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Maybe we could denote that with a B prime. We are reflecting across the X axis. Let's do another example. So here, they tell us point C prime is the image of C, which is at the coordinates negative four comma negative two, under a reflection across the Y axis. What are the coordinates of C prime? So pause this video and see if you can figure it out on your own. So there's several ways to approach it.
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Reflecting points across horizontal and vertical lines.mp3
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So here, they tell us point C prime is the image of C, which is at the coordinates negative four comma negative two, under a reflection across the Y axis. What are the coordinates of C prime? So pause this video and see if you can figure it out on your own. So there's several ways to approach it. It doesn't hurt to do a quick visual diagram. So that could be my X axis. This would be my Y axis.
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Reflecting points across horizontal and vertical lines.mp3
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So there's several ways to approach it. It doesn't hurt to do a quick visual diagram. So that could be my X axis. This would be my Y axis. And it's the point negative four comma negative two. So that might look like this. Negative four, negative two.
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Reflecting points across horizontal and vertical lines.mp3
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This would be my Y axis. And it's the point negative four comma negative two. So that might look like this. Negative four, negative two. So this is the point C right over here. And we want to reflect across the Y axis. So we want to reflect across the Y axis, which I am coloring in in red right over here.
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Reflecting points across horizontal and vertical lines.mp3
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Negative four, negative two. So this is the point C right over here. And we want to reflect across the Y axis. So we want to reflect across the Y axis, which I am coloring in in red right over here. So let's see. The point C is four to the left of the Y axis. So its reflection is going to be four to the right of the Y axis.
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Reflecting points across horizontal and vertical lines.mp3
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So we want to reflect across the Y axis, which I am coloring in in red right over here. So let's see. The point C is four to the left of the Y axis. So its reflection is going to be four to the right of the Y axis. So let me do it like this. So instead of being four to the left, we want to go four to the right. So plus four.
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Reflecting points across horizontal and vertical lines.mp3
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So its reflection is going to be four to the right of the Y axis. So let me do it like this. So instead of being four to the left, we want to go four to the right. So plus four. So where would that put our C prime? So our C prime would be right over there. And what would its coordinates be?
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Reflecting points across horizontal and vertical lines.mp3
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So plus four. So where would that put our C prime? So our C prime would be right over there. And what would its coordinates be? Well, it would have the same Y coordinates. So C prime would have a Y coordinate of negative two. But what would its X coordinate be?
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Reflecting points across horizontal and vertical lines.mp3
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And what would its coordinates be? Well, it would have the same Y coordinates. So C prime would have a Y coordinate of negative two. But what would its X coordinate be? Well, instead of it being negative four, it gets flipped over the Y axis. So now it's going to have an X coordinate of positive four. So the coordinates here would be four comma negative two.
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Reflecting points across horizontal and vertical lines.mp3
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But what would its X coordinate be? Well, instead of it being negative four, it gets flipped over the Y axis. So now it's going to have an X coordinate of positive four. So the coordinates here would be four comma negative two. Four comma negative two. You might have been able to do this in your head. Although for me, even if I tried to do it in my head, I would still have this visualization going on in my head.
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Reflecting points across horizontal and vertical lines.mp3
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So the coordinates here would be four comma negative two. Four comma negative two. You might have been able to do this in your head. Although for me, even if I tried to do it in my head, I would still have this visualization going on in my head. I was okay, negative four comma negative two. I'm sitting there in the third quadrant. If I'm flipping over the Y axis, my Y coordinate wouldn't change, but my X coordinate would become the opposite, and I would end up in the fourth quadrant.
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Reflecting points across horizontal and vertical lines.mp3
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So I have a circle here with a center at point O, and let's pick an arbitrary point that sits outside of the circle. So let me just pick this point right over here, point A. And if I have an arbitrary point outside of the circle, I can actually draw two different tangent lines that contain A that are tangent to this circle. Let me draw them. So one of them would look like this. Actually, let me just start right over here. So I wanna make it tangent to the circle so it could look like that.
