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So for example, the coordinate of the center here is for sure going to change. We go from the coordinate negative three comma zero to here we went to the coordinate negative one comma two. So the coordinates are not preserved, coordinates of the center. Let's do another example with a non-circular shape, and we'll do a different type of transformation. In this situation, let us do a reflection. So we have a quadrilateral here, quadrilateral ABCD. And we wanna think about what is preserved or not preserved as we do a reflection across the line L. So let me write that down.
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Properties perserved after rigid transformations.mp3
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Let's do another example with a non-circular shape, and we'll do a different type of transformation. In this situation, let us do a reflection. So we have a quadrilateral here, quadrilateral ABCD. And we wanna think about what is preserved or not preserved as we do a reflection across the line L. So let me write that down. We're gonna have a reflection in this situation. And we could even think about this without even doing the reflection ourselves, but let's just do the reflection really fast. So we're reflecting across the line X, Y is equal to X.
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Properties perserved after rigid transformations.mp3
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And we wanna think about what is preserved or not preserved as we do a reflection across the line L. So let me write that down. We're gonna have a reflection in this situation. And we could even think about this without even doing the reflection ourselves, but let's just do the reflection really fast. So we're reflecting across the line X, Y is equal to X. So what it essentially does to the coordinates is it swaps the X and Y coordinates, but you don't have to know that for the sake of this video. So B prime would be right over here, A prime would be right over there, D prime would be right over here. And since C is right on the line L, its image, C prime, won't change.
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Properties perserved after rigid transformations.mp3
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So we're reflecting across the line X, Y is equal to X. So what it essentially does to the coordinates is it swaps the X and Y coordinates, but you don't have to know that for the sake of this video. So B prime would be right over here, A prime would be right over there, D prime would be right over here. And since C is right on the line L, its image, C prime, won't change. And so our new, when we reflect over the line L, and you don't have to know for the sake of this video exactly how I did that fairly quickly. I really just want you to see what the reflection looks like. The real appreciation here is to think about, well, what happens with rigid transformations?
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Properties perserved after rigid transformations.mp3
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And since C is right on the line L, its image, C prime, won't change. And so our new, when we reflect over the line L, and you don't have to know for the sake of this video exactly how I did that fairly quickly. I really just want you to see what the reflection looks like. The real appreciation here is to think about, well, what happens with rigid transformations? So it's gonna look something like this. The reflection, the reflection looks something like this. So what's preserved?
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Properties perserved after rigid transformations.mp3
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The real appreciation here is to think about, well, what happens with rigid transformations? So it's gonna look something like this. The reflection, the reflection looks something like this. So what's preserved? And in general, this is good to know for any rigid transformation, what's preserved? Well, side lengths. That's actually one way that we even use to define what a rigid transformation is, a transformation that preserves the lengths between corresponding points.
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Properties perserved after rigid transformations.mp3
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So what's preserved? And in general, this is good to know for any rigid transformation, what's preserved? Well, side lengths. That's actually one way that we even use to define what a rigid transformation is, a transformation that preserves the lengths between corresponding points. Angle measures, angle measures. So for example, this angle here, the angle A, is gonna be the same as the angle A prime over here. Side lengths, the distance between A and B is gonna be the same as the distance between A prime and B prime.
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Properties perserved after rigid transformations.mp3
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That's actually one way that we even use to define what a rigid transformation is, a transformation that preserves the lengths between corresponding points. Angle measures, angle measures. So for example, this angle here, the angle A, is gonna be the same as the angle A prime over here. Side lengths, the distance between A and B is gonna be the same as the distance between A prime and B prime. Perimeter, if you have the same side lengths and the same angles, then perimeter and area are also going to be preserved, just like we saw with the rotation example. These are rigid transformations. These are the types of things that are preserved.
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Properties perserved after rigid transformations.mp3
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Side lengths, the distance between A and B is gonna be the same as the distance between A prime and B prime. Perimeter, if you have the same side lengths and the same angles, then perimeter and area are also going to be preserved, just like we saw with the rotation example. These are rigid transformations. These are the types of things that are preserved. Well, what is not preserved? Not preserved, and this just goes back to the example we just looked at. Well, coordinates are not preserved.
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Properties perserved after rigid transformations.mp3
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These are the types of things that are preserved. Well, what is not preserved? Not preserved, and this just goes back to the example we just looked at. Well, coordinates are not preserved. So as we see, the image of A, A prime, has different coordinates than A. B prime has different coordinates than B. C prime, in this case, happens to have the same coordinates as C because C happened to sit on the line that we're reflecting over.
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Properties perserved after rigid transformations.mp3
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Well, coordinates are not preserved. So as we see, the image of A, A prime, has different coordinates than A. B prime has different coordinates than B. C prime, in this case, happens to have the same coordinates as C because C happened to sit on the line that we're reflecting over. But D prime definitely does not have the same coordinates as D. So most of, or let me say, coordinates of A, B, or A, B, D, coordinates of A, B, D, not preserved after transformation, or their images, they don't have the same coordinates, after transformation. The one coordinate that happened to be preserved here is C's coordinates, because it was right on the line of reflection. And you could also look at other properties of how it might relate, how different segments might relate to lines that were not being transformed.
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Properties perserved after rigid transformations.mp3
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C prime, in this case, happens to have the same coordinates as C because C happened to sit on the line that we're reflecting over. But D prime definitely does not have the same coordinates as D. So most of, or let me say, coordinates of A, B, or A, B, D, coordinates of A, B, D, not preserved after transformation, or their images, they don't have the same coordinates, after transformation. The one coordinate that happened to be preserved here is C's coordinates, because it was right on the line of reflection. And you could also look at other properties of how it might relate, how different segments might relate to lines that were not being transformed. So for example, right over here, before transformation, C, D is parallel to the Y axis. You see that right over there. But after the transformation, C prime, D prime, so this could be C prime, D prime, is no longer parallel to the Y axis.
