about it is, B is going to be 3 5ths along the way from A to C. Now, the way that I think about it is, in order to be 3 5ths along the way from A to C, you have to be 3 5ths along the way in the X direction and 3 5ths along the way in the Y direction."}, {"video_title": "Dividing line segments according to ratio.mp3", "Sentence": "Just like that. And so the segment AC, I get my ruler tool out here, segment AC is going to look like that. And the ratio between the distance of A to B and A to C is three to five. Or another way to think about it is, B is going to be 3 5ths along the way from A to C. Now, the way that I think about it is, in order to be 3 5ths along the way from A to C, you have to be 3 5ths along the way in the X direction and 3 5ths along the way in the Y direction. So let's think about the X direction first. We are going from X equals negative one to X equals four to go from this point to that point. Our change in X is one, two, three, four, five."}, {"video_title": "Dividing line segments according to ratio.mp3", "Sentence": "Or another way to think about it is, B is going to be 3 5ths along the way from A to C. Now, the way that I think about it is, in order to be 3 5ths along the way from A to C, you have to be 3 5ths along the way in the X direction and 3 5ths along the way in the Y direction. So let's think about the X direction first. We are going from X equals negative one to X equals four to go from this point to that point. Our change in X is one, two, three, four, five. And so if we wanna go 3 5ths of that, we went a total of five, 3 5ths of that is going just three. So that is going to be B's X coordinate. And then we can look on the Y coordinate side."}, {"video_title": "Dividing line segments according to ratio.mp3", "Sentence": "Our change in X is one, two, three, four, five. And so if we wanna go 3 5ths of that, we went a total of five, 3 5ths of that is going just three. So that is going to be B's X coordinate. And then we can look on the Y coordinate side. To go from A to C, we are going from four to negative six. So we're going down by one, two, three, four, five, six, seven, eight, nine, 10. And so 3 5ths of 10 would be six."}, {"video_title": "Dividing line segments according to ratio.mp3", "Sentence": "And then we can look on the Y coordinate side. To go from A to C, we are going from four to negative six. So we're going down by one, two, three, four, five, six, seven, eight, nine, 10. And so 3 5ths of 10 would be six. So B's coordinate is going to be one, two, three, four, five, six down. So just like that, we were able to figure out the X and the Y coordinates for point B, which would be right over here. And you could look at this directly and say, look, B is going to have the coordinates."}, {"video_title": "Dividing line segments according to ratio.mp3", "Sentence": "And so 3 5ths of 10 would be six. So B's coordinate is going to be one, two, three, four, five, six down. So just like that, we were able to figure out the X and the Y coordinates for point B, which would be right over here. And you could look at this directly and say, look, B is going to have the coordinates. This looks like this is two comma negative two, which we were able to do with the graph paper. So another way you could think about it even algebraically is the coordinates of B, we could think about it as starting with the coordinates of A. So negative one comma four."}, {"video_title": "Dividing line segments according to ratio.mp3", "Sentence": "And you could look at this directly and say, look, B is going to have the coordinates. This looks like this is two comma negative two, which we were able to do with the graph paper. So another way you could think about it even algebraically is the coordinates of B, we could think about it as starting with the coordinates of A. So negative one comma four. But we're gonna move 3 5ths along the way in each of these dimensions towards C. So it's going to be plus 3 5ths times how far we've gone in the X direction. So in the X direction to go from A to C, we're going from negative one to four. And so that distance is four minus negative one."}, {"video_title": "Dividing line segments according to ratio.mp3", "Sentence": "So negative one comma four. But we're gonna move 3 5ths along the way in each of these dimensions towards C. So it's going to be plus 3 5ths times how far we've gone in the X direction. So in the X direction to go from A to C, we're going from negative one to four. And so that distance is four minus negative one. And this of course is going to be equal to five. And then on the Y dimension, this is going to be our A's Y coordinate plus 3 5ths times the distance that we travel in the Y direction. And here we're going from four to negative six."}, {"video_title": "Dividing line segments according to ratio.mp3", "Sentence": "And so that distance is four minus negative one. And this of course is going to be equal to five. And then on the Y dimension, this is going to be our A's Y coordinate plus 3 5ths times the distance that we travel in the Y direction. And here we're going from four to negative six. So we say negative six minus four, that is negative 10. And so the coordinates of B are going to be negative one plus 3 5ths times five is going to be plus three. And then four plus 3 5ths times negative 10."}, {"video_title": "Dividing line segments according to ratio.mp3", "Sentence": "And here we're going from four to negative six. So we say negative six minus four, that is negative 10. And so the coordinates of B are going to be negative one plus 3 5ths times five is going to be plus three. And then four plus 3 5ths times negative 10. Well, 3 5ths times negative 10 is negative six. And so that gets us to comma negative two. And we are done, which is exactly what we got right over there."}, {"video_title": "Writing equations of perpendicular lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "And they say that line B contains the point 6, negative 7. And they tell us lines A and B are perpendicular. So that means that slope of B must be negative inverse of slope of A. So what we'll do is we'll figure out the slope of A, then take the negative inverse of it, then we'll know the slope of B, then we can use this point right here to fill in the gaps and figure out B's y-intercept. So what's the slope of A? This is already in slope-intercept form. The slope of A is right there."}, {"video_title": "Writing equations of perpendicular lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "So what we'll do is we'll figure out the slope of A, then take the negative inverse of it, then we'll know the slope of B, then we can use this point right here to fill in the gaps and figure out B's y-intercept. So what's the slope of A? This is already in slope-intercept form. The slope of A is right there. It's the 2, mx plus b. So the slope here is equal to 2. So the slope of A is 2."}, {"video_title": "Writing equations of perpendicular lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "The slope of A is right there. It's the 2, mx plus b. So the slope here is equal to 2. So the slope of A is 2. What is the slope of B? So what is B's slope going to have to be? Well, it's perpendicular to A, so it's going to be the negative inverse of this."}, {"video_title": "Writing equations of perpendicular lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "So the slope of A is 2. What is the slope of B? So what is B's slope going to have to be? Well, it's perpendicular to A, so it's going to be the negative inverse of this. The inverse of 2 is 1 half. The negative inverse of that is negative 1 half. So B's slope is negative 1 half."}, {"video_title": "Writing equations of perpendicular lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "Well, it's perpendicular to A, so it's going to be the negative inverse of this. The inverse of 2 is 1 half. The negative inverse of that is negative 1 half. So B's slope is negative 1 half. So we know that B's equation has to be y is equal to its slope m times x plus some y-intercept. We still don't know what the y-intercept of B is, but we can use this information to figure it out. We know that y is equal to negative 7 when x is equal to 6, negative 1 half times 6 plus b. I just know that this is on the point."}, {"video_title": "Writing equations of perpendicular lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "So B's slope is negative 1 half. So we know that B's equation has to be y is equal to its slope m times x plus some y-intercept. We still don't know what the y-intercept of B is, but we can use this information to figure it out. We know that y is equal to negative 7 when x is equal to 6, negative 1 half times 6 plus b. I just know that this is on the point. So this point must satisfy the equation of line B. So let's work out what B must be. So in this, or what the B, the y-intercept, this is the lowercase b, not the line b."}, {"video_title": "Writing equations of perpendicular lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "We know that y is equal to negative 7 when x is equal to 6, negative 1 half times 6 plus b. I just know that this is on the point. So this point must satisfy the equation of line B. So let's work out what B must be. So in this, or what the B, the y-intercept, this is the lowercase b, not the line b. So we have negative 7 is equal to what's negative 1 half times 6? That's not a b there, that's a 6. What's negative 1 half times 6?"}, {"video_title": "Writing equations of perpendicular lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "So in this, or what the B, the y-intercept, this is the lowercase b, not the line b. So we have negative 7 is equal to what's negative 1 half times 6? That's not a b there, that's a 6. What's negative 1 half times 6? It's negative 3 is equal to negative 3 plus our y-intercept. Let's add 3 to both sides of this equation. So if we add 3 to both sides, I just want to get rid of this 3 right here, what do we get?"}, {"video_title": "Writing equations of perpendicular lines Mathematics I High School Math Khan Academy.mp3", "Sentence": "What's negative 1 half times 6? It's negative 3 is equal to negative 3 plus our y-intercept. Let's add 3 to both sides of this equation. So if we add 3 to both sides, I just want to get rid of this 3 right here, what do we get? The left-hand side, negative 7 plus 3 is negative 4. And that's going to be equal to, these guys cancel out, that's equal to b, our y-intercept. So this right here is a negative 4."}, {"video_title": "Constructing quadrilateral based on symmetry Transformations Geometry Khan Academy.mp3", "Sentence": "Two of the points that define a certain quadrilateral are zero comma nine and three comma four. The quadrilateral is left unchanged by a reflection over the line y is equal to three minus x. Draw and classify the quadrilateral. Now I encourage you to pause this video and try to draw and classify it on your own before I'm about to explain it. So let's at least plot the information they give us. So the point zero comma nine, that's one of the vertices of the quadrilateral. So zero comma nine, that's that point right over there."}, {"video_title": "Constructing quadrilateral based on symmetry Transformations Geometry Khan Academy.mp3", "Sentence": "Now I encourage you to pause this video and try to draw and classify it on your own before I'm about to explain it. So let's at least plot the information they give us. So the point zero comma nine, that's one of the vertices of the quadrilateral. So zero comma nine, that's that point right over there. And another one of the vertices is three comma four. Three comma four, that's that right over there. And then they tell us that the quadrilateral is left unchanged by a reflection over the line y is equal to three minus x."}, {"video_title": "Constructing quadrilateral based on symmetry Transformations Geometry Khan Academy.mp3", "Sentence": "So zero comma nine, that's that point right over there. And another one of the vertices is three comma four. Three comma four, that's that right over there. And then they tell us that the quadrilateral is left unchanged by a reflection over the line y is equal to three minus x. So when x is zero, y is three, that's our y-intercept. And it has a slope of negative one. You could view this as three minus one x."}, {"video_title": "Constructing quadrilateral based on symmetry Transformations Geometry Khan Academy.mp3", "Sentence": "And then they tell us that the quadrilateral is left unchanged by a reflection over the line y is equal to three minus x. So when x is zero, y is three, that's our y-intercept. And it has a slope of negative one. You could view this as three minus one x. So it has a slope of negative one. So the line looks like this. So every time we increase our x by one, we decrease our y by one."}, {"video_title": "Constructing quadrilateral based on symmetry Transformations Geometry Khan Academy.mp3", "Sentence": "You could view this as three minus one x. So it has a slope of negative one. So the line looks like this. So every time we increase our x by one, we decrease our y by one. So the line looks something like this. Y is equal to three minus x. Try to draw it relatively, pretty carefully."}, {"video_title": "Constructing quadrilateral based on symmetry Transformations Geometry Khan Academy.mp3", "Sentence": "So every time we increase our x by one, we decrease our y by one. So the line looks something like this. Y is equal to three minus x. Try to draw it relatively, pretty carefully. Pretty carefully, so that's what it looks like. Y is equal to three minus x. So that's my best attempt at drawing it."}, {"video_title": "Constructing quadrilateral based on symmetry Transformations Geometry Khan Academy.mp3", "Sentence": "Try to draw it relatively, pretty carefully. Pretty carefully, so that's what it looks like. Y is equal to three minus x. So that's my best attempt at drawing it. Y is equal to three minus x. So the quadrilateral is left unchanged by a reflection over this. So that means if I were to reflect each of these vertices, I would essentially end up with one of the other vertices on it, and if those get reflected, you're gonna end up with one of these, so the thing is not going to be different."}, {"video_title": "Constructing quadrilateral based on symmetry Transformations Geometry Khan Academy.mp3", "Sentence": "So that's my best attempt at drawing it. Y is equal to three minus x. So the quadrilateral is left unchanged by a reflection over this. So that means if I were to reflect each of these vertices, I would essentially end up with one of the other vertices on it, and if those get reflected, you're gonna end up with one of these, so the thing is not going to be different. So let's think about where these other two vertices of this quadrilateral need to be. So this point, let's just reflect it over this line, over y is equal to three minus x. So if we were to try to drop a perpendicular to this line, notice we have gone diagonally across one, two, three of these squares."}, {"video_title": "Constructing quadrilateral based on symmetry Transformations Geometry Khan Academy.mp3", "Sentence": "So that means if I were to reflect each of these vertices, I would essentially end up with one of the other vertices on it, and if those get reflected, you're gonna end up with one of these, so the thing is not going to be different. So let's think about where these other two vertices of this quadrilateral need to be. So this point, let's just reflect it over this line, over y is equal to three minus x. So if we were to try to drop a perpendicular to this line, notice we have gone diagonally across one, two, three of these squares. We need to go diagonally across three of them on the left-hand side. So one, two, three gets us right over there. That's the reflection of this point."}, {"video_title": "Constructing quadrilateral based on symmetry Transformations Geometry Khan Academy.mp3", "Sentence": "So if we were to try to drop a perpendicular to this line, notice we have gone diagonally across one, two, three of these squares. We need to go diagonally across three of them on the left-hand side. So one, two, three gets us right over there. That's the reflection of this point. This is the reflection of this point across that line. Now let's do the same thing for this blue point. To go, to drop a perpendicular to this line, we have to go diagonally across two of these squares."}, {"video_title": "Constructing quadrilateral based on symmetry Transformations Geometry Khan Academy.mp3", "Sentence": "That's the reflection of this point. This is the reflection of this point across that line. Now let's do the same thing for this blue point. To go, to drop a perpendicular to this line, we have to go diagonally across two of these squares. So let's go diagonally across two more of these squares, just like that, to get to that point right over there, and now we've defined our quadrilateral. Our quadrilateral looks like this. Our quadrilateral looks like this."}, {"video_title": "Constructing quadrilateral based on symmetry Transformations Geometry Khan Academy.mp3", "Sentence": "To go, to drop a perpendicular to this line, we have to go diagonally across two of these squares. So let's go diagonally across two more of these squares, just like that, to get to that point right over there, and now we've defined our quadrilateral. Our quadrilateral looks like this. Our quadrilateral looks like this. Both of these lines are perpendicular to that original line, so they're going to have the same slope, so that line is parallel to that line over there, and then we have this line. We have this line, and then we have this line. So what type of quadrilateral is this?"}, {"video_title": "Constructing circle inscribing triangle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "So this would be a circle that's inside this triangle where each of the sides of the triangle are tangent to the circle. And one way to, or probably the easiest way to think about it is the center of that circle is going to be at the in-center of the triangle. Now, what is the in-center of the triangle? The in-center of the triangle is the intersection of the angle bisectors. So if I were to make a line that perfectly splits an angle in two, so I'm eyeballing it right over here, this would be an angle bisector. But to be a little bit more precise about angle bisectors, I could actually use a compass. So let me make this a little bit smaller."}, {"video_title": "Constructing circle inscribing triangle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "The in-center of the triangle is the intersection of the angle bisectors. So if I were to make a line that perfectly splits an angle in two, so I'm eyeballing it right over here, this would be an angle bisector. But to be a little bit more precise about angle bisectors, I could actually use a compass. So let me make this a little bit smaller. And what I can do is I could put this, the center of the circle, on one of the sides of this angle right over here. Now let me get another circle. And I want to make it the same size."}, {"video_title": "Constructing circle inscribing triangle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "So let me make this a little bit smaller. And what I can do is I could put this, the center of the circle, on one of the sides of this angle right over here. Now let me get another circle. And I want to make it the same size. So let me center it there. I want to make it the exact same size. And now let me put it on the other one, on the other side of this angle."}, {"video_title": "Constructing circle inscribing triangle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "And I want to make it the same size. So let me center it there. I want to make it the exact same size. And now let me put it on the other one, on the other side of this angle. I'll put it right over here. And I want to put it so that the center of the circle is on the other side of the angle. And the circle itself, or the vertex, sits on the circle itself."}, {"video_title": "Constructing circle inscribing triangle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "And now let me put it on the other one, on the other side of this angle. I'll put it right over here. And I want to put it so that the center of the circle is on the other side of the angle. And the circle itself, or the vertex, sits on the circle itself. And what this does is I can now look at the intersection of this point, the vertex, and this point. And that's going to be an angle bisector, or the angle bisector. So let me go."}, {"video_title": "Constructing circle inscribing triangle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "And the circle itself, or the vertex, sits on the circle itself. And what this does is I can now look at the intersection of this point, the vertex, and this point. And that's going to be an angle bisector, or the angle bisector. So let me go. I'm going to go through there. And I'm going to go through there. Now let me move these circles over to here so I can take the angle bisector of this side as well."}, {"video_title": "Constructing circle inscribing triangle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "So let me go. I'm going to go through there. And I'm going to go through there. Now let me move these circles over to here so I can take the angle bisector of this side as well. So I could put this one over here. And I could put this one. See, I want to be on the side of the angle."}, {"video_title": "Constructing circle inscribing triangle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "Now let me move these circles over to here so I can take the angle bisector of this side as well. So I could put this one over here. And I could put this one. See, I want to be on the side of the angle. And I want to go right through. I want the circle to go right through the vertex. And let me add another straight edge here."}, {"video_title": "Constructing circle inscribing triangle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "See, I want to be on the side of the angle. And I want to go right through. I want the circle to go right through the vertex. And let me add another straight edge here. So I want to go through this point. And I want to bisect the angle. So I want to bisect the angle, go right through the other point of intersection of these two circles."}, {"video_title": "Constructing circle inscribing triangle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "And let me add another straight edge here. So I want to go through this point. And I want to bisect the angle. So I want to bisect the angle, go right through the other point of intersection of these two circles. Now let me get rid of one of these two circles. I don't need that anymore. And let me use this one to actually construct the circle inscribing the triangle."}, {"video_title": "Constructing circle inscribing triangle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "So I want to bisect the angle, go right through the other point of intersection of these two circles. Now let me get rid of one of these two circles. I don't need that anymore. And let me use this one to actually construct the circle inscribing the triangle. So I'm going to put it at the center right over there. And actually, this one's already pretty close in terms of dimensions. And with this tool, you don't have to be 100% precise."}, {"video_title": "Constructing circle inscribing triangle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "And let me use this one to actually construct the circle inscribing the triangle. So I'm going to put it at the center right over there. And actually, this one's already pretty close in terms of dimensions. And with this tool, you don't have to be 100% precise. It has some margin for error. It has some margin for error. And so let's just go with this."}, {"video_title": "Constructing circle inscribing triangle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "And with this tool, you don't have to be 100% precise. It has some margin for error. It has some margin for error. And so let's just go with this. This actually should be touching. But this has some margin for error. Let's see if this was good enough."}, {"video_title": "Hypotenuse of right triangle inscribed in circle Circles Geometry Khan Academy.mp3", "Sentence": "I encourage you to pause the video and try this out on your own. Well, let's think about what's going on right over here. AB is definitely the A diameter of the circle. It's a straight line. It's going through the center of the circle. O is the center of the circle right over here. And so what do we know?"}, {"video_title": "Hypotenuse of right triangle inscribed in circle Circles Geometry Khan Academy.mp3", "Sentence": "It's a straight line. It's going through the center of the circle. O is the center of the circle right over here. And so what do we know? Well, we could look at this angle right over here, angle C, and think about it as an inscribed angle, and think about the arc that it intercepts. It intercepts this arc right over here. This arc is exactly half of the circle."}, {"video_title": "Hypotenuse of right triangle inscribed in circle Circles Geometry Khan Academy.mp3", "Sentence": "And so what do we know? Well, we could look at this angle right over here, angle C, and think about it as an inscribed angle, and think about the arc that it intercepts. It intercepts this arc right over here. This arc is exactly half of the circle. This arc right over here is exactly half of the circle. Angle C is inscribed. If you take these two sides, or the two sides of the angle, it intercepts at A and B, and so it intercepts an arc, this green arc right over here."}, {"video_title": "Hypotenuse of right triangle inscribed in circle Circles Geometry Khan Academy.mp3", "Sentence": "This arc is exactly half of the circle. This arc right over here is exactly half of the circle. Angle C is inscribed. If you take these two sides, or the two sides of the angle, it intercepts at A and B, and so it intercepts an arc, this green arc right over here. So the central angle right over here is 180 degrees, and the inscribed angle is going to be half of that. It's going to be 90 degrees. Or, another way of thinking about it, it's going to be a right angle."}, {"video_title": "Hypotenuse of right triangle inscribed in circle Circles Geometry Khan Academy.mp3", "Sentence": "If you take these two sides, or the two sides of the angle, it intercepts at A and B, and so it intercepts an arc, this green arc right over here. So the central angle right over here is 180 degrees, and the inscribed angle is going to be half of that. It's going to be 90 degrees. Or, another way of thinking about it, it's going to be a right angle. And what that does for us is it tells us that triangle ACB is a right triangle. This is a right triangle, and the diameter is its hypotenuse. So we can just apply the Pythagorean theorem here."}, {"video_title": "Hypotenuse of right triangle inscribed in circle Circles Geometry Khan Academy.mp3", "Sentence": "Or, another way of thinking about it, it's going to be a right angle. And what that does for us is it tells us that triangle ACB is a right triangle. This is a right triangle, and the diameter is its hypotenuse. So we can just apply the Pythagorean theorem here. 15 squared plus 8 squared is going to be the length of side AB squared. So let me just call this side right over here, let me just call that x. That's going to be equal to x squared."}, {"video_title": "Hypotenuse of right triangle inscribed in circle Circles Geometry Khan Academy.mp3", "Sentence": "So we can just apply the Pythagorean theorem here. 15 squared plus 8 squared is going to be the length of side AB squared. So let me just call this side right over here, let me just call that x. That's going to be equal to x squared. So 15 squared, that's 225. 8 squared is 64, plus 64 is equal to x squared. I want to do that in green."}, {"video_title": "Hypotenuse of right triangle inscribed in circle Circles Geometry Khan Academy.mp3", "Sentence": "That's going to be equal to x squared. So 15 squared, that's 225. 8 squared is 64, plus 64 is equal to x squared. I want to do that in green. Is equal to x squared. 225 plus 64 is 289, is equal to x squared. And then 289 is 17 squared."}, {"video_title": "Hypotenuse of right triangle inscribed in circle Circles Geometry Khan Academy.mp3", "Sentence": "I want to do that in green. Is equal to x squared. 225 plus 64 is 289, is equal to x squared. And then 289 is 17 squared. And you could try out a few numbers if you're unsure about that. So x is equal to 17. So the diameter of the circle right over here is 17."}, {"video_title": "Mapping shapes.mp3", "Sentence": "So each of these is one unit. Which of the following sequences of transformations maps triangle PQR onto triangle ABC? So we have four different sequences of transformations. And so why don't you pause this video and figure out which of these actually does map triangle PQR, so this is PQR, onto ABC. And it could be more than one of these. So pause this video and have a go at that. All right, now let's do this together."}, {"video_title": "Mapping shapes.mp3", "Sentence": "And so why don't you pause this video and figure out which of these actually does map triangle PQR, so this is PQR, onto ABC. And it could be more than one of these. So pause this video and have a go at that. All right, now let's do this together. So let's first think about sequence A. And I will do sequence A in this purple color. So remember, we're starting with triangle PQR."}, {"video_title": "Mapping shapes.mp3", "Sentence": "All right, now let's do this together. So let's first think about sequence A. And I will do sequence A in this purple color. So remember, we're starting with triangle PQR. So first it says a rotation 90 degrees about the point R, so let's do that. And then we'll do the rest of this sequence. So if we rotate this 90 degrees, so one way to think about it is a line like that is then going to be like that."}, {"video_title": "Mapping shapes.mp3", "Sentence": "So remember, we're starting with triangle PQR. So first it says a rotation 90 degrees about the point R, so let's do that. And then we'll do the rest of this sequence. So if we rotate this 90 degrees, so one way to think about it is a line like that is then going to be like that. So we're gonna go like that. And so R is going to stay where it is. You're rotating about it."}, {"video_title": "Mapping shapes.mp3", "Sentence": "So if we rotate this 90 degrees, so one way to think about it is a line like that is then going to be like that. So we're gonna go like that. And so R is going to stay where it is. You're rotating about it. But P is now going to be right over here. One way to think about it is to go from R to P, we went down one and three to the right. Now when you do the rotation, you're going to go to the right one and then up three."}, {"video_title": "Mapping shapes.mp3", "Sentence": "You're rotating about it. But P is now going to be right over here. One way to think about it is to go from R to P, we went down one and three to the right. Now when you do the rotation, you're going to go to the right one and then up three. So P is going to be there, and you could see that. That's the rotation. So that side will look like this."}, {"video_title": "Mapping shapes.mp3", "Sentence": "Now when you do the rotation, you're going to go to the right one and then up three. So P is going to be there, and you could see that. That's the rotation. So that side will look like this. So that is P. And then Q is going to go right over here. It's going to once again also do a 90 degree rotation about R. And so after you do the 90 degree rotation, PQR is going to look like this. So that is Q."}, {"video_title": "Mapping shapes.mp3", "Sentence": "So that side will look like this. So that is P. And then Q is going to go right over here. It's going to once again also do a 90 degree rotation about R. And so after you do the 90 degree rotation, PQR is going to look like this. So that is Q. So we've done that first part. Then a translation six units to the left and seven units up. So each of these points are gonna go six units to the left and seven up."}, {"video_title": "Mapping shapes.mp3", "Sentence": "So that is Q. So we've done that first part. Then a translation six units to the left and seven units up. So each of these points are gonna go six units to the left and seven up. So if we take point P, six to the left, one, two, three, four, five, six, seven units up. One, two, three, four, five, six, seven. It'll put it right over there."}, {"video_title": "Mapping shapes.mp3", "Sentence": "So each of these points are gonna go six units to the left and seven up. So if we take point P, six to the left, one, two, three, four, five, six, seven units up. One, two, three, four, five, six, seven. It'll put it right over there. So that is point P. If we take point R, we take six units to the left, one, two, three, four, five, six, seven up. One, two, three, four, five, six, seven. It gets us right over there."}, {"video_title": "Mapping shapes.mp3", "Sentence": "It'll put it right over there. So that is point P. If we take point R, we take six units to the left, one, two, three, four, five, six, seven up. One, two, three, four, five, six, seven. It gets us right over there. And then point Q, if we go six units to the left, one, two, three, four, five, six, seven up is one, two, three, four, five, six, seven. Puts us right over there. So this looks like it worked."}, {"video_title": "Mapping shapes.mp3", "Sentence": "It gets us right over there. And then point Q, if we go six units to the left, one, two, three, four, five, six, seven up is one, two, three, four, five, six, seven. Puts us right over there. So this looks like it worked. Sequence A is good. It maps PQR onto ABC. This last one isn't an R. This is a Q right over here."}, {"video_title": "Mapping shapes.mp3", "Sentence": "So this looks like it worked. Sequence A is good. It maps PQR onto ABC. This last one isn't an R. This is a Q right over here. So that worked, sequence A. Now let's work on sequence B. I'll do this in a different color. A translation, eight units to the left and three up."}, {"video_title": "Mapping shapes.mp3", "Sentence": "This last one isn't an R. This is a Q right over here. So that worked, sequence A. Now let's work on sequence B. I'll do this in a different color. A translation, eight units to the left and three up. So let's do that first. So if we take point Q, eight to the left and three up. One, two, three, four, five, six, seven, eight."}, {"video_title": "Mapping shapes.mp3", "Sentence": "A translation, eight units to the left and three up. So let's do that first. So if we take point Q, eight to the left and three up. One, two, three, four, five, six, seven, eight. Three up, one, two, three. So this will be my red Q for now. And now if I do this point R, one, two, three, four, five, six, seven, eight."}, {"video_title": "Mapping shapes.mp3", "Sentence": "One, two, three, four, five, six, seven, eight. Three up, one, two, three. So this will be my red Q for now. And now if I do this point R, one, two, three, four, five, six, seven, eight. Let me make sure I did that right. One, two, three, four, five, six, seven, eight. Three up, one, two, three."}, {"video_title": "Mapping shapes.mp3", "Sentence": "And now if I do this point R, one, two, three, four, five, six, seven, eight. Let me make sure I did that right. One, two, three, four, five, six, seven, eight. Three up, one, two, three. So my new R is going to be there. And then last but not least, point P, eight to the left. One, two, three, four, five, six, seven, eight."}, {"video_title": "Mapping shapes.mp3", "Sentence": "Three up, one, two, three. So my new R is going to be there. And then last but not least, point P, eight to the left. One, two, three, four, five, six, seven, eight. Three up, one, two, three, goes right there. So just that translation will get us to this point. It'll get us to that point."}, {"video_title": "Mapping shapes.mp3", "Sentence": "One, two, three, four, five, six, seven, eight. Three up, one, two, three, goes right there. So just that translation will get us to this point. It'll get us to that point. So we're clearly not done mapping yet, but there's more transformation to be done. So it looks something like that. It says, then a reflection over the horizontal line through point A."}, {"video_title": "Mapping shapes.mp3", "Sentence": "It'll get us to that point. So we're clearly not done mapping yet, but there's more transformation to be done. So it looks something like that. It says, then a reflection over the horizontal line through point A. So point A is right over here. The horizontal line is right like that. So if I were to reflect, point A wouldn't change."}, {"video_title": "Mapping shapes.mp3", "Sentence": "It says, then a reflection over the horizontal line through point A. So point A is right over here. The horizontal line is right like that. So if I were to reflect, point A wouldn't change. Point R right now is three below that horizontal line. Point R will then be three above that horizontal line. So point R will then go right over there."}, {"video_title": "Mapping shapes.mp3", "Sentence": "So if I were to reflect, point A wouldn't change. Point R right now is three below that horizontal line. Point R will then be three above that horizontal line. So point R will then go right over there. Just from that, I can see that this sequence of transformations is not going to work. It's putting R in the wrong place. So I'm going to rule out sequence B. Sequence C, let me do that with another color."}, {"video_title": "Mapping shapes.mp3", "Sentence": "So point R will then go right over there. Just from that, I can see that this sequence of transformations is not going to work. It's putting R in the wrong place. So I'm going to rule out sequence B. Sequence C, let me do that with another color. I don't know how I will do it with this orange color. A reflection over the vertical point through point Q. Sorry, a reflection over the vertical line through point Q."}, {"video_title": "Mapping shapes.mp3", "Sentence": "So I'm going to rule out sequence B. Sequence C, let me do that with another color. I don't know how I will do it with this orange color. A reflection over the vertical point through point Q. Sorry, a reflection over the vertical line through point Q. So let me do that. So the vertical line through point Q looks like this. Just gonna draw that vertical line."}, {"video_title": "Mapping shapes.mp3", "Sentence": "Sorry, a reflection over the vertical line through point Q. So let me do that. So the vertical line through point Q looks like this. Just gonna draw that vertical line. So if you reflect it, Q is going to stay in place. R is one to the right of that, so now it's going to be one to the left once you do the reflection. And point P is four to the right, so now it's going to be four to the left."}, {"video_title": "Mapping shapes.mp3", "Sentence": "Just gonna draw that vertical line. So if you reflect it, Q is going to stay in place. R is one to the right of that, so now it's going to be one to the left once you do the reflection. And point P is four to the right, so now it's going to be four to the left. One, two, three, four. So P is going to be there after the reflection. And so it's going to look something like this after that first transformation."}, {"video_title": "Mapping shapes.mp3", "Sentence": "And point P is four to the right, so now it's going to be four to the left. One, two, three, four. So P is going to be there after the reflection. And so it's going to look something like this after that first transformation. I know this is getting a little bit messy, but this is what you probably have to go through as well. So I'll go through it with you. All right, so we did that first part, the reflection."}, {"video_title": "Mapping shapes.mp3", "Sentence": "And so it's going to look something like this after that first transformation. I know this is getting a little bit messy, but this is what you probably have to go through as well. So I'll go through it with you. All right, so we did that first part, the reflection. Then a translation four to the left and seven units up. So four to the left and seven up. So let me try that."}, {"video_title": "Mapping shapes.mp3", "Sentence": "All right, so we did that first part, the reflection. Then a translation four to the left and seven units up. So four to the left and seven up. So let me try that. So four to the left, one, two, three, four, seven up. One, two, three, four, five, six, seven. So it's putting Q right over here."}, {"video_title": "Mapping shapes.mp3", "Sentence": "So let me try that. So four to the left, one, two, three, four, seven up. One, two, three, four, five, six, seven. So it's putting Q right over here. I'm already suspicious of it because sequence A worked where we put P right over there. So I'm already suspicious of this, but let's keep trying. So four to the left and seven up."}, {"video_title": "Mapping shapes.mp3", "Sentence": "So it's putting Q right over here. I'm already suspicious of it because sequence A worked where we put P right over there. So I'm already suspicious of this, but let's keep trying. So four to the left and seven up. One, two, three, four, seven up. One, two, three, four, five, six, seven. So R is going to the same place that sequence A put it."}, {"video_title": "Mapping shapes.mp3", "Sentence": "So four to the left and seven up. One, two, three, four, seven up. One, two, three, four, five, six, seven. So R is going to the same place that sequence A put it. And then point P, one, two, three, four. One, two, three, four, five, six, seven. Actually, it worked."}, {"video_title": "Mapping shapes.mp3", "Sentence": "So R is going to the same place that sequence A put it. And then point P, one, two, three, four. One, two, three, four, five, six, seven. Actually, it worked. So it works because this is actually an isosceles triangle. So this one actually worked out. We were able to map PQR onto ABC with sequence C. So I like this one as well."}, {"video_title": "Mapping shapes.mp3", "Sentence": "Actually, it worked. So it works because this is actually an isosceles triangle. So this one actually worked out. We were able to map PQR onto ABC with sequence C. So I like this one as well. And then last but not least, let's try sequence D. I'll do that in black so that we can see it. So first we do a translation. Eight units to the left and three up."}, {"video_title": "Mapping shapes.mp3", "Sentence": "We were able to map PQR onto ABC with sequence C. So I like this one as well. And then last but not least, let's try sequence D. I'll do that in black so that we can see it. So first we do a translation. Eight units to the left and three up. Eight to the left and three up. So we start here. One, two, three, four, five, six, seven, eight."}, {"video_title": "Mapping shapes.mp3", "Sentence": "Eight units to the left and three up. Eight to the left and three up. So we start here. One, two, three, four, five, six, seven, eight. Three up, one, two, three. So I'll put my black Q right over there. So eight to the left."}, {"video_title": "Mapping shapes.mp3", "Sentence": "One, two, three, four, five, six, seven, eight. Three up, one, two, three. So I'll put my black Q right over there. So eight to the left. One, two, three, four, five, six, seven, eight. Three up, one, two, three. I'll put my black R right over there."}, {"video_title": "Mapping shapes.mp3", "Sentence": "So eight to the left. One, two, three, four, five, six, seven, eight. Three up, one, two, three. I'll put my black R right over there. That's actually exactly what we did in sequence B the first time. So P is going to show up right over there. So after that translation, sequence, that first translation in sequence D, it gets us right over there."}, {"video_title": "Mapping shapes.mp3", "Sentence": "I'll put my black R right over there. That's actually exactly what we did in sequence B the first time. So P is going to show up right over there. So after that translation, sequence, that first translation in sequence D, it gets us right over there. Then it says a rotation negative 270 degrees about point A. So this is point A right over here. And negative 270 degrees."}, {"video_title": "Mapping shapes.mp3", "Sentence": "So after that translation, sequence, that first translation in sequence D, it gets us right over there. Then it says a rotation negative 270 degrees about point A. So this is point A right over here. And negative 270 degrees. It's negative so it's going to go clockwise. And let's see, 180 degrees. Let's say if we were to take this line right over here."}, {"video_title": "Mapping shapes.mp3", "Sentence": "And negative 270 degrees. It's negative so it's going to go clockwise. And let's see, 180 degrees. Let's say if we were to take this line right over here. If we were to go 180 degrees, it would go to this line like that. And then if we were to go another 90 degrees, it actually does look like it would map onto that. So this is actually looking pretty good."}, {"video_title": "Mapping shapes.mp3", "Sentence": "Let's say if we were to take this line right over here. If we were to go 180 degrees, it would go to this line like that. And then if we were to go another 90 degrees, it actually does look like it would map onto that. So this is actually looking pretty good. If you were to, this line right over here, well then, if you go negative 270 degrees, will map onto this right over here. And then that point R will kind of go along for the ride is one way to think about it. And so it'll go right over there as well."}, {"video_title": "Find the volume of a triangular prism and cube Geometry Khan Academy.mp3", "Sentence": "Let's do some solid geometry volume problems. So they tell us shown is a triangular prism. And so there's a couple of types of three-dimensional figures that deal with triangles. This is what a triangular prism looks like, where it has a triangle on one, two faces. And they're kind of separated. They kind of have rectangles in between. The other types of triangular three-dimensional figures is you might see pyramids."}, {"video_title": "Find the volume of a triangular prism and cube Geometry Khan Academy.mp3", "Sentence": "This is what a triangular prism looks like, where it has a triangle on one, two faces. And they're kind of separated. They kind of have rectangles in between. The other types of triangular three-dimensional figures is you might see pyramids. This would be a rectangular pyramid because it has a rectangular or it has a square base, just like that. You could also have a triangular pyramid, where just literally every side is a triangle, so stuff like that. But this over here is a triangular prism."}, {"video_title": "Find the volume of a triangular prism and cube Geometry Khan Academy.mp3", "Sentence": "The other types of triangular three-dimensional figures is you might see pyramids. This would be a rectangular pyramid because it has a rectangular or it has a square base, just like that. You could also have a triangular pyramid, where just literally every side is a triangle, so stuff like that. But this over here is a triangular prism. I don't want to get too much into the shape classification. If the base of the triangle B is equal to 7, the height of the triangle H is equal to 3, and the length of the prism L is equal to 4, what is the total volume of the prism? So they're saying that the base is equal to 7."}, {"video_title": "Find the volume of a triangular prism and cube Geometry Khan Academy.mp3", "Sentence": "But this over here is a triangular prism. I don't want to get too much into the shape classification. If the base of the triangle B is equal to 7, the height of the triangle H is equal to 3, and the length of the prism L is equal to 4, what is the total volume of the prism? So they're saying that the base is equal to 7. So this right over here is equal to 7. The height of the triangle is equal to 3. So this right over here, this distance right over here, H is equal to 3."}, {"video_title": "Find the volume of a triangular prism and cube Geometry Khan Academy.mp3", "Sentence": "So they're saying that the base is equal to 7. So this right over here is equal to 7. The height of the triangle is equal to 3. So this right over here, this distance right over here, H is equal to 3. And the length of the prism is equal to 4. So I'm assuming it's this dimension over here is equal to 4. So length is equal to 4."}, {"video_title": "Find the volume of a triangular prism and cube Geometry Khan Academy.mp3", "Sentence": "So this right over here, this distance right over here, H is equal to 3. And the length of the prism is equal to 4. So I'm assuming it's this dimension over here is equal to 4. So length is equal to 4. So in this situation, what you really just have to do is figure out the area of this triangle right over here. We could figure out the area of this triangle, and then multiply it by how much you go deep. So multiply it by this length."}, {"video_title": "Find the volume of a triangular prism and cube Geometry Khan Academy.mp3", "Sentence": "So length is equal to 4. So in this situation, what you really just have to do is figure out the area of this triangle right over here. We could figure out the area of this triangle, and then multiply it by how much you go deep. So multiply it by this length. So the volume is going to be the area of this triangle. Let me do it in pink. The area of this triangle."}, {"video_title": "Find the volume of a triangular prism and cube Geometry Khan Academy.mp3", "Sentence": "So multiply it by this length. So the volume is going to be the area of this triangle. Let me do it in pink. The area of this triangle. We know that the area of a triangle is 1 half times the base times the height. So this area right over here is going to be 1 half times the base times the height. And then we're going to multiply it by kind of our depth of this triangular prism."}, {"video_title": "Find the volume of a triangular prism and cube Geometry Khan Academy.mp3", "Sentence": "The area of this triangle. We know that the area of a triangle is 1 half times the base times the height. So this area right over here is going to be 1 half times the base times the height. And then we're going to multiply it by kind of our depth of this triangular prism. So we have a depth of 4. So then we're going to multiply that times the 4. Times this depth."}, {"video_title": "Find the volume of a triangular prism and cube Geometry Khan Academy.mp3", "Sentence": "And then we're going to multiply it by kind of our depth of this triangular prism. So we have a depth of 4. So then we're going to multiply that times the 4. Times this depth. Times the 4. And we get, let's see, 1 half times 4 is 2. So these guys cancel out."}, {"video_title": "Find the volume of a triangular prism and cube Geometry Khan Academy.mp3", "Sentence": "Times this depth. Times the 4. And we get, let's see, 1 half times 4 is 2. So these guys cancel out. You'll just have a 2. And then 2 times 3 is 6. 6 times 7 is 42."}, {"video_title": "Find the volume of a triangular prism and cube Geometry Khan Academy.mp3", "Sentence": "So these guys cancel out. You'll just have a 2. And then 2 times 3 is 6. 6 times 7 is 42. And it would be in some type of cubic unit. So if these were in, I don't know, centimeters, it would be centimeters cubed. But they're not making us focus on the units in this problem."