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Proof Segments tangent to circle from outside point are congruent High School Math Khan Academy.mp3
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Let me draw them. So one of them would look like this. Actually, let me just start right over here. So I wanna make it tangent to the circle so it could look like that. And then the other one would look like this, would look like, would look like that. And let's say that the point that the tangent lines intersect the circle, let's say this point right over here is point B, and this point right over here is point C. This is point C right over here. What I wanna prove, what I wanna prove is that the segment AB is congruent to the segment AC, or another way of thinking about it.
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Proof Segments tangent to circle from outside point are congruent High School Math Khan Academy.mp3
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So I wanna make it tangent to the circle so it could look like that. And then the other one would look like this, would look like, would look like that. And let's say that the point that the tangent lines intersect the circle, let's say this point right over here is point B, and this point right over here is point C. This is point C right over here. What I wanna prove, what I wanna prove is that the segment AB is congruent to the segment AC, or another way of thinking about it. I wanna prove, let me do this in a new color that I haven't used yet. I wanna prove that this segment right over here is congruent to this segment, is congruent to that segment. I'd encourage you to pause the video and try to work it out, and try to work it out on your own before I work through it.
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Proof Segments tangent to circle from outside point are congruent High School Math Khan Academy.mp3
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What I wanna prove, what I wanna prove is that the segment AB is congruent to the segment AC, or another way of thinking about it. I wanna prove, let me do this in a new color that I haven't used yet. I wanna prove that this segment right over here is congruent to this segment, is congruent to that segment. I'd encourage you to pause the video and try to work it out, and try to work it out on your own before I work through it. All right, now let's try to work through this together. And to work through this together, I'm gonna actually set up two triangles, two triangles, and they're going to be right triangles as we'll see in a second. So let me draw some lines here to set up our triangles.
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Proof Segments tangent to circle from outside point are congruent High School Math Khan Academy.mp3
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I'd encourage you to pause the video and try to work it out, and try to work it out on your own before I work through it. All right, now let's try to work through this together. And to work through this together, I'm gonna actually set up two triangles, two triangles, and they're going to be right triangles as we'll see in a second. So let me draw some lines here to set up our triangles. So I'm gonna draw one line just like that, and then let me draw that, and then let me draw that. Now what do we know, what do we know about these triangles? Well, as I mentioned, we're going to be dealing with right triangles.
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Proof Segments tangent to circle from outside point are congruent High School Math Khan Academy.mp3
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So let me draw some lines here to set up our triangles. So I'm gonna draw one line just like that, and then let me draw that, and then let me draw that. Now what do we know, what do we know about these triangles? Well, as I mentioned, we're going to be dealing with right triangles. How do I know that? Well, in a previous video, we saw that if we have a radius intersecting a tangent line, that they intersect at right angles. We've proven this.
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Proof Segments tangent to circle from outside point are congruent High School Math Khan Academy.mp3
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Well, as I mentioned, we're going to be dealing with right triangles. How do I know that? Well, in a previous video, we saw that if we have a radius intersecting a tangent line, that they intersect at right angles. We've proven this. So that's a radius, that's a tangent line. They're gonna intersect at a right angle. So radius, tangent line, they intersect at a right angle.
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Proof Segments tangent to circle from outside point are congruent High School Math Khan Academy.mp3
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We've proven this. So that's a radius, that's a tangent line. They're gonna intersect at a right angle. So radius, tangent line, they intersect at a right angle. We also know since OB and OC are both radii, that they're both the length of the radius of the circle. So this side right over here, let me do some new colors here. So this side is going to be congruent to that side.
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Proof Segments tangent to circle from outside point are congruent High School Math Khan Academy.mp3
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So radius, tangent line, they intersect at a right angle. We also know since OB and OC are both radii, that they're both the length of the radius of the circle. So this side right over here, let me do some new colors here. So this side is going to be congruent to that side. And you can see that the hypotenuse of both circles is the same side, side OA. So of course, it is equal to itself. This is equal to itself.