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Properties perserved after rigid transformations.mp3
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So I have a arbitrary inscribed quadrilateral in this circle. And what I want to prove is that for any inscribed quadrilateral, that opposite angles are supplementary. So when I say they're supplementary, the measure of this angle plus the measure of this angle need to be 180 degrees. The measure of this angle plus the measure of this angle need to be 180 degrees. And the way I'm gonna prove it is we're gonna assume that this, the measure of this angle right over here, that this is x degrees. And so from that, if we can prove that the measure of this opposite angle is 180 minus x degrees, then we've proven that opposite angles for an arbitrary quadrilateral that's inscribed in a circle are supplementary, because if this is 180 minus x, 180 minus x plus x is going to be 180 degrees. So I encourage you to pause the video and see if you can do that proof.
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Inscribed quadrilaterals proof Mathematics II High School Math Khan Academy.mp3
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The measure of this angle plus the measure of this angle need to be 180 degrees. And the way I'm gonna prove it is we're gonna assume that this, the measure of this angle right over here, that this is x degrees. And so from that, if we can prove that the measure of this opposite angle is 180 minus x degrees, then we've proven that opposite angles for an arbitrary quadrilateral that's inscribed in a circle are supplementary, because if this is 180 minus x, 180 minus x plus x is going to be 180 degrees. So I encourage you to pause the video and see if you can do that proof. And I'll give you a little bit of a hint. It's going to involve the measure of the arcs that the various angles intercept. So let's think about it a little bit.
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Inscribed quadrilaterals proof Mathematics II High School Math Khan Academy.mp3
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So I encourage you to pause the video and see if you can do that proof. And I'll give you a little bit of a hint. It's going to involve the measure of the arcs that the various angles intercept. So let's think about it a little bit. This angle that has a measure of x degrees, it intercepts this arc. So we see one side of the angle goes and intercepts the circle there. The other side right over there.
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Inscribed quadrilaterals proof Mathematics II High School Math Khan Academy.mp3
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So let's think about it a little bit. This angle that has a measure of x degrees, it intercepts this arc. So we see one side of the angle goes and intercepts the circle there. The other side right over there. And so the arc that it intercepts, I am highlighting in yellow. I am highlighting it in yellow. Trying to color it in.
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Inscribed quadrilaterals proof Mathematics II High School Math Khan Academy.mp3
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The other side right over there. And so the arc that it intercepts, I am highlighting in yellow. I am highlighting it in yellow. Trying to color it in. So there you go. Not a great job at coloring it in, but you get the point. That's the arc that it intercepts.
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Inscribed quadrilaterals proof Mathematics II High School Math Khan Academy.mp3
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Trying to color it in. So there you go. Not a great job at coloring it in, but you get the point. That's the arc that it intercepts. And we've already learned in previous videos that the relationship between an inscribed angle, the vertex of this angle sits on the circle, the relationship between an inscribed angle and the measure of the arc that it intercepts is that the measure of the inscribed angle is half the measure of the arc that it intercepts. So if this angle measures x degrees, then the measure of this arc is going to be 2x. 2x degrees.
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Inscribed quadrilaterals proof Mathematics II High School Math Khan Academy.mp3
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That's the arc that it intercepts. And we've already learned in previous videos that the relationship between an inscribed angle, the vertex of this angle sits on the circle, the relationship between an inscribed angle and the measure of the arc that it intercepts is that the measure of the inscribed angle is half the measure of the arc that it intercepts. So if this angle measures x degrees, then the measure of this arc is going to be 2x. 2x degrees. All right, well that's kind of interesting, but let's keep going. If the measure of that arc is 2x degrees, what is the measure of this arc right over here, the arc that completes the circle? Well if you go all the way around the circle, that's 360 degrees.
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Inscribed quadrilaterals proof Mathematics II High School Math Khan Academy.mp3
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2x degrees. All right, well that's kind of interesting, but let's keep going. If the measure of that arc is 2x degrees, what is the measure of this arc right over here, the arc that completes the circle? Well if you go all the way around the circle, that's 360 degrees. So this blue arc that I'm showing you right now, that's going to have a measure of 360 minus 2x, minus 2x, minus 2x degrees. 360 is all the way around. The blue one is all the way around minus the yellow arc.
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Inscribed quadrilaterals proof Mathematics II High School Math Khan Academy.mp3
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Well if you go all the way around the circle, that's 360 degrees. So this blue arc that I'm showing you right now, that's going to have a measure of 360 minus 2x, minus 2x, minus 2x degrees. 360 is all the way around. The blue one is all the way around minus the yellow arc. What you have left over if you subtract out the yellow arc is you have this blue arc. Now what's the angle that intercepts this blue arc? What's the inscribed angle that intercepts this blue arc right over here?
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Inscribed quadrilaterals proof Mathematics II High School Math Khan Academy.mp3
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The blue one is all the way around minus the yellow arc. What you have left over if you subtract out the yellow arc is you have this blue arc. Now what's the angle that intercepts this blue arc? What's the inscribed angle that intercepts this blue arc right over here? What's this angle? It's the angle that we wanted to figure out in terms of x. It is, well I'm having trouble changing colors.
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Inscribed quadrilaterals proof Mathematics II High School Math Khan Academy.mp3
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What's the inscribed angle that intercepts this blue arc right over here? What's this angle? It's the angle that we wanted to figure out in terms of x. It is, well I'm having trouble changing colors. It is that angle right over there. Notice the two sides of this angle, they intercept, this angle intercepts that arc. So once again, the measure of an inscribed angle is going to be half the measure of the arc that it intercepts.
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Inscribed quadrilaterals proof Mathematics II High School Math Khan Academy.mp3
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It is, well I'm having trouble changing colors. It is that angle right over there. Notice the two sides of this angle, they intercept, this angle intercepts that arc. So once again, the measure of an inscribed angle is going to be half the measure of the arc that it intercepts. So what's 1 half? What is 1 half times 360 minus 2x? Well 1 half times 360 is 180.