}, {"video_title": "Find the volume of a triangular prism and cube Geometry Khan Academy.mp3", "Sentence": "6 times 7 is 42. And it would be in some type of cubic unit. So if these were in, I don't know, centimeters, it would be centimeters cubed. But they're not making us focus on the units in this problem. Let's do another one. Shown is a cube. If each side of the cube, or if each side is of equal length, x equals 3, what is the total volume of the cube?"}, {"video_title": "Find the volume of a triangular prism and cube Geometry Khan Academy.mp3", "Sentence": "But they're not making us focus on the units in this problem. Let's do another one. Shown is a cube. If each side of the cube, or if each side is of equal length, x equals 3, what is the total volume of the cube? So each side is equal length x, which happens to equal 3. So this side is 3. This side over here, x is equal to 3."}, {"video_title": "Find the volume of a triangular prism and cube Geometry Khan Academy.mp3", "Sentence": "If each side of the cube, or if each side is of equal length, x equals 3, what is the total volume of the cube? So each side is equal length x, which happens to equal 3. So this side is 3. This side over here, x is equal to 3. Every side, x is equal to 3. So it's actually the same exercise as the triangular prism. It's actually a little bit easier when you're dealing with a cube, where you really just want to find the area of this surface right over here."}, {"video_title": "Find the volume of a triangular prism and cube Geometry Khan Academy.mp3", "Sentence": "This side over here, x is equal to 3. Every side, x is equal to 3. So it's actually the same exercise as the triangular prism. It's actually a little bit easier when you're dealing with a cube, where you really just want to find the area of this surface right over here. Now this is pretty straightforward. This is just a square. Or it would be the base times the height."}, {"video_title": "Find the volume of a triangular prism and cube Geometry Khan Academy.mp3", "Sentence": "It's actually a little bit easier when you're dealing with a cube, where you really just want to find the area of this surface right over here. Now this is pretty straightforward. This is just a square. Or it would be the base times the height. Or since they're the same, it's just 3 times 3. So the volume is going to be the area of this surface, 3 times 3, times the depth. And so we go 3 deep."}, {"video_title": "Find the volume of a triangular prism and cube Geometry Khan Academy.mp3", "Sentence": "Or it would be the base times the height. Or since they're the same, it's just 3 times 3. So the volume is going to be the area of this surface, 3 times 3, times the depth. And so we go 3 deep. So times 3. And so we get 3 times 3 times 3. This is 27."}, {"video_title": "Find the volume of a triangular prism and cube Geometry Khan Academy.mp3", "Sentence": "And so we go 3 deep. So times 3. And so we get 3 times 3 times 3. This is 27. Or you might recognize this from exponents. This is the same thing as 3 to the third power. And that's why sometimes, if you have something to the third power, they'll say you cubed it."}, {"video_title": "Find the volume of a triangular prism and cube Geometry Khan Academy.mp3", "Sentence": "This is 27. Or you might recognize this from exponents. This is the same thing as 3 to the third power. And that's why sometimes, if you have something to the third power, they'll say you cubed it. Because literally, to find the volume of a cube, you take the length of one side, and you multiply that number by itself three times. One for each dimension. One for the length, the width, and or I guess the height, the length, and the depth, depending on how you want to define them."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "An expedition was sent to find how high the water had risen. The people measured the edge of the pyramid that's above the water and found it was 72 meters long. So this distance right over here is 72 meters. They knew that the entire length of the edge is 180 meters when it's not flooded, so this entire length is 180 meters. They also knew that the vertical height of the pyramid is 139 meters, so this is 139 meters. What is the level of the water above the ground? So the ground is right over here at the base of the pyramid, and so they want the level of the water above the ground."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "They knew that the entire length of the edge is 180 meters when it's not flooded, so this entire length is 180 meters. They also knew that the vertical height of the pyramid is 139 meters, so this is 139 meters. What is the level of the water above the ground? So the ground is right over here at the base of the pyramid, and so they want the level of the water above the ground. So that's this height right over here, so let's just call that h. We want to figure out what h is. Round your answer, if necessary, to two decimal places. So what do we know and what do we not know?"}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So the ground is right over here at the base of the pyramid, and so they want the level of the water above the ground. So that's this height right over here, so let's just call that h. We want to figure out what h is. Round your answer, if necessary, to two decimal places. So what do we know and what do we not know? So they've labeled this little angle here theta, and this, of course, is a right angle. So this angle here at the base of the pyramid, this is going to be the complement of theta. It's going to be 90 degrees minus theta."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So what do we know and what do we not know? So they've labeled this little angle here theta, and this, of course, is a right angle. So this angle here at the base of the pyramid, this is going to be the complement of theta. It's going to be 90 degrees minus theta. And using that information, we can also figure out that this angle up here is also going to be theta. If that looks a little bit strange to you, let me just draw it out here and make it a little bit clearer. If we have a right triangle where this angle right over here is 90 minus theta, and we wanted to figure out what this is up here, let's say this is x, well, we could say x plus 90 minus theta plus 90 degrees is going to be equal to, well, the sum of the angles of a triangle are going to be 180 degrees."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "It's going to be 90 degrees minus theta. And using that information, we can also figure out that this angle up here is also going to be theta. If that looks a little bit strange to you, let me just draw it out here and make it a little bit clearer. If we have a right triangle where this angle right over here is 90 minus theta, and we wanted to figure out what this is up here, let's say this is x, well, we could say x plus 90 minus theta plus 90 degrees is going to be equal to, well, the sum of the angles of a triangle are going to be 180 degrees. Well, if we subtract 180 from both sides, so that's 180 from the left, 180 from the right, we get x minus theta is equal to 0, or if you add theta to both sides, x is equal to theta. So this thing up here is going to be theta as well. So this is also going to be theta."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "If we have a right triangle where this angle right over here is 90 minus theta, and we wanted to figure out what this is up here, let's say this is x, well, we could say x plus 90 minus theta plus 90 degrees is going to be equal to, well, the sum of the angles of a triangle are going to be 180 degrees. Well, if we subtract 180 from both sides, so that's 180 from the left, 180 from the right, we get x minus theta is equal to 0, or if you add theta to both sides, x is equal to theta. So this thing up here is going to be theta as well. So this is also going to be theta. And what else do we know? Well, we know that this is 72. We know that the whole thing is 180."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So this is also going to be theta. And what else do we know? Well, we know that this is 72. We know that the whole thing is 180. So this is 72 and the whole thing is 180. The part of this edge that's below the water, this distance right over here, let me draw it without cluttering the picture too much, doing the black color, this distance right over here is going to be 108. 108 plus 72 is 180."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "We know that the whole thing is 180. So this is 72 and the whole thing is 180. The part of this edge that's below the water, this distance right over here, let me draw it without cluttering the picture too much, doing the black color, this distance right over here is going to be 108. 108 plus 72 is 180. So what does this do for us? We need to figure out this height. We know that this right over here is a right triangle."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "108 plus 72 is 180. So what does this do for us? We need to figure out this height. We know that this right over here is a right triangle. This right over here is a right triangle. I could color this in just to make it a little bit clearer. This thing in yellow right over here is a right triangle."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "We know that this right over here is a right triangle. This right over here is a right triangle. I could color this in just to make it a little bit clearer. This thing in yellow right over here is a right triangle. If we look at that right triangle and if we wanted to solve for h and solve for h using a trig ratio based on this angle theta right over here, we know that relative to this angle theta, this side of length h is an adjacent side and this length of 108 right over here along the edge, that's the hypotenuse of this yellow triangle that I just highlighted in. So which trig ratio involves an adjacent side and a hypotenuse? We just write SOH CAH TOA."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "This thing in yellow right over here is a right triangle. If we look at that right triangle and if we wanted to solve for h and solve for h using a trig ratio based on this angle theta right over here, we know that relative to this angle theta, this side of length h is an adjacent side and this length of 108 right over here along the edge, that's the hypotenuse of this yellow triangle that I just highlighted in. So which trig ratio involves an adjacent side and a hypotenuse? We just write SOH CAH TOA. Sine is opposite over hypotenuse. That would be this distance over the hypotenuse. Cosine is adjacent over hypotenuse."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "We just write SOH CAH TOA. Sine is opposite over hypotenuse. That would be this distance over the hypotenuse. Cosine is adjacent over hypotenuse. We get the cosine of theta is going to be equal to the height that we care about, that's the adjacent side of this right triangle, over the length of the hypotenuse, over 108. That doesn't help us yet because we don't know what the cosine of theta is. But there's a clue here."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Cosine is adjacent over hypotenuse. We get the cosine of theta is going to be equal to the height that we care about, that's the adjacent side of this right triangle, over the length of the hypotenuse, over 108. That doesn't help us yet because we don't know what the cosine of theta is. But there's a clue here. Theta is also sitting up here, so maybe if we can figure out what cosine of theta is based up here, then we can solve for h. If we look at this theta, what is the cosine of theta? Now we're looking at a different right triangle. We're looking at this entire right triangle now."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "But there's a clue here. Theta is also sitting up here, so maybe if we can figure out what cosine of theta is based up here, then we can solve for h. If we look at this theta, what is the cosine of theta? Now we're looking at a different right triangle. We're looking at this entire right triangle now. Based on that entire right triangle, what is cosine of theta? Well, cosine of theta, once again, is equal to adjacent over hypotenuse. The adjacent length is this length right over here."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "We're looking at this entire right triangle now. Based on that entire right triangle, what is cosine of theta? Well, cosine of theta, once again, is equal to adjacent over hypotenuse. The adjacent length is this length right over here. We already know that's 139 meters, so it's going to be equal to 139 meters. And what's the length of the hypotenuse? Well, the hypotenuse is this length right over here."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "The adjacent length is this length right over here. We already know that's 139 meters, so it's going to be equal to 139 meters. And what's the length of the hypotenuse? Well, the hypotenuse is this length right over here. It's 72 plus 108, or we already have it labeled here. It's 180. We can assume that this pyramid is an isosceles triangle."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Well, the hypotenuse is this length right over here. It's 72 plus 108, or we already have it labeled here. It's 180. We can assume that this pyramid is an isosceles triangle. So 180 on that side, 180 on that side. So the cosine is adjacent, 139, over the hypotenuse, which is 180. These thetas are the same theta."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "We can assume that this pyramid is an isosceles triangle. So 180 on that side, 180 on that side. So the cosine is adjacent, 139, over the hypotenuse, which is 180. These thetas are the same theta. We just showed that. Now we have cosine of theta is h over 108. Cosine of theta is 139 over 180."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "These thetas are the same theta. We just showed that. Now we have cosine of theta is h over 108. Cosine of theta is 139 over 180. Or we could say that h over 108, which is cosine of theta, also is equal to 139 over 180. Both of these things are equal to cosine of theta. Now to solve for h, we just multiply both sides by 108."}, {"video_title": "How much of a pyramid is submerged Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Cosine of theta is 139 over 180. Or we could say that h over 108, which is cosine of theta, also is equal to 139 over 180. Both of these things are equal to cosine of theta. Now to solve for h, we just multiply both sides by 108. So h is equal to 139 times 108 over 180. So let's get our calculator out and calculate that. So that is going to be 139 times 108 divided by 180 gets us to 83.4 meters."}, {"video_title": "Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "construct a regular hexagon inscribed inside the circle. So what I'm going to do, first I'm going to draw the diameter of the circle. Actually, I'm going to go beyond the diameter of the circle. I'm just going to draw a line that goes through the center of the circle and just keeps on going. I'm going to make it flat so it goes directly through. So this one right over here goes directly through the center. Now what I'm going to do is I'm going to construct a circle that's the exact same dimensions of this circle that's already been drawn."}, {"video_title": "Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "I'm just going to draw a line that goes through the center of the circle and just keeps on going. I'm going to make it flat so it goes directly through. So this one right over here goes directly through the center. Now what I'm going to do is I'm going to construct a circle that's the exact same dimensions of this circle that's already been drawn. So let me put this one right over here. Let me make it the same dimensions. Now what I'm going to do is I'm going to move it over so that this new circle intersects the center of the old circle."}, {"video_title": "Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "Now what I'm going to do is I'm going to construct a circle that's the exact same dimensions of this circle that's already been drawn. So let me put this one right over here. Let me make it the same dimensions. Now what I'm going to do is I'm going to move it over so that this new circle intersects the center of the old circle. These circles are the same size. Notice this center intersects the old circle and the new circle itself intersects the center of the old circle. Now the reason why this is interesting, we already know that this distance, the distance between these two centers, this is equal to a radius."}, {"video_title": "Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "Now what I'm going to do is I'm going to move it over so that this new circle intersects the center of the old circle. These circles are the same size. Notice this center intersects the old circle and the new circle itself intersects the center of the old circle. Now the reason why this is interesting, we already know that this distance, the distance between these two centers, this is equal to a radius. We also know that this distance right over here is equal to a radius. It's equal to the radius of our new circle right over here. We also know that this distance right over here is equal to a radius of our old circle and that they both have the same radius."}, {"video_title": "Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "Now the reason why this is interesting, we already know that this distance, the distance between these two centers, this is equal to a radius. We also know that this distance right over here is equal to a radius. It's equal to the radius of our new circle right over here. We also know that this distance right over here is equal to a radius of our old circle and that they both have the same radius. So this is a radius between these two points. Between these two points is a radius. And then between these two points is a radius."}, {"video_title": "Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "We also know that this distance right over here is equal to a radius of our old circle and that they both have the same radius. So this is a radius between these two points. Between these two points is a radius. And then between these two points is a radius. So now I have constructed an equilateral triangle. And essentially I have to just do this six times and then I'm going to have a hexagon inscribed inside the circle. Let me do it again."}, {"video_title": "Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "And then between these two points is a radius. So now I have constructed an equilateral triangle. And essentially I have to just do this six times and then I'm going to have a hexagon inscribed inside the circle. Let me do it again. So go from here to here. This is a radius of my new circle, which is the same as the radius of the old circle. And I could go from here to here."}, {"video_title": "Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "Let me do it again. So go from here to here. This is a radius of my new circle, which is the same as the radius of the old circle. And I could go from here to here. That's the radius of my old circle. So I have another equilateral triangle. Radius, radius, radius."}, {"video_title": "Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "And I could go from here to here. That's the radius of my old circle. So I have another equilateral triangle. Radius, radius, radius. Another equilateral triangle. I just have to do this four more times. So let me go to my original, let's see, let me make sure I can, well it's actually going to be hard for me to, let me just add another circle here to do it on the other side."}, {"video_title": "Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "Radius, radius, radius. Another equilateral triangle. I just have to do this four more times. So let me go to my original, let's see, let me make sure I can, well it's actually going to be hard for me to, let me just add another circle here to do it on the other side. So if I put the center of it right, I want to move it a little bit right over, I want to make it the same size. So that's close enough. And let's see, that looks pretty close."}, {"video_title": "Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "So let me go to my original, let's see, let me make sure I can, well it's actually going to be hard for me to, let me just add another circle here to do it on the other side. So if I put the center of it right, I want to move it a little bit right over, I want to make it the same size. So that's close enough. And let's see, that looks pretty close. That's the same size. Now let me move it over over here. Now I want this center to be on the circle, right like that."}, {"video_title": "Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "And let's see, that looks pretty close. That's the same size. Now let me move it over over here. Now I want this center to be on the circle, right like that. And now I'm ready to draw some more equilateral triangles. So add a, and really I don't have to even draw the inside of it. Now I see my six vertices for my hexagon."}, {"video_title": "Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "Now I want this center to be on the circle, right like that. And now I'm ready to draw some more equilateral triangles. So add a, and really I don't have to even draw the inside of it. Now I see my six vertices for my hexagon. Here, here, here, here, here, here. And I think you're satisfied now that you could break this up into six equilateral triangles. So let's do that."}, {"video_title": "Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "Now I see my six vertices for my hexagon. Here, here, here, here, here, here. And I think you're satisfied now that you could break this up into six equilateral triangles. So let's do that. So this would be the base of one of those equilateral triangles. And actually let me move these, let me move this one out of the way. I could move this one right over here because I really just care about the hexagon itself."}, {"video_title": "Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "So let's do that. So this would be the base of one of those equilateral triangles. And actually let me move these, let me move this one out of the way. I could move this one right over here because I really just care about the hexagon itself. And I could move this right over here. But we know that these are all the lengths of the radius anyway. Actually I'm not even having to change the length there."}, {"video_title": "Constructing regular hexagon inscribed in circle Geometric constructions Geometry Khan Academy.mp3", "Sentence": "I could move this one right over here because I really just care about the hexagon itself. And I could move this right over here. But we know that these are all the lengths of the radius anyway. Actually I'm not even having to change the length there. Then I have to just connect one more right down here. So let me add another straight edge, connect those two points, and I would have done it. I would have constructed my, I've constructed my regular hexagon inscribed in the circle."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So that's labeled right over there. AB is equal to 1. And then they tell us that BE and BD trisect angle ABC. So BE and BD trisect angle ABC. So trisect means dividing it into three equal angles. So that means that this angle is equal to this angle is equal to that angle. And what they want us to figure out is what is the perimeter of triangle BED?"}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So BE and BD trisect angle ABC. So trisect means dividing it into three equal angles. So that means that this angle is equal to this angle is equal to that angle. And what they want us to figure out is what is the perimeter of triangle BED? Triangle BED. So it's kind of this middle triangle in the rectangle right over here. So at first, it seems like a pretty hard problem."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And what they want us to figure out is what is the perimeter of triangle BED? Triangle BED. So it's kind of this middle triangle in the rectangle right over here. So at first, it seems like a pretty hard problem. Because you're like, well, what is the width of this rectangle? How can I even start on this? They've only given us one side here."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So at first, it seems like a pretty hard problem. Because you're like, well, what is the width of this rectangle? How can I even start on this? They've only given us one side here. But they've actually given us a lot of information, given that we do know that this is a rectangle. We have four sides. And then we have four angles."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "They've only given us one side here. But they've actually given us a lot of information, given that we do know that this is a rectangle. We have four sides. And then we have four angles. And the sides are all parallel to each other. And that the angles are all 90 degrees, which is more than enough information to know that this is definitely a rectangle. And so one thing we do know is that opposite sides of a rectangle are the same length."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And then we have four angles. And the sides are all parallel to each other. And that the angles are all 90 degrees, which is more than enough information to know that this is definitely a rectangle. And so one thing we do know is that opposite sides of a rectangle are the same length. So if this side is 1, then this side right over there is also 1. The other thing we know is that this angle is trisected. Now, we know what the measure of this angle is."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And so one thing we do know is that opposite sides of a rectangle are the same length. So if this side is 1, then this side right over there is also 1. The other thing we know is that this angle is trisected. Now, we know what the measure of this angle is. It was a right angle. It was a 90 degree angle. So if it's divided into three equal parts, that tells us that this angle right over here is 30 degrees."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "Now, we know what the measure of this angle is. It was a right angle. It was a 90 degree angle. So if it's divided into three equal parts, that tells us that this angle right over here is 30 degrees. This angle right over here is 30 degrees. And then this angle right over here is 30 degrees. And then we see that we're dealing with a couple of 30-60-90 triangles."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So if it's divided into three equal parts, that tells us that this angle right over here is 30 degrees. This angle right over here is 30 degrees. And then this angle right over here is 30 degrees. And then we see that we're dealing with a couple of 30-60-90 triangles. This one is 30, 90. So this other side right over here needs to be 60 degrees. So that side right over there needs to be 60 degrees."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And then we see that we're dealing with a couple of 30-60-90 triangles. This one is 30, 90. So this other side right over here needs to be 60 degrees. So that side right over there needs to be 60 degrees. This triangle right over here, you have 30, you have 90. So this one has to be 60 degrees. They have to add up to 180."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So that side right over there needs to be 60 degrees. This triangle right over here, you have 30, you have 90. So this one has to be 60 degrees. They have to add up to 180. 30-60-90 triangle. And you can also figure out the measures of this triangle, although it's not going to be a right triangle. But knowing what we know about 30-60-90 triangles, if we just have one side of them, we can actually figure out the other sides."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "They have to add up to 180. 30-60-90 triangle. And you can also figure out the measures of this triangle, although it's not going to be a right triangle. But knowing what we know about 30-60-90 triangles, if we just have one side of them, we can actually figure out the other sides. So for example, here we have the shortest side. We have the side opposite the 30 degree side. Now, if the 30 degree side is 1, then the 60 degree side is going to be square root of 3 times that."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "But knowing what we know about 30-60-90 triangles, if we just have one side of them, we can actually figure out the other sides. So for example, here we have the shortest side. We have the side opposite the 30 degree side. Now, if the 30 degree side is 1, then the 60 degree side is going to be square root of 3 times that. So this length right over here is going to be square root of 3. And that's pretty useful, because we now just figured out the length of the entire base of this rectangle right over there. And we just used our knowledge of 30-60-90 triangles."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "Now, if the 30 degree side is 1, then the 60 degree side is going to be square root of 3 times that. So this length right over here is going to be square root of 3. And that's pretty useful, because we now just figured out the length of the entire base of this rectangle right over there. And we just used our knowledge of 30-60-90 triangles. If you see that was a little bit mysterious how I came up with that, I encourage you to watch that video. We know that 30-60-90 triangles, their sides are in the ratio of 1 to square root of 3 to 2. So if this is 1, this is the 30 degree side, this is going to be square root of 3 times that."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And we just used our knowledge of 30-60-90 triangles. If you see that was a little bit mysterious how I came up with that, I encourage you to watch that video. We know that 30-60-90 triangles, their sides are in the ratio of 1 to square root of 3 to 2. So if this is 1, this is the 30 degree side, this is going to be square root of 3 times that. And then the hypotenuse right over here is going to be 2 times that. So this length right over here is going to be 2 times the side right over here. So 2 times 1 is just 2."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So if this is 1, this is the 30 degree side, this is going to be square root of 3 times that. And then the hypotenuse right over here is going to be 2 times that. So this length right over here is going to be 2 times the side right over here. So 2 times 1 is just 2. So that's pretty interesting. Let's see if we can do something similar with this side right over here. Here the 1 is not the side opposite the 30 degree side."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So 2 times 1 is just 2. So that's pretty interesting. Let's see if we can do something similar with this side right over here. Here the 1 is not the side opposite the 30 degree side. Here the 1 is the side opposite the 60 degree side. This is the 1 opposite the 60 degree side. So once again, if we multiply this side times square root of 3, we should get this side right over here."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "Here the 1 is not the side opposite the 30 degree side. Here the 1 is the side opposite the 60 degree side. This is the 1 opposite the 60 degree side. So once again, if we multiply this side times square root of 3, we should get this side right over here. Remember, this 1, so right over here, this 1, this is the 60 degree side. So this has to be 1 square root of 3's of this side. Let me write this down."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So once again, if we multiply this side times square root of 3, we should get this side right over here. Remember, this 1, so right over here, this 1, this is the 60 degree side. So this has to be 1 square root of 3's of this side. Let me write this down. 1 over the square root of 3. And the whole reason I was able to get this is, well, whatever this side, if I multiply it by the square root of 3, I should get this side right over here. I should get the 60 degree side, the side opposite the 60 degree angle."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "Let me write this down. 1 over the square root of 3. And the whole reason I was able to get this is, well, whatever this side, if I multiply it by the square root of 3, I should get this side right over here. I should get the 60 degree side, the side opposite the 60 degree angle. Or if I take the 60 degree side, if I divide it by the square root of 3, I should get the short side, the 30 degree side. So if I start with the 60 degree side and divide by the square root of 3, I get that right over there. And then the hypotenuse is always going to be twice the length of the side opposite the 30 degree angle."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "I should get the 60 degree side, the side opposite the 60 degree angle. Or if I take the 60 degree side, if I divide it by the square root of 3, I should get the short side, the 30 degree side. So if I start with the 60 degree side and divide by the square root of 3, I get that right over there. And then the hypotenuse is always going to be twice the length of the side opposite the 30 degree angle. So this is the side opposite the 30 degree angle. The hypotenuse is always twice that. So this is the side opposite the 30 degree angle."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And then the hypotenuse is always going to be twice the length of the side opposite the 30 degree angle. So this is the side opposite the 30 degree angle. The hypotenuse is always twice that. So this is the side opposite the 30 degree angle. The hypotenuse is going to be twice that. It is going to be 2. over the square root of 3. So we're doing pretty good."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So this is the side opposite the 30 degree angle. The hypotenuse is going to be twice that. It is going to be 2. over the square root of 3. So we're doing pretty good. We have to figure out the perimeter of this inner triangle right over here. We already figured out one length is 2. We figured out another length is 2 square roots of 3."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So we're doing pretty good. We have to figure out the perimeter of this inner triangle right over here. We already figured out one length is 2. We figured out another length is 2 square roots of 3. And then all we have to really figure out is what ED is. And we can do that because we know that AD is going to be the same thing as BC. We know that this entire length, because we're dealing with a rectangle, is the square root of 3."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "We figured out another length is 2 square roots of 3. And then all we have to really figure out is what ED is. And we can do that because we know that AD is going to be the same thing as BC. We know that this entire length, because we're dealing with a rectangle, is the square root of 3. This entire length. If that entire length is square root of 3, if this part, this AE, is 1 over the square root of 3, then this length right over here, ED, is going to be square root of 3 minus 1 over the square root of 3. That length minus that length right over there."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "We know that this entire length, because we're dealing with a rectangle, is the square root of 3. This entire length. If that entire length is square root of 3, if this part, this AE, is 1 over the square root of 3, then this length right over here, ED, is going to be square root of 3 minus 1 over the square root of 3. That length minus that length right over there. And now to find the perimeter is pretty straightforward. We just have to add these things up and simplify it. So it's going to be 2."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "That length minus that length right over there. And now to find the perimeter is pretty straightforward. We just have to add these things up and simplify it. So it's going to be 2. So let me write this. Perimeter of triangle BED is equal to, this is short for perimeter. I just didn't feel like writing the whole word."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So it's going to be 2. So let me write this. Perimeter of triangle BED is equal to, this is short for perimeter. I just didn't feel like writing the whole word. Is equal to 2 over the square root of 3 plus square root of 3 minus 1 over the square root of 3 plus 2. And now this just boils down to simplifying radicals. You could take a calculator out and get some type of decimal or approximation for it."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "I just didn't feel like writing the whole word. Is equal to 2 over the square root of 3 plus square root of 3 minus 1 over the square root of 3 plus 2. And now this just boils down to simplifying radicals. You could take a calculator out and get some type of decimal or approximation for it. Let's see, if we have 2 square roots of 3 minus 1 square root of 3, that'll leave us with 1 over the square root of 3. 2 square root of thirds, I should say. 2 over the square root of 3 minus 1 over the square root of 3 is 1 over the square root of 3."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "You could take a calculator out and get some type of decimal or approximation for it. Let's see, if we have 2 square roots of 3 minus 1 square root of 3, that'll leave us with 1 over the square root of 3. 2 square root of thirds, I should say. 2 over the square root of 3 minus 1 over the square root of 3 is 1 over the square root of 3. And then you have plus the square root of 3 plus 2. And let's see, I can rationalize this. If I multiply the numerator and the denominator by the square root of 3, this gives me the square root of 3 over 3 plus the square root of 3, which I can rewrite that as plus 3 square roots of 3 over 3."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "2 over the square root of 3 minus 1 over the square root of 3 is 1 over the square root of 3. And then you have plus the square root of 3 plus 2. And let's see, I can rationalize this. If I multiply the numerator and the denominator by the square root of 3, this gives me the square root of 3 over 3 plus the square root of 3, which I can rewrite that as plus 3 square roots of 3 over 3. I just multiplied this times 3 over 3 plus 2. And so this gives us, this is the drum roll part now. So 1 square root of 3 plus 3 square roots of 3, and all of that over 3 gives us 4 square roots of 3 over 3 plus 2."}, {"video_title": "30-60-90 triangle example problem Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "If I multiply the numerator and the denominator by the square root of 3, this gives me the square root of 3 over 3 plus the square root of 3, which I can rewrite that as plus 3 square roots of 3 over 3. I just multiplied this times 3 over 3 plus 2. And so this gives us, this is the drum roll part now. So 1 square root of 3 plus 3 square roots of 3, and all of that over 3 gives us 4 square roots of 3 over 3 plus 2. Or you could have put the 2 first. Some people like to write the non-irrational part before the irrational part. But we're done."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Sort the expressions according to their values. You can put any number of cards in a category or leave a category empty. And so we have this diagram right over here. And then we have these cards that have these expressions. And we're supposed to sort these into different buckets. So we're trying to say, well, what is the length of segment AC over the length of segment BC equal to? Which of these expressions is it equal to?"}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And then we have these cards that have these expressions. And we're supposed to sort these into different buckets. So we're trying to say, well, what is the length of segment AC over the length of segment BC equal to? Which of these expressions is it equal to? And then we should drag it into the appropriate buckets. So to figure these out, I've actually already redrawn this problem on my little, I guess you'd call it, scratch pad or blackboard, whatever you want to call it. This right over here is that same diagram blown up a little bit."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Which of these expressions is it equal to? And then we should drag it into the appropriate buckets. So to figure these out, I've actually already redrawn this problem on my little, I guess you'd call it, scratch pad or blackboard, whatever you want to call it. This right over here is that same diagram blown up a little bit. Here are the expressions that we need to drag into things. And here are the buckets that we need to see which of these expressions are equal to which of these expressions. So let's first look at this."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "This right over here is that same diagram blown up a little bit. Here are the expressions that we need to drag into things. And here are the buckets that we need to see which of these expressions are equal to which of these expressions. So let's first look at this. The length of segment AC over the length of segment BC. So let's think about what AC is. The length of segment AC."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So let's first look at this. The length of segment AC over the length of segment BC. So let's think about what AC is. The length of segment AC. AC is this right over here. So it's this length right over here in purple over the length of segment BC. Over this length right over here."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "The length of segment AC. AC is this right over here. So it's this length right over here in purple over the length of segment BC. Over this length right over here. So it's the ratio of the lengths of two sides of a right triangle. This is clearly a right triangle. Triangle ABC."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Over this length right over here. So it's the ratio of the lengths of two sides of a right triangle. This is clearly a right triangle. Triangle ABC. And I could color that in just so you know what triangle I'm talking about. Triangle ABC is this entire triangle that we could focus on. So you could imagine that it's reasonable that the ratio of two sides of a right triangle are going to be the sine of one of its angles."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Triangle ABC. And I could color that in just so you know what triangle I'm talking about. Triangle ABC is this entire triangle that we could focus on. So you could imagine that it's reasonable that the ratio of two sides of a right triangle are going to be the sine of one of its angles. And they give us one of the angles right over here. They give us this angle right over here. You say, well, no, all they did is mark that angle."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So you could imagine that it's reasonable that the ratio of two sides of a right triangle are going to be the sine of one of its angles. And they give us one of the angles right over here. They give us this angle right over here. You say, well, no, all they did is mark that angle. But notice, one arc is here. One arc is here. So anywhere we see only one arc, that's going to be 30 degrees."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "You say, well, no, all they did is mark that angle. But notice, one arc is here. One arc is here. So anywhere we see only one arc, that's going to be 30 degrees. So this is 30 degrees as well. You have two arcs here. That's 41 degrees."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So anywhere we see only one arc, that's going to be 30 degrees. So this is 30 degrees as well. You have two arcs here. That's 41 degrees. Two arcs here. This is going to be congruent to that. This over here is going to be 41 degrees."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "That's 41 degrees. Two arcs here. This is going to be congruent to that. This over here is going to be 41 degrees. This is three arcs. They don't tell us how many degrees that is. But this angle with the three arcs is congruent to this angle with the three arcs right over there."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "This over here is going to be 41 degrees. This is three arcs. They don't tell us how many degrees that is. But this angle with the three arcs is congruent to this angle with the three arcs right over there. So anyway, this yellow triangle, triangle ABC, we know the measure of this angle is 30 degrees. And then they give us these two sides. So how do these sides relate to this 30 degree angle?"}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "But this angle with the three arcs is congruent to this angle with the three arcs right over there. So anyway, this yellow triangle, triangle ABC, we know the measure of this angle is 30 degrees. And then they give us these two sides. So how do these sides relate to this 30 degree angle? Well, side AC is adjacent to it. It's literally one of the sides of the angle that is not the hypotenuse. So let me write that down."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So how do these sides relate to this 30 degree angle? Well, side AC is adjacent to it. It's literally one of the sides of the angle that is not the hypotenuse. So let me write that down. This is adjacent. And what is BC? Well, BC is the hypotenuse of this right triangle."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So let me write that down. This is adjacent. And what is BC? Well, BC is the hypotenuse of this right triangle. It's the side opposite the 90 degrees. So this is the hypotenuse. So what trig function, when applied to 30 degrees, is equal to the adjacent side over the hypotenuse?"}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Well, BC is the hypotenuse of this right triangle. It's the side opposite the 90 degrees. So this is the hypotenuse. So what trig function, when applied to 30 degrees, is equal to the adjacent side over the hypotenuse? Let's write down SOH CAH TOA just to remind ourselves. So SOH CAH TOA, sine of an angle is opposite over a hypotenuse. Cosine of an angle is adjacent over a hypotenuse."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So what trig function, when applied to 30 degrees, is equal to the adjacent side over the hypotenuse? Let's write down SOH CAH TOA just to remind ourselves. So SOH CAH TOA, sine of an angle is opposite over a hypotenuse. Cosine of an angle is adjacent over a hypotenuse. So cosine, let's write this down, cosine of 30 degrees is going to be equal to the length of the adjacent side, so that is AC, over the length of the hypotenuse, which is equal to BC. So this right over here is the same thing as the cosine of 30 degrees. So let's drag it in there."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Cosine of an angle is adjacent over a hypotenuse. So cosine, let's write this down, cosine of 30 degrees is going to be equal to the length of the adjacent side, so that is AC, over the length of the hypotenuse, which is equal to BC. So this right over here is the same thing as the cosine of 30 degrees. So let's drag it in there. This is equal to the cosine of 30 degrees. Now let's look at the next one. Cosine of angle DEC. Where is DEC?"}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So let's drag it in there. This is equal to the cosine of 30 degrees. Now let's look at the next one. Cosine of angle DEC. Where is DEC? So DEC. So that's this angle right over here. I'll put four arcs here so we don't get it confused."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Cosine of angle DEC. Where is DEC? So DEC. So that's this angle right over here. I'll put four arcs here so we don't get it confused. So this is angle DEC. So what is the cosine of DEC? Well, once again, cosine is adjacent over a hypotenuse."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "I'll put four arcs here so we don't get it confused. So this is angle DEC. So what is the cosine of DEC? Well, once again, cosine is adjacent over a hypotenuse. So cosine of angle DEC, the adjacent side to this, well, that's this right over here. You might say, well, isn't this side adjacent? Well, that side, side DE, that is the actual hypotenuse."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Well, once again, cosine is adjacent over a hypotenuse. So cosine of angle DEC, the adjacent side to this, well, that's this right over here. You might say, well, isn't this side adjacent? Well, that side, side DE, that is the actual hypotenuse. So that's not going to be the adjacent side. So the adjacent side is E. The adjacent side is, I could call it EC. It's the length of segment EC."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Well, that side, side DE, that is the actual hypotenuse. So that's not going to be the adjacent side. So the adjacent side is E. The adjacent side is, I could call it EC. It's the length of segment EC. And then the hypotenuse is this right over here. It's the length of the hypotenuse. The hypotenuse is side DE or ED, however you want to call it."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "It's the length of segment EC. And then the hypotenuse is this right over here. It's the length of the hypotenuse. The hypotenuse is side DE or ED, however you want to call it. And so the length of it is, we could just write it as DE. Now what is this also equal to? We don't see this choice over here."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "The hypotenuse is side DE or ED, however you want to call it. And so the length of it is, we could just write it as DE. Now what is this also equal to? We don't see this choice over here. We don't have the ratio EC over DE as one of these choices here. But what we do have is we do get one of the angles here. They give us this 41 degrees."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "We don't see this choice over here. We don't have the ratio EC over DE as one of these choices here. But what we do have is we do get one of the angles here. They give us this 41 degrees. And the ratio of this green side, the length of this green side over this orange side, what would that be in terms of if we wanted to apply a trig function to this angle? Well, relative to this angle, the green side is the opposite side. And the orange side is still the hypotenuse."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "They give us this 41 degrees. And the ratio of this green side, the length of this green side over this orange side, what would that be in terms of if we wanted to apply a trig function to this angle? Well, relative to this angle, the green side is the opposite side. And the orange side is still the hypotenuse. So relative to 41 degrees, so let's write this down. Relative to 41 degrees, this ratio is the opposite over the hypotenuse. It's the cosine of this angle."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And the orange side is still the hypotenuse. So relative to 41 degrees, so let's write this down. Relative to 41 degrees, this ratio is the opposite over the hypotenuse. It's the cosine of this angle. But it's the sine of this angle right over here. So sine is opposite over hypotenuse. So this is equal to the sine of this angle right over here."