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Proof Segments tangent to circle from outside point are congruent High School Math Khan Academy.mp3
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So this side is going to be congruent to that side. And you can see that the hypotenuse of both circles is the same side, side OA. So of course, it is equal to itself. This is equal to itself. And so we see triangle ABO and triangle ACO, they're both right triangles that have two sides in common. In particular, they both have a hypotenuse that are equal to each other, and they both have a base or a leg. And we know from hypotenuse leg congruence that if you have two right triangles where the hypotenuses are equal and you have a leg that are equal, then the two triangles are going to be congruent.
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Proof Segments tangent to circle from outside point are congruent High School Math Khan Academy.mp3
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This is equal to itself. And so we see triangle ABO and triangle ACO, they're both right triangles that have two sides in common. In particular, they both have a hypotenuse that are equal to each other, and they both have a base or a leg. And we know from hypotenuse leg congruence that if you have two right triangles where the hypotenuses are equal and you have a leg that are equal, then the two triangles are going to be congruent. So triangle ABO is congruent to triangle ACO. And in that proof where we prove it is, well, the Pythagorean theorem tells us if you know two sides of a right triangle, that determines what the third side is. So the third side, so it's length of AB is going to be the same thing as the length of AC.
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Proof Segments tangent to circle from outside point are congruent High School Math Khan Academy.mp3
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And we know from hypotenuse leg congruence that if you have two right triangles where the hypotenuses are equal and you have a leg that are equal, then the two triangles are going to be congruent. So triangle ABO is congruent to triangle ACO. And in that proof where we prove it is, well, the Pythagorean theorem tells us if you know two sides of a right triangle, that determines what the third side is. So the third side, so it's length of AB is going to be the same thing as the length of AC. And that's, once again, it just comes out of these are both right triangles. If two sides, if two corresponding sides of these two right triangles are congruent, then the third side has to. That comes straight from the Pythagorean theorem.
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Proof Segments tangent to circle from outside point are congruent High School Math Khan Academy.mp3
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So the third side, so it's length of AB is going to be the same thing as the length of AC. And that's, once again, it just comes out of these are both right triangles. If two sides, if two corresponding sides of these two right triangles are congruent, then the third side has to. That comes straight from the Pythagorean theorem. And there you go. We've hopefully feel good about the fact that AB is going to be congruent to AC. Or another way to think about it, if I take a point outside of a circle and I construct segments that are tangent to the circle, that those two segments are going to be congruent to each other.
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Proof Segments tangent to circle from outside point are congruent High School Math Khan Academy.mp3
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So that's a square base, that's a square, that's a square, and that's a square. All of the figures except for C are prisms. Yes, C is a pyramid right over here. All of the figures except for D are right. You can see D right over here is a little bit skewed or you could view it as oblique. All of the figures except for E have the same base area. The base of figure E is a dilation of the base of figure A by a scale factor of 1.5.
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Using related volumes Solid geometry High school geometry Khan Academy.mp3
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All of the figures except for D are right. You can see D right over here is a little bit skewed or you could view it as oblique. All of the figures except for E have the same base area. The base of figure E is a dilation of the base of figure A by a scale factor of 1.5. All right. So they tell us if figure A has a volume of 28 cubic centimeters, what are the volumes of the other figures? So pause this video and see if you can have a go at that.
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Using related volumes Solid geometry High school geometry Khan Academy.mp3
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The base of figure E is a dilation of the base of figure A by a scale factor of 1.5. All right. So they tell us if figure A has a volume of 28 cubic centimeters, what are the volumes of the other figures? So pause this video and see if you can have a go at that. All right, now let's work through this together. Now, they're telling us about the bases and the heights, that a lot of these have the same base area, figure E is going to be different, and they also tell us that they all have the same height. So one way to think about volume is it's going to deal with base and height.