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Inscribed quadrilaterals proof Mathematics II High School Math Khan Academy.mp3
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So once again, the measure of an inscribed angle is going to be half the measure of the arc that it intercepts. So what's 1 half? What is 1 half times 360 minus 2x? Well 1 half times 360 is 180. 1 half times 2x is x. So the measure of this angle is going to be 180 minus x degrees. 180 minus x degrees.
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Inscribed quadrilaterals proof Mathematics II High School Math Khan Academy.mp3
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Well 1 half times 360 is 180. 1 half times 2x is x. So the measure of this angle is going to be 180 minus x degrees. 180 minus x degrees. And just like that, we've proven that these opposite sides for this arbitrary inscribed quadrilateral, that they are supplementary. You add these together, x plus 180 minus x, you're going to get 180 degrees. So they are supplementary.
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Inscribed quadrilaterals proof Mathematics II High School Math Khan Academy.mp3
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And they want us to make a line that goes right in between that angle, that divides that angle into two angles that have equal measure, that have half the measure of the first angle. So let's first find two points that are equidistant from this point right over here on each of these rays. So to do that, let's draw one circle here. And I could make this of any radius. Wherever this intersects with the rays, that's where I'm going to put a point. So let's say here and here. Notice both of these points, since they're both on this circle, are going to be equidistant from this point, which is the center of the circle.
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Constructing an angle bisector using a compass and straightedge Geometry Khan Academy.mp3
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And I could make this of any radius. Wherever this intersects with the rays, that's where I'm going to put a point. So let's say here and here. Notice both of these points, since they're both on this circle, are going to be equidistant from this point, which is the center of the circle. Now, what I want to do is construct a line that is equidistant from both of these points. And we've done that already when we looked at perpendicular bisectors for lines in this construction module. So let's do that.
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Constructing an angle bisector using a compass and straightedge Geometry Khan Academy.mp3
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Notice both of these points, since they're both on this circle, are going to be equidistant from this point, which is the center of the circle. Now, what I want to do is construct a line that is equidistant from both of these points. And we've done that already when we looked at perpendicular bisectors for lines in this construction module. So let's do that. So let's add a compass. And so what I want to do, this circle is centered at this point. And it has a radius equal to the distance between this point and that point.
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Constructing an angle bisector using a compass and straightedge Geometry Khan Academy.mp3
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So let's do that. So let's add a compass. And so what I want to do, this circle is centered at this point. And it has a radius equal to the distance between this point and that point. And then I do that again. So this circle is centered at this point and has a radius equal to the distance between that point and that point. And then the two places where they intersect are equidistant to both of these points.
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Constructing an angle bisector using a compass and straightedge Geometry Khan Academy.mp3
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And it has a radius equal to the distance between this point and that point. And then I do that again. So this circle is centered at this point and has a radius equal to the distance between that point and that point. And then the two places where they intersect are equidistant to both of these points. They're equidistant to both of these points. And so we can now draw our angle bisector just like that. And you might say, well, how do we really know that this angle is equal to this angle?
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Constructing an angle bisector using a compass and straightedge Geometry Khan Academy.mp3
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And then the two places where they intersect are equidistant to both of these points. They're equidistant to both of these points. And so we can now draw our angle bisector just like that. And you might say, well, how do we really know that this angle is equal to this angle? Well, there's a couple of ways we can tell. We know this distance right over here is equal to this distance right over there. We know that this distance over here is equal to this distance over here.
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Constructing an angle bisector using a compass and straightedge Geometry Khan Academy.mp3
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And you might say, well, how do we really know that this angle is equal to this angle? Well, there's a couple of ways we can tell. We know this distance right over here is equal to this distance right over there. We know that this distance over here is equal to this distance over here. And both of these triangles share this line. So essentially, if you look at this point, this point, and this point, that forms a triangle. And if you look at this point, this point, and this point, that forms a triangle.
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Constructing an angle bisector using a compass and straightedge Geometry Khan Academy.mp3
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We know that this distance over here is equal to this distance over here. And both of these triangles share this line. So essentially, if you look at this point, this point, and this point, that forms a triangle. And if you look at this point, this point, and this point, that forms a triangle. We know those two triangles are congruent. So this angle must be equal to this angle. These are the corresponding angles.
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Constructing an angle bisector using a compass and straightedge Geometry Khan Academy.mp3
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In past videos, we've thought about whether segment lengths or angle measures are preserved with a transformation. What we're now going to think about is what is preserved with a sequence of transformations. And in particular, we're gonna think about angle measure, angle measure, and segment lengths. So if you're transforming some type of a shape, segment, segment lengths. So let's look at this first example. We say a sequence of transformations is described below. So we first do a translation, then we do a reflection over a horizontal line, PQ, then we do a vertical stretch about PQ.
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Shape properties after a sequence of transformations.mp3
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So if you're transforming some type of a shape, segment, segment lengths. So let's look at this first example. We say a sequence of transformations is described below. So we first do a translation, then we do a reflection over a horizontal line, PQ, then we do a vertical stretch about PQ. What is this going to do? Is this going to preserve angle measures, and is this going to preserve segment lengths? Well, a translation is a rigid transformation, and so that will preserve both angle measures and segment lengths.
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Shape properties after a sequence of transformations.mp3
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So we first do a translation, then we do a reflection over a horizontal line, PQ, then we do a vertical stretch about PQ. What is this going to do? Is this going to preserve angle measures, and is this going to preserve segment lengths? Well, a translation is a rigid transformation, and so that will preserve both angle measures and segment lengths. So after that, angle measures and segment lengths are still going to be the same. A reflection over a horizontal line, PQ. Well, a reflection is also a rigid transformation, and so we will continue to preserve angle measure and segment lengths.
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Shape properties after a sequence of transformations.mp3
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Well, a translation is a rigid transformation, and so that will preserve both angle measures and segment lengths. So after that, angle measures and segment lengths are still going to be the same. A reflection over a horizontal line, PQ. Well, a reflection is also a rigid transformation, and so we will continue to preserve angle measure and segment lengths. Then they say a vertical stretch about PQ. Well, let's just think about what a vertical stretch does. So if I have some triangle right over here, if I have some triangle that looks like this, this is triangle ABC, and if you were to do a vertical stretch, what's going to happen?