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "It's the cosine of this angle. But it's the sine of this angle right over here. So sine is opposite over hypotenuse. So this is equal to the sine of this angle right over here. It's equal to the sine of 41 degrees. So that is this one right over here, the sine of 41 degrees. So let's drag that into the appropriate bucket."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So this is equal to the sine of this angle right over here. It's equal to the sine of 41 degrees. So that is this one right over here, the sine of 41 degrees. So let's drag that into the appropriate bucket. So let's sine of 41 degrees is the same thing as the cosine of angle DEC. Only have two left. So now we have to figure out what the sine of angle CDA is. So let's see, where is CDA?"}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So let's drag that into the appropriate bucket. So let's sine of 41 degrees is the same thing as the cosine of angle DEC. Only have two left. So now we have to figure out what the sine of angle CDA is. So let's see, where is CDA? CDA is this entire angle. It's this entire angle right over here. So I could put a bunch of arcs here if I want, just to show that it's different than all the other ones."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So let's see, where is CDA? CDA is this entire angle. It's this entire angle right over here. So I could put a bunch of arcs here if I want, just to show that it's different than all the other ones. So that's that angle right over there. So now we're really dealing with this larger right triangle. Let me highlight that in some."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So I could put a bunch of arcs here if I want, just to show that it's different than all the other ones. So that's that angle right over there. So now we're really dealing with this larger right triangle. Let me highlight that in some. Let me highlight it in this pink color. So we're now dealing with this larger right triangle right over here. We care about the sine of this whole thing."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Let me highlight that in some. Let me highlight it in this pink color. So we're now dealing with this larger right triangle right over here. We care about the sine of this whole thing. Remember, sine is opposite over hypotenuse. Sine is opposite over hypotenuse. So the opposite side is going to be side CA."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "We care about the sine of this whole thing. Remember, sine is opposite over hypotenuse. Sine is opposite over hypotenuse. So the opposite side is going to be side CA. So this is going to be equal to the length of CA over the hypotenuse, which is AD. So that is going to be over AD. Now once again, we don't see that as a choice here."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So the opposite side is going to be side CA. So this is going to be equal to the length of CA over the hypotenuse, which is AD. So that is going to be over AD. Now once again, we don't see that as a choice here. But maybe we can express this ratio. Maybe this ratio is a trig function applied to one of the other angles. And they give us one of the angles."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Now once again, we don't see that as a choice here. But maybe we can express this ratio. Maybe this ratio is a trig function applied to one of the other angles. And they give us one of the angles. They give us this angle right over here. I guess we could call this angle DAC. This is 30 degrees."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And they give us one of the angles. They give us this angle right over here. I guess we could call this angle DAC. This is 30 degrees. So relative to this angle, what two sides are we taking the ratio of? We're taking now the ratio of, relative to this angle, the adjacent side over the hypotenuse. So this is the adjacent side over the hypotenuse."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "This is 30 degrees. So relative to this angle, what two sides are we taking the ratio of? We're taking now the ratio of, relative to this angle, the adjacent side over the hypotenuse. So this is the adjacent side over the hypotenuse. What deals with adjacent over hypotenuse? Well, cosine. So this is equal to the cosine of this angle."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So this is the adjacent side over the hypotenuse. What deals with adjacent over hypotenuse? Well, cosine. So this is equal to the cosine of this angle. So this is equal to cosine of 30 degrees. Sine of CDA is equal to the cosine of this angle right over here. So this one is equal to this right over here."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So this is equal to the cosine of this angle. So this is equal to cosine of 30 degrees. Sine of CDA is equal to the cosine of this angle right over here. So this one is equal to this right over here. So let me drag that in. So this one is equal to, so you can see that I just dragged it in, equal to that. And now we have one left."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So this one is equal to this right over here. So let me drag that in. So this one is equal to, so you can see that I just dragged it in, equal to that. And now we have one left. We have one left. Home stretch. We should be getting excited."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And now we have one left. We have one left. Home stretch. We should be getting excited. AE over EB. AE, let me use this color. Length of segment AE."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "We should be getting excited. AE over EB. AE, let me use this color. Length of segment AE. That's this length right over here. Let me make that stand out more. Let me do it in this red."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Length of segment AE. That's this length right over here. Let me make that stand out more. Let me do it in this red. This color right over here. That's length of segment AE over length of segment EB. This is EB right over here."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Let me do it in this red. This color right over here. That's length of segment AE over length of segment EB. This is EB right over here. This is EB. So now we are focused on this triangle, this right triangle right over here. Well, we know the measure of this angle over here."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "This is EB right over here. This is EB. So now we are focused on this triangle, this right triangle right over here. Well, we know the measure of this angle over here. We have double arcs. We have double arcs right over here. And they say this is 41 degrees."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Well, we know the measure of this angle over here. We have double arcs. We have double arcs right over here. And they say this is 41 degrees. So we have double marks over here. And this is also going to be 41 degrees. So relative to this angle, what ratio is this?"}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And they say this is 41 degrees. So we have double marks over here. And this is also going to be 41 degrees. So relative to this angle, what ratio is this? This is the opposite over the hypotenuse. Opposite over the hypotenuse. This right over here is going to be sine of that angle, sine of 41 degrees."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So relative to this angle, what ratio is this? This is the opposite over the hypotenuse. Opposite over the hypotenuse. This right over here is going to be sine of that angle, sine of 41 degrees. So it's equal to this first one right over there. So let's drag it. So this is going to be equal to sine of 41 degrees."}, {"video_title": "Example relating trig function to side ratios Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "This right over here is going to be sine of that angle, sine of 41 degrees. So it's equal to this first one right over there. So let's drag it. So this is going to be equal to sine of 41 degrees. So none of the ones actually ended up being equal to the tangent of 41 degrees. Now let's see if we actually got this right. I hope I did."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And then we're asked, what is the sine of 32 degrees? And I encourage you to pause this video and try it on your own. And a hint is to look at this right triangle, one of the angles is already labeled 32 degrees, figure out what all of the angles are, and then use the fundamental definitions, your SOH CAH TOA definitions to see if you could figure out what sine of 32 degrees is. So I'm assuming you've given a go at it. Let's work it through now. So we know that the sum of the angles of a triangle add up to 180. Now in a right angle, one of the angles is 90 degrees."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So I'm assuming you've given a go at it. Let's work it through now. So we know that the sum of the angles of a triangle add up to 180. Now in a right angle, one of the angles is 90 degrees. So that means that the other two must add up to 90. These two add up to 90, plus another 90 is going to be 180 degrees. Or another way to think about it is that the other two non-right angles are going to be complementary."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Now in a right angle, one of the angles is 90 degrees. So that means that the other two must add up to 90. These two add up to 90, plus another 90 is going to be 180 degrees. Or another way to think about it is that the other two non-right angles are going to be complementary. So what plus 32 is equal to 90? Well, 90 minus 32 is 58. So this right over here is going to be 58 degrees."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Or another way to think about it is that the other two non-right angles are going to be complementary. So what plus 32 is equal to 90? Well, 90 minus 32 is 58. So this right over here is going to be 58 degrees. Well, why is that interesting? Well, we already know what the cosine of 58 degrees is equal to. But let's think about it in terms of ratios of the lengths of sides of this right triangle."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So this right over here is going to be 58 degrees. Well, why is that interesting? Well, we already know what the cosine of 58 degrees is equal to. But let's think about it in terms of ratios of the lengths of sides of this right triangle. Let's just write down SOH CAH TOA. So sine is opposite over hypotenuse. CAH, cosine is adjacent over hypotenuse."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "But let's think about it in terms of ratios of the lengths of sides of this right triangle. Let's just write down SOH CAH TOA. So sine is opposite over hypotenuse. CAH, cosine is adjacent over hypotenuse. TOA, tangent is opposite over adjacent. So we could write down the cosine of 58 degrees, which we already know. If we think about it in terms of these fundamental ratios, cosine is adjacent over hypotenuse."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "CAH, cosine is adjacent over hypotenuse. TOA, tangent is opposite over adjacent. So we could write down the cosine of 58 degrees, which we already know. If we think about it in terms of these fundamental ratios, cosine is adjacent over hypotenuse. This is the 58 degree angle. The side that is adjacent to it is, let me do it in this color, is side BC, right over here. It's one of the sides of the angle, the side of the angle that is not the hypotenuse."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "If we think about it in terms of these fundamental ratios, cosine is adjacent over hypotenuse. This is the 58 degree angle. The side that is adjacent to it is, let me do it in this color, is side BC, right over here. It's one of the sides of the angle, the side of the angle that is not the hypotenuse. The other side, this over here, is the hypotenuse. So this is going to be the adjacent, the length of the adjacent side, BC, over the length of the hypotenuse. Over the length of the hypotenuse."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "It's one of the sides of the angle, the side of the angle that is not the hypotenuse. The other side, this over here, is the hypotenuse. So this is going to be the adjacent, the length of the adjacent side, BC, over the length of the hypotenuse. Over the length of the hypotenuse. The length of the hypotenuse, well that is AB. Now let's think about what the sine of 32 degrees would be. So the sine of 32 degrees, well sine is opposite over hypotenuse."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Over the length of the hypotenuse. The length of the hypotenuse, well that is AB. Now let's think about what the sine of 32 degrees would be. So the sine of 32 degrees, well sine is opposite over hypotenuse. So now we're looking at this 32 degree angle. What side is opposite? Well it opens up onto BC."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So the sine of 32 degrees, well sine is opposite over hypotenuse. So now we're looking at this 32 degree angle. What side is opposite? Well it opens up onto BC. It opens up onto BC, just like that. And what's the hypotenuse? Well we've already, or the length of the hypotenuse, it's AB."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Well it opens up onto BC. It opens up onto BC, just like that. And what's the hypotenuse? Well we've already, or the length of the hypotenuse, it's AB. It's AB. Notice, the sine of 32 degrees is BC over AB. The cosine of 58 degrees is BC over AB."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Well we've already, or the length of the hypotenuse, it's AB. It's AB. Notice, the sine of 32 degrees is BC over AB. The cosine of 58 degrees is BC over AB. Or another way of thinking about it, the sine of this angle is the same thing as the cosine of this angle. So we could literally write the sine, I'm gonna do that in that pink color, the sine, the sine of 32 degrees is equal to the cosine, cosine of 58 degrees, which is roughly equal to 0.53. And this is a really, really useful property."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "The cosine of 58 degrees is BC over AB. Or another way of thinking about it, the sine of this angle is the same thing as the cosine of this angle. So we could literally write the sine, I'm gonna do that in that pink color, the sine, the sine of 32 degrees is equal to the cosine, cosine of 58 degrees, which is roughly equal to 0.53. And this is a really, really useful property. The sine of an angle is equal to the cosine of its complement. So we could write this in general terms. We could write that the sine of some angle is equal to the cosine of its complement, is equal to the cosine of 90 minus theta."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And this is a really, really useful property. The sine of an angle is equal to the cosine of its complement. So we could write this in general terms. We could write that the sine of some angle is equal to the cosine of its complement, is equal to the cosine of 90 minus theta. Think about it. I could have done, I could change this entire problem. Instead of making this the sine of 32 degrees, I could make this the sine of 25 degrees."}, {"video_title": "Sine and cosine of complements example Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "We could write that the sine of some angle is equal to the cosine of its complement, is equal to the cosine of 90 minus theta. Think about it. I could have done, I could change this entire problem. Instead of making this the sine of 32 degrees, I could make this the sine of 25 degrees. And if someone gave you the cosine of what's 90 minus 25, if someone gave you the cosine of 65 degrees, then you could think about this as 25. The complement is going to be right over here. This would be 65 degrees."}, {"video_title": "Area density.mp3", "Sentence": "And the simplest way of thinking about it is density is going to be some quantity per unit area. So for example, let's say that I have a football field right over here, and I have another identical football field right over here. Now they have the same area, but if I have, let's say, five people on this football field, actually six people on this football field, and I only have three people on this football field, the density of people per average unit area, or the density of people, I should say, per football field, is going to be higher in this left example. So it's always going to be quantity per area. Now with that out of the way, let's do a worked example that helps us understand this idea a little bit better. So here we're told the town of Tigersville has a population density of 13 cats per square kilometer. So they're giving us the density."}, {"video_title": "Area density.mp3", "Sentence": "So it's always going to be quantity per area. Now with that out of the way, let's do a worked example that helps us understand this idea a little bit better. So here we're told the town of Tigersville has a population density of 13 cats per square kilometer. So they're giving us the density. I'll rewrite that, 13 cats, so the quantity is quantity in cats, per square kilometer. That's the density right over there. The town is shaped like a perfect isosceles trapezoid."}, {"video_title": "Area density.mp3", "Sentence": "So they're giving us the density. I'll rewrite that, 13 cats, so the quantity is quantity in cats, per square kilometer. That's the density right over there. The town is shaped like a perfect isosceles trapezoid. So it looks something like this. It's a perfect isosceles trapezoid. It's gonna look something like that, with two parallel boundaries 12 kilometers apart."}, {"video_title": "Area density.mp3", "Sentence": "The town is shaped like a perfect isosceles trapezoid. So it looks something like this. It's a perfect isosceles trapezoid. It's gonna look something like that, with two parallel boundaries 12 kilometers apart. So this distance right over here is 12 kilometers, one measuring eight kilometers. So this side over here is eight kilometers. The other is 16, that's the longer one over there."}, {"video_title": "Area density.mp3", "Sentence": "It's gonna look something like that, with two parallel boundaries 12 kilometers apart. So this distance right over here is 12 kilometers, one measuring eight kilometers. So this side over here is eight kilometers. The other is 16, that's the longer one over there. How many cats are in Tigersville? So they give us the density here, and they give us, I think, enough information to figure out the area, and they want us to figure out how many cats we have, so what is the quantity? So pause this video and see if you can figure that out."}, {"video_title": "Area density.mp3", "Sentence": "The other is 16, that's the longer one over there. How many cats are in Tigersville? So they give us the density here, and they give us, I think, enough information to figure out the area, and they want us to figure out how many cats we have, so what is the quantity? So pause this video and see if you can figure that out. Well, just as we said, the density is equal to quantity divided by area. If we multiply both sides of this equation by area, you get area times density is going to be equal to quantity. And we know the density, it's 13 cats per square kilometer, and we can figure out the area, and then just multiply the two."}, {"video_title": "Area density.mp3", "Sentence": "So pause this video and see if you can figure that out. Well, just as we said, the density is equal to quantity divided by area. If we multiply both sides of this equation by area, you get area times density is going to be equal to quantity. And we know the density, it's 13 cats per square kilometer, and we can figure out the area, and then just multiply the two. So what's the area of this right over here? Well, the area of a trapezoid is going to be, let me write it here, area is going to be 12 kilometers, the height of the trapezoid, times the average of the two parallel sides, I guess you could say. So the average of those is going to be eight kilometers plus 16 kilometers over two."}, {"video_title": "Area density.mp3", "Sentence": "And we know the density, it's 13 cats per square kilometer, and we can figure out the area, and then just multiply the two. So what's the area of this right over here? Well, the area of a trapezoid is going to be, let me write it here, area is going to be 12 kilometers, the height of the trapezoid, times the average of the two parallel sides, I guess you could say. So the average of those is going to be eight kilometers plus 16 kilometers over two. So this is going to be equal to 12 kilometers times eight plus 16 is 24, divided by two is 12, so times 12 kilometers. So this gives us 144 square kilometers. Now, we know we have 13 cats per square kilometer, so let me do this here in another color."}, {"video_title": "Area density.mp3", "Sentence": "So the average of those is going to be eight kilometers plus 16 kilometers over two. So this is going to be equal to 12 kilometers times eight plus 16 is 24, divided by two is 12, so times 12 kilometers. So this gives us 144 square kilometers. Now, we know we have 13 cats per square kilometer, so let me do this here in another color. So if I multiply 13 cats per kilometer squared, and I multiply that times this business right over here, times 144 square kilometers, and you might also notice that the units cancel out the same way that variables might, so that cancels out with that. You're going to get 13 times 144, and then the units that you're left with is just cats. So 144 times 13, three times four is 12, three times four is 12, that gives us to 13, three times 100, 300 plus another 100 is 400."}, {"video_title": "Area density.mp3", "Sentence": "Now, we know we have 13 cats per square kilometer, so let me do this here in another color. So if I multiply 13 cats per kilometer squared, and I multiply that times this business right over here, times 144 square kilometers, and you might also notice that the units cancel out the same way that variables might, so that cancels out with that. You're going to get 13 times 144, and then the units that you're left with is just cats. So 144 times 13, three times four is 12, three times four is 12, that gives us to 13, three times 100, 300 plus another 100 is 400. Now I'm just going to multiply 144 essentially by 10, which is just going to be 1440. And so if I add up all of that together, I'm going to jump down to here, I get 1872. So this is 1872 cats in total, and we are done."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "So when they say solve the right triangle, we can assume that they're saying, hey, figure out the lengths of all the sides, so whatever a is equal to, whatever b is equal to, and also what are all the angles of the right triangle? They've given two of them, we might have to figure out this third right over here. So there's multiple ways to tackle this, but we'll just go and we'll just try to tackle side xw first, try to figure out what a is, and I'll give you a hint. You can use a calculator, and using a calculator, you can use your trigonometric functions that we've looked at a good bit now. So I'll give you a few seconds to think about how to figure out what a is. Well, what do we know? We know this angle y right over here, we know the side adjacent to angle y, and length a, this is the side, that's the length of the side that is opposite, that is opposite to angle y."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "You can use a calculator, and using a calculator, you can use your trigonometric functions that we've looked at a good bit now. So I'll give you a few seconds to think about how to figure out what a is. Well, what do we know? We know this angle y right over here, we know the side adjacent to angle y, and length a, this is the side, that's the length of the side that is opposite, that is opposite to angle y. So what trigonometric ratio, what trigonometric ratio deals with the opposite and the adjacent? So if we're looking at angle y, relative to angle y, this is the opposite, and this right over here is the adjacent. Well, if we don't remember, we can go back to SOHCAHTOA."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "We know this angle y right over here, we know the side adjacent to angle y, and length a, this is the side, that's the length of the side that is opposite, that is opposite to angle y. So what trigonometric ratio, what trigonometric ratio deals with the opposite and the adjacent? So if we're looking at angle y, relative to angle y, this is the opposite, and this right over here is the adjacent. Well, if we don't remember, we can go back to SOHCAHTOA. SOHCAHTOA. Sine deals with opposite and hypotenuse, cosine deals with adjacent and hypotenuse, tangent deals with opposite over adjacent, opposite over adjacent. So we can say that the tangent, the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a, over the length of the adjacent side, which they gave us in the diagram, which has length, which has length five."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "Well, if we don't remember, we can go back to SOHCAHTOA. SOHCAHTOA. Sine deals with opposite and hypotenuse, cosine deals with adjacent and hypotenuse, tangent deals with opposite over adjacent, opposite over adjacent. So we can say that the tangent, the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a, over the length of the adjacent side, which they gave us in the diagram, which has length, which has length five. And you might say, well, how do I figure out a? Well, we can use our calculator to evaluate what the tangent of 65 degrees are, and then we can solve for a. And actually, if we just wanna get the expression, explicitly solving for a, we can just multiply both sides of this equation times five."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "So we can say that the tangent, the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a, over the length of the adjacent side, which they gave us in the diagram, which has length, which has length five. And you might say, well, how do I figure out a? Well, we can use our calculator to evaluate what the tangent of 65 degrees are, and then we can solve for a. And actually, if we just wanna get the expression, explicitly solving for a, we can just multiply both sides of this equation times five. So let's do that. Five times, times five, these cancel out, and we are left with, if we flip the equal around, we're left with a is equal to five times the tangent of 65, of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest 10th."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "And actually, if we just wanna get the expression, explicitly solving for a, we can just multiply both sides of this equation times five. So let's do that. Five times, times five, these cancel out, and we are left with, if we flip the equal around, we're left with a is equal to five times the tangent of 65, of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest 10th. My handy TI-85 out, and I have five times the, not the, the tangent, no, I didn't need to press that second right over there, just the regular tangent, of 65 degrees. And I am, I get, if I round to the nearest 10th like they asked me to, I get 10.7. So this is, so a is approximately equal to 10.7."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "So now we can get our calculator out and figure out what this is to the nearest 10th. My handy TI-85 out, and I have five times the, not the, the tangent, no, I didn't need to press that second right over there, just the regular tangent, of 65 degrees. And I am, I get, if I round to the nearest 10th like they asked me to, I get 10.7. So this is, so a is approximately equal to 10.7. I say approximately, because I rounded it, I rounded it down. I didn't, this is not the exact number, but a is equal to 10.7. So we now know that this has length 10.7 approximately."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "So this is, so a is approximately equal to 10.7. I say approximately, because I rounded it, I rounded it down. I didn't, this is not the exact number, but a is equal to 10.7. So we now know that this has length 10.7 approximately. There's several ways that we can try to tackle b, and I'll let you pick the way you want to, but then I'll just do it the way I would like to. So my next question to you is, what is the length of the side yw, or what is the value of b? Well, there's several ways to do it."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "So we now know that this has length 10.7 approximately. There's several ways that we can try to tackle b, and I'll let you pick the way you want to, but then I'll just do it the way I would like to. So my next question to you is, what is the length of the side yw, or what is the value of b? Well, there's several ways to do it. This is a hypotenuse, so we could use trigonometric functions that deal with adjacent over hypotenuse or opposite over hypotenuse, or we could just use the Pythagorean theorem. We know two sides of a right triangle. We can come up with the third side."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "Well, there's several ways to do it. This is a hypotenuse, so we could use trigonometric functions that deal with adjacent over hypotenuse or opposite over hypotenuse, or we could just use the Pythagorean theorem. We know two sides of a right triangle. We can come up with the third side. I will go with using trigonometric ratios, since that's what we've been working on a good bit. So this length b, that's the length of the hypotenuse. So this side wy is the hypotenuse."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "We can come up with the third side. I will go with using trigonometric ratios, since that's what we've been working on a good bit. So this length b, that's the length of the hypotenuse. So this side wy is the hypotenuse. And so what trigonometric ratios, and we can decide what we want to use. We could use opposite and hypotenuse. We could use adjacent and hypotenuse."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "So this side wy is the hypotenuse. And so what trigonometric ratios, and we can decide what we want to use. We could use opposite and hypotenuse. We could use adjacent and hypotenuse. Since we know that xy is exactly five, we don't have to deal with this approximation. Let's use that side. So what trigonometric ratios deal with adjacent and hypotenuse?"}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "We could use adjacent and hypotenuse. Since we know that xy is exactly five, we don't have to deal with this approximation. Let's use that side. So what trigonometric ratios deal with adjacent and hypotenuse? Well, we see from Sohcahtoa, cosine deals with adjacent over hypotenuse. So we could say that the cosine of 65 degrees, cosine of 65 degrees, is equal to the length of the adjacent side, which is five, over the length of the hypotenuse, which has length b. And then we can try to solve for b."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "So what trigonometric ratios deal with adjacent and hypotenuse? Well, we see from Sohcahtoa, cosine deals with adjacent over hypotenuse. So we could say that the cosine of 65 degrees, cosine of 65 degrees, is equal to the length of the adjacent side, which is five, over the length of the hypotenuse, which has length b. And then we can try to solve for b. You multiply both sides times b. You're left with b times cosine of 65 degrees is equal to five. And then to solve for b, you can divide both sides by cosine of 65 degrees."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "And then we can try to solve for b. You multiply both sides times b. You're left with b times cosine of 65 degrees is equal to five. And then to solve for b, you can divide both sides by cosine of 65 degrees. This is just a number here. So we're just dividing. We have to figure it out with our calculator, but this is just going to evaluate to some number."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "And then to solve for b, you can divide both sides by cosine of 65 degrees. This is just a number here. So we're just dividing. We have to figure it out with our calculator, but this is just going to evaluate to some number. So we can divide both sides by that, by cosine of 65 degrees, cosine of 65 degrees. And we're left with b is equal to five over the cosine of 65 degrees. So let us now use our calculator to figure out the length of b."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "We have to figure it out with our calculator, but this is just going to evaluate to some number. So we can divide both sides by that, by cosine of 65 degrees, cosine of 65 degrees. And we're left with b is equal to five over the cosine of 65 degrees. So let us now use our calculator to figure out the length of b. Length of b is five divided by cosine of 65 degrees. And I get, if I round to the nearest tenth, 11.8. So b is approximately equal to, rounded to the nearest tenth, 11.8."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "So let us now use our calculator to figure out the length of b. Length of b is five divided by cosine of 65 degrees. And I get, if I round to the nearest tenth, 11.8. So b is approximately equal to, rounded to the nearest tenth, 11.8. So b is equal to 11.8. And then we're almost done solving this right triangle. And you could have figured this out using the Pythagorean theorem as well, saying that five squared plus 10.7 squared should be equal to b squared."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "So b is approximately equal to, rounded to the nearest tenth, 11.8. So b is equal to 11.8. And then we're almost done solving this right triangle. And you could have figured this out using the Pythagorean theorem as well, saying that five squared plus 10.7 squared should be equal to b squared. And hopefully you would get the exact same answer. And the last thing we have to figure out is the measure of angle w. Angle w right over here. So I'll give you a few seconds to think about what the measure of angle w is."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "And you could have figured this out using the Pythagorean theorem as well, saying that five squared plus 10.7 squared should be equal to b squared. And hopefully you would get the exact same answer. And the last thing we have to figure out is the measure of angle w. Angle w right over here. So I'll give you a few seconds to think about what the measure of angle w is. Well, here we just have to remember that the sum of the angles of a triangle add up to 180 degrees. So angle w plus 65 degrees, that's this angle right up here, plus the right angle, this is a right triangle, they're going to add up to 180 degrees. So all we need to do is, well, we could simplify the left-hand side right over here."}, {"video_title": "Example Trig to solve the sides and angles of a right triangle Trigonometry Khan Academy.mp3", "Sentence": "So I'll give you a few seconds to think about what the measure of angle w is. Well, here we just have to remember that the sum of the angles of a triangle add up to 180 degrees. So angle w plus 65 degrees, that's this angle right up here, plus the right angle, this is a right triangle, they're going to add up to 180 degrees. So all we need to do is, well, we could simplify the left-hand side right over here. 65 plus 90 is 155. So angle w plus 155 degrees is equal to 180 degrees. And then we get angle w, if we subtract 155 from both sides, angle w is equal to 25 degrees."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "We also know that FG is a perpendicular bisector of BC. So we've already shown that it's perpendicular, that this is a 90 degree angle, but it also bisects BC. So this length is equal to that length right over there. Then they tell us that arc AC, that's a curve on top, arc AC is part of circle B. So this is a circle centered at B. So if this is the center of the circle, this is part of that circle. It's really kind of the bottom left quarter of that circle."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "Then they tell us that arc AC, that's a curve on top, arc AC is part of circle B. So this is a circle centered at B. So if this is the center of the circle, this is part of that circle. It's really kind of the bottom left quarter of that circle. And then given that information, they want us to find what the measure of angle BED is. So what is BED? So it's BED."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "It's really kind of the bottom left quarter of that circle. And then given that information, they want us to find what the measure of angle BED is. So what is BED? So it's BED. So we need to figure out the measure of this angle right over here. And I encourage you to pause it and try it out. And you might imagine, well, you could pause it and try it without any hints."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "So it's BED. So we need to figure out the measure of this angle right over here. And I encourage you to pause it and try it out. And you might imagine, well, you could pause it and try it without any hints. And now I will give you a hint if you tried it the first time and you weren't able to do it and you should pause again after this hint, is try to draw some triangles that maybe split up this angle into a couple of different angles and it might be a little bit easier. You might be able to use some of what we know about triangles. Now, with that said, I will try to solve it."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "And you might imagine, well, you could pause it and try it without any hints. And now I will give you a hint if you tried it the first time and you weren't able to do it and you should pause again after this hint, is try to draw some triangles that maybe split up this angle into a couple of different angles and it might be a little bit easier. You might be able to use some of what we know about triangles. Now, with that said, I will try to solve it. And you should pause at any point where you think that you know exactly how to do this and try to do it yourself. So the trick here is to realize that this is a circle. And so any line that goes between B and any point on this arc is going to be equal to the radius of the circle."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "Now, with that said, I will try to solve it. And you should pause at any point where you think that you know exactly how to do this and try to do it yourself. So the trick here is to realize that this is a circle. And so any line that goes between B and any point on this arc is going to be equal to the radius of the circle. So AB is equal to the radius of the circle. BE is equal to the radius of the circle. And we can keep drawing other things that are equal to the radius of the circle."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "And so any line that goes between B and any point on this arc is going to be equal to the radius of the circle. So AB is equal to the radius of the circle. BE is equal to the radius of the circle. And we can keep drawing other things that are equal to the radius of the circle. BC is equal to the radius of the circle. So let's think about it a little bit. If we were to draw, and a lot of trickier geometry problems really all revolve around drawing the right lines or visualizing the right triangles."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "And we can keep drawing other things that are equal to the radius of the circle. BC is equal to the radius of the circle. So let's think about it a little bit. If we were to draw, and a lot of trickier geometry problems really all revolve around drawing the right lines or visualizing the right triangles. And I'll do one right here that might open up a lot for you in terms of thinking about how to do this problem. So let me draw segment EC. I'll draw that as straight as possible."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "If we were to draw, and a lot of trickier geometry problems really all revolve around drawing the right lines or visualizing the right triangles. And I'll do one right here that might open up a lot for you in terms of thinking about how to do this problem. So let me draw segment EC. I'll draw that as straight as possible. I can draw a better job of that. So segment EC. Now something becomes interesting."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "I'll draw that as straight as possible. I can draw a better job of that. So segment EC. Now something becomes interesting. Because what is the relationship between triangle EBG and triangle ECG? Well, they both definitely share this side right over here. They both share side EG."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "Now something becomes interesting. Because what is the relationship between triangle EBG and triangle ECG? Well, they both definitely share this side right over here. They both share side EG. And then BG is equal to GC. And they both have 90 degree angles. You have a 90 degree angle here."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "They both share side EG. And then BG is equal to GC. And they both have 90 degree angles. You have a 90 degree angle here. You have a 90 degree angle there. So you see by side, angle, side, side, angle, side, these two triangles are going to be congruent. So we know that triangle EBG is congruent to triangle ECG by side, angle, side, congruency."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "You have a 90 degree angle here. You have a 90 degree angle there. So you see by side, angle, side, side, angle, side, these two triangles are going to be congruent. So we know that triangle EBG is congruent to triangle ECG by side, angle, side, congruency. And that also tells us that all of the corresponding angles and sides are going to be the same. So that tells us right there that EC is equal to EB. So we know that EB is equal to EC."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "So we know that triangle EBG is congruent to triangle ECG by side, angle, side, congruency. And that also tells us that all of the corresponding angles and sides are going to be the same. So that tells us right there that EC is equal to EB. So we know that EB is equal to EC. And what also is equal to that length? Well, once again, this is the radius of the circle. BE is one radius of the circle going from the center to the arc."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "So we know that EB is equal to EC. And what also is equal to that length? Well, once again, this is the radius of the circle. BE is one radius of the circle going from the center to the arc. But so is BC. It is also a radius of the circle going from the center to the arc. So this is also equal to BC."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "BE is one radius of the circle going from the center to the arc. But so is BC. It is also a radius of the circle going from the center to the arc. So this is also equal to BC. So I could draw this out of the three things right here. I'm referring to the entire thing, not just one of the segments. All of BC."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "So this is also equal to BC. So I could draw this out of the three things right here. I'm referring to the entire thing, not just one of the segments. All of BC. So what kind of a triangle is this right over here? Triangle BEC is equilateral. And we know that because all three sides are equal."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "All of BC. So what kind of a triangle is this right over here? Triangle BEC is equilateral. And we know that because all three sides are equal. So that tells us that all of its angles are equal. So that tells us that the measure of angle BEC, we're not done yet, but it gets us close, is 60 degrees. So the measure of angle BEC right over there is 60 degrees."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "And we know that because all three sides are equal. So that tells us that all of its angles are equal. So that tells us that the measure of angle BEC, we're not done yet, but it gets us close, is 60 degrees. So the measure of angle BEC right over there is 60 degrees. So that gives us part of the problem. BEC is part of the angle BED. If we can just figure out the measure of angle CED now, if we can figure out this angle right over here, we just add that to 60 degrees and we're done."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "So the measure of angle BEC right over there is 60 degrees. So that gives us part of the problem. BEC is part of the angle BED. If we can just figure out the measure of angle CED now, if we can figure out this angle right over here, we just add that to 60 degrees and we're done. We've figured out the entire BED. Now let's think about how we can do this right over here. So there's a couple of interesting things that we already do know."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "If we can just figure out the measure of angle CED now, if we can figure out this angle right over here, we just add that to 60 degrees and we're done. We've figured out the entire BED. Now let's think about how we can do this right over here. So there's a couple of interesting things that we already do know. We know that this right over here is equal to the radius of the circle. And we also know that this length down here, this is a square. We know that this length down here is the same as this length up here."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "So there's a couple of interesting things that we already do know. We know that this right over here is equal to the radius of the circle. And we also know that this length down here, this is a square. We know that this length down here is the same as this length up here. That these are the exact same length. And this is equal to the radius of the circle. We already put these three slashes here."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "We know that this length down here is the same as this length up here. That these are the exact same length. And this is equal to the radius of the circle. We already put these three slashes here. BC is the same as that length. This is the same as that length. And so all four sides are going to be that same length because this is a square."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "We already put these three slashes here. BC is the same as that length. This is the same as that length. And so all four sides are going to be that same length because this is a square. So let me write this down. Because it's a square, I'll just write it this way, because it's a square, we know that CD is equal to BC, which is equal to, and we already established, which is equal to EC, which is equal to EB. But the important thing here is to realize that this and this are the same length."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "And so all four sides are going to be that same length because this is a square. So let me write this down. Because it's a square, I'll just write it this way, because it's a square, we know that CD is equal to BC, which is equal to, and we already established, which is equal to EC, which is equal to EB. But the important thing here is to realize that this and this are the same length. And the reason why that is interesting is it lets us know that this is an isosceles triangle. So whatever isosceles triangle, if you have your two legs of it, the two base angles are going to be congruent. So whatever angle this green angle is, this angle is going to be as well."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "But the important thing here is to realize that this and this are the same length. And the reason why that is interesting is it lets us know that this is an isosceles triangle. So whatever isosceles triangle, if you have your two legs of it, the two base angles are going to be congruent. So whatever angle this green angle is, this angle is going to be as well. So somehow we can figure out this angle right over here. We can subtract that from 180 and then divide by 2 to figure out these two because we know that they're the same. So how can we figure out this angle?"}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "So whatever angle this green angle is, this angle is going to be as well. So somehow we can figure out this angle right over here. We can subtract that from 180 and then divide by 2 to figure out these two because we know that they're the same. So how can we figure out this angle? Well, we know all of the angles of this thing up. We can figure out the angles of all of this larger one up here. We know this is an equilateral triangle."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "So how can we figure out this angle? Well, we know all of the angles of this thing up. We can figure out the angles of all of this larger one up here. We know this is an equilateral triangle. So this over here has to be 60 degrees as well. That's 60 degrees, and that is also 60 degrees. In fact, I could write it over here, which is equal to the measure of angle BCE."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "We know this is an equilateral triangle. So this over here has to be 60 degrees as well. That's 60 degrees, and that is also 60 degrees. In fact, I could write it over here, which is equal to the measure of angle BCE. Measure of angle BCE. So if this is 60 degrees, we know we're dealing with a square, so this whole angle over here is a right angle. What is the measure of angle ECD?"}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "In fact, I could write it over here, which is equal to the measure of angle BCE. Measure of angle BCE. So if this is 60 degrees, we know we're dealing with a square, so this whole angle over here is a right angle. What is the measure of angle ECD? What is this angle right over here? Let me do this in a new color. This right over here is going to have to be 30 degrees."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "What is the measure of angle ECD? What is this angle right over here? Let me do this in a new color. This right over here is going to have to be 30 degrees. So that is going to be 30 degrees. And now we're ready to solve for these two base angles. If we call these x, and we know they have to be the same, we have x plus x plus 30 degrees is going to be equal to 180 degrees."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "This right over here is going to have to be 30 degrees. So that is going to be 30 degrees. And now we're ready to solve for these two base angles. If we call these x, and we know they have to be the same, we have x plus x plus 30 degrees is going to be equal to 180 degrees. That's the sum of all of the interior angles of a triangle. So you get 2x plus 30 is equal to 180 degrees. Now you can subtract 30 from both sides, and we are left with 2x is equal to 150."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "If we call these x, and we know they have to be the same, we have x plus x plus 30 degrees is going to be equal to 180 degrees. That's the sum of all of the interior angles of a triangle. So you get 2x plus 30 is equal to 180 degrees. Now you can subtract 30 from both sides, and we are left with 2x is equal to 150. Divide both sides by 2, you get x is equal to 75. So we've figured out that x is equal to 75, and now we're at the home stretch. We need to figure out angle BED."