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Using related volumes Solid geometry High school geometry Khan Academy.mp3
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So pause this video and see if you can have a go at that. All right, now let's work through this together. Now, they're telling us about the bases and the heights, that a lot of these have the same base area, figure E is going to be different, and they also tell us that they all have the same height. So one way to think about volume is it's going to deal with base and height. And so for figure A, it's pretty straightforward. If we call this area right over here, let's call that B for the area of the base, and then it has some height, H, right over here, we know that the base area times the height is going to be the volume. So we could say that based on figure A, base times height is going to be equal to 28 cubic centimeters.
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Using related volumes Solid geometry High school geometry Khan Academy.mp3
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So one way to think about volume is it's going to deal with base and height. And so for figure A, it's pretty straightforward. If we call this area right over here, let's call that B for the area of the base, and then it has some height, H, right over here, we know that the base area times the height is going to be the volume. So we could say that based on figure A, base times height is going to be equal to 28 cubic centimeters. Fair enough. Now, what's going on over here with figure B? Well, it's a cylinder.
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Using related volumes Solid geometry High school geometry Khan Academy.mp3
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So we could say that based on figure A, base times height is going to be equal to 28 cubic centimeters. Fair enough. Now, what's going on over here with figure B? Well, it's a cylinder. Now, for a cylinder, what is the volume of a cylinder? Well, it too is going to be base times height. So it's going to be the area of the base times the height.
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Using related volumes Solid geometry High school geometry Khan Academy.mp3
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Well, it's a cylinder. Now, for a cylinder, what is the volume of a cylinder? Well, it too is going to be base times height. So it's going to be the area of the base times the height. And if you're wondering how is that possible that it's the same as the volume of a rectangular prism over here, it's actually Cavalieri's principle. If they have the same height, and if at any point along that height, they have the same cross-sectional area, then you're going to have the same volume. So this volume is also going to be base times height.
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Using related volumes Solid geometry High school geometry Khan Academy.mp3
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So it's going to be the area of the base times the height. And if you're wondering how is that possible that it's the same as the volume of a rectangular prism over here, it's actually Cavalieri's principle. If they have the same height, and if at any point along that height, they have the same cross-sectional area, then you're going to have the same volume. So this volume is also going to be base times height. So let me say this is figure A. Figure B, right over here, let me draw those dots a little better, these colons a little bit better. Figure B, the volume is also going to be base times height, which is equal to 28 cubic centimeters.
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Using related volumes Solid geometry High school geometry Khan Academy.mp3
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So this volume is also going to be base times height. So let me say this is figure A. Figure B, right over here, let me draw those dots a little better, these colons a little bit better. Figure B, the volume is also going to be base times height, which is equal to 28 cubic centimeters. Let me make that clear. That's the volume is equal to that. Volume is equal to that.
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Using related volumes Solid geometry High school geometry Khan Academy.mp3
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Figure B, the volume is also going to be base times height, which is equal to 28 cubic centimeters. Let me make that clear. That's the volume is equal to that. Volume is equal to that. Now, what about for figure C? What is the volume going to be? What is the formula for the volume of a pyramid?
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Using related volumes Solid geometry High school geometry Khan Academy.mp3
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Volume is equal to that. Now, what about for figure C? What is the volume going to be? What is the formula for the volume of a pyramid? And we've gotten the intuition and proven this to ourselves in other videos. Well, we know that for a pyramid, the volume is going to be equal to 1 3rd times base times height. And we know that it has the same base area as these other characters here, it has the same height.
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Using related volumes Solid geometry High school geometry Khan Academy.mp3
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What is the formula for the volume of a pyramid? And we've gotten the intuition and proven this to ourselves in other videos. Well, we know that for a pyramid, the volume is going to be equal to 1 3rd times base times height. And we know that it has the same base area as these other characters here, it has the same height. And so we know what base times height is, is 28 cubic centimeters. So this is going to be 1 3rd times 28 cubic centimeters. So this is going to be equal to 1 3rd times 28 cubic centimeters, which we could rewrite as 28 over three cubic centimeters.