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Shape properties after a sequence of transformations.mp3
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Well, a reflection is also a rigid transformation, and so we will continue to preserve angle measure and segment lengths. Then they say a vertical stretch about PQ. Well, let's just think about what a vertical stretch does. So if I have some triangle right over here, if I have some triangle that looks like this, this is triangle ABC, and if you were to do a vertical stretch, what's going to happen? Well, let's just imagine that we take these sides and we stretch them out so that we now have A is over here, or A prime, I should say, is over there. Let's say that B prime is now over here. This isn't going to be exact.
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Shape properties after a sequence of transformations.mp3
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So if I have some triangle right over here, if I have some triangle that looks like this, this is triangle ABC, and if you were to do a vertical stretch, what's going to happen? Well, let's just imagine that we take these sides and we stretch them out so that we now have A is over here, or A prime, I should say, is over there. Let's say that B prime is now over here. This isn't going to be exact. Well, what just happened to my triangle? Well, the measure of angle C is for sure going to be different now, and my segment lengths are for sure going to be different now. A prime, C prime is going to be different than AC in terms of segment length.
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Shape properties after a sequence of transformations.mp3
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This isn't going to be exact. Well, what just happened to my triangle? Well, the measure of angle C is for sure going to be different now, and my segment lengths are for sure going to be different now. A prime, C prime is going to be different than AC in terms of segment length. So a vertical stretch, if we're talking about a stretch in general, this is going to preserve neither. So neither preserved. Neither preserved.
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Shape properties after a sequence of transformations.mp3
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A prime, C prime is going to be different than AC in terms of segment length. So a vertical stretch, if we're talking about a stretch in general, this is going to preserve neither. So neither preserved. Neither preserved. So in general, if you're doing rigid transformation after rigid transformation, you're gonna preserve both angles and segment lengths, but if you throw a stretch in there, then all bets are off. You're not going to preserve either of them. Let's do another example.
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Shape properties after a sequence of transformations.mp3
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Neither preserved. So in general, if you're doing rigid transformation after rigid transformation, you're gonna preserve both angles and segment lengths, but if you throw a stretch in there, then all bets are off. You're not going to preserve either of them. Let's do another example. A sequence of transformations is described below, and so they give three transformations. So pause this video and think about whether angle measures, segment lengths, or will either both or neither or only one of them be preserved? All right, so first we have a rotation about a point P. That's a rigid transformation.
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Shape properties after a sequence of transformations.mp3
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Let's do another example. A sequence of transformations is described below, and so they give three transformations. So pause this video and think about whether angle measures, segment lengths, or will either both or neither or only one of them be preserved? All right, so first we have a rotation about a point P. That's a rigid transformation. It would preserve both segment lengths and angle measures. Then you have a translation, which is also a rigid transformation, and so that would preserve both again. Then we have a rotation about point P. So once again, another rigid transformation.
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Shape properties after a sequence of transformations.mp3
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All right, so first we have a rotation about a point P. That's a rigid transformation. It would preserve both segment lengths and angle measures. Then you have a translation, which is also a rigid transformation, and so that would preserve both again. Then we have a rotation about point P. So once again, another rigid transformation. So in this situation, everything is going to be preserved. So both angle measure, angle measure, and segment length are going to be preserved in this example. Let's do one more example.
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Shape properties after a sequence of transformations.mp3
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Then we have a rotation about point P. So once again, another rigid transformation. So in this situation, everything is going to be preserved. So both angle measure, angle measure, and segment length are going to be preserved in this example. Let's do one more example. So here, once again, we have a sequence of transformations, and so pause this video again and see if you can figure out whether angle measures, segment lengths, both or neither are going to be preserved. So the first transformation is a dilation. So a dilation is a non-rigid transformation.
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Shape properties after a sequence of transformations.mp3
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Let's do one more example. So here, once again, we have a sequence of transformations, and so pause this video again and see if you can figure out whether angle measures, segment lengths, both or neither are going to be preserved. So the first transformation is a dilation. So a dilation is a non-rigid transformation. So segment lengths not preserved. Segment lengths not preserved. And we've seen this in multiple videos already.
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Shape properties after a sequence of transformations.mp3
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So a dilation is a non-rigid transformation. So segment lengths not preserved. Segment lengths not preserved. And we've seen this in multiple videos already. But in a dilation, angles are preserved. Angles preserved. So already we've lost our segment lengths, but we still got our angles.
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Shape properties after a sequence of transformations.mp3
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And we've seen this in multiple videos already. But in a dilation, angles are preserved. Angles preserved. So already we've lost our segment lengths, but we still got our angles. Then we have a rotation about another point Q. So this is a rigid transformation. It would preserve both, but we've already lost our segment lengths, but angles are going to continue to be preserved.
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Shape properties after a sequence of transformations.mp3
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Construct a circle circumscribing the triangle. So that would be a circle that touches the vertices, the three vertices of this triangle. So we can construct it using a compass and a straight edge, or a virtual compass and a virtual straight edge. So what we want to do is center the circle at the perpendicular bisectors of the sides, or sometimes that's called the circumcenter of this triangle. So let's do that. And so let's think about, let's try to construct where the perpendicular bisectors of the sides are. So let me put a circle right over here whose radius is longer than this side right over here.
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Constructing circumscribing circle Geometric constructions Geometry Khan Academy.mp3
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So what we want to do is center the circle at the perpendicular bisectors of the sides, or sometimes that's called the circumcenter of this triangle. So let's do that. And so let's think about, let's try to construct where the perpendicular bisectors of the sides are. So let me put a circle right over here whose radius is longer than this side right over here. Now let me get one that has the same size. So let me make it the same size as the one I just did. And let me put it right over here.