}, {"video_title": "Problem involving angle derived from square and circle Congruence Geometry Khan Academy.mp3", "Sentence": "Now you can subtract 30 from both sides, and we are left with 2x is equal to 150. Divide both sides by 2, you get x is equal to 75. So we've figured out that x is equal to 75, and now we're at the home stretch. We need to figure out angle BED. Well, that's going to be angle C, so x is equal to the measure of angle CED. So BED is CED plus BEC. So the 60 degrees plus 75 degrees."}, {"video_title": "Identifying transformation described with other algebra and geometry concepts.mp3", "Sentence": "We're told that a certain mapping in the xy-plane has the following two properties. Each point on the line y is equal to three x minus two maps to itself. Any point p not on the line maps to a new point p prime in such a way that the perpendicular bisector of the segment p, p prime is the line y is equal to three x minus two. Which of the following statements is true? So is this describing a reflection, a rotation, or a translation? So pause this video and see if you can work through it on your own. All right, so let me just try to visualize this."}, {"video_title": "Identifying transformation described with other algebra and geometry concepts.mp3", "Sentence": "Which of the following statements is true? So is this describing a reflection, a rotation, or a translation? So pause this video and see if you can work through it on your own. All right, so let me just try to visualize this. So, and I'll just do a very quick, so if that's my y-axis, and that this right over here is my x-axis, three x minus two might look something like this. The line three x minus two would look something like that. And so what we're saying is, or what they're telling us, is any point on this after the transformation maps to itself."}, {"video_title": "Identifying transformation described with other algebra and geometry concepts.mp3", "Sentence": "All right, so let me just try to visualize this. So, and I'll just do a very quick, so if that's my y-axis, and that this right over here is my x-axis, three x minus two might look something like this. The line three x minus two would look something like that. And so what we're saying is, or what they're telling us, is any point on this after the transformation maps to itself. Now that by itself is a pretty good clue that we're likely dealing with a reflection. Because remember, with a reflection, you reflect over a line. But if a point sits on the line, well it's just gonna continue to sit on the line."}, {"video_title": "Identifying transformation described with other algebra and geometry concepts.mp3", "Sentence": "And so what we're saying is, or what they're telling us, is any point on this after the transformation maps to itself. Now that by itself is a pretty good clue that we're likely dealing with a reflection. Because remember, with a reflection, you reflect over a line. But if a point sits on the line, well it's just gonna continue to sit on the line. But let's just make sure that the second point is consistent with it being a reflection. So any point p not on the line, so let's see, point p right over here, it maps to a new point p prime in such a way that the perpendicular bisector of p p prime is the line y equals three x minus two. So I need to connect, so the line three x minus two, y is equal to three x minus two, would be the perpendicular bisector of the segment between p and what?"}, {"video_title": "Identifying transformation described with other algebra and geometry concepts.mp3", "Sentence": "But if a point sits on the line, well it's just gonna continue to sit on the line. But let's just make sure that the second point is consistent with it being a reflection. So any point p not on the line, so let's see, point p right over here, it maps to a new point p prime in such a way that the perpendicular bisector of p p prime is the line y equals three x minus two. So I need to connect, so the line three x minus two, y is equal to three x minus two, would be the perpendicular bisector of the segment between p and what? Well let's see, I'd have to draw a perpendicular line. It would have to have the same length on both sides of the line y equals three x minus two. So p prime would have to be right over there."}, {"video_title": "Identifying transformation described with other algebra and geometry concepts.mp3", "Sentence": "So I need to connect, so the line three x minus two, y is equal to three x minus two, would be the perpendicular bisector of the segment between p and what? Well let's see, I'd have to draw a perpendicular line. It would have to have the same length on both sides of the line y equals three x minus two. So p prime would have to be right over there. So once again, this is consistent with being a reflection. P prime is equidistant on the other side of the line as p. So I definitely feel good that this is going to be a reflection right over here. Let's do another example."}, {"video_title": "Identifying transformation described with other algebra and geometry concepts.mp3", "Sentence": "So p prime would have to be right over there. So once again, this is consistent with being a reflection. P prime is equidistant on the other side of the line as p. So I definitely feel good that this is going to be a reflection right over here. Let's do another example. So here we are told, and I'll switch my colors up, a certain mapping in the plane has the following two properties. Point O maps to itself. Every point V on a circle C centered at O, maps to a new point W on a circle, on circle C, so that the counterclockwise angle from segment OV to OW measures 137 degrees."}, {"video_title": "Identifying transformation described with other algebra and geometry concepts.mp3", "Sentence": "Let's do another example. So here we are told, and I'll switch my colors up, a certain mapping in the plane has the following two properties. Point O maps to itself. Every point V on a circle C centered at O, maps to a new point W on a circle, on circle C, so that the counterclockwise angle from segment OV to OW measures 137 degrees. So is this a reflection, rotation, or translation? Pause this video and try to figure it out on your own. All right, so let's see."}, {"video_title": "Identifying transformation described with other algebra and geometry concepts.mp3", "Sentence": "Every point V on a circle C centered at O, maps to a new point W on a circle, on circle C, so that the counterclockwise angle from segment OV to OW measures 137 degrees. So is this a reflection, rotation, or translation? Pause this video and try to figure it out on your own. All right, so let's see. We're talking about circle centered at O. So let's see, let me just say, so I have this point O. It maps to itself on its transformation."}, {"video_title": "Identifying transformation described with other algebra and geometry concepts.mp3", "Sentence": "All right, so let's see. We're talking about circle centered at O. So let's see, let me just say, so I have this point O. It maps to itself on its transformation. Now every point V on circle C centered at O, so let's see, let's say this is circle C centered at point O. So I'm gonna try to draw a decent-looking circle here. You get the idea."}, {"video_title": "Identifying transformation described with other algebra and geometry concepts.mp3", "Sentence": "It maps to itself on its transformation. Now every point V on circle C centered at O, so let's see, let's say this is circle C centered at point O. So I'm gonna try to draw a decent-looking circle here. You get the idea. This is not the best hand-drawn circle ever, all right? So every point, let's just pick a point V here. So let's say that that is the point V on a circle centered at O, maps to a new point W on the circle C. So maybe it maps to a new point W on, actually, let me keep reading, W on circle C, so that the counterclockwise angle from OV to OW measures 137 degrees."}, {"video_title": "Identifying transformation described with other algebra and geometry concepts.mp3", "Sentence": "You get the idea. This is not the best hand-drawn circle ever, all right? So every point, let's just pick a point V here. So let's say that that is the point V on a circle centered at O, maps to a new point W on the circle C. So maybe it maps to a new point W on, actually, let me keep reading, W on circle C, so that the counterclockwise angle from OV to OW measures 137 degrees. Okay, so we wanna know the angle from OV to OW, going counterclockwise, is 137 degrees. So this right over here is 137 degrees. And so this would be the segment OW."}, {"video_title": "Identifying transformation described with other algebra and geometry concepts.mp3", "Sentence": "So let's say that that is the point V on a circle centered at O, maps to a new point W on the circle C. So maybe it maps to a new point W on, actually, let me keep reading, W on circle C, so that the counterclockwise angle from OV to OW measures 137 degrees. Okay, so we wanna know the angle from OV to OW, going counterclockwise, is 137 degrees. So this right over here is 137 degrees. And so this would be the segment OW. W would go right over there. And so what this looks like is, well, if we're talking about angles and we are rotating something, this point corresponds to this point. It's essentially the point has been rotated by 137 degrees around point O."}, {"video_title": "Identifying transformation described with other algebra and geometry concepts.mp3", "Sentence": "And so this would be the segment OW. W would go right over there. And so what this looks like is, well, if we're talking about angles and we are rotating something, this point corresponds to this point. It's essentially the point has been rotated by 137 degrees around point O. So this right over here is clearly a rotation. This is a rotation. Sometimes reading this language at first is a little bit daunting."}, {"video_title": "Identifying transformation described with other algebra and geometry concepts.mp3", "Sentence": "It's essentially the point has been rotated by 137 degrees around point O. So this right over here is clearly a rotation. This is a rotation. Sometimes reading this language at first is a little bit daunting. It was a little bit daunting to me when I first read it. But when you actually just break it down and you actually try to visualize what's going on, you'll say, okay, well, look, they're just taking point V and they're rotating it by 137 degrees around point O. And so this would be a rotation."}, {"video_title": "Identifying transformation described with other algebra and geometry concepts.mp3", "Sentence": "Sometimes reading this language at first is a little bit daunting. It was a little bit daunting to me when I first read it. But when you actually just break it down and you actually try to visualize what's going on, you'll say, okay, well, look, they're just taking point V and they're rotating it by 137 degrees around point O. And so this would be a rotation. Let's do one more example. So here we are told, so they're talking about, again, a certain mapping in the XY plane. Each circle O with radius R and centered at X, Y is mapped to a circle O prime with radius R and centered at X plus 11 and then Y minus seven."}, {"video_title": "Identifying transformation described with other algebra and geometry concepts.mp3", "Sentence": "And so this would be a rotation. Let's do one more example. So here we are told, so they're talking about, again, a certain mapping in the XY plane. Each circle O with radius R and centered at X, Y is mapped to a circle O prime with radius R and centered at X plus 11 and then Y minus seven. So once again, pause this video. What is this, reflection, rotation, or translation? All right, so you might be tempted, if they're talking about circles like we did in the last example, maybe they're talking about a rotation."}, {"video_title": "Identifying transformation described with other algebra and geometry concepts.mp3", "Sentence": "Each circle O with radius R and centered at X, Y is mapped to a circle O prime with radius R and centered at X plus 11 and then Y minus seven. So once again, pause this video. What is this, reflection, rotation, or translation? All right, so you might be tempted, if they're talking about circles like we did in the last example, maybe they're talking about a rotation. But look, what they're really saying is is that if I have a circle, let's say I have a circle right over here, centered right over here, after, so this is X comma Y, centered at X comma Y, it's mapped to a new circle O prime with the same radius. So if this is the radius, it's mapped to a new circle with the same radius, but now it is centered at, now it is centered at X plus 11, so our new X coordinate is gonna be 11 larger, X plus 11, and our Y coordinate is gonna be seven less. But we have the exact same radius."}, {"video_title": "Identifying transformation described with other algebra and geometry concepts.mp3", "Sentence": "All right, so you might be tempted, if they're talking about circles like we did in the last example, maybe they're talking about a rotation. But look, what they're really saying is is that if I have a circle, let's say I have a circle right over here, centered right over here, after, so this is X comma Y, centered at X comma Y, it's mapped to a new circle O prime with the same radius. So if this is the radius, it's mapped to a new circle with the same radius, but now it is centered at, now it is centered at X plus 11, so our new X coordinate is gonna be 11 larger, X plus 11, and our Y coordinate is gonna be seven less. But we have the exact same radius. We have the exact same radius. So our circle would still, so we have the exact same radius right over here. So what just happened to this circle?"}, {"video_title": "Identifying transformation described with other algebra and geometry concepts.mp3", "Sentence": "But we have the exact same radius. We have the exact same radius. So our circle would still, so we have the exact same radius right over here. So what just happened to this circle? Well, we kept the radius the same and we just shifted, we just shifted our center to the right by 11, plus 11, and we shifted it down by seven. We shifted it down by seven. So this is clearly a translation."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "We've got two right triangles here. Let's say we also know that they both have an angle whose measure is equal to theta. So angle A is congruent to angle D. What do we now know about these two triangles? Well, for any right, or any triangle, if you know two of the angles, you're going to know the third angle because the sum of the angles of a triangle add up to 180 degrees. So if you have two angles in common, that means you're going to have three angles in common. And if you have three angles in common, you are dealing with similar triangles. Let me make that a little bit clearer."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Well, for any right, or any triangle, if you know two of the angles, you're going to know the third angle because the sum of the angles of a triangle add up to 180 degrees. So if you have two angles in common, that means you're going to have three angles in common. And if you have three angles in common, you are dealing with similar triangles. Let me make that a little bit clearer. So if this angle is theta, this is 90, they all have to add up to 180 degrees. That means that this angle plus this angle up here have to add up to 90. We've already used up 90 right over here."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Let me make that a little bit clearer. So if this angle is theta, this is 90, they all have to add up to 180 degrees. That means that this angle plus this angle up here have to add up to 90. We've already used up 90 right over here. So this angle, angle A and angle B need to be complements. So this angle right over here needs to be 90 minus theta. Well, we can use the same logic over here."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "We've already used up 90 right over here. So this angle, angle A and angle B need to be complements. So this angle right over here needs to be 90 minus theta. Well, we can use the same logic over here. We already used up 90 degrees over here. So we have a remaining 90 degrees between theta and that angle. So this angle is going to be 90 degrees minus theta."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Well, we can use the same logic over here. We already used up 90 degrees over here. So we have a remaining 90 degrees between theta and that angle. So this angle is going to be 90 degrees minus theta. 90 degrees minus theta. You have three congruent, three corresponding angles being congruent, you are dealing with similar triangles. Now, why is that interesting?"}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So this angle is going to be 90 degrees minus theta. 90 degrees minus theta. You have three congruent, three corresponding angles being congruent, you are dealing with similar triangles. Now, why is that interesting? Well, we know from geometry that the ratio of corresponding sides of a similar triangles are always going to be the same. So let's explore the corresponding sides here. Well, the side that jumps out when you're dealing with the right triangles the most is always the hypotenuse."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Now, why is that interesting? Well, we know from geometry that the ratio of corresponding sides of a similar triangles are always going to be the same. So let's explore the corresponding sides here. Well, the side that jumps out when you're dealing with the right triangles the most is always the hypotenuse. So this right over here is the hypotenuse. This hypotenuse is going to correspond to this hypotenuse right over here. And we could write that down."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Well, the side that jumps out when you're dealing with the right triangles the most is always the hypotenuse. So this right over here is the hypotenuse. This hypotenuse is going to correspond to this hypotenuse right over here. And we could write that down. This is the hypotenuse of this triangle. This is the hypotenuse of that triangle. Now, this side right over here, side BC, what side does that correspond to?"}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And we could write that down. This is the hypotenuse of this triangle. This is the hypotenuse of that triangle. Now, this side right over here, side BC, what side does that correspond to? Well, if you look at this triangle, you can kind of view it as the side that is opposite this angle theta. So it's opposite. If you go across a triangle, you get there."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Now, this side right over here, side BC, what side does that correspond to? Well, if you look at this triangle, you can kind of view it as the side that is opposite this angle theta. So it's opposite. If you go across a triangle, you get there. So let's go opposite angle D. If you go opposite angle A, you get to BC. Opposite angle D, you get to EF. You get EF, so it corresponds to this side right over here."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "If you go across a triangle, you get there. So let's go opposite angle D. If you go opposite angle A, you get to BC. Opposite angle D, you get to EF. You get EF, so it corresponds to this side right over here. And then finally, side AC is the one remaining one. We could view it as, well, there's two sides that make up this angle A. One of them is a hypotenuse."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "You get EF, so it corresponds to this side right over here. And then finally, side AC is the one remaining one. We could view it as, well, there's two sides that make up this angle A. One of them is a hypotenuse. We could call this maybe the adjacent side to it. And so D corresponds to A. And so this would be the side that corresponds."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "One of them is a hypotenuse. We could call this maybe the adjacent side to it. And so D corresponds to A. And so this would be the side that corresponds. Now, the whole reason I did that is to leverage that corresponding sides, the ratio between corresponding sides of similar triangles is always going to be the same. So for example, the ratio between BC and the hypotenuse BA, so let me write that down. BC over BA is going to be equal to EF over ED."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And so this would be the side that corresponds. Now, the whole reason I did that is to leverage that corresponding sides, the ratio between corresponding sides of similar triangles is always going to be the same. So for example, the ratio between BC and the hypotenuse BA, so let me write that down. BC over BA is going to be equal to EF over ED. EF, the length of segment EF over the length of segment ED. Over the length of segment ED. Or we could also write that the length of segment AC, so AC over the hypotenuse, over this triangle's hypotenuse, over AB is equal to DF over DE."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "BC over BA is going to be equal to EF over ED. EF, the length of segment EF over the length of segment ED. Over the length of segment ED. Or we could also write that the length of segment AC, so AC over the hypotenuse, over this triangle's hypotenuse, over AB is equal to DF over DE. Once again, this green side over the orange side, these are similar triangles. They're corresponding to each other. So this is equal to, this is equal to DF over DE."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Or we could also write that the length of segment AC, so AC over the hypotenuse, over this triangle's hypotenuse, over AB is equal to DF over DE. Once again, this green side over the orange side, these are similar triangles. They're corresponding to each other. So this is equal to, this is equal to DF over DE. Over DE. Or we could also say, we could keep going, but I'll just do another one. Or we could say that the ratio of this side, right over this blue side to the green side, so of this triangle, BC, the length of BC over CA, over CA is going to be the same as the blue, the ratio between these two corresponding sides, the blue over the green."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So this is equal to, this is equal to DF over DE. Over DE. Or we could also say, we could keep going, but I'll just do another one. Or we could say that the ratio of this side, right over this blue side to the green side, so of this triangle, BC, the length of BC over CA, over CA is going to be the same as the blue, the ratio between these two corresponding sides, the blue over the green. EF over DF. Over DF. And we got all of this from the fact that these are similar triangles."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Or we could say that the ratio of this side, right over this blue side to the green side, so of this triangle, BC, the length of BC over CA, over CA is going to be the same as the blue, the ratio between these two corresponding sides, the blue over the green. EF over DF. Over DF. And we got all of this from the fact that these are similar triangles. So if this is true for all similar triangles, or this is true for any right triangle that has an angle theta, then those two triangles are going to be similar, and all of these ratios are going to be the same. Well, maybe we can give names to these ratios relative to the angle theta. So from angle theta's point of view, from theta's point of view, I'll write theta right over here, or we can just remember that."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And we got all of this from the fact that these are similar triangles. So if this is true for all similar triangles, or this is true for any right triangle that has an angle theta, then those two triangles are going to be similar, and all of these ratios are going to be the same. Well, maybe we can give names to these ratios relative to the angle theta. So from angle theta's point of view, from theta's point of view, I'll write theta right over here, or we can just remember that. What are these, what is the ratio of these two sides? Well, from theta's point of view, that blue side is the opposite side, it's opposite, so the opposite side of the right triangle. And then the orange side, we've already labeled the hypotenuse."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So from angle theta's point of view, from theta's point of view, I'll write theta right over here, or we can just remember that. What are these, what is the ratio of these two sides? Well, from theta's point of view, that blue side is the opposite side, it's opposite, so the opposite side of the right triangle. And then the orange side, we've already labeled the hypotenuse. So from theta's point of view, this is the opposite side over the hypotenuse. And I keep stating from theta's point of view, because that wouldn't be the case for this other angle, for angle B. From angle B's point of view, this is the adjacent side over the hypotenuse."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And then the orange side, we've already labeled the hypotenuse. So from theta's point of view, this is the opposite side over the hypotenuse. And I keep stating from theta's point of view, because that wouldn't be the case for this other angle, for angle B. From angle B's point of view, this is the adjacent side over the hypotenuse. And we'll think about that relationship later on. But let's just all think of it from theta's point of view right over here. So from theta's point of view, what is this?"}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "From angle B's point of view, this is the adjacent side over the hypotenuse. And we'll think about that relationship later on. But let's just all think of it from theta's point of view right over here. So from theta's point of view, what is this? Well, theta's right over here. Clearly, A, B, and D, E, A, B, and D, E are still the hypotenuses, hypotenai, I don't know how to say that in plural again. And what is AC?"}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So from theta's point of view, what is this? Well, theta's right over here. Clearly, A, B, and D, E, A, B, and D, E are still the hypotenuses, hypotenai, I don't know how to say that in plural again. And what is AC? Or what is AC and what are DF? Well, these are adjacent to it. They're one of the sides, two sides that make up this angle that is not the hypotenuse."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And what is AC? Or what is AC and what are DF? Well, these are adjacent to it. They're one of the sides, two sides that make up this angle that is not the hypotenuse. So this we can view as the ratio in either of these triangles between the adjacent side. So this is relative, once again, this is opposite angle B, but we're only thinking about angle A right here, or the angle that measures theta, or angle D right over here. Relative to angle A, AC is adjacent."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "They're one of the sides, two sides that make up this angle that is not the hypotenuse. So this we can view as the ratio in either of these triangles between the adjacent side. So this is relative, once again, this is opposite angle B, but we're only thinking about angle A right here, or the angle that measures theta, or angle D right over here. Relative to angle A, AC is adjacent. Relative to angle D, DF is adjacent. So this ratio right over here is the adjacent over the hypotenuse. And it's going to be the same for any right triangle that has an angle theta in it."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Relative to angle A, AC is adjacent. Relative to angle D, DF is adjacent. So this ratio right over here is the adjacent over the hypotenuse. And it's going to be the same for any right triangle that has an angle theta in it. And then finally, this over here, this is going to be the opposite side. Once again, this was the opposite side over here. This ratio for either right triangle is going to be the opposite side over the adjacent side."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And it's going to be the same for any right triangle that has an angle theta in it. And then finally, this over here, this is going to be the opposite side. Once again, this was the opposite side over here. This ratio for either right triangle is going to be the opposite side over the adjacent side. Over the adjacent side. And I really wanna stress the importance, and we're gonna do many, many more examples of this to make this very concrete. But for any right triangle that has an angle theta, the ratio between its opposite side and its hypotenuse is going to be the same."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "This ratio for either right triangle is going to be the opposite side over the adjacent side. Over the adjacent side. And I really wanna stress the importance, and we're gonna do many, many more examples of this to make this very concrete. But for any right triangle that has an angle theta, the ratio between its opposite side and its hypotenuse is going to be the same. That comes out of similar triangles, we just explored that. The ratio between the adjacent side to that angle that is theta and the hypotenuse is going to be the same for any of these triangles as long as it has that angle theta in it. And the ratio relative to the angle theta between the opposite side and the adjacent side, between the blue side and the green side, is always going to be the same."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "But for any right triangle that has an angle theta, the ratio between its opposite side and its hypotenuse is going to be the same. That comes out of similar triangles, we just explored that. The ratio between the adjacent side to that angle that is theta and the hypotenuse is going to be the same for any of these triangles as long as it has that angle theta in it. And the ratio relative to the angle theta between the opposite side and the adjacent side, between the blue side and the green side, is always going to be the same. These are similar triangles. So given that, mathematicians decided to give these things names. Relative to the angle theta, this ratio is always going to be the same."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And the ratio relative to the angle theta between the opposite side and the adjacent side, between the blue side and the green side, is always going to be the same. These are similar triangles. So given that, mathematicians decided to give these things names. Relative to the angle theta, this ratio is always going to be the same. So they call this, the opposite of our hypotenuse, they call this the sine of the angle theta. So this is the sine, let me do this in a new color. This is by definition, and we're going to extend this definition in the future, this is sine of theta."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Relative to the angle theta, this ratio is always going to be the same. So they call this, the opposite of our hypotenuse, they call this the sine of the angle theta. So this is the sine, let me do this in a new color. This is by definition, and we're going to extend this definition in the future, this is sine of theta. This right over here, by definition, is the cosine of theta. And this right over here, by definition, is the tangent. That, by definition, is the tangent of theta."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "This is by definition, and we're going to extend this definition in the future, this is sine of theta. This right over here, by definition, is the cosine of theta. And this right over here, by definition, is the tangent. That, by definition, is the tangent of theta. And a mnemonic that will help you remember this, and these really are just definitions. People realize, wow, by similar triangles, for any angle theta, this ratio is always going to be the same. Because of similar triangles, for any angle theta, this ratio is always going to be same."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "That, by definition, is the tangent of theta. And a mnemonic that will help you remember this, and these really are just definitions. People realize, wow, by similar triangles, for any angle theta, this ratio is always going to be the same. Because of similar triangles, for any angle theta, this ratio is always going to be same. This ratio is always going to be same, so let's make these definitions. And to help us remember it, there's the mnemonic SOH CAH TOA. So I'll write it like this."}, {"video_title": "Similarity to define sine, cosine, and tangent Basic trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Because of similar triangles, for any angle theta, this ratio is always going to be same. This ratio is always going to be same, so let's make these definitions. And to help us remember it, there's the mnemonic SOH CAH TOA. So I'll write it like this. SOH, SOH, SOH is sine is opposite of our hypotenuse, CAH, cosine, cosine is adjacent over hypotenuse, cosine is adjacent over hypotenuse. And then finally, tangent, tangent is opposite over adjacent. Tangent is opposite, opposite over, opposite over adjacent."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "What we have drawn over here is five different triangles. What I want to do in this video is figure out which of these triangles are congruent to which other of these triangles. And to figure that out, I'm just over here going to write our triangle congruency postulate. So we know that two triangles are congruent if all of their sides are the same, so side, side, side. We also know they are congruent if we have a side and then an angle between the sides, and then another side that is congruent, so side, angle, side. If we reverse the angles on the sides, we know that's also a congruence postulate. So if we have an angle and then another angle, and then the side in between them is congruent, then we also have two congruent triangles."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "So we know that two triangles are congruent if all of their sides are the same, so side, side, side. We also know they are congruent if we have a side and then an angle between the sides, and then another side that is congruent, so side, angle, side. If we reverse the angles on the sides, we know that's also a congruence postulate. So if we have an angle and then another angle, and then the side in between them is congruent, then we also have two congruent triangles. And then finally, if we have an angle and then another angle and then a side, then that is also. Any of these imply congruency. So let's see our congruent triangles."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "So if we have an angle and then another angle, and then the side in between them is congruent, then we also have two congruent triangles. And then finally, if we have an angle and then another angle and then a side, then that is also. Any of these imply congruency. So let's see our congruent triangles. So let's see what we can figure out right over here for these triangles. So right in this triangle ABC over here, we're given this length 7, then 60 degrees, and then 40 degrees. Or another way to think about it, we're given an angle, an angle, and a side."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "So let's see our congruent triangles. So let's see what we can figure out right over here for these triangles. So right in this triangle ABC over here, we're given this length 7, then 60 degrees, and then 40 degrees. Or another way to think about it, we're given an angle, an angle, and a side. 40 degrees, then 60 degrees, then 7. And in order for something to be congruent here, they would have to have an angle, angle, side given, at least unless maybe we have to figure it out some other way. But I'm guessing for this problem, they'll just already give us the angle."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "Or another way to think about it, we're given an angle, an angle, and a side. 40 degrees, then 60 degrees, then 7. And in order for something to be congruent here, they would have to have an angle, angle, side given, at least unless maybe we have to figure it out some other way. But I'm guessing for this problem, they'll just already give us the angle. So they'll have to have an angle, an angle, side. And it can't just be any angle, angle, and side. It has to be 40, 60, and 7."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "But I'm guessing for this problem, they'll just already give us the angle. So they'll have to have an angle, an angle, side. And it can't just be any angle, angle, and side. It has to be 40, 60, and 7. And it has to be in the same order. It can't be 60 and then 40 and then 7. If the 40 degree side has, if one of its side has the length 7, then that is not the same thing here."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "It has to be 40, 60, and 7. And it has to be in the same order. It can't be 60 and then 40 and then 7. If the 40 degree side has, if one of its side has the length 7, then that is not the same thing here. Here the 60 degree side has length 7. So let's see if any of these other triangles have this kind of 40, 60 degrees, and then the 7 right over here. So this has the 40 degrees and the 60 degrees, but the 7 is in between them."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "If the 40 degree side has, if one of its side has the length 7, then that is not the same thing here. Here the 60 degree side has length 7. So let's see if any of these other triangles have this kind of 40, 60 degrees, and then the 7 right over here. So this has the 40 degrees and the 60 degrees, but the 7 is in between them. So this looks like it might be congruent to some other triangle, maybe closer to something like angle, side, angle, because they have an angle, side, angle. So it wouldn't be that one. This one looks interesting."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "So this has the 40 degrees and the 60 degrees, but the 7 is in between them. So this looks like it might be congruent to some other triangle, maybe closer to something like angle, side, angle, because they have an angle, side, angle. So it wouldn't be that one. This one looks interesting. This is also angle, side, angle. So maybe these are congruent, but we'll check back on that. We're still focused on this one right over here."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "This one looks interesting. This is also angle, side, angle. So maybe these are congruent, but we'll check back on that. We're still focused on this one right over here. This one we have a 60 degrees, then a 40 degrees, and a 7. This is tempting. We have an angle, an angle, and a side."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "We're still focused on this one right over here. This one we have a 60 degrees, then a 40 degrees, and a 7. This is tempting. We have an angle, an angle, and a side. But the angles are in a different order. Here it's 40, 60, 7. Here it's 60, 40, 7."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "We have an angle, an angle, and a side. But the angles are in a different order. Here it's 40, 60, 7. Here it's 60, 40, 7. So it's an angle, an angle, and side, but the side is not on the 60 degree angle. It's on the 40 degree angle over here. So this doesn't look right either."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "Here it's 60, 40, 7. So it's an angle, an angle, and side, but the side is not on the 60 degree angle. It's on the 40 degree angle over here. So this doesn't look right either. Here we have 40 degrees, 60 degrees, and then 7. So this is looking pretty good. We have this side right over here is congruent to this side right over here."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "So this doesn't look right either. Here we have 40 degrees, 60 degrees, and then 7. So this is looking pretty good. We have this side right over here is congruent to this side right over here. Then you have your 60 degree angle right over here. 60 degree angle over here. It might not be obvious, because it's flipped, and they're drawn a little bit different."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "We have this side right over here is congruent to this side right over here. Then you have your 60 degree angle right over here. 60 degree angle over here. It might not be obvious, because it's flipped, and they're drawn a little bit different. But you should never assume that just the drawing tells you what's going on. And then finally, you have your 40 degree angle here, which is your 40 degree angle here. So we can say, we can write down that, and I'll do it."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "It might not be obvious, because it's flipped, and they're drawn a little bit different. But you should never assume that just the drawing tells you what's going on. And then finally, you have your 40 degree angle here, which is your 40 degree angle here. So we can say, we can write down that, and I'll do it. Let me think of a good place to do it. I'll write it right over here. We can write down that triangle ABC is congruent to triangle."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "So we can say, we can write down that, and I'll do it. Let me think of a good place to do it. I'll write it right over here. We can write down that triangle ABC is congruent to triangle. Now we have to be very careful with how we name this. We have to make sure that we have the corresponding vertices map up together. So for example, we started this triangle at vertex A."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "We can write down that triangle ABC is congruent to triangle. Now we have to be very careful with how we name this. We have to make sure that we have the corresponding vertices map up together. So for example, we started this triangle at vertex A. So point A right over here, that's where we have the 60 degree angle. That's the vertex of the 60 degree angle. So the vertex of the 60 degree angle over here is point N. So I'm going to go to N. And then we went from A to B."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "So for example, we started this triangle at vertex A. So point A right over here, that's where we have the 60 degree angle. That's the vertex of the 60 degree angle. So the vertex of the 60 degree angle over here is point N. So I'm going to go to N. And then we went from A to B. B was the side, was the vertex that we did not have any angle for. And we could figure it out. If these two guys add up to 100, then this is going to be the 80 degree angle."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "So the vertex of the 60 degree angle over here is point N. So I'm going to go to N. And then we went from A to B. B was the side, was the vertex that we did not have any angle for. And we could figure it out. If these two guys add up to 100, then this is going to be the 80 degree angle. So over here, the 80 degree angle is going to be M, the one that we don't have any label for. It's kind of the other side. It's the thing that shares the 7 length side right over here."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "If these two guys add up to 100, then this is going to be the 80 degree angle. So over here, the 80 degree angle is going to be M, the one that we don't have any label for. It's kind of the other side. It's the thing that shares the 7 length side right over here. So then we want to go to N, then M, and then finish up the triangle in O. And I want to really stress this, that we have to make sure we get the order of these right. Because then we're kind of referring to, we're not showing the corresponding vertices in each triangle."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "It's the thing that shares the 7 length side right over here. So then we want to go to N, then M, and then finish up the triangle in O. And I want to really stress this, that we have to make sure we get the order of these right. Because then we're kind of referring to, we're not showing the corresponding vertices in each triangle. So now we see vertex A, or point A, maps to point N on this congruent triangle. Vertex B maps to point M. And so you can say, look, AB, the length of AB, is congruent to NM. So it all matches up."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "Because then we're kind of referring to, we're not showing the corresponding vertices in each triangle. So now we see vertex A, or point A, maps to point N on this congruent triangle. Vertex B maps to point M. And so you can say, look, AB, the length of AB, is congruent to NM. So it all matches up. And we can say that these two are congruent by angle, angle side, by AAS. So we did this one is, this one right over here is congruent to this one right over there. Now let's look at these two characters."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "So it all matches up. And we can say that these two are congruent by angle, angle side, by AAS. So we did this one is, this one right over here is congruent to this one right over there. Now let's look at these two characters. So here we have an angle, 40 degrees, a side in between, and then another angle. So it looks like ASA is going to be involved. We look at this one right over here."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "Now let's look at these two characters. So here we have an angle, 40 degrees, a side in between, and then another angle. So it looks like ASA is going to be involved. We look at this one right over here. We have a 40 degrees, 40 degrees, 7, and then 60. And you might say, wait, here the 40 degrees is on the bottom, and here it's on the top. But remember, things can be congruent if you can flip them."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "We look at this one right over here. We have a 40 degrees, 40 degrees, 7, and then 60. And you might say, wait, here the 40 degrees is on the bottom, and here it's on the top. But remember, things can be congruent if you can flip them. If you can flip them, rotate them, shift them, whatever. So if you flip this guy over, you will get this one over here. And that would not have happened if you had flipped this one to get this one over here."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "But remember, things can be congruent if you can flip them. If you can flip them, rotate them, shift them, whatever. So if you flip this guy over, you will get this one over here. And that would not have happened if you had flipped this one to get this one over here. So you see, these two by, let me just make it clear. You have the 60 degree angle is congruent to this 60 degree angle. You have the side of length 7 is congruent to this side of length 7."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "And that would not have happened if you had flipped this one to get this one over here. So you see, these two by, let me just make it clear. You have the 60 degree angle is congruent to this 60 degree angle. You have the side of length 7 is congruent to this side of length 7. And then you have the 40 degree angle is congruent to this 40 degree angle. So once again, these two characters are congruent to each other. And we can write, I'll write it right over here, we can say triangle DEF is congruent to triangle."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "You have the side of length 7 is congruent to this side of length 7. And then you have the 40 degree angle is congruent to this 40 degree angle. So once again, these two characters are congruent to each other. And we can write, I'll write it right over here, we can say triangle DEF is congruent to triangle. And here we have to be careful again. Point D is the vertex for the 60 degree side. So I'm going to start at H, which is the vertex of the 60 degree side over here."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "And we can write, I'll write it right over here, we can say triangle DEF is congruent to triangle. And here we have to be careful again. Point D is the vertex for the 60 degree side. So I'm going to start at H, which is the vertex of the 60 degree side over here. It's congruent to triangle H. And then we went from D to E. E is the vertex on the 40 degree side, kind of the other vertex that shares the 7 length segment right over here. We want to go from H to G. HGI. And we know that from angle side angle."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "So I'm going to start at H, which is the vertex of the 60 degree side over here. It's congruent to triangle H. And then we went from D to E. E is the vertex on the 40 degree side, kind of the other vertex that shares the 7 length segment right over here. We want to go from H to G. HGI. And we know that from angle side angle. By angle side angle. So that gives us that character right over there is congruent to this character right over here. And then finally, we're left with this poor chap."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "And we know that from angle side angle. By angle side angle. So that gives us that character right over there is congruent to this character right over here. And then finally, we're left with this poor chap. And it looks like it is not congruent to any of them. It is tempting to try to match it up to this one, especially because the angles here on the bottom, and you have the 7 side over here, angles here on the bottom, and you have the 7 side over here. But it doesn't match up."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "And then finally, we're left with this poor chap. And it looks like it is not congruent to any of them. It is tempting to try to match it up to this one, especially because the angles here on the bottom, and you have the 7 side over here, angles here on the bottom, and you have the 7 side over here. But it doesn't match up. Because the order of the angles aren't the same. You don't have the same corresponding angles. If you try to do this little exercise where you map everything to each other, you wouldn't be able to do it right over here."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "But it doesn't match up. Because the order of the angles aren't the same. You don't have the same corresponding angles. If you try to do this little exercise where you map everything to each other, you wouldn't be able to do it right over here. And this over here, it might have been a trick question where maybe if you did the math, if this was like a 40 or 60 degree angle, then maybe you could have matched this to some of the other triangles, or maybe even some of them to each other. But this last angle in all of these cases, 40 plus 60 is 100. This is going to be an 80 degree angle right over here."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "If you try to do this little exercise where you map everything to each other, you wouldn't be able to do it right over here. And this over here, it might have been a trick question where maybe if you did the math, if this was like a 40 or 60 degree angle, then maybe you could have matched this to some of the other triangles, or maybe even some of them to each other. But this last angle in all of these cases, 40 plus 60 is 100. This is going to be an 80 degree angle right over here. To add up to 180, this is an 80 degree angle. If this ended up by the math being a 40 or 60 degree angle, then it could have been a little bit more interesting. There might have been other congruent pairs."