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Using related volumes Solid geometry High school geometry Khan Academy.mp3
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And we know that it has the same base area as these other characters here, it has the same height. And so we know what base times height is, is 28 cubic centimeters. So this is going to be 1 3rd times 28 cubic centimeters. So this is going to be equal to 1 3rd times 28 cubic centimeters, which we could rewrite as 28 over three cubic centimeters. You could also write that as 9 1 3rd cubic centimeters. So that's for figure C. Now let's think about figure D. I'll do that right over here. Well, for this oblique prism, I guess we could say, you're going to have the same idea, and it comes from Cavalier's principle again.
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Using related volumes Solid geometry High school geometry Khan Academy.mp3
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So this is going to be equal to 1 3rd times 28 cubic centimeters, which we could rewrite as 28 over three cubic centimeters. You could also write that as 9 1 3rd cubic centimeters. So that's for figure C. Now let's think about figure D. I'll do that right over here. Well, for this oblique prism, I guess we could say, you're going to have the same idea, and it comes from Cavalier's principle again. It's gonna have the same formula for volume as figure A. It's going to be the base times the height right over here. So I could write volume is going to be equal to base times height.
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Using related volumes Solid geometry High school geometry Khan Academy.mp3
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Well, for this oblique prism, I guess we could say, you're going to have the same idea, and it comes from Cavalier's principle again. It's gonna have the same formula for volume as figure A. It's going to be the base times the height right over here. So I could write volume is going to be equal to base times height. And we already know what that is. They tell us the base times height is going to be the same as figure A. It's gonna be 28 cubic centimeters.
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Using related volumes Solid geometry High school geometry Khan Academy.mp3
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So I could write volume is going to be equal to base times height. And we already know what that is. They tell us the base times height is going to be the same as figure A. It's gonna be 28 cubic centimeters. Now let's go to figure E. This is an interesting one because it has a different base area. What is going to be the area right over here? Well, they tell us that the base of figure E is a dilation of the base of figure A by a scale factor of 1.5.
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Using related volumes Solid geometry High school geometry Khan Academy.mp3
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It's gonna be 28 cubic centimeters. Now let's go to figure E. This is an interesting one because it has a different base area. What is going to be the area right over here? Well, they tell us that the base of figure E is a dilation of the base of figure A by a scale factor of 1.5. And these are both squares. So figure A will say X by X. This one over here is going to be 1.5X by 1.5X.
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Using related volumes Solid geometry High school geometry Khan Academy.mp3
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Well, they tell us that the base of figure E is a dilation of the base of figure A by a scale factor of 1.5. And these are both squares. So figure A will say X by X. This one over here is going to be 1.5X by 1.5X. So let me write that down. 1.5X times 1.5X. Or another way to think about it, let me do it over here where I have some free space.
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Using related volumes Solid geometry High school geometry Khan Academy.mp3
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This one over here is going to be 1.5X by 1.5X. So let me write that down. 1.5X times 1.5X. Or another way to think about it, let me do it over here where I have some free space. We know that B, which we know is the area of figure A, that would be equal to X times X. Now what's the area of the base of figure E? Well, it's going to be 1.5X times 1.5X or 1.5X squared, which is the same thing as 1.5 squared is 2.25X squared.
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Using related volumes Solid geometry High school geometry Khan Academy.mp3
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Or another way to think about it, let me do it over here where I have some free space. We know that B, which we know is the area of figure A, that would be equal to X times X. Now what's the area of the base of figure E? Well, it's going to be 1.5X times 1.5X or 1.5X squared, which is the same thing as 1.5 squared is 2.25X squared. And we know X squared or X times X, that is equal to B. That is equal to our original base area and all of these other figures. So the area over here, this area right over here, is going to be 2.25 times B.
