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Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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? |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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? |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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? |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Cylinder volume and surface area Perimeter, area, and volume Geometry Khan Academy.mp3 | 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. |
Proof perpendicular radius bisects chord.mp3 | 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. |
Proof perpendicular radius bisects chord.mp3 | 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. |
Proof perpendicular radius bisects chord.mp3 | 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. |
Proof perpendicular radius bisects chord.mp3 | 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. |
Proof perpendicular radius bisects chord.mp3 | 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. |
Proof perpendicular radius bisects chord.mp3 | 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. |
Proof perpendicular radius bisects chord.mp3 | 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. |
Proof perpendicular radius bisects chord.mp3 | 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. |
Proof perpendicular radius bisects chord.mp3 | 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. |
Proof perpendicular radius bisects chord.mp3 | 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. |
Proof perpendicular radius bisects chord.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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? |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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? |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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? |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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? |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Similarity postulates Similarity Geometry Khan Academy.mp3 | 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. |
Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3 | 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. |
Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3 | 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. |
Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3 | 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. |
Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3 | 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. |
Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3 | 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. |
Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3 | 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. |
Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3 | 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. |
Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3 | 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. |
Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3 | 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. |
Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3 | 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? |
Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3 | 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. |
Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3 | 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. |
Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3 | 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. |
Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3 | 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. |
Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3 | 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. |
Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3 | 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. |
Proof Radius is perpendicular to tangent line Mathematics II High School Math Khan Academy.mp3 | 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. |
Constructing a perpendicular line using a compass and straightedge Geometry Khan Academy.mp3 | 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. |
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