}, {"video_title": "Finding congruent triangles Congruence Geometry Khan Academy.mp3", "Sentence": "This is going to be an 80 degree angle right over here. To add up to 180, this is an 80 degree angle. If this ended up by the math being a 40 or 60 degree angle, then it could have been a little bit more interesting. There might have been other congruent pairs. But this is an 80 degree angle in every case. The other angle is 80 degrees. So this is just a loan, unfortunately, for him."}, {"video_title": "Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3", "Sentence": "We're going to pick whether it's a square, rhombus, rectangle, parallelogram, trapezoid, none of the above. And I'm assuming we're going to pick the most specific one possible, because obviously all squares are rhombuses, or rhombi, I guess you'd say. Not all rhombi are squares. All squares are also rectangles. All squares, rhombi, and rectangles are parallelograms. So we want to be as specific as possible in picking this. So let's see, point A is at 1, 6."}, {"video_title": "Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3", "Sentence": "All squares are also rectangles. All squares, rhombi, and rectangles are parallelograms. So we want to be as specific as possible in picking this. So let's see, point A is at 1, 6. So 1, 6. I encourage you to pause this video and actually try this on your own before seeing how I do it, but I'll just proceed. So that's point A right over there."}, {"video_title": "Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So let's see, point A is at 1, 6. So 1, 6. I encourage you to pause this video and actually try this on your own before seeing how I do it, but I'll just proceed. So that's point A right over there. Point B is at negative 5, 2. Negative 5, 2. That's point B."}, {"video_title": "Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So that's point A right over there. Point B is at negative 5, 2. Negative 5, 2. That's point B. Point C is at some carbonated water, so some air is coming out. Point C is at negative 7, 8. Negative 7, 8."}, {"video_title": "Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3", "Sentence": "That's point B. Point C is at some carbonated water, so some air is coming out. Point C is at negative 7, 8. Negative 7, 8. So that is point C right over there. And then finally, point D is at 2, 11. 2, 11."}, {"video_title": "Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Negative 7, 8. So that is point C right over there. And then finally, point D is at 2, 11. 2, 11. And actually, that kind of goes off the screen. This is 10, 11 would be right like this. So that would be 2, 11."}, {"video_title": "Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3", "Sentence": "2, 11. And actually, that kind of goes off the screen. This is 10, 11 would be right like this. So that would be 2, 11. If we were to extend this, this is 10, and this is 11 right up here. 2, 11. So let's see what this quadrilateral looks like."}, {"video_title": "Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So that would be 2, 11. If we were to extend this, this is 10, and this is 11 right up here. 2, 11. So let's see what this quadrilateral looks like. You have this line right over here, this line right over there, that line right over there. And then you have this line like this, like this. And then you have this like this."}, {"video_title": "Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So let's see what this quadrilateral looks like. You have this line right over here, this line right over there, that line right over there. And then you have this line like this, like this. And then you have this like this. So right off the bat, it's definitely a quadrilateral. I have four sides. But the key question, are any of these sides parallel to any of the other sides?"}, {"video_title": "Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And then you have this like this. So right off the bat, it's definitely a quadrilateral. I have four sides. But the key question, are any of these sides parallel to any of the other sides? So just looking at it, side CB is clearly not parallel to AD. You just can look at it. Now, it also looks like CD is not parallel to BA."}, {"video_title": "Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3", "Sentence": "But the key question, are any of these sides parallel to any of the other sides? So just looking at it, side CB is clearly not parallel to AD. You just can look at it. Now, it also looks like CD is not parallel to BA. But maybe I just drew it badly. Maybe they actually are parallel. So let's see if we can verify that."}, {"video_title": "Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Now, it also looks like CD is not parallel to BA. But maybe I just drew it badly. Maybe they actually are parallel. So let's see if we can verify that. So the way to tell whether two things are parallel is to actually figure out their slopes. So let's first figure out the slope. Let's figure out our slope of AB or BA."}, {"video_title": "Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So let's see if we can verify that. So the way to tell whether two things are parallel is to actually figure out their slopes. So let's first figure out the slope. Let's figure out our slope of AB or BA. So let's figure out our slope here. Your slope is going to be the change in y over your change in x. And in this case, you could think of it as we're starting at the point negative 5, 2, and we're ending at the point 1, 6."}, {"video_title": "Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Let's figure out our slope of AB or BA. So let's figure out our slope here. Your slope is going to be the change in y over your change in x. And in this case, you could think of it as we're starting at the point negative 5, 2, and we're ending at the point 1, 6. So what's our change in y? We go from 2 all the way to 6. Or you could say it's 6 minus 2."}, {"video_title": "Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And in this case, you could think of it as we're starting at the point negative 5, 2, and we're ending at the point 1, 6. So what's our change in y? We go from 2 all the way to 6. Or you could say it's 6 minus 2. Our change in y is 4. What's our change in x? Well, we go from negative 5 to 1."}, {"video_title": "Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Or you could say it's 6 minus 2. Our change in y is 4. What's our change in x? Well, we go from negative 5 to 1. So we increased by 6. Or another way of thinking about it, 1 minus negative 5. Well, that's going to be equal to 6."}, {"video_title": "Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Well, we go from negative 5 to 1. So we increased by 6. Or another way of thinking about it, 1 minus negative 5. Well, that's going to be equal to 6. So our slope here is 2 thirds. It's 2 over 3. Every time we move 3 in the x direction, we go up 2 in the y direction."}, {"video_title": "Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Well, that's going to be equal to 6. So our slope here is 2 thirds. It's 2 over 3. Every time we move 3 in the x direction, we go up 2 in the y direction. Move 3 in the x direction, go up 2 in the y direction. Now let's think about line CD up here. What is our slope?"}, {"video_title": "Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Every time we move 3 in the x direction, we go up 2 in the y direction. Move 3 in the x direction, go up 2 in the y direction. Now let's think about line CD up here. What is our slope? Our change in y over change in x is going to be equal to, let's see, our change in, let's figure out our change in x first. Our change in x, we're going from negative 7, 8 all the way to 2, 11. So our change in x, we're going from negative 7 to 2."}, {"video_title": "Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3", "Sentence": "What is our slope? Our change in y over change in x is going to be equal to, let's see, our change in, let's figure out our change in x first. Our change in x, we're going from negative 7, 8 all the way to 2, 11. So our change in x, we're going from negative 7 to 2. Or we could say 2 minus negative 7. So we are increasing by 9. That's our change in x."}, {"video_title": "Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So our change in x, we're going from negative 7 to 2. Or we could say 2 minus negative 7. So we are increasing by 9. That's our change in x. So that's going to be equal to 9. And then our change in y, well, it looks like we've gone up. We've gone from 8 to 11."}, {"video_title": "Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3", "Sentence": "That's our change in x. So that's going to be equal to 9. And then our change in y, well, it looks like we've gone up. We've gone from 8 to 11. So we've gone up 3. Or we could say 11 minus 8. Notice, end point minus start point."}, {"video_title": "Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3", "Sentence": "We've gone from 8 to 11. So we've gone up 3. Or we could say 11 minus 8. Notice, end point minus start point. You have to do that on both the top and the bottom. Otherwise, you're not going to actually be calculating your change in y over change in x. But you notice our change, when our x increases by 9, our y increased by 3."}, {"video_title": "Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Notice, end point minus start point. You have to do that on both the top and the bottom. Otherwise, you're not going to actually be calculating your change in y over change in x. But you notice our change, when our x increases by 9, our y increased by 3. So the slope here is equal to 1 third. So these actually have different slopes. So none of these lines are parallel to each other."}, {"video_title": "Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3", "Sentence": "But you notice our change, when our x increases by 9, our y increased by 3. So the slope here is equal to 1 third. So these actually have different slopes. So none of these lines are parallel to each other. So this isn't even a parallelogram. This isn't even a trapezoid. Parallelogram, you have to have two pairs of parallel sides."}, {"video_title": "Classifying a quadrilateral on the coordinate plane Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So none of these lines are parallel to each other. So this isn't even a parallelogram. This isn't even a trapezoid. Parallelogram, you have to have two pairs of parallel sides. Trapezoid, you have to have one pair of parallel sides. This isn't the case for any of these. Or none of these sides are parallel."}, {"video_title": "Geometric constructions parallel line Congruence High school geometry Khan Academy.mp3", "Sentence": "How would we do that? Well, the way that we can approach it is by creating what will eventually be a transversal between the two parallel lines. So let me draw that. So I'm just drawing a line that goes through my point and intersects my original line. I'm doing that, so it's going to look like that. And then I'm really just going to use the idea of corresponding angle congruence for parallel lines. So what I can do is now take my compass and think about this angle right over here."}, {"video_title": "Geometric constructions parallel line Congruence High school geometry Khan Academy.mp3", "Sentence": "So I'm just drawing a line that goes through my point and intersects my original line. I'm doing that, so it's going to look like that. And then I'm really just going to use the idea of corresponding angle congruence for parallel lines. So what I can do is now take my compass and think about this angle right over here. So I'll draw it like that and say, all right, if I draw an arc of the same radius over here, can I reconstruct that angle? And so where should the point be on this left end? Well, to do that, I can just measure the distance between these two points using my compass."}, {"video_title": "Geometric constructions parallel line Congruence High school geometry Khan Academy.mp3", "Sentence": "So what I can do is now take my compass and think about this angle right over here. So I'll draw it like that and say, all right, if I draw an arc of the same radius over here, can I reconstruct that angle? And so where should the point be on this left end? Well, to do that, I can just measure the distance between these two points using my compass. So I'm adjusting it a little bit to get the distance between those two points. And then I can use that up over here to figure out, I got a little bit shaky, I can figure out that point right over there. And just like that, I now have two corresponding angles defined by a transversal and parallel lines."}, {"video_title": "Geometric constructions parallel line Congruence High school geometry Khan Academy.mp3", "Sentence": "Well, to do that, I can just measure the distance between these two points using my compass. So I'm adjusting it a little bit to get the distance between those two points. And then I can use that up over here to figure out, I got a little bit shaky, I can figure out that point right over there. And just like that, I now have two corresponding angles defined by a transversal and parallel lines. So what I could do is take my straight edge and make it go through those points that I just created. So let's see, make sure I'm going through them. And it would look like that."}, {"video_title": "Geometric constructions parallel line Congruence High school geometry Khan Academy.mp3", "Sentence": "And just like that, I now have two corresponding angles defined by a transversal and parallel lines. So what I could do is take my straight edge and make it go through those points that I just created. So let's see, make sure I'm going through them. And it would look like that. And I have just constructed two parallel lines. And once again, how do I know that this line is parallel to this line? Because we have a transversal that intersects both of them."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "we're asked to convert 150 degrees and negative 45 degrees to radians. So let's think about the relationship between degrees and radians. And to do that, let me just draw a little circle here. So that's the center of the circle. And then do my best shot, best attempt to freehand draw a reasonable looking circle. Yeah, that's not, I've done worse. I've done worse than that."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "So that's the center of the circle. And then do my best shot, best attempt to freehand draw a reasonable looking circle. Yeah, that's not, I've done worse. I've done worse than that. All right. Now, if we were to go in degrees, if we were to go one time around the circle like that, how many degrees is that? Well, we know that that would be 360 degrees."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "I've done worse than that. All right. Now, if we were to go in degrees, if we were to go one time around the circle like that, how many degrees is that? Well, we know that that would be 360 degrees. Well, if we did the same thing, how many radians is that? If we were to go all the way around the circle? Well, we just have to remember, when we're measuring in terms of radians, we're really talking about the arc that subtends that angle."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "Well, we know that that would be 360 degrees. Well, if we did the same thing, how many radians is that? If we were to go all the way around the circle? Well, we just have to remember, when we're measuring in terms of radians, we're really talking about the arc that subtends that angle. So if you go all the way around, you're really talking about the arc length of the entire circle, or essentially the circumference of the circle. And you're essentially saying how many radiuses this is, or radii, or how many radii is the circumference of the circle? You know a circumference of a circle is two pi, is two pi times the radius."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "Well, we just have to remember, when we're measuring in terms of radians, we're really talking about the arc that subtends that angle. So if you go all the way around, you're really talking about the arc length of the entire circle, or essentially the circumference of the circle. And you're essentially saying how many radiuses this is, or radii, or how many radii is the circumference of the circle? You know a circumference of a circle is two pi, is two pi times the radius. Or you could say that the length of the circumference of the circle is two pi radii. Two pi radii. If you wanna know the exact length, you just have to get the length of the radius and multiply it by two pi."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "You know a circumference of a circle is two pi, is two pi times the radius. Or you could say that the length of the circumference of the circle is two pi radii. Two pi radii. If you wanna know the exact length, you just have to get the length of the radius and multiply it by two pi. That just comes from, really actually the definition of pi, but it comes from what we know as the formula for the circumference of a circle. So if we were to go all the way around this, this is also two pi radians. This is also two pi, two pi radians."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "If you wanna know the exact length, you just have to get the length of the radius and multiply it by two pi. That just comes from, really actually the definition of pi, but it comes from what we know as the formula for the circumference of a circle. So if we were to go all the way around this, this is also two pi radians. This is also two pi, two pi radians. So that tells us that two pi radians, as an angle measure, is the exact same thing, and I'm gonna write it out as 360, 360 degrees. And then we can take all of this relationship and manipulate it in different ways. If we wanna simplify a little bit, we can divide both sides of this equation by two, in which case you are left with, if you divide both sides by two, you are left with pi radians, pi radians is equal to 180 degrees, is equal to 180 degrees, 180 degrees."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "This is also two pi, two pi radians. So that tells us that two pi radians, as an angle measure, is the exact same thing, and I'm gonna write it out as 360, 360 degrees. And then we can take all of this relationship and manipulate it in different ways. If we wanna simplify a little bit, we can divide both sides of this equation by two, in which case you are left with, if you divide both sides by two, you are left with pi radians, pi radians is equal to 180 degrees, is equal to 180 degrees, 180 degrees. So how can we use this relationship now to figure out what 150 degrees is? Well, this relationship we could write in different ways. We could divide both sides by 180 degrees, and we could get pi radians, pi radians over 180 degrees, over 180 degrees is equal to one, which is just another way of saying that there are pi radians for every 180 degrees, or you could say pi over 180 radians per degree."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "If we wanna simplify a little bit, we can divide both sides of this equation by two, in which case you are left with, if you divide both sides by two, you are left with pi radians, pi radians is equal to 180 degrees, is equal to 180 degrees, 180 degrees. So how can we use this relationship now to figure out what 150 degrees is? Well, this relationship we could write in different ways. We could divide both sides by 180 degrees, and we could get pi radians, pi radians over 180 degrees, over 180 degrees is equal to one, which is just another way of saying that there are pi radians for every 180 degrees, or you could say pi over 180 radians per degree. The other option, you could divide both sides of this by pi radians. You could say, you would get, on the left-hand side, you get one, and you would also get, on the right-hand side, you would get 180 degrees for every pi radians, 180 degrees for every pi radians, for every pi radians, or you could interpret this as 180 over pi degrees per radian. So how would we figure out, how would we do what they asked us?"}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "We could divide both sides by 180 degrees, and we could get pi radians, pi radians over 180 degrees, over 180 degrees is equal to one, which is just another way of saying that there are pi radians for every 180 degrees, or you could say pi over 180 radians per degree. The other option, you could divide both sides of this by pi radians. You could say, you would get, on the left-hand side, you get one, and you would also get, on the right-hand side, you would get 180 degrees for every pi radians, 180 degrees for every pi radians, for every pi radians, or you could interpret this as 180 over pi degrees per radian. So how would we figure out, how would we do what they asked us? So let's convert 150 degrees to radians. Let me write the word out. So 150 degrees, 150 degrees."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "So how would we figure out, how would we do what they asked us? So let's convert 150 degrees to radians. Let me write the word out. So 150 degrees, 150 degrees. Well, we wanna convert this to radians, so we really care about how many radians there are, how many radians there are per degree. Actually, let me do that in that color. So we care about how many radians, radians there are per degree, per, let me do that same green color, per degree."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "So 150 degrees, 150 degrees. Well, we wanna convert this to radians, so we really care about how many radians there are, how many radians there are per degree. Actually, let me do that in that color. So we care about how many radians, radians there are per degree, per, let me do that same green color, per degree. Well, how many radians are there per degree? We already know, there's pi radians for every 180 degrees, or there are pi, pi, let me do that yellow color, there are pi over 180, pi over 180 radians per degree. And so if we multiply, and this all works out, because you have degrees in the numerator, degrees in the denominator, these cancel out, and so you are left with 150 times pi divided by 180 radians."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "So we care about how many radians, radians there are per degree, per, let me do that same green color, per degree. Well, how many radians are there per degree? We already know, there's pi radians for every 180 degrees, or there are pi, pi, let me do that yellow color, there are pi over 180, pi over 180 radians per degree. And so if we multiply, and this all works out, because you have degrees in the numerator, degrees in the denominator, these cancel out, and so you are left with 150 times pi divided by 180 radians. So what do we get? Well, this becomes, let me just rewrite it, 150 times pi, times pi, all of that over 180, all of that over 180, so this is equal to, and we get it in radians, radians. And so if we simplify it, let's see, we can divide the numerator and the denominator both by, looks like 30."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "And so if we multiply, and this all works out, because you have degrees in the numerator, degrees in the denominator, these cancel out, and so you are left with 150 times pi divided by 180 radians. So what do we get? Well, this becomes, let me just rewrite it, 150 times pi, times pi, all of that over 180, all of that over 180, so this is equal to, and we get it in radians, radians. And so if we simplify it, let's see, we can divide the numerator and the denominator both by, looks like 30. So if you divide the numerator by 30, you get five. You divide the denominator by 30, you get six. So you get five pi over six radians, or five six pi radians, depending how you want to do it."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "And so if we simplify it, let's see, we can divide the numerator and the denominator both by, looks like 30. So if you divide the numerator by 30, you get five. You divide the denominator by 30, you get six. So you get five pi over six radians, or five six pi radians, depending how you want to do it. Now let's do the same thing for negative 45 degrees. What do you get for negative 45 degrees if you were to convert that to radians? Well, same exact process."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "So you get five pi over six radians, or five six pi radians, depending how you want to do it. Now let's do the same thing for negative 45 degrees. What do you get for negative 45 degrees if you were to convert that to radians? Well, same exact process. You have negative, and I'll do this one a little quicker, for negative 45 degrees, I'll write down the word, times pi radians, pi radians for every 180 degrees, for every 180 degrees, the degrees cancel out, and you are left with negative 45 pi over 180 radians. So this is equal to negative 45 pi, pi over 180, over 180 radians, radians. So how can we simplify this?"}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "Well, same exact process. You have negative, and I'll do this one a little quicker, for negative 45 degrees, I'll write down the word, times pi radians, pi radians for every 180 degrees, for every 180 degrees, the degrees cancel out, and you are left with negative 45 pi over 180 radians. So this is equal to negative 45 pi, pi over 180, over 180 radians, radians. So how can we simplify this? Well, it looks like they're both, at minimum, divisible by nine. Nine times five is 45, this is nine times 20, so actually it's gonna be divisible by more than just, let's see, nine, nine, they're both, well, I will just, actually, they're both divisible by 45, what am I doing? Okay, so if you divide the numerator by 45, you get one."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "So how can we simplify this? Well, it looks like they're both, at minimum, divisible by nine. Nine times five is 45, this is nine times 20, so actually it's gonna be divisible by more than just, let's see, nine, nine, they're both, well, I will just, actually, they're both divisible by 45, what am I doing? Okay, so if you divide the numerator by 45, you get one. You divide the denominator by 45, 45 goes into 180 four times, four times. You're left with negative pi over four radians. This is equal to negative pi over four radians, and we are done."}, {"video_title": "Determining a translation between points Transformations Geometry Khan Academy.mp3", "Sentence": "And I encourage you to pause this video and try to think about it on your own. So let's think about what the change in x that this translation does and the change in y that this translation does. So our change in x, our change in x is going to be our ending point, our ending point. So negative 203 minus our starting point minus negative 169, minus negative 169. And this is the same thing as, I'll just write it in yellow, negative 203 plus 169, which is the same thing as 169 minus 203. I'm just writing it this way for my brain to process it, which is the same thing as the negative of 203 minus 169. And let's see, this one is the easiest for my brain to process."}, {"video_title": "Determining a translation between points Transformations Geometry Khan Academy.mp3", "Sentence": "So negative 203 minus our starting point minus negative 169, minus negative 169. And this is the same thing as, I'll just write it in yellow, negative 203 plus 169, which is the same thing as 169 minus 203. I'm just writing it this way for my brain to process it, which is the same thing as the negative of 203 minus 169. And let's see, this one is the easiest for my brain to process. This is going to be 34, so this is going to be negative 34. That's our change in x. Now let's think about our change in y."}, {"video_title": "Determining a translation between points Transformations Geometry Khan Academy.mp3", "Sentence": "And let's see, this one is the easiest for my brain to process. This is going to be 34, so this is going to be negative 34. That's our change in x. Now let's think about our change in y. Our change in y is, our ending point is negative 68, negative 68, and our starting point is 434. So our ending point minus our starting point, minus 434, is going to be equal to, so this is just going to be the same thing as the negative, of 68 plus 434. And let's see, 430 plus 60 is 490 plus another eight plus four, plus another 12, so that's going to get us to 502."}, {"video_title": "Determining a translation between points Transformations Geometry Khan Academy.mp3", "Sentence": "Now let's think about our change in y. Our change in y is, our ending point is negative 68, negative 68, and our starting point is 434. So our ending point minus our starting point, minus 434, is going to be equal to, so this is just going to be the same thing as the negative, of 68 plus 434. And let's see, 430 plus 60 is 490 plus another eight plus four, plus another 12, so that's going to get us to 502. This is going to be negative 502. So this translation, what it does is, it shifts us in the x direction by negative 34, and it shifts us in the y direction by negative 502. So if we're starting at 31 comma negative 529, we're going to end up at, let me write it this way."}, {"video_title": "Determining a translation between points Transformations Geometry Khan Academy.mp3", "Sentence": "And let's see, 430 plus 60 is 490 plus another eight plus four, plus another 12, so that's going to get us to 502. This is going to be negative 502. So this translation, what it does is, it shifts us in the x direction by negative 34, and it shifts us in the y direction by negative 502. So if we're starting at 31 comma negative 529, we're going to end up at, let me write it this way. So if we're starting at this point, if we're starting at 31 comma negative 529, we're going to end up at 31 minus 34, minus 34, that's how much we're going to shift it, and negative 529, negative 529, so negative 529 minus 502, minus 502. So what's this going to be? 31 minus 34 is negative three, and then if we have negative 529 minus 502, that's going to be negative 1031."}, {"video_title": "Axis of symmetry Transformations Geometry Khan Academy.mp3", "Sentence": "For each of these diagrams, I want to think about whether this blue line represents an axis of symmetry. And the way we can tell is if on both sides of the blue line, we essentially have mirror images. So let's imagine, let's take this top part of this polygon, the part that is above this blue line here. And let's reflect it across the blue line. You could almost imagine that it's a reflection over some type of a lake or something and see if we get exactly what we have below. Then this would be an axis of symmetry. So this point right over here, this distance to the blue line, let's go the same amount on the other side would get you right there."}, {"video_title": "Axis of symmetry Transformations Geometry Khan Academy.mp3", "Sentence": "And let's reflect it across the blue line. You could almost imagine that it's a reflection over some type of a lake or something and see if we get exactly what we have below. Then this would be an axis of symmetry. So this point right over here, this distance to the blue line, let's go the same amount on the other side would get you right there. And so you immediately see, we start ending up with a point that is off what's actually here in black, the actual bottom part of the polygon. So this is a pretty good clue that this is not an axis of symmetry. But let's just continue it, just to go through the exercise."}, {"video_title": "Axis of symmetry Transformations Geometry Khan Academy.mp3", "Sentence": "So this point right over here, this distance to the blue line, let's go the same amount on the other side would get you right there. And so you immediately see, we start ending up with a point that is off what's actually here in black, the actual bottom part of the polygon. So this is a pretty good clue that this is not an axis of symmetry. But let's just continue it, just to go through the exercise. So this point, if you reflected it across this blue line, would get you here. This point, we'll do it in a different color. This point, if you were to reflect it across this blue line, it would get you."}, {"video_title": "Axis of symmetry Transformations Geometry Khan Academy.mp3", "Sentence": "But let's just continue it, just to go through the exercise. So this point, if you reflected it across this blue line, would get you here. This point, we'll do it in a different color. This point, if you were to reflect it across this blue line, it would get you. Make sure I can do that relatively straight. I can do a straighter job than that. So if you go about that distance about it, and I want to go straight down into the blue line, and say I'm going to go the same distance on the other side, it gets me to right around there."}, {"video_title": "Axis of symmetry Transformations Geometry Khan Academy.mp3", "Sentence": "This point, if you were to reflect it across this blue line, it would get you. Make sure I can do that relatively straight. I can do a straighter job than that. So if you go about that distance about it, and I want to go straight down into the blue line, and say I'm going to go the same distance on the other side, it gets me to right around there. And then this point over here, if I were to drop it straight down, then if I were to go the same distance on the other side, it gets me right around there. And then finally, this point gets me right around there. So its mirror image of this top part would look something like this."}, {"video_title": "Axis of symmetry Transformations Geometry Khan Academy.mp3", "Sentence": "So if you go about that distance about it, and I want to go straight down into the blue line, and say I'm going to go the same distance on the other side, it gets me to right around there. And then this point over here, if I were to drop it straight down, then if I were to go the same distance on the other side, it gets me right around there. And then finally, this point gets me right around there. So its mirror image of this top part would look something like this. My best attempt to draw it would look something like this, which is very different than the part of the polygon that's actually on the other side of this blue line. So in this case, the blue line is not an axis of symmetry. So this is no, no."}, {"video_title": "Axis of symmetry Transformations Geometry Khan Academy.mp3", "Sentence": "So its mirror image of this top part would look something like this. My best attempt to draw it would look something like this, which is very different than the part of the polygon that's actually on the other side of this blue line. So in this case, the blue line is not an axis of symmetry. So this is no, no. This blue line is not an axis of symmetry. Now let's look at it over here. And our eyes pick this out very naturally."}, {"video_title": "Axis of symmetry Transformations Geometry Khan Academy.mp3", "Sentence": "So this is no, no. This blue line is not an axis of symmetry. Now let's look at it over here. And our eyes pick this out very naturally. Here you can see that it looks like this blue line really divides this polygon in half. It really does look like mirror images. It really does look like, if you imagine that this is some type of a lake, a still lake, so I shouldn't actually draw waves, but this is some type of a lake, that this is the reflection."}, {"video_title": "Axis of symmetry Transformations Geometry Khan Academy.mp3", "Sentence": "And our eyes pick this out very naturally. Here you can see that it looks like this blue line really divides this polygon in half. It really does look like mirror images. It really does look like, if you imagine that this is some type of a lake, a still lake, so I shouldn't actually draw waves, but this is some type of a lake, that this is the reflection. And we can even go point by point here. So this point right over here is the same distance from if we dropped a perpendicular to this point as this one right over here. This one over here, same distance as this point right over here."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "Well, we're going to prove in this video, it's a couple of fairly straightforward parallelogram related proofs. And this first one we're gonna say, hey, if we have this parallelogram ABCD, let's prove that the opposite sides have the same length. So prove that AB is equal to DC, and that AD is equal to BC. So let me draw a diagonal here. So I'm gonna draw a diagonal. And this diagonal, depending on how you view it, is intersecting two sets of parallel lines. So you could also consider it to be a transversal."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So let me draw a diagonal here. So I'm gonna draw a diagonal. And this diagonal, depending on how you view it, is intersecting two sets of parallel lines. So you could also consider it to be a transversal. Actually, let me draw it a little bit neater than that. I can do a better job. So, nope, that's not any better."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So you could also consider it to be a transversal. Actually, let me draw it a little bit neater than that. I can do a better job. So, nope, that's not any better. That is about as good as I can do. So if we view DB, this diagonal DB, we can view it as a transversal for the parallel lines AB and DC. And if you view it that way, you can pick out that angle ABD is going to be congruent."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So, nope, that's not any better. That is about as good as I can do. So if we view DB, this diagonal DB, we can view it as a transversal for the parallel lines AB and DC. And if you view it that way, you can pick out that angle ABD is going to be congruent. So angle ABD, that's that angle right there, is going to be congruent to angle BDC because they're alternate interior angles. You have a transversal, parallel lines. So we know that angle ABD is going to be congruent to angle BDC."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "And if you view it that way, you can pick out that angle ABD is going to be congruent. So angle ABD, that's that angle right there, is going to be congruent to angle BDC because they're alternate interior angles. You have a transversal, parallel lines. So we know that angle ABD is going to be congruent to angle BDC. Angle BDC. Now, you could also view this diagonal DB, you could view it as a transversal of these two parallel lines, of the other pair of parallel lines, AD and BC. And if you look at it that way, then you immediately see that angle DBC, angle DBC right over here, angle DBC is going to be congruent to angle ADB, to angle ADB, angle ADB, for the exact same reason."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So we know that angle ABD is going to be congruent to angle BDC. Angle BDC. Now, you could also view this diagonal DB, you could view it as a transversal of these two parallel lines, of the other pair of parallel lines, AD and BC. And if you look at it that way, then you immediately see that angle DBC, angle DBC right over here, angle DBC is going to be congruent to angle ADB, to angle ADB, angle ADB, for the exact same reason. They are alternate interior angles of a transversal intersecting these two parallel lines. So I could write this. This is alternate interior angles, interior angles are congruent when you have a transversal intersecting two parallel lines."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "And if you look at it that way, then you immediately see that angle DBC, angle DBC right over here, angle DBC is going to be congruent to angle ADB, to angle ADB, angle ADB, for the exact same reason. They are alternate interior angles of a transversal intersecting these two parallel lines. So I could write this. This is alternate interior angles, interior angles are congruent when you have a transversal intersecting two parallel lines. And we also see that both of these triangles, triangle ADB and triangle CDB, both share this side over here. It's obviously equal to itself. Now, why is this useful?"}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "This is alternate interior angles, interior angles are congruent when you have a transversal intersecting two parallel lines. And we also see that both of these triangles, triangle ADB and triangle CDB, both share this side over here. It's obviously equal to itself. Now, why is this useful? Well, you might realize that we've just shown that both of these triangles, they have this pink angle, then they have this side in common, and then they have the green angle. Pink angle, side in common, and then the green angle. So we've just shown by angle, side, angle, that these two triangles are congruent."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "Now, why is this useful? Well, you might realize that we've just shown that both of these triangles, they have this pink angle, then they have this side in common, and then they have the green angle. Pink angle, side in common, and then the green angle. So we've just shown by angle, side, angle, that these two triangles are congruent. So let me write this down. We have shown that triangle, I'll go from non-labeled to pink to green. ADB is congruent to triangle non-labeled to pink to green."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So we've just shown by angle, side, angle, that these two triangles are congruent. So let me write this down. We have shown that triangle, I'll go from non-labeled to pink to green. ADB is congruent to triangle non-labeled to pink to green. Non-labeled to pink to green. C, CBD, CBD. And this comes out of angle, side, angle congruency."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "ADB is congruent to triangle non-labeled to pink to green. Non-labeled to pink to green. C, CBD, CBD. And this comes out of angle, side, angle congruency. So this is from angle, side, angle, angle, side, angle, congruency. Well, what does that do for us? Well, if two triangles are congruent, then all of the corresponding features of the two triangles are going to be congruent."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "And this comes out of angle, side, angle congruency. So this is from angle, side, angle, angle, side, angle, congruency. Well, what does that do for us? Well, if two triangles are congruent, then all of the corresponding features of the two triangles are going to be congruent. In particular, side DC, side DC corresponds to side BA, side DC on this bottom triangle corresponds to side BA on that top triangle. So they need to be congruent. So DC, so we get DC is going to be equal to BA, and that's because they are corresponding sides, corresponding sides of congruent, congruent triangles."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "Well, if two triangles are congruent, then all of the corresponding features of the two triangles are going to be congruent. In particular, side DC, side DC corresponds to side BA, side DC on this bottom triangle corresponds to side BA on that top triangle. So they need to be congruent. So DC, so we get DC is going to be equal to BA, and that's because they are corresponding sides, corresponding sides of congruent, congruent triangles. So this is going to be equal to that, and by that exact same logic, AD, AD corresponds to CB. AD corresponds to CB. AD is equal to CB, and for the exact same reason, corresponding sides of congruent triangles."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So DC, so we get DC is going to be equal to BA, and that's because they are corresponding sides, corresponding sides of congruent, congruent triangles. So this is going to be equal to that, and by that exact same logic, AD, AD corresponds to CB. AD corresponds to CB. AD is equal to CB, and for the exact same reason, corresponding sides of congruent triangles. And then we're done. We've proven that opposite sides are congruent. Now let's go the other way."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "AD is equal to CB, and for the exact same reason, corresponding sides of congruent triangles. And then we're done. We've proven that opposite sides are congruent. Now let's go the other way. Let's go the other way. Let's say that we have some type of a quadrilateral, and we know that the opposite sides are congruent. Can we prove to ourselves that this is a parallelogram?"}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "Now let's go the other way. Let's go the other way. Let's say that we have some type of a quadrilateral, and we know that the opposite sides are congruent. Can we prove to ourselves that this is a parallelogram? Well, it's kind of the same proof in reverse. So let's draw a diagonal here, since we know a lot about triangles. So let me draw, there we go."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "Can we prove to ourselves that this is a parallelogram? Well, it's kind of the same proof in reverse. So let's draw a diagonal here, since we know a lot about triangles. So let me draw, there we go. That's the hardest part. Let's see, draw it, that's pretty good. All right, so we obviously know that CB is going to be equal to itself."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So let me draw, there we go. That's the hardest part. Let's see, draw it, that's pretty good. All right, so we obviously know that CB is going to be equal to itself. So I'll draw it like that. We all, obviously, because it's the same line. And then we have something interesting."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "All right, so we obviously know that CB is going to be equal to itself. So I'll draw it like that. We all, obviously, because it's the same line. And then we have something interesting. We've split this quadrilateral into two triangles, triangle ACB and triangle DBC. And notice, they have, all three sides of these two triangles are equal to each other. So we know by side, side, side, that they are congruent."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "And then we have something interesting. We've split this quadrilateral into two triangles, triangle ACB and triangle DBC. And notice, they have, all three sides of these two triangles are equal to each other. So we know by side, side, side, that they are congruent. So we know that triangle, triangle A, and we're starting A, and then I'm going to the one-half side. So ACB is congruent to triangle DBC. DBC, and this is by side, side, side, side, side, side, side, congruency."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So we know by side, side, side, that they are congruent. So we know that triangle, triangle A, and we're starting A, and then I'm going to the one-half side. So ACB is congruent to triangle DBC. DBC, and this is by side, side, side, side, side, side, side, congruency. Well, what does that do for us? Well, it tells us that all of the corresponding angles are going to be congruent. So for example, ABC, angle ABC is going to be, so let me mark that."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "DBC, and this is by side, side, side, side, side, side, side, congruency. Well, what does that do for us? Well, it tells us that all of the corresponding angles are going to be congruent. So for example, ABC, angle ABC is going to be, so let me mark that. Angle ABC is going to be congruent, and you can say ABC is going to be congruent to DCB. DCB, angle DCB, and you could say by, you could say corresponding angles congruent of congruent triangles. I'm just using some shorthand here to save some time."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So for example, ABC, angle ABC is going to be, so let me mark that. Angle ABC is going to be congruent, and you can say ABC is going to be congruent to DCB. DCB, angle DCB, and you could say by, you could say corresponding angles congruent of congruent triangles. I'm just using some shorthand here to save some time. So ABC is going to be congruent to DCB. So these two angles are going to be congruent. Well, this is interesting because here you have a line, and it's intersecting AB and CD, and we clearly see that these things that could be alternate angles, alternate interior angles, are congruent."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "I'm just using some shorthand here to save some time. So ABC is going to be congruent to DCB. So these two angles are going to be congruent. Well, this is interesting because here you have a line, and it's intersecting AB and CD, and we clearly see that these things that could be alternate angles, alternate interior angles, are congruent. And because we have these congruent alternate interior angles, we know that AB must be parallel to CD. So this must be parallel to that. So we know that AB is parallel to CD by alternate interior angles of a transversal intersecting parallel lines."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "Well, this is interesting because here you have a line, and it's intersecting AB and CD, and we clearly see that these things that could be alternate angles, alternate interior angles, are congruent. And because we have these congruent alternate interior angles, we know that AB must be parallel to CD. So this must be parallel to that. So we know that AB is parallel to CD by alternate interior angles of a transversal intersecting parallel lines. Now, we can use that exact same logic. We also know that angle, let me get this right, angle ACB is congruent to angle DBC. And we know that by corresponding angles congruent of congruent triangles."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So we know that AB is parallel to CD by alternate interior angles of a transversal intersecting parallel lines. Now, we can use that exact same logic. We also know that angle, let me get this right, angle ACB is congruent to angle DBC. And we know that by corresponding angles congruent of congruent triangles. So we're just saying that this angle is equal to that angle. Well, once again, these could be alternate interior angles. They look like they could be."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "And we know that by corresponding angles congruent of congruent triangles. So we're just saying that this angle is equal to that angle. Well, once again, these could be alternate interior angles. They look like they could be. This is a transversal, and here's two lines here, which we're not sure whether they're parallel. But because the alternate interior angles are congruent, we know that they are parallel. So this is parallel to that."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "They look like they could be. This is a transversal, and here's two lines here, which we're not sure whether they're parallel. But because the alternate interior angles are congruent, we know that they are parallel. So this is parallel to that. So we know that AC is parallel to BD by alternate interior angles. By alternate interior angles. And we're done."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "So this is parallel to that. So we know that AC is parallel to BD by alternate interior angles. By alternate interior angles. And we're done. So what we've done is, it's interesting, we've shown if you have a parallelogram, opposite sides have the same length. And if opposite sides have the same length, then you have a parallelogram. And so we've actually proven it in both directions."