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Using related volumes Solid geometry High school geometry Khan Academy.mp3
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Well, it's going to be 1.5X times 1.5X or 1.5X squared, which is the same thing as 1.5 squared is 2.25X squared. And we know X squared or X times X, that is equal to B. That is equal to our original base area and all of these other figures. So the area over here, this area right over here, is going to be 2.25 times B. 2.25 times B. I didn't, that wasn't so easy to read. Let me write that a little bit clearer. So 2.25B is the base area right over here.
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Using related volumes Solid geometry High school geometry Khan Academy.mp3
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So the area over here, this area right over here, is going to be 2.25 times B. 2.25 times B. I didn't, that wasn't so easy to read. Let me write that a little bit clearer. So 2.25B is the base area right over here. And so what's the volume of this figure? The volume is going to be the area of the base, which is 2.25 times the area of all these other figures' bases, times the height, which is the same, times H. Now we know what base time, we know what B times H is, where B is the area of figure, the base area of figure A. We know that B times H is 28 cubic centimeters.
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Using related volumes Solid geometry High school geometry Khan Academy.mp3
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So 2.25B is the base area right over here. And so what's the volume of this figure? The volume is going to be the area of the base, which is 2.25 times the area of all these other figures' bases, times the height, which is the same, times H. Now we know what base time, we know what B times H is, where B is the area of figure, the base area of figure A. We know that B times H is 28 cubic centimeters. So the volume for figure E is going to be 2.25 times 28 cubic centimeters, times 28 cubic centimeters. And I don't have a calculator here in front of me and I could do it by hand, but I think you get the general point. You just have to multiply 2.25 times 28 to get the cubic, to get the volume of figure E. And that's because its base has been scaled in each dimension by 1.5.
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Using related volumes Solid geometry High school geometry Khan Academy.mp3
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One, all of their corresponding angles are the same. So the angle right here, angle BAC, is congruent to angle YXZ. Angle BCA is congruent to angle YZX. And angle ABC is congruent to angle XYZ. So all of their angles, their corresponding angles, are the same. And we also see that the sides are just scaled up versions of each other. So to go from the length of XZ to AC, we can multiply by 3.
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Similar triangle basics Similarity Geometry Khan Academy.mp3
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And angle ABC is congruent to angle XYZ. So all of their angles, their corresponding angles, are the same. And we also see that the sides are just scaled up versions of each other. So to go from the length of XZ to AC, we can multiply by 3. We multiplied by 3 there. To go from XY, the length of XY, to the length of AB, which is the corresponding side, we are multiplying by 3. We had to multiply by 3.
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Similar triangle basics Similarity Geometry Khan Academy.mp3
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So to go from the length of XZ to AC, we can multiply by 3. We multiplied by 3 there. To go from XY, the length of XY, to the length of AB, which is the corresponding side, we are multiplying by 3. We had to multiply by 3. And then to go from the length of YZ to the length of BC, we also multiplied by 3. So essentially, triangle ABC is just a scaled up version of triangle XYZ. If they were the same scale, they would be the exact same triangles.
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Similar triangle basics Similarity Geometry Khan Academy.mp3
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We had to multiply by 3. And then to go from the length of YZ to the length of BC, we also multiplied by 3. So essentially, triangle ABC is just a scaled up version of triangle XYZ. If they were the same scale, they would be the exact same triangles. But one is just a bigger, a blown up version of the other one. Or this is a miniaturized version of that one over there. If you just multiply all the sides by 3, you get to this triangle.
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Similar triangle basics Similarity Geometry Khan Academy.mp3
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If they were the same scale, they would be the exact same triangles. But one is just a bigger, a blown up version of the other one. Or this is a miniaturized version of that one over there. If you just multiply all the sides by 3, you get to this triangle. And so we can't call them congruent. But this does seem to be a bit of a special relationship. So we call this special relationship similarity.