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Constructing circumscribing circle Geometric constructions Geometry Khan Academy.mp3
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So let me put a circle right over here whose radius is longer than this side right over here. Now let me get one that has the same size. So let me make it the same size as the one I just did. And let me put it right over here. And this allows us to construct a perpendicular bisector. So if I go through that point and this point right over here, this bisects this side over here, and it's at a right angle. So now let's do that for the other sides.
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Constructing circumscribing circle Geometric constructions Geometry Khan Academy.mp3
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And let me put it right over here. And this allows us to construct a perpendicular bisector. So if I go through that point and this point right over here, this bisects this side over here, and it's at a right angle. So now let's do that for the other sides. So if I move this over here, and I really just have to do it for one of the other sides because the intersection of two lines is gonna give me a point. So I can do it for this side right over here. Let me scroll down so you can see a little bit clearer.
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Constructing circumscribing circle Geometric constructions Geometry Khan Academy.mp3
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So now let's do that for the other sides. So if I move this over here, and I really just have to do it for one of the other sides because the intersection of two lines is gonna give me a point. So I can do it for this side right over here. Let me scroll down so you can see a little bit clearer. So let me add another straight edge right over here. So I'm gonna go through that point, and I'm gonna go through this point. So that's the perpendicular bisector of this side right over here.
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Constructing circumscribing circle Geometric constructions Geometry Khan Academy.mp3
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Let me scroll down so you can see a little bit clearer. So let me add another straight edge right over here. So I'm gonna go through that point, and I'm gonna go through this point. So that's the perpendicular bisector of this side right over here. Now I could do the third side, and it should intersect at that point. I'm not ultra, ultra precise, but I'm close enough. And now I just have to center one of these circles.
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Constructing circumscribing circle Geometric constructions Geometry Khan Academy.mp3
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So that's the perpendicular bisector of this side right over here. Now I could do the third side, and it should intersect at that point. I'm not ultra, ultra precise, but I'm close enough. And now I just have to center one of these circles. Let me move one of these away. So let me just get rid of this one. And I just have to move this circle to the circumcenter and adjust its radius so that it gets pretty close to touching the three sides, the three vertices of this triangle.
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Constructing circumscribing circle Geometric constructions Geometry Khan Academy.mp3
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And now I just have to center one of these circles. Let me move one of these away. So let me just get rid of this one. And I just have to move this circle to the circumcenter and adjust its radius so that it gets pretty close to touching the three sides, the three vertices of this triangle. It doesn't have to be perfect. I think this exercise has some margin for error. But they really want to see that you've made an attempt at drawing the perpendicular bisectors of the sides to find the circumcenter, and then you put a circle right over there.
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Constructing circumscribing circle Geometric constructions Geometry Khan Academy.mp3
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And a compass is something that allows you to draw circles centered where you want them to be centered of different radii. And your typical compass would be like a metal thing. It has a pin on one end and it's kind of shaped like an angle and then you have a pencil at the other end. Now I don't have real physical rulers and pencils in front of me, but I have the virtual equivalent. I can say add a compass and now I can draw a circle. I can pick where I want to center it and I can change the radius. And I can draw a straight line segment.
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Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3
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Now I don't have real physical rulers and pencils in front of me, but I have the virtual equivalent. I can say add a compass and now I can draw a circle. I can pick where I want to center it and I can change the radius. And I can draw a straight line segment. So I can move it around and I can draw, this is equivalent to having a straight edge. And so using these tools, I want to construct a line going through P that is tangent to the circle. Now it might be tempting, oh, I could just, let me just try to draw a line.
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Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3
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And I can draw a straight line segment. So I can move it around and I can draw, this is equivalent to having a straight edge. And so using these tools, I want to construct a line going through P that is tangent to the circle. Now it might be tempting, oh, I could just, let me just try to draw a line. Maybe it looks something like this. Do you remember a tangent? A tangent line will touch the circle at exactly one point and that point, since it's going through P, should be the point P. Another way to think about a tangent line is it's going to be perpendicular to the radius between that point and the center.
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Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3
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Now it might be tempting, oh, I could just, let me just try to draw a line. Maybe it looks something like this. Do you remember a tangent? A tangent line will touch the circle at exactly one point and that point, since it's going through P, should be the point P. Another way to think about a tangent line is it's going to be perpendicular to the radius between that point and the center. Now what I just drew actually looks pretty good, but it's not so precise. I don't know if it's exactly perpendicular to the radius. I don't know if it's touching at exactly one point right over there.
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Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3
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A tangent line will touch the circle at exactly one point and that point, since it's going through P, should be the point P. Another way to think about a tangent line is it's going to be perpendicular to the radius between that point and the center. Now what I just drew actually looks pretty good, but it's not so precise. I don't know if it's exactly perpendicular to the radius. I don't know if it's touching at exactly one point right over there. So what we're going to do is use our virtual compass and our virtual straight edge to try to do a more precise drawing. So let's do that. So the first thing I'm going to do is I'm going to set up P as the midpoint of a line where the center of the circle is one other end of the line.
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Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3
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I don't know if it's touching at exactly one point right over there. So what we're going to do is use our virtual compass and our virtual straight edge to try to do a more precise drawing. So let's do that. So the first thing I'm going to do is I'm going to set up P as the midpoint of a line where the center of the circle is one other end of the line. And the way I can do that, let me add a compass here, and let me construct a circle that has the same, okay, so I've set up my compass to have the same radius as my original circle, but now let me move it over right over here. So now it's centered at P. Well, why is this useful? Well, now a diameter of this new circle is going to be a segment that is centered at P. So I'm going to have a segment where P is the midpoint, and then my center of my original circle is going to be one of the endpoints.
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Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3
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So the first thing I'm going to do is I'm going to set up P as the midpoint of a line where the center of the circle is one other end of the line. And the way I can do that, let me add a compass here, and let me construct a circle that has the same, okay, so I've set up my compass to have the same radius as my original circle, but now let me move it over right over here. So now it's centered at P. Well, why is this useful? Well, now a diameter of this new circle is going to be a segment that is centered at P. So I'm going to have a segment where P is the midpoint, and then my center of my original circle is going to be one of the endpoints. So let's do that. So I'm going to add a straight edge. So I'll make that one of the endpoints.