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "And we're done. So what we've done is, it's interesting, we've shown if you have a parallelogram, opposite sides have the same length. And if opposite sides have the same length, then you have a parallelogram. And so we've actually proven it in both directions. And so we can actually make what you call an if and only if statement. You can say if opposite sides are parallel of a quadrilateral, or you could say opposite sides of a quadrilateral are parallel if and only if their lengths are equal. And you say if and only if."}, {"video_title": "Proof Opposite sides of parallelogram congruent Quadrilaterals Geometry Khan Academy.mp3", "Sentence": "And so we've actually proven it in both directions. And so we can actually make what you call an if and only if statement. You can say if opposite sides are parallel of a quadrilateral, or you could say opposite sides of a quadrilateral are parallel if and only if their lengths are equal. And you say if and only if. So if they are parallel, then you could say their lengths are equal. And only if, only if their lengths are equal are they parallel. We've proven it in both directions."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "You didn't have to be told it's a hexagon. But the regular part lets us know that all of the sides, all six sides have the same length and all of the interior angles have the same measure. Fair enough. And then they give us the length of one of the sides. And since this is a regular hexagon, they're actually giving us the length of all of the sides. They say it's two squares of three. So this side right over here is two squares of three."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And then they give us the length of one of the sides. And since this is a regular hexagon, they're actually giving us the length of all of the sides. They say it's two squares of three. So this side right over here is two squares of three. This side over here is two squares of three. And I could just go around the hexagon. Every one of their sides is two squares of three."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So this side right over here is two squares of three. This side over here is two squares of three. And I could just go around the hexagon. Every one of their sides is two squares of three. They want us to find the area of this hexagon. Find the area of A, B, C, D, E, F. And the best way to find the area, especially of regular polygons, is try to split it up into triangles. And hexagons are a bit of a special case."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "Every one of their sides is two squares of three. They want us to find the area of this hexagon. Find the area of A, B, C, D, E, F. And the best way to find the area, especially of regular polygons, is try to split it up into triangles. And hexagons are a bit of a special case. Maybe in future videos we'll think about the more general case of any polygon. But with a hexagon, what you could think about is if we take this point right over here, and let's call this point G, and let's say it's the center of the hexagon. And when I'm talking about a center of a hexagon, I'm talking about a point."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And hexagons are a bit of a special case. Maybe in future videos we'll think about the more general case of any polygon. But with a hexagon, what you could think about is if we take this point right over here, and let's call this point G, and let's say it's the center of the hexagon. And when I'm talking about a center of a hexagon, I'm talking about a point. It can't be equidistant from everything over here because this isn't a circle. But we could say it's equidistant from all of the vertices. So GD is the same thing as GC, is the same thing as GB, which is the same thing as GA, which is the same thing as GF, which is the same thing as GE."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And when I'm talking about a center of a hexagon, I'm talking about a point. It can't be equidistant from everything over here because this isn't a circle. But we could say it's equidistant from all of the vertices. So GD is the same thing as GC, is the same thing as GB, which is the same thing as GA, which is the same thing as GF, which is the same thing as GE. So let me draw some of those that I just talked about. So that is GE. There's GD."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So GD is the same thing as GC, is the same thing as GB, which is the same thing as GA, which is the same thing as GF, which is the same thing as GE. So let me draw some of those that I just talked about. So that is GE. There's GD. There's GC. All of these lengths are going to be the same. So there's a point G, which we can call the center, the center of this polygon."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "There's GD. There's GC. All of these lengths are going to be the same. So there's a point G, which we can call the center, the center of this polygon. And we know that this length is equal to that length, which is equal to that length, which is equal to that length, which is equal to that length, which is equal to that length. We also know that if we go all the way around the circle, if we go all the way around a circle like that, we've gone 360 degrees. And we know that these triangles, these triangles are all going to be congruent to each other."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So there's a point G, which we can call the center, the center of this polygon. And we know that this length is equal to that length, which is equal to that length, which is equal to that length, which is equal to that length, which is equal to that length. We also know that if we go all the way around the circle, if we go all the way around a circle like that, we've gone 360 degrees. And we know that these triangles, these triangles are all going to be congruent to each other. And there's multiple ways that we could show it. But the easiest way is, look, they have two sides. All of them have this side and this side be congruent to each other because G is in the center."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And we know that these triangles, these triangles are all going to be congruent to each other. And there's multiple ways that we could show it. But the easiest way is, look, they have two sides. All of them have this side and this side be congruent to each other because G is in the center. And they all have this third common side of 2 squared is 3. So all of them by side, side, side, they are all congruent. What that tells us is if they're all congruent, then this angle, this interior angle right over here, is going to be the same for all 6 of these triangles over here."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "All of them have this side and this side be congruent to each other because G is in the center. And they all have this third common side of 2 squared is 3. So all of them by side, side, side, they are all congruent. What that tells us is if they're all congruent, then this angle, this interior angle right over here, is going to be the same for all 6 of these triangles over here. And maybe I call that x. That's angle x. That's x."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "What that tells us is if they're all congruent, then this angle, this interior angle right over here, is going to be the same for all 6 of these triangles over here. And maybe I call that x. That's angle x. That's x. That's x. That's x. That's x."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "That's x. That's x. That's x. That's x. And if you add them all up, we've gone around a circle. We've gone 360 degrees. And we have 6 of these x's."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "That's x. And if you add them all up, we've gone around a circle. We've gone 360 degrees. And we have 6 of these x's. So you get 6x is equal to 360 degrees. You divide both sides by 6. You get x is equal to 60 degrees."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And we have 6 of these x's. So you get 6x is equal to 360 degrees. You divide both sides by 6. You get x is equal to 60 degrees. All of these are equal to 60 degrees. Now there's something interesting. We know that these triangles, for example, triangle GBC, and we could do that for any of these 6 triangles."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "You get x is equal to 60 degrees. All of these are equal to 60 degrees. Now there's something interesting. We know that these triangles, for example, triangle GBC, and we could do that for any of these 6 triangles. It looks kind of like a trivial pursuit piece. We know that they're definitely isosceles triangles, that this distance is equal to this distance. So we can use that information to figure out what the other angles are."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "We know that these triangles, for example, triangle GBC, and we could do that for any of these 6 triangles. It looks kind of like a trivial pursuit piece. We know that they're definitely isosceles triangles, that this distance is equal to this distance. So we can use that information to figure out what the other angles are. Because these 2 base angles, it's an isosceles triangle. The 2 legs are the same, so our 2 base angles, this angle is going to be congruent to that angle. We could call that y right over there."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So we can use that information to figure out what the other angles are. Because these 2 base angles, it's an isosceles triangle. The 2 legs are the same, so our 2 base angles, this angle is going to be congruent to that angle. We could call that y right over there. So you have y plus y, which is 2y, plus 60 degrees is going to be equal to 180. Because the interior angles of any triangle, they add up to 180. And so subtract 60 from both sides, you get 2y is equal to 120."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "We could call that y right over there. So you have y plus y, which is 2y, plus 60 degrees is going to be equal to 180. Because the interior angles of any triangle, they add up to 180. And so subtract 60 from both sides, you get 2y is equal to 120. Divide both sides by 2, you get y is equal to 60 degrees. Now this is interesting. I could have done this with any of these triangles."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And so subtract 60 from both sides, you get 2y is equal to 120. Divide both sides by 2, you get y is equal to 60 degrees. Now this is interesting. I could have done this with any of these triangles. All of these triangles are 60-60-60 triangles, which tells us, and we've proven this earlier on when we first started studying equilateral triangles, we know that all of the angles of a triangle are 60 degrees, and we're dealing with an equilateral triangle, which means that all the sides have the same length. So if this is 2 square roots of 3, then so is this. This is also 2 square roots of 3, and this is also 2 square roots of 3."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "I could have done this with any of these triangles. All of these triangles are 60-60-60 triangles, which tells us, and we've proven this earlier on when we first started studying equilateral triangles, we know that all of the angles of a triangle are 60 degrees, and we're dealing with an equilateral triangle, which means that all the sides have the same length. So if this is 2 square roots of 3, then so is this. This is also 2 square roots of 3, and this is also 2 square roots of 3. So pretty much all of these green lines are 2 square roots of 3, and we already knew, because it's a regular hexagon, that each side of the hexagon itself is also 2 square roots of 3. So now we can essentially use that information to figure out, actually we don't even have to figure this part out, I'll show you in a second, to figure out the area of any one of these triangles, and then we can just multiply by 6. So let's focus on, let me focus on this triangle right over here, and think about how we can find its area."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "This is also 2 square roots of 3, and this is also 2 square roots of 3. So pretty much all of these green lines are 2 square roots of 3, and we already knew, because it's a regular hexagon, that each side of the hexagon itself is also 2 square roots of 3. So now we can essentially use that information to figure out, actually we don't even have to figure this part out, I'll show you in a second, to figure out the area of any one of these triangles, and then we can just multiply by 6. So let's focus on, let me focus on this triangle right over here, and think about how we can find its area. We know that length of DC is 2 square roots of 3. We can drop an altitude over here. We can drop an altitude just like that."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So let's focus on, let me focus on this triangle right over here, and think about how we can find its area. We know that length of DC is 2 square roots of 3. We can drop an altitude over here. We can drop an altitude just like that. And then if we drop an altitude, we know that this is an equilateral triangle, and we can show very easily that these two triangles are symmetric. These are both 90 degree angles. We know that these two are 60 degree angles already, and then if you look at each of these two independent triangles, you'd have to just say, well, they have to add up to 180, so this has to be 30 degrees, this has to be 30 degrees."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "We can drop an altitude just like that. And then if we drop an altitude, we know that this is an equilateral triangle, and we can show very easily that these two triangles are symmetric. These are both 90 degree angles. We know that these two are 60 degree angles already, and then if you look at each of these two independent triangles, you'd have to just say, well, they have to add up to 180, so this has to be 30 degrees, this has to be 30 degrees. All the angles are the same. They also share a side in common, so these two are congruent triangles. So if we want to find the area of this little slice of the pie right over here, we can just find the area of this slice or this sub-slice and then multiply by 2, or we can just find this area and multiply by 12 for the entire hexagon."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "We know that these two are 60 degree angles already, and then if you look at each of these two independent triangles, you'd have to just say, well, they have to add up to 180, so this has to be 30 degrees, this has to be 30 degrees. All the angles are the same. They also share a side in common, so these two are congruent triangles. So if we want to find the area of this little slice of the pie right over here, we can just find the area of this slice or this sub-slice and then multiply by 2, or we can just find this area and multiply by 12 for the entire hexagon. So how do we figure out the area of this thing? Well, this is going to be half of this base length, so this length right over here, let me call this point H. DH is going to be the square root of 3. And we, well, hopefully we've already recognized that this is a 30-60-90 triangle."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So if we want to find the area of this little slice of the pie right over here, we can just find the area of this slice or this sub-slice and then multiply by 2, or we can just find this area and multiply by 12 for the entire hexagon. So how do we figure out the area of this thing? Well, this is going to be half of this base length, so this length right over here, let me call this point H. DH is going to be the square root of 3. And we, well, hopefully we've already recognized that this is a 30-60-90 triangle. Let me draw it over here. So this is a 30-60-90 triangle. We know that this length over here is square root of 3."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And we, well, hopefully we've already recognized that this is a 30-60-90 triangle. Let me draw it over here. So this is a 30-60-90 triangle. We know that this length over here is square root of 3. We know, and we already actually did calculate that this is 2 square roots of 3, although we don't really need it. What we really need to figure out is this altitude height. And from 30-60-90 triangles, we know that the side opposite the 60-degree side is a square root of 3 times the side opposite the 30-degree side."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "We know that this length over here is square root of 3. We know, and we already actually did calculate that this is 2 square roots of 3, although we don't really need it. What we really need to figure out is this altitude height. And from 30-60-90 triangles, we know that the side opposite the 60-degree side is a square root of 3 times the side opposite the 30-degree side. So this is going to be square root of 3 times this square root of 3. Times the square root of 3. Square root of 3 times the square root of 3 is obviously just 3."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "And from 30-60-90 triangles, we know that the side opposite the 60-degree side is a square root of 3 times the side opposite the 30-degree side. So this is going to be square root of 3 times this square root of 3. Times the square root of 3. Square root of 3 times the square root of 3 is obviously just 3. So this altitude right over here is just going to be 3. So if we want the area of this triangle right over here, which is this triangle right over here, it's just 1 half base times height. So the area of this little sub-slice is just 1 half times our base, just the base over here."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "Square root of 3 times the square root of 3 is obviously just 3. So this altitude right over here is just going to be 3. So if we want the area of this triangle right over here, which is this triangle right over here, it's just 1 half base times height. So the area of this little sub-slice is just 1 half times our base, just the base over here. Actually, let's take a step back. We don't even have to worry about this thing. Let's just go straight to the larger triangle, GDC."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "So the area of this little sub-slice is just 1 half times our base, just the base over here. Actually, let's take a step back. We don't even have to worry about this thing. Let's just go straight to the larger triangle, GDC. So let me rewind this a little bit, because now we have the base and the height of the whole thing. If we care about the area of triangle GDC, so now I'm looking at this entire triangle right over here. This is equal to 1 half times base times height, which is equal to 1 half."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "Let's just go straight to the larger triangle, GDC. So let me rewind this a little bit, because now we have the base and the height of the whole thing. If we care about the area of triangle GDC, so now I'm looking at this entire triangle right over here. This is equal to 1 half times base times height, which is equal to 1 half. What's our base? Our base, we already know, it's one of the sides of our hexagon. It's 2 square roots of 3."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "This is equal to 1 half times base times height, which is equal to 1 half. What's our base? Our base, we already know, it's one of the sides of our hexagon. It's 2 square roots of 3. It's this whole thing right over here. So times 2 square roots of 3, and then we want to multiply that times our height, and that's what we just figured out using 30-60-90 triangles. Our height is 3."}, {"video_title": "Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3", "Sentence": "It's 2 square roots of 3. It's this whole thing right over here. So times 2 square roots of 3, and then we want to multiply that times our height, and that's what we just figured out using 30-60-90 triangles. Our height is 3. So times 3, 1 half and 2 cancel out. We're left with 3 square roots of 3. That's just the area of one of these little wedges right over here."}, {"video_title": "Proving the SSS triangle congruence criterion using transformations Geometry Khan Academy.mp3", "Sentence": "What we're going to do in this video is see that if we have two different triangles where the corresponding sides have the same measure, so this orange side has the same length as this orange side, this blue side has the same length as this blue side, this gray side has the same length as this gray side, then we can deduce that these two triangles are congruent to each other based on the rigid transformation definition of congruence. And to show that, we just have to show that there's always a series of rigid transformations that maps triangle ABC onto triangle EDF. So how do we do that? Well, first of all, in other videos, we showed that if we have two line segments that have the same measure, they are congruent. You can map one onto the other using rigid transformations. So let's do a series of rigid transformations that maps AB onto ED, and you could imagine how to do that. You would translate point A, you would translate this entire left triangle so that point A coincides with point E, and then side AB would be moving, and this would be on this direction over here, and then you would rotate around this point right over here, you could call that A prime, so this is going to be equal to A prime."}, {"video_title": "Proving the SSS triangle congruence criterion using transformations Geometry Khan Academy.mp3", "Sentence": "Well, first of all, in other videos, we showed that if we have two line segments that have the same measure, they are congruent. You can map one onto the other using rigid transformations. So let's do a series of rigid transformations that maps AB onto ED, and you could imagine how to do that. You would translate point A, you would translate this entire left triangle so that point A coincides with point E, and then side AB would be moving, and this would be on this direction over here, and then you would rotate around this point right over here, you could call that A prime, so this is going to be equal to A prime. You rotate around that so that side AB coincides with side ED, and we've talked about that in other videos. So at that point, D would be equal to B prime, the point to which B is mapped. But the question is, where is C?"}, {"video_title": "Proving the SSS triangle congruence criterion using transformations Geometry Khan Academy.mp3", "Sentence": "You would translate point A, you would translate this entire left triangle so that point A coincides with point E, and then side AB would be moving, and this would be on this direction over here, and then you would rotate around this point right over here, you could call that A prime, so this is going to be equal to A prime. You rotate around that so that side AB coincides with side ED, and we've talked about that in other videos. So at that point, D would be equal to B prime, the point to which B is mapped. But the question is, where is C? If we can show that for sure C is either at point F or with another rigid transformation, we can get C to point F, then we would have completed our proof. We would have been able to show that with a series of rigid transformations, you can go from this triangle, you can map this triangle onto that triangle. And to think about where point C is, this is where this compass is going to prove useful."}, {"video_title": "Proving the SSS triangle congruence criterion using transformations Geometry Khan Academy.mp3", "Sentence": "But the question is, where is C? If we can show that for sure C is either at point F or with another rigid transformation, we can get C to point F, then we would have completed our proof. We would have been able to show that with a series of rigid transformations, you can go from this triangle, you can map this triangle onto that triangle. And to think about where point C is, this is where this compass is going to prove useful. We know that point C is exactly this far away from point A. I will measure that with my compass, so I could do it this way as well. Point C is exactly that far from point A. And so that means that point C needs to be someplace, someplace on this curve right over here, on this arc that I'm doing."}, {"video_title": "Proving the SSS triangle congruence criterion using transformations Geometry Khan Academy.mp3", "Sentence": "And to think about where point C is, this is where this compass is going to prove useful. We know that point C is exactly this far away from point A. I will measure that with my compass, so I could do it this way as well. Point C is exactly that far from point A. And so that means that point C needs to be someplace, someplace on this curve right over here, on this arc that I'm doing. These are some of the points that are exactly that far away. I could do a complete circle, but you see where this is going. So point C, I guess we could say C prime, where C will be mapped to some point on that circle, if you take it from A's perspective, because that's how far C is from A."}, {"video_title": "Proving the SSS triangle congruence criterion using transformations Geometry Khan Academy.mp3", "Sentence": "And so that means that point C needs to be someplace, someplace on this curve right over here, on this arc that I'm doing. These are some of the points that are exactly that far away. I could do a complete circle, but you see where this is going. So point C, I guess we could say C prime, where C will be mapped to some point on that circle, if you take it from A's perspective, because that's how far C is from A. But then we also know that C is this far from B. So let me adjust my compass again. C is that far from B."}, {"video_title": "Proving the SSS triangle congruence criterion using transformations Geometry Khan Academy.mp3", "Sentence": "So point C, I guess we could say C prime, where C will be mapped to some point on that circle, if you take it from A's perspective, because that's how far C is from A. But then we also know that C is this far from B. So let me adjust my compass again. C is that far from B. And so if B is mapped to this point, this is where B prime is, then C prime, where C is mapped, is going to be someplace along this curve. And so you could view those two curves as constraints, so we know that C prime has to sit on both of these curves. So it's either going to sit right over here where F is."}, {"video_title": "Proving the SSS triangle congruence criterion using transformations Geometry Khan Academy.mp3", "Sentence": "C is that far from B. And so if B is mapped to this point, this is where B prime is, then C prime, where C is mapped, is going to be someplace along this curve. And so you could view those two curves as constraints, so we know that C prime has to sit on both of these curves. So it's either going to sit right over here where F is. And so if my rigid transformation got us to a point where C sits exactly where F is, well, then our proof is complete. We have come up with a rigid transformation. Now, another possibility is when we do that transformation, C prime ends up right over here."}, {"video_title": "Proving the SSS triangle congruence criterion using transformations Geometry Khan Academy.mp3", "Sentence": "So it's either going to sit right over here where F is. And so if my rigid transformation got us to a point where C sits exactly where F is, well, then our proof is complete. We have come up with a rigid transformation. Now, another possibility is when we do that transformation, C prime ends up right over here. So what could we then do to continue to transform rigidly so that C prime ends up with F? Remember, the other two points have already coincided on with E and D, so we just have to get C prime to coincide with F. Well, one way to think about it is, if we think about it, E, point E is equidistant to C prime and F. We see this is going to be equal to, we could put three hashtags there, because once again, that defined the radius of this arc. And we know that point C prime in this case, point C prime in this case, is the same distance from D as F is, as F is."}, {"video_title": "Proving the SSS triangle congruence criterion using transformations Geometry Khan Academy.mp3", "Sentence": "Now, another possibility is when we do that transformation, C prime ends up right over here. So what could we then do to continue to transform rigidly so that C prime ends up with F? Remember, the other two points have already coincided on with E and D, so we just have to get C prime to coincide with F. Well, one way to think about it is, if we think about it, E, point E is equidistant to C prime and F. We see this is going to be equal to, we could put three hashtags there, because once again, that defined the radius of this arc. And we know that point C prime in this case, point C prime in this case, is the same distance from D as F is, as F is. And so one way to think about it, imagine a line between F and, I could get it, let me get my straight edge here so it looks a little bit neater. Imagine a line that connects F and this C prime. And once again, we're in the case where C prime immediately didn't go to F, where C prime ended up being on this side, so to speak."}, {"video_title": "Proving the SSS triangle congruence criterion using transformations Geometry Khan Academy.mp3", "Sentence": "And we know that point C prime in this case, point C prime in this case, is the same distance from D as F is, as F is. And so one way to think about it, imagine a line between F and, I could get it, let me get my straight edge here so it looks a little bit neater. Imagine a line that connects F and this C prime. And once again, we're in the case where C prime immediately didn't go to F, where C prime ended up being on this side, so to speak. And you can see that point E, because it is equidistant to C prime and F, it must sit on the perpendicular bisector of the segment F, C. Same thing about point D or B prime. This must be the perpendicular bisector, because this point is equidistant to F as it is to C, as to C prime. This point is equidistant to F as it is to C prime."}, {"video_title": "Proving the SSS triangle congruence criterion using transformations Geometry Khan Academy.mp3", "Sentence": "And once again, we're in the case where C prime immediately didn't go to F, where C prime ended up being on this side, so to speak. And you can see that point E, because it is equidistant to C prime and F, it must sit on the perpendicular bisector of the segment F, C. Same thing about point D or B prime. This must be the perpendicular bisector, because this point is equidistant to F as it is to C, as to C prime. This point is equidistant to F as it is to C prime. The set of points whose distance is equal to F and C prime, they will form the perpendicular bisector of F, C. So we know that this orange line is a perpendicular bisector of F, C. Why is that helpful? Well, that tells us is if when we do that first transformation to make A, B coincide with E, F, if C prime doesn't end up here and it ends up there, we just have to do one more transformation. We just have to do a reflection about ED or about A prime, B prime, however you wanna view it, about this orange line, and then C will coincide with F, because orange is a perpendicular bisector, so I could do something like this."}, {"video_title": "Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3", "Sentence": "So we have a trapezoid here on the coordinate plane. And what we want to do is find the area of this trapezoid just given this diagram. And like always, pause this video and see if you can figure it out. Well, we know how to figure out the area of a trapezoid. We have videos where we derive this formula. But the area of a trapezoid, just put simply, is equal to the average of the lengths of the bases. We could say base one plus base two times the height."}, {"video_title": "Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3", "Sentence": "Well, we know how to figure out the area of a trapezoid. We have videos where we derive this formula. But the area of a trapezoid, just put simply, is equal to the average of the lengths of the bases. We could say base one plus base two times the height. And so what are our bases here? And what is going to be our height over here? Well, we could call base one, we could call that segment CL."}, {"video_title": "Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3", "Sentence": "We could say base one plus base two times the height. And so what are our bases here? And what is going to be our height over here? Well, we could call base one, we could call that segment CL. So it would be the length of segment CL right over here. I'll do that in magenta. That is going to be base one."}, {"video_title": "Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3", "Sentence": "Well, we could call base one, we could call that segment CL. So it would be the length of segment CL right over here. I'll do that in magenta. That is going to be base one. Base two, that could, let me do that in a different color. Base two would be the length of segment OW, or B2 would be the length of segment OW, right over there. And then our height, our height H, well, that would just be an altitude."}, {"video_title": "Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3", "Sentence": "That is going to be base one. Base two, that could, let me do that in a different color. Base two would be the length of segment OW, or B2 would be the length of segment OW, right over there. And then our height, our height H, well, that would just be an altitude. And they did one in a dotted line here. Notice it intersects the base one. I guess you could say segment CL at a right angle here."}, {"video_title": "Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3", "Sentence": "And then our height, our height H, well, that would just be an altitude. And they did one in a dotted line here. Notice it intersects the base one. I guess you could say segment CL at a right angle here. And so this would be the height. So if we know the lengths of each of these, if we know each of these values, which are the lengths of these segments, then we can evaluate the area of this actual trapezoid. And once again, if this is completely unfamiliar to you, or if you're curious, we have multiple videos talking about the proofs, or how we came up with this formula."}, {"video_title": "Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3", "Sentence": "I guess you could say segment CL at a right angle here. And so this would be the height. So if we know the lengths of each of these, if we know each of these values, which are the lengths of these segments, then we can evaluate the area of this actual trapezoid. And once again, if this is completely unfamiliar to you, or if you're curious, we have multiple videos talking about the proofs, or how we came up with this formula. You can even break down a trapezoid into two triangles and a rectangle, which is one way to think about it. But anyway, let's see how we can figure this out. So the first one is what is B1 going to be?"}, {"video_title": "Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3", "Sentence": "And once again, if this is completely unfamiliar to you, or if you're curious, we have multiple videos talking about the proofs, or how we came up with this formula. You can even break down a trapezoid into two triangles and a rectangle, which is one way to think about it. But anyway, let's see how we can figure this out. So the first one is what is B1 going to be? B1 is the length of segment CL. And you could say, well, look, we know what the coordinates of these points are. You could say, let's use the distance formula."}, {"video_title": "Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3", "Sentence": "So the first one is what is B1 going to be? B1 is the length of segment CL. And you could say, well, look, we know what the coordinates of these points are. You could say, let's use the distance formula. And you could say, well, the distance formula is just an application of the Pythagorean theorem. So this is just going to be the square root of our change in X squared. So our change in X is going to be this right over here."}, {"video_title": "Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3", "Sentence": "You could say, let's use the distance formula. And you could say, well, the distance formula is just an application of the Pythagorean theorem. So this is just going to be the square root of our change in X squared. So our change in X is going to be this right over here. And notice we're going from X equals negative four to X equals eight as we go from C to L. So our change in X is equal to eight minus negative four, which is equal to 12. And our change in Y, we're going from Y equals negative one to Y equals five. So we could say our change in Y is equal to five minus negative one, which of course is equal to six."}, {"video_title": "Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3", "Sentence": "So our change in X is going to be this right over here. And notice we're going from X equals negative four to X equals eight as we go from C to L. So our change in X is equal to eight minus negative four, which is equal to 12. And our change in Y, we're going from Y equals negative one to Y equals five. So we could say our change in Y is equal to five minus negative one, which of course is equal to six. And you see that here, one, two, three, four, five, six. And the segment that we care about, its length that we care about, that's just the hypotenuse of this right triangle that has one side 12 and one side six. So the length of that hypotenuse from the Pythagorean theorem, and as I mentioned, the distance formula is just an application of the Pythagorean theorem, this is going to be our change in X squared, 12 squared, plus our change in Y squared, so plus six squared."}, {"video_title": "Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3", "Sentence": "So we could say our change in Y is equal to five minus negative one, which of course is equal to six. And you see that here, one, two, three, four, five, six. And the segment that we care about, its length that we care about, that's just the hypotenuse of this right triangle that has one side 12 and one side six. So the length of that hypotenuse from the Pythagorean theorem, and as I mentioned, the distance formula is just an application of the Pythagorean theorem, this is going to be our change in X squared, 12 squared, plus our change in Y squared, so plus six squared. And this is going to be equal to 144 plus 36. So the square root of 144 plus 36 is 100, 180, which is equal to, let's see, 180 is 36 times five. So that is six square roots, six square roots of five."}, {"video_title": "Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3", "Sentence": "So the length of that hypotenuse from the Pythagorean theorem, and as I mentioned, the distance formula is just an application of the Pythagorean theorem, this is going to be our change in X squared, 12 squared, plus our change in Y squared, so plus six squared. And this is going to be equal to 144 plus 36. So the square root of 144 plus 36 is 100, 180, which is equal to, let's see, 180 is 36 times five. So that is six square roots, six square roots of five. All right, let me not skip some steps. So this is the square root of 36 times five, which is equal to, the square root of 36 is six, so six square roots of five. Now let's figure out B2."}, {"video_title": "Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3", "Sentence": "So that is six square roots, six square roots of five. All right, let me not skip some steps. So this is the square root of 36 times five, which is equal to, the square root of 36 is six, so six square roots of five. Now let's figure out B2. So B2, once again, change in X squared plus the square root of change in X squared plus change in Y squared. Well, let's see, if we're going from, we could set up a right triangle if you like like this to figure those things out. So our change in X, we're going from X is at negative two, X is going from negative two to positive four, so our change in X is six."}, {"video_title": "Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3", "Sentence": "Now let's figure out B2. So B2, once again, change in X squared plus the square root of change in X squared plus change in Y squared. Well, let's see, if we're going from, we could set up a right triangle if you like like this to figure those things out. So our change in X, we're going from X is at negative two, X is going from negative two to positive four, so our change in X is six. Our change in Y, we are going from, we are going from Y equals, Y equals five to Y equals eight, so our change in Y is equal to three. So just applying the Pythagorean theorem to find the length of the hypotenuse here, it's going to be the square root of change in X squared, six squared, plus change in Y squared, plus three squared, which is going to be equal to, it's going to be equal to 36 plus, 36 plus nine, which is 45, so square root of 45, which is equal to the square root of nine times five, which is equal to three square roots of five, and so we only have one left to figure out. We have to figure out H. We have to figure out the length of H. So H is going to be equal to, and so what is our, if we're going from W to N, our change in X is two."}, {"video_title": "Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3", "Sentence": "So our change in X, we're going from X is at negative two, X is going from negative two to positive four, so our change in X is six. Our change in Y, we are going from, we are going from Y equals, Y equals five to Y equals eight, so our change in Y is equal to three. So just applying the Pythagorean theorem to find the length of the hypotenuse here, it's going to be the square root of change in X squared, six squared, plus change in Y squared, plus three squared, which is going to be equal to, it's going to be equal to 36 plus, 36 plus nine, which is 45, so square root of 45, which is equal to the square root of nine times five, which is equal to three square roots of five, and so we only have one left to figure out. We have to figure out H. We have to figure out the length of H. So H is going to be equal to, and so what is our, if we're going from W to N, our change in X is two. Change in X is equal to two. We're going from X equals four to X equals six. If you wanted to do that purely numerically, you would say, okay, our end point, our X value is six, our starting point, our X value is four."}, {"video_title": "Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3", "Sentence": "We have to figure out H. We have to figure out the length of H. So H is going to be equal to, and so what is our, if we're going from W to N, our change in X is two. Change in X is equal to two. We're going from X equals four to X equals six. If you wanted to do that purely numerically, you would say, okay, our end point, our X value is six, our starting point, our X value is four. Six minus four is two. You see that visually here. So it's going to be the square root of two squared, plus our, let me write that radical a little bit better."}, {"video_title": "Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3", "Sentence": "If you wanted to do that purely numerically, you would say, okay, our end point, our X value is six, our starting point, our X value is four. Six minus four is two. You see that visually here. So it's going to be the square root of two squared, plus our, let me write that radical a little bit better. So it's the square root of our change in X squared, plus our change in Y squared. Our change in Y is negative four. Change in Y is negative four, but we're going to square it, so it's going to become a positive 16."}, {"video_title": "Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3", "Sentence": "So it's going to be the square root of two squared, plus our, let me write that radical a little bit better. So it's the square root of our change in X squared, plus our change in Y squared. Our change in Y is negative four. Change in Y is negative four, but we're going to square it, so it's going to become a positive 16. So this is going to be equal to the square root of four plus 16 square root of 20, which is equal to the square root of four times five, which is equal to two times the square root of five. It's nice that the square root of five keeps popping up. And so now we just substitute into our original expression."}, {"video_title": "Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3", "Sentence": "Change in Y is negative four, but we're going to square it, so it's going to become a positive 16. So this is going to be equal to the square root of four plus 16 square root of 20, which is equal to the square root of four times five, which is equal to two times the square root of five. It's nice that the square root of five keeps popping up. And so now we just substitute into our original expression. And so our area of our trapezoid is going to be 1 half times six square roots of five, six square roots of five, plus three square roots of five, plus three square roots of five, let me close that parentheses, times two square roots of five, times two square root of five. And let's see how we can simplify this. So six square roots of five plus three square roots of five, that is nine square roots of five."}, {"video_title": "Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3", "Sentence": "And so now we just substitute into our original expression. And so our area of our trapezoid is going to be 1 half times six square roots of five, six square roots of five, plus three square roots of five, plus three square roots of five, let me close that parentheses, times two square roots of five, times two square root of five. And let's see how we can simplify this. So six square roots of five plus three square roots of five, that is nine square roots of five. Let's see, the 1 half times a two, those cancel out to just be one. And so we're left with nine square roots of five times the square root of five. Well, the square root of five times the square root of five is just going to be five."}, {"video_title": "Measure of circumscribed angle Circles Geometry Khan Academy.mp3", "Sentence": "So this is angle A right over here. Then when I say it's a circumscribed angle, that means that the two sides of the angle are tangent to the circle. So AC is tangent to the circle at point C. AB is tangent to the circle at point B. What is the measure of angle A? Now I encourage you to pause the video now and to try this out on your own. And I'll give you a hint. It will leverage the fact that this is a circumscribed angle as you could imagine."}, {"video_title": "Measure of circumscribed angle Circles Geometry Khan Academy.mp3", "Sentence": "What is the measure of angle A? Now I encourage you to pause the video now and to try this out on your own. And I'll give you a hint. It will leverage the fact that this is a circumscribed angle as you could imagine. So I'm assuming you've given a go at it. So the other piece of information they give us is that angle D, which is an inscribed angle, is 48 degrees. And it intercepts the same arc."}, {"video_title": "Measure of circumscribed angle Circles Geometry Khan Academy.mp3", "Sentence": "It will leverage the fact that this is a circumscribed angle as you could imagine. So I'm assuming you've given a go at it. So the other piece of information they give us is that angle D, which is an inscribed angle, is 48 degrees. And it intercepts the same arc. So this is the arc that it intercepts, arc CB I guess you could call it. It intercepts this arc right over here. It's the inscribed angle."}, {"video_title": "Measure of circumscribed angle Circles Geometry Khan Academy.mp3", "Sentence": "And it intercepts the same arc. So this is the arc that it intercepts, arc CB I guess you could call it. It intercepts this arc right over here. It's the inscribed angle. So the central angle that intersects that same arc is going to be twice the inscribed angle. So this is going to be 96 degrees. I could put three markers here just because we've already used a double marker."}, {"video_title": "Measure of circumscribed angle Circles Geometry Khan Academy.mp3", "Sentence": "It's the inscribed angle. So the central angle that intersects that same arc is going to be twice the inscribed angle. So this is going to be 96 degrees. I could put three markers here just because we've already used a double marker. Notice, they both intercept arc CB. So you could, some people would say the measure of arc CB is 96 degrees. The central angle is 96 degrees."}, {"video_title": "Measure of circumscribed angle Circles Geometry Khan Academy.mp3", "Sentence": "I could put three markers here just because we've already used a double marker. Notice, they both intercept arc CB. So you could, some people would say the measure of arc CB is 96 degrees. The central angle is 96 degrees. The inscribed angle is going to be half of that, 48 degrees. So how does this help us? Well, a key clue is that angle is a circumscribed angle."}, {"video_title": "Measure of circumscribed angle Circles Geometry Khan Academy.mp3", "Sentence": "The central angle is 96 degrees. The inscribed angle is going to be half of that, 48 degrees. So how does this help us? Well, a key clue is that angle is a circumscribed angle. So that means AC and AB are each tangent to the circle. Well, a line that is tangent to the circle is going to be perpendicular to the radius of the circle that intersects the circle at the same point. So this right over here, this right over here is going to be a 90 degree angle."}, {"video_title": "Measure of circumscribed angle Circles Geometry Khan Academy.mp3", "Sentence": "Well, a key clue is that angle is a circumscribed angle. So that means AC and AB are each tangent to the circle. Well, a line that is tangent to the circle is going to be perpendicular to the radius of the circle that intersects the circle at the same point. So this right over here, this right over here is going to be a 90 degree angle. And this right over here is going to be a 90 degree angle. OC is perpendicular to CA. OB, which is a radius, is perpendicular to BA, which is a tangent line, and they both intersect right over here at B."}, {"video_title": "Measure of circumscribed angle Circles Geometry Khan Academy.mp3", "Sentence": "So this right over here, this right over here is going to be a 90 degree angle. And this right over here is going to be a 90 degree angle. OC is perpendicular to CA. OB, which is a radius, is perpendicular to BA, which is a tangent line, and they both intersect right over here at B. Now this might jump out at you. We have a quadrilateral going on here. A, B, O, C is a quadrilateral."}, {"video_title": "Measure of circumscribed angle Circles Geometry Khan Academy.mp3", "Sentence": "OB, which is a radius, is perpendicular to BA, which is a tangent line, and they both intersect right over here at B. Now this might jump out at you. We have a quadrilateral going on here. A, B, O, C is a quadrilateral. So its sides are going to add up to 360 degrees. So we could know, we know, we could write it this way. We could write the measure of angle A plus 90 degrees, plus 90 degrees, plus another 90 degrees, plus another 90 degrees, plus 96 degrees, plus 96 degrees is going to be equal to 360 degrees."