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Similar triangle basics Similarity Geometry Khan Academy.mp3
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If you just multiply all the sides by 3, you get to this triangle. And so we can't call them congruent. But this does seem to be a bit of a special relationship. So we call this special relationship similarity. So we can write that triangle ABC is similar to triangle. And we want to make sure we get the corresponding sides right. ABC is going to be similar to XYZ.
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Similar triangle basics Similarity Geometry Khan Academy.mp3
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So we call this special relationship similarity. So we can write that triangle ABC is similar to triangle. And we want to make sure we get the corresponding sides right. ABC is going to be similar to XYZ. And so based on what we just saw, there's actually kind of three ideas here. And they're all equivalent ways of thinking about similarity. One way to think about it is that one is a scaled up version of the other.
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Similar triangle basics Similarity Geometry Khan Academy.mp3
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ABC is going to be similar to XYZ. And so based on what we just saw, there's actually kind of three ideas here. And they're all equivalent ways of thinking about similarity. One way to think about it is that one is a scaled up version of the other. So scaled up or down version of the other. When we talked about congruency, they had to be exactly the same. You could rotate it.
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Similar triangle basics Similarity Geometry Khan Academy.mp3
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One way to think about it is that one is a scaled up version of the other. So scaled up or down version of the other. When we talked about congruency, they had to be exactly the same. You could rotate it. You could shift it. You could flip it. But when you do all those things, they would have to essentially be identical.
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Similar triangle basics Similarity Geometry Khan Academy.mp3
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You could rotate it. You could shift it. You could flip it. But when you do all those things, they would have to essentially be identical. With similarity, you can rotate it. You can shift it. You can flip it.
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Similar triangle basics Similarity Geometry Khan Academy.mp3
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But when you do all those things, they would have to essentially be identical. With similarity, you can rotate it. You can shift it. You can flip it. And you can also scale it up and down in order for something to be similar. So for example, if you say if something is congruent, if let me say triangle CDE, if we know that triangle CDE is congruent to triangle FGH, then we definitely know that they are similar. They're scaled up by a factor of 1.
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Similar triangle basics Similarity Geometry Khan Academy.mp3
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You can flip it. And you can also scale it up and down in order for something to be similar. So for example, if you say if something is congruent, if let me say triangle CDE, if we know that triangle CDE is congruent to triangle FGH, then we definitely know that they are similar. They're scaled up by a factor of 1. Then we know for a fact that CDE is also similar to triangle FGH. But we can't say it the other way around. If triangle ABC is similar to XYZ, we can't say that it's necessarily congruent.
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Similar triangle basics Similarity Geometry Khan Academy.mp3
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They're scaled up by a factor of 1. Then we know for a fact that CDE is also similar to triangle FGH. But we can't say it the other way around. If triangle ABC is similar to XYZ, we can't say that it's necessarily congruent. And we see for this particular example, they definitely are not congruent. So this is one way to think about similarity. The other way to think about similarity is that all of the corresponding angles will be equal.
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Similar triangle basics Similarity Geometry Khan Academy.mp3
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If triangle ABC is similar to XYZ, we can't say that it's necessarily congruent. And we see for this particular example, they definitely are not congruent. So this is one way to think about similarity. The other way to think about similarity is that all of the corresponding angles will be equal. So if something is similar, then all of the corresponding angles are going to be congruent. I always have trouble spelling this. It is two Rs, one S. Corresponding angles are congruent.
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Similar triangle basics Similarity Geometry Khan Academy.mp3
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The other way to think about similarity is that all of the corresponding angles will be equal. So if something is similar, then all of the corresponding angles are going to be congruent. I always have trouble spelling this. It is two Rs, one S. Corresponding angles are congruent. So if we say that triangle ABC is similar to triangle XYZ, that is equivalent to saying that angle ABC is congruent, or we can say that their measures are equal to angle XYZ. That angle BAC is going to be congruent to angle YXZ. YXZ to angle YXZ.