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Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3
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Well, now a diameter of this new circle is going to be a segment that is centered at P. So I'm going to have a segment where P is the midpoint, and then my center of my original circle is going to be one of the endpoints. So let's do that. So I'm going to add a straight edge. So I'll make that one of the endpoints. And I'm going to go through P all the way to the other side of my new circle. Now, what's the whole point of me, let me try to do it as well as I can, what's the whole point of me doing that? Well, now I've made P the midpoint of a segment.
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Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3
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So I'll make that one of the endpoints. And I'm going to go through P all the way to the other side of my new circle. Now, what's the whole point of me, let me try to do it as well as I can, what's the whole point of me doing that? Well, now I've made P the midpoint of a segment. So if I can construct a perpendicular bisector of the segment, it will go through P because P is the midpoint, and then that thing, it's going to be exactly perpendicular to the radius because the radius, the original radius, let me be careful here, the original radius is part of this segment. So let's see how I can do this. So what I could do is, I'm going to draw another circle.
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Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3
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Well, now I've made P the midpoint of a segment. So if I can construct a perpendicular bisector of the segment, it will go through P because P is the midpoint, and then that thing, it's going to be exactly perpendicular to the radius because the radius, the original radius, let me be careful here, the original radius is part of this segment. So let's see how I can do this. So what I could do is, I'm going to draw another circle. I'm going to center this one at the original circle. And I'm going to make it have a different radius, maybe a radius something like that. And now I'm going to construct another circle of this larger size, but I'm going to center it at this point right over here.
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Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3
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So what I could do is, I'm going to draw another circle. I'm going to center this one at the original circle. And I'm going to make it have a different radius, maybe a radius something like that. And now I'm going to construct another circle of this larger size, but I'm going to center it at this point right over here. And I think you'll see quickly what this will accomplish. So I'm going to construct another circle of that same larger radius, of that same larger radius, just like that. And now I'm going to move it over here.
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Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3
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And now I'm going to construct another circle of this larger size, but I'm going to center it at this point right over here. And I think you'll see quickly what this will accomplish. So I'm going to construct another circle of that same larger radius, of that same larger radius, just like that. And now I'm going to move it over here. Over here. So what's interesting about the intersection of these two larger circles? Well this point right over here is equidistant to this end of the segment and to this end of the segment.
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Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3
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And now I'm going to move it over here. Over here. So what's interesting about the intersection of these two larger circles? Well this point right over here is equidistant to this end of the segment and to this end of the segment. Because remember, these two larger circles have the same radius. So if I'm sitting on both of them, then I am that distance away from this point, and I am sitting that distance away from this point. So something that is equidistant from the two end points of a segment, it's going to sit on the perpendicular bisector.
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Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3
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Well this point right over here is equidistant to this end of the segment and to this end of the segment. Because remember, these two larger circles have the same radius. So if I'm sitting on both of them, then I am that distance away from this point, and I am sitting that distance away from this point. So something that is equidistant from the two end points of a segment, it's going to sit on the perpendicular bisector. So this point is going to sit on the perpendicular bisector, and this point is going to be sit on the perpendicular bisector. So now we can draw a perpendicular bisector. We can go from this point right over here, the intersection of our two larger circles, this point that is equidistant from my two centers of the large circle, to this point that is equidistant to the two centers of the large circle.
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Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3
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So something that is equidistant from the two end points of a segment, it's going to sit on the perpendicular bisector. So this point is going to sit on the perpendicular bisector, and this point is going to be sit on the perpendicular bisector. So now we can draw a perpendicular bisector. We can go from this point right over here, the intersection of our two larger circles, this point that is equidistant from my two centers of the large circle, to this point that is equidistant to the two centers of the large circle. And once again, it's equidistant to the two centers of the large circle, but those points are also the end points of this segment. So these two points are on the perpendicular bisector. You just need two points for a line.
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Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3
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We can go from this point right over here, the intersection of our two larger circles, this point that is equidistant from my two centers of the large circle, to this point that is equidistant to the two centers of the large circle. And once again, it's equidistant to the two centers of the large circle, but those points are also the end points of this segment. So these two points are on the perpendicular bisector. You just need two points for a line. So I've just constructed a perpendicular bisector to P. And it's perpendicular, once again, to the radius from the center to P of our original circle. Now, that is going to be a tangent line. Because if we go only through P, or we go through P, and we are exactly perpendicular to the radius from P to the center, then this line that we've just constructed is actually tangent.
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Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3
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You just need two points for a line. So I've just constructed a perpendicular bisector to P. And it's perpendicular, once again, to the radius from the center to P of our original circle. Now, that is going to be a tangent line. Because if we go only through P, or we go through P, and we are exactly perpendicular to the radius from P to the center, then this line that we've just constructed is actually tangent. So anyway, it might seem like a lot of work to do all of this. You know, I could have started just eyeballing it. But when we do it like this, we can feel really, really, really, really, really good that we're being precise.
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Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3
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Because if we go only through P, or we go through P, and we are exactly perpendicular to the radius from P to the center, then this line that we've just constructed is actually tangent. So anyway, it might seem like a lot of work to do all of this. You know, I could have started just eyeballing it. But when we do it like this, we can feel really, really, really, really, really good that we're being precise. Imagine if you were trying to do this in a larger scale, or if you're trying to engineer some very precise instrument. You would want to do it this way. You want to draw a very precise drawing, maybe an architectural drawing.
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Constructing a tangent line using compass and straightedge Geometry Khan Academy.mp3
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Construct a square inscribed inside the circle. In order to do this, we just have to remember that a square, well, we know all squares, all four sides are congruent and they intersect at right angles, and we also have to remember that the two diagonals of the square are going to be perpendicular bisectors of each other. So let's see if we can construct two lines that are perpendicular bisectors of each other, and actually where those two lines intersect our bigger circle, those are going to be the vertices of our square. So let's throw a straight edge right over here, and let's make a diameter. So that's a diameter right over here. It just goes through the circle, goes through the center of the circle, and two sides of the circle. And now let's think about how we can construct a perpendicular bisector of this, and other compass construction or construction videos.