}, {"video_title": "Measure of circumscribed angle Circles Geometry Khan Academy.mp3", "Sentence": "A, B, O, C is a quadrilateral. So its sides are going to add up to 360 degrees. So we could know, we know, we could write it this way. We could write the measure of angle A plus 90 degrees, plus 90 degrees, plus another 90 degrees, plus another 90 degrees, plus 96 degrees, plus 96 degrees is going to be equal to 360 degrees. Is going to be equal to 360 degrees. Or another way of thinking about it, if we subtract 180 from both sides, if we subtract that from both sides, we get the measure of angle A plus 96 degrees, plus 96 degrees is going to be equal to 180 degrees. Or another way of thinking about it is the measure of angle A, or that angle A and angle O right over here, you could call it angle COB, that these are going to be supplementary angles, that they add up to 180 degrees."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "So we're starting off with triangle ABC here, and we see from the drawing that we already know that the length of AB is equal to the length of AC, or line segment AB is congruent to line segment AC. And since this is a triangle and two sides of this triangle are congruent, or they have the same length, we can say that this is an isosceles triangle. Isosceles triangle, one of the hardest words for me to spell. I think I got it right. And that just means that two of the sides are equal to each other. Now, what I want to do in this video is show what I want to prove. So what I want to prove here is that these two, and they're sometimes referred to as base angles, these angles that are between one of the sides and the side that isn't necessarily equal to it, and the other side that is equal and the side that's not equal to it, I want to show that they're congruent."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "I think I got it right. And that just means that two of the sides are equal to each other. Now, what I want to do in this video is show what I want to prove. So what I want to prove here is that these two, and they're sometimes referred to as base angles, these angles that are between one of the sides and the side that isn't necessarily equal to it, and the other side that is equal and the side that's not equal to it, I want to show that they're congruent. So I want to prove that angle ABC, I want to prove that that is congruent to angle ACB. And so for an isosceles triangle, those two angles are often called base angles, and this might be called the vertex angle over here. And these are often called the sides, and these are the legs of the isosceles triangle, and these are obviously their sides."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "So what I want to prove here is that these two, and they're sometimes referred to as base angles, these angles that are between one of the sides and the side that isn't necessarily equal to it, and the other side that is equal and the side that's not equal to it, I want to show that they're congruent. So I want to prove that angle ABC, I want to prove that that is congruent to angle ACB. And so for an isosceles triangle, those two angles are often called base angles, and this might be called the vertex angle over here. And these are often called the sides, and these are the legs of the isosceles triangle, and these are obviously their sides. These are the legs of the isosceles triangle, and this one down here that isn't necessarily the same as the other two you would call the base. So let's see if we can prove that. So there's not a lot of information here, just that these two sides are equal, but we have in our toolkit a lot that we know about triangle congruency."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "And these are often called the sides, and these are the legs of the isosceles triangle, and these are obviously their sides. These are the legs of the isosceles triangle, and this one down here that isn't necessarily the same as the other two you would call the base. So let's see if we can prove that. So there's not a lot of information here, just that these two sides are equal, but we have in our toolkit a lot that we know about triangle congruency. So maybe we can construct two triangles here that are congruent, and then we can use that information to figure out whether this angle is congruent to that angle there. And the first step, if we're going to use triangle congruency, is to actually construct two triangles. So one way to construct two triangles is let's set up another point right over here."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "So there's not a lot of information here, just that these two sides are equal, but we have in our toolkit a lot that we know about triangle congruency. So maybe we can construct two triangles here that are congruent, and then we can use that information to figure out whether this angle is congruent to that angle there. And the first step, if we're going to use triangle congruency, is to actually construct two triangles. So one way to construct two triangles is let's set up another point right over here. Let's set up another point D, and let's just say that D is the midpoint of B and C. So it's the midpoint. So the distance from B to D is going to be the same thing as the distance. Let me do a double slash here to show you it's not the same as that distance."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "So one way to construct two triangles is let's set up another point right over here. Let's set up another point D, and let's just say that D is the midpoint of B and C. So it's the midpoint. So the distance from B to D is going to be the same thing as the distance. Let me do a double slash here to show you it's not the same as that distance. So the distance from B to D is going to be the same thing as the distance from D to C. And obviously, between any two points, you have a midpoint. And so let me draw a segment AD. And what's useful about that is that we have now constructed two triangles."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "Let me do a double slash here to show you it's not the same as that distance. So the distance from B to D is going to be the same thing as the distance from D to C. And obviously, between any two points, you have a midpoint. And so let me draw a segment AD. And what's useful about that is that we have now constructed two triangles. And what's even cooler is that triangle ABD and triangle ACD, they have this side is congruent, this side is congruent, and they actually share this side right over here. They actually share that side right over there. So we know that triangle ABD, we know that it is congruent to triangle ACD."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "And what's useful about that is that we have now constructed two triangles. And what's even cooler is that triangle ABD and triangle ACD, they have this side is congruent, this side is congruent, and they actually share this side right over here. They actually share that side right over there. So we know that triangle ABD, we know that it is congruent to triangle ACD. And we know it because of SSS, side, side, side. You have two triangles that have three sides that are congruent or they have the same length, then the two triangles are congruent. And what's useful about that is if these two triangles are congruent, then their corresponding angles are congruent."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "So we know that triangle ABD, we know that it is congruent to triangle ACD. And we know it because of SSS, side, side, side. You have two triangles that have three sides that are congruent or they have the same length, then the two triangles are congruent. And what's useful about that is if these two triangles are congruent, then their corresponding angles are congruent. And so we've actually now proved our result. Because the corresponding angle to ABC in this triangle is angle ACD in this triangle right over here. So that we then know that angle ABC is congruent to angle ACB."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "And what's useful about that is if these two triangles are congruent, then their corresponding angles are congruent. And so we've actually now proved our result. Because the corresponding angle to ABC in this triangle is angle ACD in this triangle right over here. So that we then know that angle ABC is congruent to angle ACB. So that's a pretty neat result. If you have an isosceles triangle, a triangle where two of the sides are congruent, then their base angles are also going to be congruent. Now let's think about it the other way."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "So that we then know that angle ABC is congruent to angle ACB. So that's a pretty neat result. If you have an isosceles triangle, a triangle where two of the sides are congruent, then their base angles are also going to be congruent. Now let's think about it the other way. Can we make the other statement? If the base angles are congruent, do we know that these two legs are going to be congruent? So let's try to construct a triangle and see if we can prove it the other way."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "Now let's think about it the other way. Can we make the other statement? If the base angles are congruent, do we know that these two legs are going to be congruent? So let's try to construct a triangle and see if we can prove it the other way. So I'll do another triangle right over here. We draw another one just like that. That's not that pretty of a triangle, so let me draw it a little nicer."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "So let's try to construct a triangle and see if we can prove it the other way. So I'll do another triangle right over here. We draw another one just like that. That's not that pretty of a triangle, so let me draw it a little nicer. I'm going to draw it like this. I'm going to call this, let me do it in a different color. So I'll call that A. I will call this B. I will call that C right over there."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "That's not that pretty of a triangle, so let me draw it a little nicer. I'm going to draw it like this. I'm going to call this, let me do it in a different color. So I'll call that A. I will call this B. I will call that C right over there. And now we're going to start off with the idea that this angle, angle ABC, is congruent to angle ACB. So this is where they have the same exact measure. And what we want to do in this case, we want to prove."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "So I'll call that A. I will call this B. I will call that C right over there. And now we're going to start off with the idea that this angle, angle ABC, is congruent to angle ACB. So this is where they have the same exact measure. And what we want to do in this case, we want to prove. So let me draw a little line here to show that we're doing a different idea. Here we're saying if these two sides are the same, then the base angles are going to be the same. We proved that."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "And what we want to do in this case, we want to prove. So let me draw a little line here to show that we're doing a different idea. Here we're saying if these two sides are the same, then the base angles are going to be the same. We proved that. Now let's go the other way. If the base angles are the same, do we know that the two sides are the same? So we want to prove that segment AB is congruent to segment AC, or AC is congruent to AB."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "We proved that. Now let's go the other way. If the base angles are the same, do we know that the two sides are the same? So we want to prove that segment AB is congruent to segment AC, or AC is congruent to AB. Or you could say that the length of segment AC, which we would denote that way, is equal to the length of segment AB. These are essentially equivalent statements. So let's see."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "So we want to prove that segment AB is congruent to segment AC, or AC is congruent to AB. Or you could say that the length of segment AC, which we would denote that way, is equal to the length of segment AB. These are essentially equivalent statements. So let's see. Once again, in our toolkit, we have our congruency theorems. But in order to apply them, you really do need to have two triangles. So let's construct two triangles here."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "So let's see. Once again, in our toolkit, we have our congruency theorems. But in order to apply them, you really do need to have two triangles. So let's construct two triangles here. And this time, instead of defining another point as a midpoint, I'm going to define D this time as the point that if I were to go straight up, the point that is essentially, if we view BC as straight horizontal, the point that goes straight down from A. And the reason why I say that is there's some point, you could call it an altitude, that intersects BC at a right angle. And there will definitely be some point like that."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "So let's construct two triangles here. And this time, instead of defining another point as a midpoint, I'm going to define D this time as the point that if I were to go straight up, the point that is essentially, if we view BC as straight horizontal, the point that goes straight down from A. And the reason why I say that is there's some point, you could call it an altitude, that intersects BC at a right angle. And there will definitely be some point like that. And so if it's a right angle on that side, if that's 90 degrees, then we know that this is 90 degrees as well. Now what's interesting about this? And let me write this down."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "And there will definitely be some point like that. And so if it's a right angle on that side, if that's 90 degrees, then we know that this is 90 degrees as well. Now what's interesting about this? And let me write this down. So I've constructed AD such that AD is perpendicular to BC. And you can always construct an altitude. Essentially, you just have to make BC lie flat on the ground."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "And let me write this down. So I've constructed AD such that AD is perpendicular to BC. And you can always construct an altitude. Essentially, you just have to make BC lie flat on the ground. And then you just have to drop something from A. And that'll give you point D. You can always do that with a triangle like this. So what does this give us?"}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "Essentially, you just have to make BC lie flat on the ground. And then you just have to drop something from A. And that'll give you point D. You can always do that with a triangle like this. So what does this give us? So over here, we have an angle, an angle, and then a side in common. And over here, you have an angle that corresponds to that angle, an angle that corresponds to this angle, and the same side in common. And so we know that these triangles are congruent by AAS, angle-angle-side, which we've shown is a valid congruent postulate."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "So what does this give us? So over here, we have an angle, an angle, and then a side in common. And over here, you have an angle that corresponds to that angle, an angle that corresponds to this angle, and the same side in common. And so we know that these triangles are congruent by AAS, angle-angle-side, which we've shown is a valid congruent postulate. So we can say now that triangle ABD is congruent to triangle ACD. And we know that by angle-angle-side. This angle, then this angle, and this side."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "And so we know that these triangles are congruent by AAS, angle-angle-side, which we've shown is a valid congruent postulate. So we can say now that triangle ABD is congruent to triangle ACD. And we know that by angle-angle-side. This angle, then this angle, and this side. This angle, then this angle, then this side. And once we know these two triangles are congruent, we know that every corresponding angle or side of the two triangles are also going to be congruent. So then we know that AB is a corresponding side to AC."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "This angle, then this angle, and this side. This angle, then this angle, then this side. And once we know these two triangles are congruent, we know that every corresponding angle or side of the two triangles are also going to be congruent. So then we know that AB is a corresponding side to AC. So these two sides must be congruent. And so you get AB is going to be congruent to AC. And that's because these are congruent triangles."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "So then we know that AB is a corresponding side to AC. So these two sides must be congruent. And so you get AB is going to be congruent to AC. And that's because these are congruent triangles. And we've proven what we wanted to show. If the base angles are equal, then the two legs are going to be equal. If the two legs are equal, then the base angles are equal."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "And that's because these are congruent triangles. And we've proven what we wanted to show. If the base angles are equal, then the two legs are going to be equal. If the two legs are equal, then the base angles are equal. So it's a very, very, very useful tool in geometry. And in case you're curious for this specific isosceles triangle, over here we set up D so it was the midpoint. Over here we set up D so it went directly below A."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "If the two legs are equal, then the base angles are equal. So it's a very, very, very useful tool in geometry. And in case you're curious for this specific isosceles triangle, over here we set up D so it was the midpoint. Over here we set up D so it went directly below A. We didn't say whether it was the midpoint. But here we can actually show that it is the midpoint, just as a little bit of a bonus result. Because we know that since these two triangles are congruent, BD is going to be congruent to DC, because they are the corresponding sides."}, {"video_title": "Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3", "Sentence": "Over here we set up D so it went directly below A. We didn't say whether it was the midpoint. But here we can actually show that it is the midpoint, just as a little bit of a bonus result. Because we know that since these two triangles are congruent, BD is going to be congruent to DC, because they are the corresponding sides. So it actually turns out that point D for an isosceles triangle not only is it the midpoint, but it is the place where it is a point at which AD, or we could say that AD is a perpendicular bisector of BC. So not only is AD perpendicular to BC, but it bisects it. That D is the midpoint of that entire base."}, {"video_title": "Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3", "Sentence": "And we know that the length of AC is equal to the length of CB. So this is an isosceles triangle. We have two of its legs are equal to each other. And then they also tell us that this line up here, they didn't put another label there. Let me put another label there just for fun. Let's call this, you could even call this a ray, because it's starting at C. That line or ray CD is parallel to this segment AB over here. And that's interesting."}, {"video_title": "Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3", "Sentence": "And then they also tell us that this line up here, they didn't put another label there. Let me put another label there just for fun. Let's call this, you could even call this a ray, because it's starting at C. That line or ray CD is parallel to this segment AB over here. And that's interesting. And they give us these two angles right over here, these adjacent angles. They give it to us in terms of x. And what I want to do in this video is try to figure out what x is."}, {"video_title": "Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3", "Sentence": "And that's interesting. And they give us these two angles right over here, these adjacent angles. They give it to us in terms of x. And what I want to do in this video is try to figure out what x is. And so given that they told us that this line and this line are parallel, and we can turn this into a line CD, so it's not just a ray anymore. So it just keeps going on and on in both directions. The fact that they've given us a parallel line tells us that maybe we can use some of what we know about transversals and parallel lines to figure out some of the angles here."}, {"video_title": "Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3", "Sentence": "And what I want to do in this video is try to figure out what x is. And so given that they told us that this line and this line are parallel, and we can turn this into a line CD, so it's not just a ray anymore. So it just keeps going on and on in both directions. The fact that they've given us a parallel line tells us that maybe we can use some of what we know about transversals and parallel lines to figure out some of the angles here. And you might recognize that this right over here, this line, let me do that in a better color, you might recognize that line CB is a transversal for those two parallel lines. Let me draw both of the parallel lines a little bit more so that you can recognize that as a transversal. And then a few things might jump out."}, {"video_title": "Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3", "Sentence": "The fact that they've given us a parallel line tells us that maybe we can use some of what we know about transversals and parallel lines to figure out some of the angles here. And you might recognize that this right over here, this line, let me do that in a better color, you might recognize that line CB is a transversal for those two parallel lines. Let me draw both of the parallel lines a little bit more so that you can recognize that as a transversal. And then a few things might jump out. You have this x plus 10 right over here. And its corresponding angle is right down here. This would also be x plus 10."}, {"video_title": "Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3", "Sentence": "And then a few things might jump out. You have this x plus 10 right over here. And its corresponding angle is right down here. This would also be x plus 10. And if this is x plus 10, then you have a vertical angle right over here that would also be x plus 10. Or you could say that you have alternate interior angles that would also be congruent. Either way, this base angle is going to be x plus 10."}, {"video_title": "Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3", "Sentence": "This would also be x plus 10. And if this is x plus 10, then you have a vertical angle right over here that would also be x plus 10. Or you could say that you have alternate interior angles that would also be congruent. Either way, this base angle is going to be x plus 10. Well, it's an isosceles triangle. So your two base angles are going to be congruent. So if this is x plus 10, then this is going to be x plus 10 as well."}, {"video_title": "Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3", "Sentence": "Either way, this base angle is going to be x plus 10. Well, it's an isosceles triangle. So your two base angles are going to be congruent. So if this is x plus 10, then this is going to be x plus 10 as well. And now we have the three angles of a triangle expressed in terms of x. So when we take their sum, they need to be equal to 180. And then we can actually solve for x."}, {"video_title": "Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3", "Sentence": "So if this is x plus 10, then this is going to be x plus 10 as well. And now we have the three angles of a triangle expressed in terms of x. So when we take their sum, they need to be equal to 180. And then we can actually solve for x. So we get 2x plus x plus 10 is going to be equal to 180 degrees. And then we can add up the x's. So we have a 2x there plus an x plus another x."}, {"video_title": "Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3", "Sentence": "And then we can actually solve for x. So we get 2x plus x plus 10 is going to be equal to 180 degrees. And then we can add up the x's. So we have a 2x there plus an x plus another x. That gives us 4x. And then we have a plus 10 and another plus 10. So that gives us a plus 20 is equal to 180."}, {"video_title": "Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3", "Sentence": "So we have a 2x there plus an x plus another x. That gives us 4x. And then we have a plus 10 and another plus 10. So that gives us a plus 20 is equal to 180. And we can subtract 20 from both sides of that. And we get 4x is equal to 160. Divide both sides by 4."}, {"video_title": "Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3", "Sentence": "So that gives us a plus 20 is equal to 180. And we can subtract 20 from both sides of that. And we get 4x is equal to 160. Divide both sides by 4. And we get x is equal to 40. And we're done. We figured out what x is."}, {"video_title": "Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3", "Sentence": "Divide both sides by 4. And we get x is equal to 40. And we're done. We figured out what x is. And then we can actually figure out what these angles are. If this is x plus 10, then you have 40 plus 10. This right over here is going to be a 50 degree angle."}, {"video_title": "Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3", "Sentence": "We figured out what x is. And then we can actually figure out what these angles are. If this is x plus 10, then you have 40 plus 10. This right over here is going to be a 50 degree angle. This is 2x, so 2 times 40, this is an 80 degree angle. It doesn't look at it the way I've drawn it. And that's why you should never assume anything based on how a diagram is drawn."}, {"video_title": "Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3", "Sentence": "This right over here is going to be a 50 degree angle. This is 2x, so 2 times 40, this is an 80 degree angle. It doesn't look at it the way I've drawn it. And that's why you should never assume anything based on how a diagram is drawn. So this right over here is going to be an 80 degree angle. And then these two base angles right over here are also going to be 50 degrees. So you have 50 degrees, 50 degrees, and 80."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "So that's negative four comma negative two and zero comma five. And zero comma five, so that's zero comma five right over there. The quadrilateral is left unchanged by a reflection over the line y is equal to x over two. So what does that line look like? Y is equal to x over two. I'll do that in blue. Y is equal to x over two."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "So what does that line look like? Y is equal to x over two. I'll do that in blue. Y is equal to x over two. So when x is equal to zero, y is zero. The y-intercept is zero here, and the slope is one half. Every time x increases by one, y will increase by one half, or when x increases by two, y will increase by one."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "Y is equal to x over two. So when x is equal to zero, y is zero. The y-intercept is zero here, and the slope is one half. Every time x increases by one, y will increase by one half, or when x increases by two, y will increase by one. So x increases by two, y increases by one. x increases by two, y increases by one. Another way to think about it, y is always one half of x."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "Every time x increases by one, y will increase by one half, or when x increases by two, y will increase by one. So x increases by two, y increases by one. x increases by two, y increases by one. Another way to think about it, y is always one half of x. When x is four, y is two. When x is six, y is three. When x is eight, y is four."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "Another way to think about it, y is always one half of x. When x is four, y is two. When x is six, y is three. When x is eight, y is four. So we could connect these. Let me try my best attempt to draw these in a relatively straight line, and then I can keep going. When x is negative two, y is negative one."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "When x is eight, y is four. So we could connect these. Let me try my best attempt to draw these in a relatively straight line, and then I can keep going. When x is negative two, y is negative one. When x is negative four, y is negative two. So it actually goes through that point right there, and it just keeps going with a slope of one half. So this line, and I can draw it a little bit thicker now, now that I've dotted it out."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "When x is negative two, y is negative one. When x is negative four, y is negative two. So it actually goes through that point right there, and it just keeps going with a slope of one half. So this line, and I can draw it a little bit thicker now, now that I've dotted it out. This is the line y is equal to x over two. And they also say that the quadrilateral is left unchanged by reflection over the line y is equal to negative two x plus five. So the y-intercept here is five."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "So this line, and I can draw it a little bit thicker now, now that I've dotted it out. This is the line y is equal to x over two. And they also say that the quadrilateral is left unchanged by reflection over the line y is equal to negative two x plus five. So the y-intercept here is five. When x is zero, y is five. So it actually goes through that point, and the slope is negative two. Every time we increase x by one, we decrease y by two."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "So the y-intercept here is five. When x is zero, y is five. So it actually goes through that point, and the slope is negative two. Every time we increase x by one, we decrease y by two. So we go there, we go there, and we keep going at a slope of negative two. So it's going to look something like this. It's going to look something like this."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "Every time we increase x by one, we decrease y by two. So we go there, we go there, and we keep going at a slope of negative two. So it's going to look something like this. It's going to look something like this. It actually goes through that point and just keeps going on and on. So this is my best attempt at drawing that line. So that is y is equal to negative two x plus five."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "It's going to look something like this. It actually goes through that point and just keeps going on and on. So this is my best attempt at drawing that line. So that is y is equal to negative two x plus five. Now let's think about it. Let's see if we can draw this quadrilateral. So let's first reflect the quadrilateral, or let's reflect the points we have over the line y is equal to x over two."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "So that is y is equal to negative two x plus five. Now let's think about it. Let's see if we can draw this quadrilateral. So let's first reflect the quadrilateral, or let's reflect the points we have over the line y is equal to x over two. So this is the line y is equal to x over two. This magenta point, the point negative four, two, is already on that line. So it's its own reflection, I guess you could say, or it's on the mirror, one way is to think about it."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "So let's first reflect the quadrilateral, or let's reflect the points we have over the line y is equal to x over two. So this is the line y is equal to x over two. This magenta point, the point negative four, two, is already on that line. So it's its own reflection, I guess you could say, or it's on the mirror, one way is to think about it. But we can easily reflect this line over here. This line, if we were to drop a perpendicular, and actually this line right over here, y is equal to negative two x plus five, is perpendicular to y is equal to x over two. How do we know?"}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "So it's its own reflection, I guess you could say, or it's on the mirror, one way is to think about it. But we can easily reflect this line over here. This line, if we were to drop a perpendicular, and actually this line right over here, y is equal to negative two x plus five, is perpendicular to y is equal to x over two. How do we know? Well, if one line has a slope of m, then the line that's perpendicular would be the negative reciprocal of this. It would be negative one over m. So this first line has a slope of one half. What's the negative reciprocal of one half?"}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "How do we know? Well, if one line has a slope of m, then the line that's perpendicular would be the negative reciprocal of this. It would be negative one over m. So this first line has a slope of one half. What's the negative reciprocal of one half? The reciprocal of one half is two over one, and you make that negative. So it is equal to negative two. So this slope is the negative reciprocal of this slope."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "What's the negative reciprocal of one half? The reciprocal of one half is two over one, and you make that negative. So it is equal to negative two. So this slope is the negative reciprocal of this slope. So these lines are indeed perpendicular. So we literally could drop a perpendicular, literally go along this line right over here in our attempt to reflect, and we see that we're going down two over one, down two over one twice. So let's go down two over one, down two over one twice again."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "So this slope is the negative reciprocal of this slope. So these lines are indeed perpendicular. So we literally could drop a perpendicular, literally go along this line right over here in our attempt to reflect, and we see that we're going down two over one, down two over one twice. So let's go down two over one, down two over one twice again. The reflection of this point across y is equal to x over two is this point right over there. So now we have three points of our quadrilateral. Let's see if we can get a fourth."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "So let's go down two over one, down two over one twice again. The reflection of this point across y is equal to x over two is this point right over there. So now we have three points of our quadrilateral. Let's see if we can get a fourth. So let's go to the magenta point. The magenta point we've already seen. It's sitting on top of y equals x over two."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "Let's see if we can get a fourth. So let's go to the magenta point. The magenta point we've already seen. It's sitting on top of y equals x over two. So trying to reflect it doesn't help us much, but we could try to reflect it across y is equal to negative 2x plus 5. So once again, these lines are perpendicular to each other. Actually, let me mark that off."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "It's sitting on top of y equals x over two. So trying to reflect it doesn't help us much, but we could try to reflect it across y is equal to negative 2x plus 5. So once again, these lines are perpendicular to each other. Actually, let me mark that off. These lines are perpendicular, so we can drop a perpendicular and try to find its reflection. So we're going to the right two and up one. We're doing that once, twice, three times on the left side."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "Actually, let me mark that off. These lines are perpendicular, so we can drop a perpendicular and try to find its reflection. So we're going to the right two and up one. We're doing that once, twice, three times on the left side. So let's do that once, twice, three times on the right side. So the reflection is right there. We essentially want to go to that line, and however far we were to the left of it, we want to go that bottom left direction."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "We're doing that once, twice, three times on the left side. So let's do that once, twice, three times on the right side. So the reflection is right there. We essentially want to go to that line, and however far we were to the left of it, we want to go that bottom left direction. We want to go in the same direction to the top right, the same distance to get the reflection. So there you have it. There is our other point."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "We essentially want to go to that line, and however far we were to the left of it, we want to go that bottom left direction. We want to go in the same direction to the top right, the same distance to get the reflection. So there you have it. There is our other point. So now we have the four points of this quadrilateral. Four points of this quadrilateral are the four sides. Let me actually just draw the quadrilateral."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "There is our other point. So now we have the four points of this quadrilateral. Four points of this quadrilateral are the four sides. Let me actually just draw the quadrilateral. We have our four points. This is one side right over here. This is one side right over here."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "Let me actually just draw the quadrilateral. We have our four points. This is one side right over here. This is one side right over here. This is another side right over here. And you can verify that these are parallel. How would you verify that they're parallel?"}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "This is one side right over here. This is another side right over here. And you can verify that these are parallel. How would you verify that they're parallel? Well, they have the same slope. If you get from this point to that point, you have to go over, so your run has to be four, and you have to rise one, two, three, four, five, six, seven. So the slope here is 7 4ths."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "How would you verify that they're parallel? Well, they have the same slope. If you get from this point to that point, you have to go over, so your run has to be four, and you have to rise one, two, three, four, five, six, seven. So the slope here is 7 4ths. So slope here, rise over run, or change in y over change in x is 7 4ths. And over here, you go one, two, three, four. So you run four, and then you rise one, two, three, four, five, six, seven."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "So the slope here is 7 4ths. So slope here, rise over run, or change in y over change in x is 7 4ths. And over here, you go one, two, three, four. So you run four, and then you rise one, two, three, four, five, six, seven. So the slope here is also 7 over 4. So these two lines are going to be parallel. And then we could draw these lines over here."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "So you run four, and then you rise one, two, three, four, five, six, seven. So the slope here is also 7 over 4. So these two lines are going to be parallel. And then we could draw these lines over here. So this one at the top right over there. And what's its slope? Well, let's see."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "And then we could draw these lines over here. So this one at the top right over there. And what's its slope? Well, let's see. We go from x equals 0 to x equals 8. So we go down. Our change in y is negative 1 every time we increase x by 8."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "Well, let's see. We go from x equals 0 to x equals 8. So we go down. Our change in y is negative 1 every time we increase x by 8. So this is a slope of slope is equal to negative 1 over 8. And that's the exact same slope that we have right over here, negative 1 over 8. So these two lines are parallel as well."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "Our change in y is negative 1 every time we increase x by 8. So this is a slope of slope is equal to negative 1 over 8. And that's the exact same slope that we have right over here, negative 1 over 8. So these two lines are parallel as well. This line is parallel to this line as well. So at minimum, we're dealing with a parallelogram. Let's see if we can go even more specific because this kind of looks like a rhombus."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "So these two lines are parallel as well. This line is parallel to this line as well. So at minimum, we're dealing with a parallelogram. Let's see if we can go even more specific because this kind of looks like a rhombus. It looks like a parallelogram where all four sides have the same length. So there's a couple of ways that you could verify that this parallelogram is a rhombus. One way is you could actually find the distance between the points."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "Let's see if we can go even more specific because this kind of looks like a rhombus. It looks like a parallelogram where all four sides have the same length. So there's a couple of ways that you could verify that this parallelogram is a rhombus. One way is you could actually find the distance between the points. You could use that. We know the coordinates, so you could use the distance formula, which really comes straight out of the Pythagorean theorem. Or even better, you could look at the diagonals of this rhombus."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "One way is you could actually find the distance between the points. You could use that. We know the coordinates, so you could use the distance formula, which really comes straight out of the Pythagorean theorem. Or even better, you could look at the diagonals of this rhombus. You could look at the diagonals of this parallelogram. We're trying to figure out if it's a rhombus. And if the diagonals are perpendicular, then you're dealing with a rhombus."}, {"video_title": "Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3", "Sentence": "Or even better, you could look at the diagonals of this rhombus. You could look at the diagonals of this parallelogram. We're trying to figure out if it's a rhombus. And if the diagonals are perpendicular, then you're dealing with a rhombus. And we've already shown that these diagonals, that this diagonal, this diagonal, and this diagonal are perpendicular. They intersect at right angles. And so this must be a rhombus."}, {"video_title": "Arc length as fraction of circumference Trigonometry Khan Academy.mp3", "Sentence": "It's a little bit of a hairy circle, but you get the idea. So this is a circle, this is the center of the circle. And let's say that I have an arc along this circle. So I'll do the arc in green. So I have an arc that is part of the circle, and it subtends an angle. So that's my arc right over there. And it subtends an angle, and the angle that it subtends, so what I mean by subtends, you take each of the end points of the arc, go to the center of the circle, go to the center of the circle, just like this."}, {"video_title": "Arc length as fraction of circumference Trigonometry Khan Academy.mp3", "Sentence": "So I'll do the arc in green. So I have an arc that is part of the circle, and it subtends an angle. So that's my arc right over there. And it subtends an angle, and the angle that it subtends, so what I mean by subtends, you take each of the end points of the arc, go to the center of the circle, go to the center of the circle, just like this. And so it subtends angle theta right over here. So it subtends angle theta. And let's say that we know that angle theta is equal to two radians."}, {"video_title": "Arc length as fraction of circumference Trigonometry Khan Academy.mp3", "Sentence": "And it subtends an angle, and the angle that it subtends, so what I mean by subtends, you take each of the end points of the arc, go to the center of the circle, go to the center of the circle, just like this. And so it subtends angle theta right over here. So it subtends angle theta. And let's say that we know that angle theta is equal to two radians. So my question to you is, what fraction of the entire circumference is this green arc? What fraction of the entire circumference is this green arc? And like always, pause the video and give it a go."}, {"video_title": "Arc length as fraction of circumference Trigonometry Khan Academy.mp3", "Sentence": "And let's say that we know that angle theta is equal to two radians. So my question to you is, what fraction of the entire circumference is this green arc? What fraction of the entire circumference is this green arc? And like always, pause the video and give it a go. All right, so let's think through it a little bit. So you might say, well, how do I know that? I don't know what the radius of this thing is."}, {"video_title": "Arc length as fraction of circumference Trigonometry Khan Academy.mp3", "Sentence": "And like always, pause the video and give it a go. All right, so let's think through it a little bit. So you might say, well, how do I know that? I don't know what the radius of this thing is. I don't, I, you know, how do I think through this? And we just have to remind ourselves what radians mean. What radians mean."}, {"video_title": "Arc length as fraction of circumference Trigonometry Khan Academy.mp3", "Sentence": "I don't know what the radius of this thing is. I don't, I, you know, how do I think through this? And we just have to remind ourselves what radians mean. What radians mean. If an arc subtends an angle of two radians, that means that the arc itself is two radiuses long. So this right over here, so let me make this a little clearer. If the radius is r, if this radius, I already used that color."}, {"video_title": "Arc length as fraction of circumference Trigonometry Khan Academy.mp3", "Sentence": "What radians mean. If an arc subtends an angle of two radians, that means that the arc itself is two radiuses long. So this right over here, so let me make this a little clearer. If the radius is r, if this radius, I already used that color. If this radius, I have trouble switching colors. All right. If this radius is length r, then the length, if this angle is two radians, then the arc that subtends it is going to be two radiuses long."}, {"video_title": "Arc length as fraction of circumference Trigonometry Khan Academy.mp3", "Sentence": "If the radius is r, if this radius, I already used that color. If this radius, I have trouble switching colors. All right. If this radius is length r, then the length, if this angle is two radians, then the arc that subtends it is going to be two radiuses long. So this length right over here is two radiuses. Now, what fraction of the entire circumference is that? Well, the entire circumference, we know this from basic geometry, the entire circumference is two pi times the radius, or you could say it's two pi radii, two pi radiuses."}, {"video_title": "Arc length as fraction of circumference Trigonometry Khan Academy.mp3", "Sentence": "If this radius is length r, then the length, if this angle is two radians, then the arc that subtends it is going to be two radiuses long. So this length right over here is two radiuses. Now, what fraction of the entire circumference is that? Well, the entire circumference, we know this from basic geometry, the entire circumference is two pi times the radius, or you could say it's two pi radii, two pi radiuses. Two pi radii is the correct way to say it. So what fraction is it? It's two radii, it's two radii over two pi radii, over two pi radii."}, {"video_title": "Finding arc length from radian angle measure Trigonometry Khan Academy.mp3", "Sentence": "The measure of angle P is 0.4 radians, and the length of the radius is five units. Length of the radius is five units, that's this length right over here. Find the length of the green arc. So just to kind of conceptualize this a little bit, P, you can imagine, is the center of this larger circle, is the larger circle, and this angle right over here that has a measure of 0.4 radians, it intercepts this green arc right over here. In order to figure this out, and actually, I encourage you to pause this video now and try to think about this question on your own. How long is this arc, given the information that they've given us? Well, all we have to do is remind ourselves what a radian is."}, {"video_title": "Finding arc length from radian angle measure Trigonometry Khan Academy.mp3", "Sentence": "So just to kind of conceptualize this a little bit, P, you can imagine, is the center of this larger circle, is the larger circle, and this angle right over here that has a measure of 0.4 radians, it intercepts this green arc right over here. In order to figure this out, and actually, I encourage you to pause this video now and try to think about this question on your own. How long is this arc, given the information that they've given us? Well, all we have to do is remind ourselves what a radian is. One way to think about a radian is, if you look at the arc that the angle intercepts, which is this green arc, and you think about its length, the length of this green arc is going to be 0.4 radii. One way to think about radians is, if this angle is 0.4 radians, that means that the arc that it intercepts is going to be 0.4 radii long. So this length, we could write as 0.4 radii, or radiuses, but radii is the proper term."}, {"video_title": "Finding arc length from radian angle measure Trigonometry Khan Academy.mp3", "Sentence": "Well, all we have to do is remind ourselves what a radian is. One way to think about a radian is, if you look at the arc that the angle intercepts, which is this green arc, and you think about its length, the length of this green arc is going to be 0.4 radii. One way to think about radians is, if this angle is 0.4 radians, that means that the arc that it intercepts is going to be 0.4 radii long. So this length, we could write as 0.4 radii, or radiuses, but radii is the proper term. Now, we don't want our length in terms of radii, we want our length in terms of whatever units the radius is, these kind of five units. Well, we know that each radius has a length of five, that our radius of the circle has a length of five. So this is going to be 0.4 radii times five, and you know, they just call it units right over here, times five, I'll put it in quotes because it's kind of a generic term, five units per radii."}, {"video_title": "Finding arc length from radian angle measure Trigonometry Khan Academy.mp3", "Sentence": "So this length, we could write as 0.4 radii, or radiuses, but radii is the proper term. Now, we don't want our length in terms of radii, we want our length in terms of whatever units the radius is, these kind of five units. Well, we know that each radius has a length of five, that our radius of the circle has a length of five. So this is going to be 0.4 radii times five, and you know, they just call it units right over here, times five, I'll put it in quotes because it's kind of a generic term, five units per radii. So the radii cancel out, we're left with just the units, which is what we want, so 0.4 times five is two. So this is going to give us, this is going to be equal to two. So just as a refresher again, when the angle measure in radians, one way to think about it is the arc that it intercepts, that's going to be this many radii long."