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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. | Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3 |
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. | Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3 |
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. | Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3 |
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. | Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3 |
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. | Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3 |
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. | Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3 |
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. | Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3 |
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. | Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3 |
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. | Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3 |
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. | Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3 |
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. | Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3 |
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. | Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3 |
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. | Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3 |
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. | Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3 |
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. | Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3 |
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. | Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3 |
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. | Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3 |
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. | Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3 |
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. | Area of a regular hexagon Right triangles and trigonometry Geometry Khan Academy.mp3 |
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. | Proving the SSS triangle congruence criterion using transformations Geometry Khan Academy.mp3 |
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? | Proving the SSS triangle congruence criterion using transformations Geometry Khan Academy.mp3 |
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. | Proving the SSS triangle congruence criterion using transformations Geometry Khan Academy.mp3 |
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. | Proving the SSS triangle congruence criterion using transformations Geometry Khan Academy.mp3 |
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. | Proving the SSS triangle congruence criterion using transformations Geometry Khan Academy.mp3 |
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. | Proving the SSS triangle congruence criterion using transformations Geometry Khan Academy.mp3 |
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. | Proving the SSS triangle congruence criterion using transformations Geometry Khan Academy.mp3 |
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. | Proving the SSS triangle congruence criterion using transformations Geometry Khan Academy.mp3 |
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. | Proving the SSS triangle congruence criterion using transformations Geometry Khan Academy.mp3 |
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. | Proving the SSS triangle congruence criterion using transformations Geometry Khan Academy.mp3 |
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. | Proving the SSS triangle congruence criterion using transformations Geometry Khan Academy.mp3 |
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. | Proving the SSS triangle congruence criterion using transformations Geometry Khan Academy.mp3 |
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. | Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3 |
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. | Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3 |
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. | Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3 |
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. | Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3 |
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. | Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3 |
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. | Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3 |
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? | Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3 |
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. | Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3 |
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. | Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3 |
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. | Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3 |
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. | Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3 |
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. | Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3 |
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. | Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3 |
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. | Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3 |
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. | Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3 |
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. | Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3 |
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. | Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3 |
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. | Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3 |
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. | Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3 |
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. | Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3 |
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. | Area of trapezoid on the coordinate plane High School Math Khan Academy.mp3 |
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. | Measure of circumscribed angle Circles Geometry Khan Academy.mp3 |
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. | Measure of circumscribed angle Circles Geometry Khan Academy.mp3 |
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. | Measure of circumscribed angle Circles Geometry Khan Academy.mp3 |
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. | Measure of circumscribed angle Circles Geometry Khan Academy.mp3 |
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. | Measure of circumscribed angle Circles Geometry Khan Academy.mp3 |
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. | Measure of circumscribed angle Circles Geometry Khan Academy.mp3 |
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. | Measure of circumscribed angle Circles Geometry Khan Academy.mp3 |
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. | Measure of circumscribed angle Circles Geometry Khan Academy.mp3 |
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. | Measure of circumscribed angle Circles Geometry Khan Academy.mp3 |
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. | Measure of circumscribed angle Circles Geometry Khan Academy.mp3 |
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. | Measure of circumscribed angle Circles Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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? | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Congruent legs and base angles of isosceles triangles Congruence Geometry Khan Academy.mp3 |
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. | Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3 |
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. | Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3 |
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. | Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3 |
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. | Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3 |
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. | Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3 |
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. | Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3 |
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. | Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3 |