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Constructing square inscribed in circle Geometric constructions Geometry Khan Academy.mp3
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So let's throw a straight edge right over here, and let's make a diameter. So that's a diameter right over here. It just goes through the circle, goes through the center of the circle, and two sides of the circle. And now let's think about how we can construct a perpendicular bisector of this, and other compass construction or construction videos. But what we can do is we can put a circle, let's throw a circle right over here. We've got to make its radius bigger than the center. What we're going to do is we're going to reuse this, we're going to make another circle that's the exact same size, put it there, and where they intersect is going to be exactly along, those two points of intersection are going to be along a perpendicular bisector.
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Constructing square inscribed in circle Geometric constructions Geometry Khan Academy.mp3
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And now let's think about how we can construct a perpendicular bisector of this, and other compass construction or construction videos. But what we can do is we can put a circle, let's throw a circle right over here. We've got to make its radius bigger than the center. What we're going to do is we're going to reuse this, we're going to make another circle that's the exact same size, put it there, and where they intersect is going to be exactly along, those two points of intersection are going to be along a perpendicular bisector. So that's one of them. Let's do another one. So I'll draw a circle of the exact same dimensions, so I'll center it at the same place.
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Constructing square inscribed in circle Geometric constructions Geometry Khan Academy.mp3
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What we're going to do is we're going to reuse this, we're going to make another circle that's the exact same size, put it there, and where they intersect is going to be exactly along, those two points of intersection are going to be along a perpendicular bisector. So that's one of them. Let's do another one. So I'll draw a circle of the exact same dimensions, so I'll center it at the same place. I'll drag it out there. That looks pretty good. I'll move it on to this side, the other side of my diameter.
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Constructing square inscribed in circle Geometric constructions Geometry Khan Academy.mp3
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So I'll draw a circle of the exact same dimensions, so I'll center it at the same place. I'll drag it out there. That looks pretty good. I'll move it on to this side, the other side of my diameter. So that looks pretty good. And notice, if I connect that point to that point, I have now, I will have constructed a perpendicular bisector of this original segment. So let's do that.
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Constructing square inscribed in circle Geometric constructions Geometry Khan Academy.mp3
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I'll move it on to this side, the other side of my diameter. So that looks pretty good. And notice, if I connect that point to that point, I have now, I will have constructed a perpendicular bisector of this original segment. So let's do that. Let's connect those two points. So that point and that point, and then we could just keep going all the way to, so keep going all the way to the end of the circle. Go all the way over, all the way over there.
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Constructing square inscribed in circle Geometric constructions Geometry Khan Academy.mp3
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So let's do that. Let's connect those two points. So that point and that point, and then we could just keep going all the way to, so keep going all the way to the end of the circle. Go all the way over, all the way over there. That looks pretty good. And now we just have to connect these four points to have a square. So let's do that.
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Constructing square inscribed in circle Geometric constructions Geometry Khan Academy.mp3
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Go all the way over, all the way over there. That looks pretty good. And now we just have to connect these four points to have a square. So let's do that. So I'll connect that and that, and then I will connect, draw another straight edge there, I will connect that with that, and then two more to go. I'll connect this with that, and then one more. I can connect this with that, and there you go.
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Constructing square inscribed in circle Geometric constructions Geometry Khan Academy.mp3
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What we're going to do in this video is study a proof of the Pythagorean Theorem that was first discovered, first discovered, or as far as we know, first discovered by James Garfield in 1876. And what's exciting about this is he was not a professional mathematician. You might know James Garfield as the 20th president of the United States. He was elected president, he was elected four years after, in 1880, and then he became president in 1881. And he did this proof while he was a sitting member of the United States House of Representatives. And what's exciting about that is it shows that Abraham Lincoln was not the only U.S. politician, or not the only U.S. president who was really into geometry. And what Garfield realized is if you construct a right triangle, so I'm going to do my best attempt to construct one.
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Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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He was elected president, he was elected four years after, in 1880, and then he became president in 1881. And he did this proof while he was a sitting member of the United States House of Representatives. And what's exciting about that is it shows that Abraham Lincoln was not the only U.S. politician, or not the only U.S. president who was really into geometry. And what Garfield realized is if you construct a right triangle, so I'm going to do my best attempt to construct one. So let me construct one right here. So let's say that this side right over here is length b. Let's say this side is length a.
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Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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And what Garfield realized is if you construct a right triangle, so I'm going to do my best attempt to construct one. So let me construct one right here. So let's say that this side right over here is length b. Let's say this side is length a. And let's say that this side, the hypotenuse of my right triangle, has length c. So I've just constructed a right triangle, and let me make it clear, it is a right triangle. He essentially flipped and rotated this right triangle to construct another one that is congruent to the first one. So let me construct that.
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Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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Let's say this side is length a. And let's say that this side, the hypotenuse of my right triangle, has length c. So I've just constructed a right triangle, and let me make it clear, it is a right triangle. He essentially flipped and rotated this right triangle to construct another one that is congruent to the first one. So let me construct that. So we're going to have length b, and it's collinear with length a. It's along the same line, I should say. They don't overlap with each other.
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Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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So let me construct that. So we're going to have length b, and it's collinear with length a. It's along the same line, I should say. They don't overlap with each other. So this is side of length b. And then you have a side of length, let me draw it, this should be a little bit taller, side of length b. Then you have your side of length a at a right angle.
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Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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They don't overlap with each other. So this is side of length b. And then you have a side of length, let me draw it, this should be a little bit taller, side of length b. Then you have your side of length a at a right angle. Your side of length a comes in at a right angle. And then you have your side of length c. You have your side of length c. So the first thing we need to think about is what's the angle between these two sides? What's this mystery angle?