}, {"video_title": "Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And this is true for all triangles. And so to set up this proof, I've put an arbitrary triangle here, but I've put one vertex at the origin. That'll simplify the math. And then I put another vertex on the x-axis. And I've given them coordinates. So this one right over here is at zero, zero. This one over here, we're just saying that the x-coordinate is a, and so it's a comma zero."}, {"video_title": "Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And then I put another vertex on the x-axis. And I've given them coordinates. So this one right over here is at zero, zero. This one over here, we're just saying that the x-coordinate is a, and so it's a comma zero. And then this one up here has some x and some y-coordinate. We're just calling them b and c. This is some arbitrary triangle. And if you had some other triangle that had the same dimensions as this one, and this one can have any dimensions because we haven't defined a, b, and c, you could go from this triangle to any of those other ones using rigid transformations."}, {"video_title": "Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3", "Sentence": "This one over here, we're just saying that the x-coordinate is a, and so it's a comma zero. And then this one up here has some x and some y-coordinate. We're just calling them b and c. This is some arbitrary triangle. And if you had some other triangle that had the same dimensions as this one, and this one can have any dimensions because we haven't defined a, b, and c, you could go from this triangle to any of those other ones using rigid transformations. So if we can prove that the medians of this triangle, this general triangle, always intersect at one point, this will be true for all triangles. So let's do a little bit more. Let's draw the medians."}, {"video_title": "Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And if you had some other triangle that had the same dimensions as this one, and this one can have any dimensions because we haven't defined a, b, and c, you could go from this triangle to any of those other ones using rigid transformations. So if we can prove that the medians of this triangle, this general triangle, always intersect at one point, this will be true for all triangles. So let's do a little bit more. Let's draw the medians. So what we're going to do is draw lines from each of the vertex to the midpoint of the opposite side. So if we do that, we've drawn all the medians, and it for sure looks like they intersect at one point, but to prove that, let's think about what the coordinates are of the midpoints of each of these sides. So what is the coordinate right over here?"}, {"video_title": "Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Let's draw the medians. So what we're going to do is draw lines from each of the vertex to the midpoint of the opposite side. So if we do that, we've drawn all the medians, and it for sure looks like they intersect at one point, but to prove that, let's think about what the coordinates are of the midpoints of each of these sides. So what is the coordinate right over here? Pause this video and think about that. Well, this is going to be the midpoint of this top point and this bottom right point. So this length is equal to that length."}, {"video_title": "Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So what is the coordinate right over here? Pause this video and think about that. Well, this is going to be the midpoint of this top point and this bottom right point. So this length is equal to that length. And for the midpoint, you can really just think about it as you're taking the average of each of the coordinates. So the x-coordinate here is going to be the average of b and a. So you could just write that as a plus b over two."}, {"video_title": "Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So this length is equal to that length. And for the midpoint, you can really just think about it as you're taking the average of each of the coordinates. So the x-coordinate here is going to be the average of b and a. So you could just write that as a plus b over two. And then the y-coordinate is going to be the average of c and zero. That would be c plus zero over two, or just c over two. And we could do that for each of these points."}, {"video_title": "Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So you could just write that as a plus b over two. And then the y-coordinate is going to be the average of c and zero. That would be c plus zero over two, or just c over two. And we could do that for each of these points. So this point right over here, its x-coordinate is going to be the average of zero and a. So that's just a over two. And its y-coordinate is going to be the average of zero and zero."}, {"video_title": "Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And we could do that for each of these points. So this point right over here, its x-coordinate is going to be the average of zero and a. So that's just a over two. And its y-coordinate is going to be the average of zero and zero. You can see that it sits on the x-axis, so its y-coordinate is zero. And then last but not least, what's the coordinate of this point? Pause the video and try to figure that out."}, {"video_title": "Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And its y-coordinate is going to be the average of zero and zero. You can see that it sits on the x-axis, so its y-coordinate is zero. And then last but not least, what's the coordinate of this point? Pause the video and try to figure that out. All right, well, the x-coordinate is going to be the average of b and zero, which is just going to be b over two. And then the y-coordinate is going to be the average of c and zero, which is just going to be c over two. So the way that I'm gonna prove that all three of these medians intersect at a unique point is by showing you a coordinate that sits on all three lines."}, {"video_title": "Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Pause the video and try to figure that out. All right, well, the x-coordinate is going to be the average of b and zero, which is just going to be b over two. And then the y-coordinate is going to be the average of c and zero, which is just going to be c over two. So the way that I'm gonna prove that all three of these medians intersect at a unique point is by showing you a coordinate that sits on all three lines. If it sits on all three lines, that must be the point of intersection. And that interesting point is 2 3rds along the way of any one of the medians. So one way to think about it is the distance between the vertex and that point is 2 3rds of the length of the median."}, {"video_title": "Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So the way that I'm gonna prove that all three of these medians intersect at a unique point is by showing you a coordinate that sits on all three lines. If it sits on all three lines, that must be the point of intersection. And that interesting point is 2 3rds along the way of any one of the medians. So one way to think about it is the distance between the vertex and that point is 2 3rds of the length of the median. So if we just look at this blue median, the coordinate of this point that is twice as far away from the vertex as it is from the opposite side, it will be based on a weighted average of the x and y-coordinates. When we did a midpoint and things were equally far away, you equally weighted the coordinates. So you just took their average."}, {"video_title": "Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So one way to think about it is the distance between the vertex and that point is 2 3rds of the length of the median. So if we just look at this blue median, the coordinate of this point that is twice as far away from the vertex as it is from the opposite side, it will be based on a weighted average of the x and y-coordinates. When we did a midpoint and things were equally far away, you equally weighted the coordinates. So you just took their average. But if you are closer to this side, you will take a weighted average accordingly. So it's going to be 2 3rds times a plus b over two plus 1 3rd times zero. And then the y-coordinate is going to be 2 3rds times c over two plus 1 3rd times the y-coordinate here, which is just going to be zero."}, {"video_title": "Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So you just took their average. But if you are closer to this side, you will take a weighted average accordingly. So it's going to be 2 3rds times a plus b over two plus 1 3rd times zero. And then the y-coordinate is going to be 2 3rds times c over two plus 1 3rd times the y-coordinate here, which is just going to be zero. Now once again, why do we have this 2 3rd and this 1 3rd weighting? Because we are twice as close to this point as we are to that point. Now if we wanted to simplify it, what would we get?"}, {"video_title": "Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3", "Sentence": "And then the y-coordinate is going to be 2 3rds times c over two plus 1 3rd times the y-coordinate here, which is just going to be zero. Now once again, why do we have this 2 3rd and this 1 3rd weighting? Because we are twice as close to this point as we are to that point. Now if we wanted to simplify it, what would we get? So this two would cancel with that two, so and this is zero, so we would get a plus b over three for the x-coordinate. And for the y-coordinate, this is zero. That two cancels with that two, and we get c over three."}, {"video_title": "Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Now if we wanted to simplify it, what would we get? So this two would cancel with that two, so and this is zero, so we would get a plus b over three for the x-coordinate. And for the y-coordinate, this is zero. That two cancels with that two, and we get c over three. So we just found a point that for sure sits on this blue median. Now let's do a similar exercise with this pink-colored median. So that pink-colored median, what is the coordinate of the point that sits on that median that is twice as far from the vertex as it is from the opposite side?"}, {"video_title": "Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3", "Sentence": "That two cancels with that two, and we get c over three. So we just found a point that for sure sits on this blue median. Now let's do a similar exercise with this pink-colored median. So that pink-colored median, what is the coordinate of the point that sits on that median that is twice as far from the vertex as it is from the opposite side? Well, it would be the exact same exercise. We would doubly weight these coordinates, so the x-coordinate would be 2 3rds times b over two plus 1 3rd times a, 1 3rd times a, and then the y-coordinate would be 2 3rds times c over two, 2 3rds times c over two plus 1 3rd times zero, 1 3rd times zero, and what does that get us? Well, let's see, this two cancels out with that two, and we are left with b over three plus a over three, so that's the same thing as a plus b over three, and over here, that's zero, that cancels out."}, {"video_title": "Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3", "Sentence": "So that pink-colored median, what is the coordinate of the point that sits on that median that is twice as far from the vertex as it is from the opposite side? Well, it would be the exact same exercise. We would doubly weight these coordinates, so the x-coordinate would be 2 3rds times b over two plus 1 3rd times a, 1 3rd times a, and then the y-coordinate would be 2 3rds times c over two, 2 3rds times c over two plus 1 3rd times zero, 1 3rd times zero, and what does that get us? Well, let's see, this two cancels out with that two, and we are left with b over three plus a over three, so that's the same thing as a plus b over three, and over here, that's zero, that cancels out. That is c over three. So notice, this exact coordinate sits on both the blue median and this pink median, so that must be the place that they intersect. Let's see if that's also true for this orange median."}, {"video_title": "Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Well, let's see, this two cancels out with that two, and we are left with b over three plus a over three, so that's the same thing as a plus b over three, and over here, that's zero, that cancels out. That is c over three. So notice, this exact coordinate sits on both the blue median and this pink median, so that must be the place that they intersect. Let's see if that's also true for this orange median. So in the orange median, same exact exercise, and I encourage you to pause this video and try to calculate the value of this point on your own. What's the coordinate of that point on the orange line where this distance is twice as large as this distance? Well, same idea, we will doubly weight these points right over here, so the x coordinate would be 2 3rds times a over two plus 1 3rd times b."}, {"video_title": "Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Let's see if that's also true for this orange median. So in the orange median, same exact exercise, and I encourage you to pause this video and try to calculate the value of this point on your own. What's the coordinate of that point on the orange line where this distance is twice as large as this distance? Well, same idea, we will doubly weight these points right over here, so the x coordinate would be 2 3rds times a over two plus 1 3rd times b. The y coordinate would be 2 3rds times zero, 2 3rds times zero plus 1 3rd times c, 1 3rd times c, and what does that get us? Let's see, that cancels with that, so we have a over three plus b over three. Well, that's the same thing as a plus b over three, and then over here, that's just zero, and we're left with c over three."}, {"video_title": "Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3", "Sentence": "Well, same idea, we will doubly weight these points right over here, so the x coordinate would be 2 3rds times a over two plus 1 3rd times b. The y coordinate would be 2 3rds times zero, 2 3rds times zero plus 1 3rd times c, 1 3rd times c, and what does that get us? Let's see, that cancels with that, so we have a over three plus b over three. Well, that's the same thing as a plus b over three, and then over here, that's just zero, and we're left with c over three. So notice, we have just shown that this exact coordinate sits on all three medians, and so therefore, all three medians must intersect at that point because that point exists on all the lines. We've just shown that, and that's true for this arbitrary triangle. You can make this triangle have arbitrary dimensions by changing your value for a, b, or c, and if you see a triangle that has the same dimensions but it's shifted or it's in a different orientation, you can do a rigid transformation which doesn't change any of the dimensions, and you can show that that would be true for that triangle as well."}, {"video_title": "Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3", "Sentence": "And the point is called the focus of the parabola, and the line is called the directrix of the parabola. What I want to do in this video, it's gonna get a little bit of hairy algebra, but given that definition, I want to see, and given that definition and given a focus at the point x equals a, y equals b, and align a directrix at y equals k to figure out what is the equation of that parabola actually going to be. And it's going to be based on a's, b's, and k's. So let's do that. So let's take a arbitrary point on the parabola. Let's say we take this point right over here, and its x-coordinate is x and its y-coordinate is y. And by definition, in order for this to be a parabola, it has to be equidistant to its focus and its directrix."}, {"video_title": "Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3", "Sentence": "So let's do that. So let's take a arbitrary point on the parabola. Let's say we take this point right over here, and its x-coordinate is x and its y-coordinate is y. And by definition, in order for this to be a parabola, it has to be equidistant to its focus and its directrix. So what does that mean? That means that the distance to the directrix, which I'm drawing here in blue, has to be the same as the distance to the focus, which I am drawing in magenta. And when we take the distance to the directrix, we literally just drop a perpendicular, I guess you could say."}, {"video_title": "Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3", "Sentence": "And by definition, in order for this to be a parabola, it has to be equidistant to its focus and its directrix. So what does that mean? That means that the distance to the directrix, which I'm drawing here in blue, has to be the same as the distance to the focus, which I am drawing in magenta. And when we take the distance to the directrix, we literally just drop a perpendicular, I guess you could say. That is the, that's going to be the shortest distance to that line. And then when we, but the distance to the focus, well that's, we see that's at a bit of an angle, and we might have to use the distance formula, which is really just the Pythagorean theorem. So let's do that."}, {"video_title": "Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3", "Sentence": "And when we take the distance to the directrix, we literally just drop a perpendicular, I guess you could say. That is the, that's going to be the shortest distance to that line. And then when we, but the distance to the focus, well that's, we see that's at a bit of an angle, and we might have to use the distance formula, which is really just the Pythagorean theorem. So let's do that. This distance has to be the same as that distance. So what's this blue distance? Well that's just going to be our change in y."}, {"video_title": "Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3", "Sentence": "So let's do that. This distance has to be the same as that distance. So what's this blue distance? Well that's just going to be our change in y. It's going to be this y, it's going to be this y minus k. It's just this distance. So it's going to be y minus k. Now we have to be careful. The way I've just drawn it, yes, y is greater, y is greater than k, so this is going to give us a positive value, and you need a non-negative value if you're talking about distances."}, {"video_title": "Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3", "Sentence": "Well that's just going to be our change in y. It's going to be this y, it's going to be this y minus k. It's just this distance. So it's going to be y minus k. Now we have to be careful. The way I've just drawn it, yes, y is greater, y is greater than k, so this is going to give us a positive value, and you need a non-negative value if you're talking about distances. But there are, you can definitely have a parabola where the y coordinate of the focus is lower than the y coordinate of the directrix, in which case this would be negative. So what we really want is the absolute value of this, or you could say we could square it, and then we could take the square root, the principal root, which would be equivalent to taking the absolute value of y minus k. So that's this distance right over here. And by the definition of a parabola, in order for x comma y to be sitting on the parabola, that distance needs to be the same as the distance from x comma y to a comma b, to the focus."}, {"video_title": "Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3", "Sentence": "The way I've just drawn it, yes, y is greater, y is greater than k, so this is going to give us a positive value, and you need a non-negative value if you're talking about distances. But there are, you can definitely have a parabola where the y coordinate of the focus is lower than the y coordinate of the directrix, in which case this would be negative. So what we really want is the absolute value of this, or you could say we could square it, and then we could take the square root, the principal root, which would be equivalent to taking the absolute value of y minus k. So that's this distance right over here. And by the definition of a parabola, in order for x comma y to be sitting on the parabola, that distance needs to be the same as the distance from x comma y to a comma b, to the focus. So what's that going to be? Well, we just apply the distance formula, or really just the Pythagorean theorem. It's going to be our change in x, so x minus a squared plus the change in y, y minus b squared, and the square root of that whole thing."}, {"video_title": "Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3", "Sentence": "And by the definition of a parabola, in order for x comma y to be sitting on the parabola, that distance needs to be the same as the distance from x comma y to a comma b, to the focus. So what's that going to be? Well, we just apply the distance formula, or really just the Pythagorean theorem. It's going to be our change in x, so x minus a squared plus the change in y, y minus b squared, and the square root of that whole thing. The square root of all of that business. Now, this right over here is an equation of a parabola. It doesn't look like it."}, {"video_title": "Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3", "Sentence": "It's going to be our change in x, so x minus a squared plus the change in y, y minus b squared, and the square root of that whole thing. The square root of all of that business. Now, this right over here is an equation of a parabola. It doesn't look like it. It looks really hairy, but it is the equation of a parabola. And to show you that, we just have to simplify this. And if you get inspired, I encourage you to try to simplify this on your own."}, {"video_title": "Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3", "Sentence": "It doesn't look like it. It looks really hairy, but it is the equation of a parabola. And to show you that, we just have to simplify this. And if you get inspired, I encourage you to try to simplify this on your own. It's just going to be a little bit of hairy algebra, but it really is not too bad. You're going to get an equation for a parabola that you might recognize, and it's going to be in terms of kind of a general focus, a comma b, and a general directrix, y equals k. So let's do that. So the simplest thing to start here is let's just square both sides so we get rid of the radicals."}, {"video_title": "Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3", "Sentence": "And if you get inspired, I encourage you to try to simplify this on your own. It's just going to be a little bit of hairy algebra, but it really is not too bad. You're going to get an equation for a parabola that you might recognize, and it's going to be in terms of kind of a general focus, a comma b, and a general directrix, y equals k. So let's do that. So the simplest thing to start here is let's just square both sides so we get rid of the radicals. So if you square both sides, on the left-hand side, you're going to get y minus k squared. You're going to get y minus k squared is equal to, is equal to x minus a squared plus y minus b, plus y minus b squared. Fair enough."}, {"video_title": "Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3", "Sentence": "So the simplest thing to start here is let's just square both sides so we get rid of the radicals. So if you square both sides, on the left-hand side, you're going to get y minus k squared. You're going to get y minus k squared is equal to, is equal to x minus a squared plus y minus b, plus y minus b squared. Fair enough. Now, what I want to do is I just want to end up with just a y on the left-hand side and just x's, a, b's, and k's on the right-hand side. So the first thing I might want to do is let's expand each of these expressions that involve with y. So this blue one on the left-hand side, that is going to be, that's going to be y squared minus two yk plus k squared, and that is going to be equal to, I'm going to keep this first one the same, so it's going to be x minus a squared."}, {"video_title": "Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3", "Sentence": "Fair enough. Now, what I want to do is I just want to end up with just a y on the left-hand side and just x's, a, b's, and k's on the right-hand side. So the first thing I might want to do is let's expand each of these expressions that involve with y. So this blue one on the left-hand side, that is going to be, that's going to be y squared minus two yk plus k squared, and that is going to be equal to, I'm going to keep this first one the same, so it's going to be x minus a squared. And now let me expand, let me expand, I'm going to find a color, expand this in green, so plus y squared minus two yb plus b squared. All I did is I multiplied y minus b times y minus b. Now let's see if we can simplify things."}, {"video_title": "Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3", "Sentence": "So this blue one on the left-hand side, that is going to be, that's going to be y squared minus two yk plus k squared, and that is going to be equal to, I'm going to keep this first one the same, so it's going to be x minus a squared. And now let me expand, let me expand, I'm going to find a color, expand this in green, so plus y squared minus two yb plus b squared. All I did is I multiplied y minus b times y minus b. Now let's see if we can simplify things. So I have a y squared on the left, I have a y squared on the right. Well, if I subtract y squared from both sides, so I can do that. Well, that simplified things a little bit."}, {"video_title": "Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3", "Sentence": "Now let's see if we can simplify things. So I have a y squared on the left, I have a y squared on the right. Well, if I subtract y squared from both sides, so I can do that. Well, that simplified things a little bit. And now I can, I can, let's see what I can do. So let's get the k squared on this side, so let's subtract k squared from both sides. So subtract k squared, subtract k squared from both sides, so that's going to get rid of it on the left-hand side."}, {"video_title": "Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3", "Sentence": "Well, that simplified things a little bit. And now I can, I can, let's see what I can do. So let's get the k squared on this side, so let's subtract k squared from both sides. So subtract k squared, subtract k squared from both sides, so that's going to get rid of it on the left-hand side. And now let's add two yb to both sides, so we have all the y's on the left-hand side. So plus two yb, that's going to move, that's going to give us a two yb on the left-hand side, plus two yb. So what is this going to be equal to?"}, {"video_title": "Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3", "Sentence": "So subtract k squared, subtract k squared from both sides, so that's going to get rid of it on the left-hand side. And now let's add two yb to both sides, so we have all the y's on the left-hand side. So plus two yb, that's going to move, that's going to give us a two yb on the left-hand side, plus two yb. So what is this going to be equal to? I'm starting to run into my graph, so let me give myself a little bit more real estate over here. So on the left-hand side, what am I going to have? This is the same thing as two yb minus two yk, which is the same thing, actually let me just write that down."}, {"video_title": "Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3", "Sentence": "So what is this going to be equal to? I'm starting to run into my graph, so let me give myself a little bit more real estate over here. So on the left-hand side, what am I going to have? This is the same thing as two yb minus two yk, which is the same thing, actually let me just write that down. That's going to be two y, do it in green. Actually, well, yeah, why not green? That's going to be, actually let me start a new color."}, {"video_title": "Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3", "Sentence": "This is the same thing as two yb minus two yk, which is the same thing, actually let me just write that down. That's going to be two y, do it in green. Actually, well, yeah, why not green? That's going to be, actually let me start a new color. That's going to be two yb minus two yk. You can factor out a two y, and it's going to be two y times b minus k. So let's do that. So we could write this as two, or we could write it two times b minus ky, if you factor out a two and a y."}, {"video_title": "Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3", "Sentence": "That's going to be, actually let me start a new color. That's going to be two yb minus two yk. You can factor out a two y, and it's going to be two y times b minus k. So let's do that. So we could write this as two, or we could write it two times b minus ky, if you factor out a two and a y. So that's the left-hand side, so that's that piece right over there. These things cancel out. Now on our right-hand side, I promised you a little bit of hairy algebra, so hopefully you see that I'm delivering on that promise."}, {"video_title": "Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3", "Sentence": "So we could write this as two, or we could write it two times b minus ky, if you factor out a two and a y. So that's the left-hand side, so that's that piece right over there. These things cancel out. Now on our right-hand side, I promised you a little bit of hairy algebra, so hopefully you see that I'm delivering on that promise. On the right-hand side, you have x minus a squared, and then, let's see, these characters cancel out, and you're left with b squared minus k squared. So these two are going to be b squared minus k squared, plus b squared minus k squared. Now, I said all I want is a y on the left-hand side, so let's divide everything by two times b minus k. So let's divide everything times two times b minus k. So two times b minus k, and I'm actually going to divide this whole thing by two times b minus k. Now, obviously, on the left-hand side, this all cancels out."}, {"video_title": "Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3", "Sentence": "Now on our right-hand side, I promised you a little bit of hairy algebra, so hopefully you see that I'm delivering on that promise. On the right-hand side, you have x minus a squared, and then, let's see, these characters cancel out, and you're left with b squared minus k squared. So these two are going to be b squared minus k squared, plus b squared minus k squared. Now, I said all I want is a y on the left-hand side, so let's divide everything by two times b minus k. So let's divide everything times two times b minus k. So two times b minus k, and I'm actually going to divide this whole thing by two times b minus k. Now, obviously, on the left-hand side, this all cancels out. You're left with just a y, and then it's going to be y equals, y is equal to one over two times b minus k, and notice b minus k is the difference between the y-coordinate of the focus and the y-coordinate, I guess you could say, of the line, y equals k. So it's one over two times that, times x minus a squared. x minus a squared. So if you knew what b minus k was, this would just simplify to some number, some number that's being multiplied times x minus a squared."}, {"video_title": "Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3", "Sentence": "Now, I said all I want is a y on the left-hand side, so let's divide everything by two times b minus k. So let's divide everything times two times b minus k. So two times b minus k, and I'm actually going to divide this whole thing by two times b minus k. Now, obviously, on the left-hand side, this all cancels out. You're left with just a y, and then it's going to be y equals, y is equal to one over two times b minus k, and notice b minus k is the difference between the y-coordinate of the focus and the y-coordinate, I guess you could say, of the line, y equals k. So it's one over two times that, times x minus a squared. x minus a squared. So if you knew what b minus k was, this would just simplify to some number, some number that's being multiplied times x minus a squared. So hopefully this is starting to look like the parabolas that you remember from your childhood, if you do remember parabolas from your childhood. All right, so then let's see if we could simplify this thing on the right, and you might recognize b squared minus k squared, that's the difference of squares. That's the same thing as, that's the same thing as b plus k times b minus k. Times b, whoops, times b minus k. So the b minus k's cancel out, and we are just left with, and we deserve a little bit of a drum roll, we're just left with 1 1 2 times b plus k. So there you go."}, {"video_title": "Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3", "Sentence": "So if you knew what b minus k was, this would just simplify to some number, some number that's being multiplied times x minus a squared. So hopefully this is starting to look like the parabolas that you remember from your childhood, if you do remember parabolas from your childhood. All right, so then let's see if we could simplify this thing on the right, and you might recognize b squared minus k squared, that's the difference of squares. That's the same thing as, that's the same thing as b plus k times b minus k. Times b, whoops, times b minus k. So the b minus k's cancel out, and we are just left with, and we deserve a little bit of a drum roll, we're just left with 1 1 2 times b plus k. So there you go. Given a focus at the point a comma b, and a directrix at y equals k, we now know what the formula of the parabola is actually going to be. So for example, so for example, if I had a focus, if I had a focus at the point, I don't know, let's say the point one comma two, and I had a directrix, I had a directrix at y, y is equal to, I don't know, let's make it y is equal to negative one, what would the formula, or what would the equation of this parabola be? Well, it would be y is equal to one over two times b minus k. So two minus negative one, that's the same thing as two plus one, so that's two times, that's just three, two minus negative one is three, times x minus, times x minus one, whoops, x minus one squared plus 1 1 2 times b plus k. Two plus negative one is one."}, {"video_title": "Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3", "Sentence": "That's the same thing as, that's the same thing as b plus k times b minus k. Times b, whoops, times b minus k. So the b minus k's cancel out, and we are just left with, and we deserve a little bit of a drum roll, we're just left with 1 1 2 times b plus k. So there you go. Given a focus at the point a comma b, and a directrix at y equals k, we now know what the formula of the parabola is actually going to be. So for example, so for example, if I had a focus, if I had a focus at the point, I don't know, let's say the point one comma two, and I had a directrix, I had a directrix at y, y is equal to, I don't know, let's make it y is equal to negative one, what would the formula, or what would the equation of this parabola be? Well, it would be y is equal to one over two times b minus k. So two minus negative one, that's the same thing as two plus one, so that's two times, that's just three, two minus negative one is three, times x minus, times x minus one, whoops, x minus one squared plus 1 1 2 times b plus k. Two plus negative one is one. So one. And so what's this going to be? You're going to get y is equal to one over 1 6th x minus one squared plus 1 1 2."}, {"video_title": "Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3", "Sentence": "Well, it would be y is equal to one over two times b minus k. So two minus negative one, that's the same thing as two plus one, so that's two times, that's just three, two minus negative one is three, times x minus, times x minus one, whoops, x minus one squared plus 1 1 2 times b plus k. Two plus negative one is one. So one. And so what's this going to be? You're going to get y is equal to one over 1 6th x minus one squared plus 1 1 2. There you go. That is the parabola with a focus at one comma two and a directrix at y equals negative one. Fascinating."}, {"video_title": "Graphing circles from features Mathematics II High School Math Khan Academy.mp3", "Sentence": "And it's a pretty neat little widget here, because what I can do is I can take this dot and I can move it around to redefine the center of the circle. So it's centered at 3, negative 2. So x is 3 and y is negative 2, so that's its center. It has to have a radius of 5. The way it's drawn right now, it has a radius of 1. The distance between the center and the actual circle, the points that define the circle, right now it's 1. I need to make this radius equal to 5."}, {"video_title": "Graphing circles from features Mathematics II High School Math Khan Academy.mp3", "Sentence": "It has to have a radius of 5. The way it's drawn right now, it has a radius of 1. The distance between the center and the actual circle, the points that define the circle, right now it's 1. I need to make this radius equal to 5. So let's see if I take that. So now the radius is equal to 2, 3, 4, and 5. There you go."}, {"video_title": "Graphing circles from features Mathematics II High School Math Khan Academy.mp3", "Sentence": "I need to make this radius equal to 5. So let's see if I take that. So now the radius is equal to 2, 3, 4, and 5. There you go. Centered at 3, negative 2, radius of 5. Notice, go from the center to the actual circle, it's 5, no matter where you go. Let's do one more of these."}, {"video_title": "Graphing circles from features Mathematics II High School Math Khan Academy.mp3", "Sentence": "There you go. Centered at 3, negative 2, radius of 5. Notice, go from the center to the actual circle, it's 5, no matter where you go. Let's do one more of these. Graph the circle which is centered at negative 4, 1 and which has the point 0, 4 on it. So once again, let's drag the center. So it's going to be negative 4, x is negative 4, y is 1, so that's the center."}, {"video_title": "Graphing circles from features Mathematics II High School Math Khan Academy.mp3", "Sentence": "Let's do one more of these. Graph the circle which is centered at negative 4, 1 and which has the point 0, 4 on it. So once again, let's drag the center. So it's going to be negative 4, x is negative 4, y is 1, so that's the center. It has the point 0, 4 on it. So x is 0, y is 4. So I have to drag, I have to increase the radius of the circle."}, {"video_title": "Graphing circles from features Mathematics II High School Math Khan Academy.mp3", "Sentence": "So it's going to be negative 4, x is negative 4, y is 1, so that's the center. It has the point 0, 4 on it. So x is 0, y is 4. So I have to drag, I have to increase the radius of the circle. Let's see, whoops, nope, I want to make sure I don't change the center. I want to increase the radius of the circle until it includes this point right over here, 0, 4. So I'm not there quite yet."}, {"video_title": "Graphing circles from features Mathematics II High School Math Khan Academy.mp3", "Sentence": "So I have to drag, I have to increase the radius of the circle. Let's see, whoops, nope, I want to make sure I don't change the center. I want to increase the radius of the circle until it includes this point right over here, 0, 4. So I'm not there quite yet. There you go. I am now including the point 0, 4. And if we're curious what the radius is, we could just go along the x-axis."}, {"video_title": "Graphing circles from features Mathematics II High School Math Khan Academy.mp3", "Sentence": "So I'm not there quite yet. There you go. I am now including the point 0, 4. And if we're curious what the radius is, we could just go along the x-axis. x equals negative 4 is the x-coordinate for the center and we see that this point, so this is negative 4, 1 and we see that 1, 1 is actually on the circle. So the distance here is, you go 4 and then another one, it's 5. So this has a radius of 5."}, {"video_title": "Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3", "Sentence": "So first of all, there's this idea of rigid transformations, which we've talked about in other videos. But just as a refresher, these are transformations that preserve distance between points. So for example, if I have points A and B, a rigid transformation could be something like a translation because after I've translated them, notice the distance between the points is still the same. It could be like that. It includes rotation. Let's say I rotated about point A as the center of rotation. That still would not change, that still would not change my distance between points A and B."}, {"video_title": "Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3", "Sentence": "It could be like that. It includes rotation. Let's say I rotated about point A as the center of rotation. That still would not change, that still would not change my distance between points A and B. It could even be things like taking the mirror image. Once again, that's not going to change the distance between A and B. What's not a rigid transformation?"}, {"video_title": "Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3", "Sentence": "That still would not change, that still would not change my distance between points A and B. It could even be things like taking the mirror image. Once again, that's not going to change the distance between A and B. What's not a rigid transformation? Well, one thing you might imagine is dilating, scaling it up or down. That is going to change the distance. So rigid transformation are any transformations that preserve the distance between points."}, {"video_title": "Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3", "Sentence": "What's not a rigid transformation? Well, one thing you might imagine is dilating, scaling it up or down. That is going to change the distance. So rigid transformation are any transformations that preserve the distance between points. Now another idea is congruence. And in the context of this video, we're going to be viewing the definition of congruence as two figures are congruent if and only if there exists a series of rigid transformations which will map one figure on to the other. You might see other definitions of congruence in your life, but this is the rigid transformation definition of congruence that we will use."}, {"video_title": "Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3", "Sentence": "So rigid transformation are any transformations that preserve the distance between points. Now another idea is congruence. And in the context of this video, we're going to be viewing the definition of congruence as two figures are congruent if and only if there exists a series of rigid transformations which will map one figure on to the other. You might see other definitions of congruence in your life, but this is the rigid transformation definition of congruence that we will use. And we're going to use these two definitions to prove the following. To prove that saying two segments are congruent is equivalent to saying that they have the same length. So let me get some space here to do that in."}, {"video_title": "Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3", "Sentence": "You might see other definitions of congruence in your life, but this is the rigid transformation definition of congruence that we will use. And we're going to use these two definitions to prove the following. To prove that saying two segments are congruent is equivalent to saying that they have the same length. So let me get some space here to do that in. So first let me prove that if segment AB is congruent to segment CD, then the length of segment AB, which we'll just denote as AB without the line over it, is equal to the length of segment CD. How do we do that? Well, the first thing to realize is if AB, if AB is congruent to CD, then AB can be mapped onto CD with rigid transformations, rigid transformations."}, {"video_title": "Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3", "Sentence": "So let me get some space here to do that in. So first let me prove that if segment AB is congruent to segment CD, then the length of segment AB, which we'll just denote as AB without the line over it, is equal to the length of segment CD. How do we do that? Well, the first thing to realize is if AB, if AB is congruent to CD, then AB can be mapped onto CD with rigid transformations, rigid transformations. That comes out of the definition of congruence. And then we could say since the transformations are rigid, distance is preserved, preserved, and so that would imply that the distance between the points are going to be the same. AB, the distance between points AB or the length of segment AB is equal to the length of segment CD."}, {"video_title": "Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3", "Sentence": "Well, the first thing to realize is if AB, if AB is congruent to CD, then AB can be mapped onto CD with rigid transformations, rigid transformations. That comes out of the definition of congruence. And then we could say since the transformations are rigid, distance is preserved, preserved, and so that would imply that the distance between the points are going to be the same. AB, the distance between points AB or the length of segment AB is equal to the length of segment CD. That might almost seem too intuitive for you, but that's all we're talking about. So now let's see if we can prove the other way. Let's see if we can prove that if the length of segment AB is equal to the length of segment CD, then segment AB is congruent to segment CD."}, {"video_title": "Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3", "Sentence": "AB, the distance between points AB or the length of segment AB is equal to the length of segment CD. That might almost seem too intuitive for you, but that's all we're talking about. So now let's see if we can prove the other way. Let's see if we can prove that if the length of segment AB is equal to the length of segment CD, then segment AB is congruent to segment CD. And let me draw them right over here just to, so let's say I have segment AB right over there, and I'll draw another segment that has the same length. So maybe it looks something like this, and this is obviously hand-drawn. So then let's call this CD."}, {"video_title": "Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3", "Sentence": "Let's see if we can prove that if the length of segment AB is equal to the length of segment CD, then segment AB is congruent to segment CD. And let me draw them right over here just to, so let's say I have segment AB right over there, and I'll draw another segment that has the same length. So maybe it looks something like this, and this is obviously hand-drawn. So then let's call this CD. So in order to prove this, I have to show, hey, if I have two segments with the same length, that there's always a set of rigid transformations that will map one segment onto the other, which means by definition they are congruent. So let me just construct those transformations. So my first rigid transformation that I could do is to translate, translate, and I'll underline the name of the transformation, segment AB so that point A is on top of point C, or A is mapped onto C. And you could see that there's always a translation to do that."}, {"video_title": "Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3", "Sentence": "So then let's call this CD. So in order to prove this, I have to show, hey, if I have two segments with the same length, that there's always a set of rigid transformations that will map one segment onto the other, which means by definition they are congruent. So let me just construct those transformations. So my first rigid transformation that I could do is to translate, translate, and I'll underline the name of the transformation, segment AB so that point A is on top of point C, or A is mapped onto C. And you could see that there's always a translation to do that. It would be doing that. And of course we would translate, B would end up like that. And so after this translation, it's going to be A right over there."}, {"video_title": "Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3", "Sentence": "So my first rigid transformation that I could do is to translate, translate, and I'll underline the name of the transformation, segment AB so that point A is on top of point C, or A is mapped onto C. And you could see that there's always a translation to do that. It would be doing that. And of course we would translate, B would end up like that. And so after this translation, it's going to be A right over there. A is going to be there. And then B is going to be right over there. Now the second step I would do is then rotate AB about A, so A is the center of rotation, so I'm gonna rotate it, so that point B lies on ray CD."}, {"video_title": "Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3", "Sentence": "And so after this translation, it's going to be A right over there. A is going to be there. And then B is going to be right over there. Now the second step I would do is then rotate AB about A, so A is the center of rotation, so I'm gonna rotate it, so that point B lies on ray CD. Well, what does this transformation do? Well, since point A is the center of rotation, A is going to stay mapped on top of C from our first translation, but now B is rotated, so it sits on top of the ray that starts at C and goes through D and keeps going. And where will B be on that ray?"}, {"video_title": "Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3", "Sentence": "Now the second step I would do is then rotate AB about A, so A is the center of rotation, so I'm gonna rotate it, so that point B lies on ray CD. Well, what does this transformation do? Well, since point A is the center of rotation, A is going to stay mapped on top of C from our first translation, but now B is rotated, so it sits on top of the ray that starts at C and goes through D and keeps going. And where will B be on that ray? Well, since the distance between B and A is the same as the distance between D and C, and A and C are at the same point, and now B sits on that ray, B will now sit right on top of D because AB is equal to CD, B will be mapped onto D. And just like that, we've shown that if the segment lengths are equal, there is always a set of rigid transformations that will map one segment onto the other. Therefore, since A and B have been mapped onto C and D, we know that segment AB is congruent to segment CD, and we are done. We have proven what we set out to prove both ways."}, {"video_title": "Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3", "Sentence": "And where will B be on that ray? Well, since the distance between B and A is the same as the distance between D and C, and A and C are at the same point, and now B sits on that ray, B will now sit right on top of D because AB is equal to CD, B will be mapped onto D. And just like that, we've shown that if the segment lengths are equal, there is always a set of rigid transformations that will map one segment onto the other. Therefore, since A and B have been mapped onto C and D, we know that segment AB is congruent to segment CD, and we are done. We have proven what we set out to prove both ways."}, {"video_title": "Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3", "Sentence": "We have proven what we set out to prove both ways."}]