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Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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Then you have your side of length a at a right angle. Your side of length a comes in at a right angle. And then you have your side of length c. You have your side of length c. So the first thing we need to think about is what's the angle between these two sides? What's this mystery angle? What's that mystery angle going to be? Well, it looks like something, but let's see if we can prove to ourselves it really is what we think it looks like. If we look at this original triangle, and we call this angle theta, what's this angle over here, the angle that's between sides of length a and length c, what's the measure of this angle going to be?
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Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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What's this mystery angle? What's that mystery angle going to be? Well, it looks like something, but let's see if we can prove to ourselves it really is what we think it looks like. If we look at this original triangle, and we call this angle theta, what's this angle over here, the angle that's between sides of length a and length c, what's the measure of this angle going to be? Well, theta plus this angle have to add up to 90, because you add those two together, they add up to 90, and then you have another 90, it's going to get 180 degrees for the interior angles of this triangle. So these two have to add up to 90, this angle's going to be 90 minus theta. Well, if this triangle up here is congruent, and we've constructed it so it is congruent, the corresponding angle to this one is this angle right over here, so this is also going to be theta, and this right over here is going to be 90 minus theta.
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Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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If we look at this original triangle, and we call this angle theta, what's this angle over here, the angle that's between sides of length a and length c, what's the measure of this angle going to be? Well, theta plus this angle have to add up to 90, because you add those two together, they add up to 90, and then you have another 90, it's going to get 180 degrees for the interior angles of this triangle. So these two have to add up to 90, this angle's going to be 90 minus theta. Well, if this triangle up here is congruent, and we've constructed it so it is congruent, the corresponding angle to this one is this angle right over here, so this is also going to be theta, and this right over here is going to be 90 minus theta. So given that this is theta, this is 90 minus theta, what is our angle going to be? Well, they all collectively kind of go 180 degrees, so you have theta plus 90 minus theta, plus our mystery angle is going to be equal to 180 degrees, the thetas cancel out, theta minus theta, and you have 90 plus our mystery angle is 180 degrees, subtract 90 from both sides, and you are left with your mystery angle equaling 90 degrees. So that all worked out well, so let me make that clear.
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Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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Well, if this triangle up here is congruent, and we've constructed it so it is congruent, the corresponding angle to this one is this angle right over here, so this is also going to be theta, and this right over here is going to be 90 minus theta. So given that this is theta, this is 90 minus theta, what is our angle going to be? Well, they all collectively kind of go 180 degrees, so you have theta plus 90 minus theta, plus our mystery angle is going to be equal to 180 degrees, the thetas cancel out, theta minus theta, and you have 90 plus our mystery angle is 180 degrees, subtract 90 from both sides, and you are left with your mystery angle equaling 90 degrees. So that all worked out well, so let me make that clear. That's going to be useful for us in a second. So we can now say definitively that this is 90 degrees, this is a right angle. Now what we are going to do is we are going to construct a trapezoid.
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Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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So that all worked out well, so let me make that clear. That's going to be useful for us in a second. So we can now say definitively that this is 90 degrees, this is a right angle. Now what we are going to do is we are going to construct a trapezoid. This side A is parallel to side B down here, the way it's been constructed, this is just one side right over here, this goes straight up, and now let's just connect these two sides right over there. So there's a couple of ways to think about the area of this trapezoid. One is we can just think of it as a trapezoid and come up with its area, and then we can think about it as the sum of the areas of its components.
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Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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Now what we are going to do is we are going to construct a trapezoid. This side A is parallel to side B down here, the way it's been constructed, this is just one side right over here, this goes straight up, and now let's just connect these two sides right over there. So there's a couple of ways to think about the area of this trapezoid. One is we can just think of it as a trapezoid and come up with its area, and then we can think about it as the sum of the areas of its components. So let's just first think of it as a trapezoid. So what do we know about the area of a trapezoid? Well the area of a trapezoid is going to be the height of the trapezoid, which is A plus B, A plus B, that's the height of the trapezoid, times the mean of the top and the bottom, or the average of the top and the bottom.
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Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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One is we can just think of it as a trapezoid and come up with its area, and then we can think about it as the sum of the areas of its components. So let's just first think of it as a trapezoid. So what do we know about the area of a trapezoid? Well the area of a trapezoid is going to be the height of the trapezoid, which is A plus B, A plus B, that's the height of the trapezoid, times the mean of the top and the bottom, or the average of the top and the bottom. So this times 1 half times A plus B. And the intuition there, you're taking the height times the average of this bottom and the top. The average of the bottom and the top gives you the area of the trapezoid.
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Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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Well the area of a trapezoid is going to be the height of the trapezoid, which is A plus B, A plus B, that's the height of the trapezoid, times the mean of the top and the bottom, or the average of the top and the bottom. So this times 1 half times A plus B. And the intuition there, you're taking the height times the average of this bottom and the top. The average of the bottom and the top gives you the area of the trapezoid. Now, how can we also figure out the area with its component parts? So this should be, regardless of how we calculate the area, as long as we do correct things, we should come up with the same result. So how else can we come up with this area?
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Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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The average of the bottom and the top gives you the area of the trapezoid. Now, how can we also figure out the area with its component parts? So this should be, regardless of how we calculate the area, as long as we do correct things, we should come up with the same result. So how else can we come up with this area? Well, we could say it's the area of the two right triangles. The area of each of them is 1 half A times B. But there's two of them.
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Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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So how else can we come up with this area? Well, we could say it's the area of the two right triangles. The area of each of them is 1 half A times B. But there's two of them. Let me do that B in that same blue color. But there's two of these right triangles, so let's multiply by two. So this 2 times 1 half AB, that takes into consideration this bottom right triangle and this top one.
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Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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But there's two of them. Let me do that B in that same blue color. But there's two of these right triangles, so let's multiply by two. So this 2 times 1 half AB, that takes into consideration this bottom right triangle and this top one. And what's the area of this large one that I will color in in green? What's the area of this large one? Well, that's pretty straightforward.
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Garfield's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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