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Similar triangle basics Similarity Geometry Khan Academy.mp3
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It is two Rs, one S. Corresponding angles are congruent. So if we say that triangle ABC is similar to triangle XYZ, that is equivalent to saying that angle ABC is congruent, or we can say that their measures are equal to angle XYZ. That angle BAC is going to be congruent to angle YXZ. YXZ to angle YXZ. And then finally, angle ACB is going to be congruent to angle XZY. So if you have two triangles, all of their angles are the same, then you can say that they're similar. Or if you find two triangles and you're told that they are similar triangles, then you know that all of their corresponding angles are the same.
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Similar triangle basics Similarity Geometry Khan Academy.mp3
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YXZ to angle YXZ. And then finally, angle ACB is going to be congruent to angle XZY. So if you have two triangles, all of their angles are the same, then you can say that they're similar. Or if you find two triangles and you're told that they are similar triangles, then you know that all of their corresponding angles are the same. And the last way to think about it is that the sides are all just scaled up versions of each other. So the sides scaled by the same factor. In the example we did here, the scaling factor was 3.
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Similar triangle basics Similarity Geometry Khan Academy.mp3
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Or if you find two triangles and you're told that they are similar triangles, then you know that all of their corresponding angles are the same. And the last way to think about it is that the sides are all just scaled up versions of each other. So the sides scaled by the same factor. In the example we did here, the scaling factor was 3. It doesn't have to be 3. It just has to be the same scaling factor for every side. If we scale this side up by 3 and we only scale this side up by 2, then we would not be dealing with a similar triangle.
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Similar triangle basics Similarity Geometry Khan Academy.mp3
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In the example we did here, the scaling factor was 3. It doesn't have to be 3. It just has to be the same scaling factor for every side. If we scale this side up by 3 and we only scale this side up by 2, then we would not be dealing with a similar triangle. But if we scaled all of these sides up by 7, then that's still a similar, as long as you have all of them scaled up or scaled down by the exact same factor. So one way to think about it is, and I want to keep having, well, I want to still visualize those triangles. Let me redraw them right over here a little bit simpler, because I'm not talking now in general terms, not even for that specific case.
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Similar triangle basics Similarity Geometry Khan Academy.mp3
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If we scale this side up by 3 and we only scale this side up by 2, then we would not be dealing with a similar triangle. But if we scaled all of these sides up by 7, then that's still a similar, as long as you have all of them scaled up or scaled down by the exact same factor. So one way to think about it is, and I want to keep having, well, I want to still visualize those triangles. Let me redraw them right over here a little bit simpler, because I'm not talking now in general terms, not even for that specific case. So if we say that this is A, B, and C, and this right over here is X, Y, and Z. I just redrew them so I can refer to them when we write down here. If we're saying that these two things right over here are similar, that means that corresponding sides are scaled up versions of each other. So we could say that the length of AB is equal to some scaling factor, and this thing could be less than 1, times the length of XY, the corresponding sides.
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Similar triangle basics Similarity Geometry Khan Academy.mp3
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Let me redraw them right over here a little bit simpler, because I'm not talking now in general terms, not even for that specific case. So if we say that this is A, B, and C, and this right over here is X, Y, and Z. I just redrew them so I can refer to them when we write down here. If we're saying that these two things right over here are similar, that means that corresponding sides are scaled up versions of each other. So we could say that the length of AB is equal to some scaling factor, and this thing could be less than 1, times the length of XY, the corresponding sides. And I know that AB corresponds to XY because of the order in which I wrote this similarity statement. So some scaling factor times XY. We know that BC, the length of BC, needs to be that same scaling factor times the length of YZ.
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Similar triangle basics Similarity Geometry Khan Academy.mp3
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So we could say that the length of AB is equal to some scaling factor, and this thing could be less than 1, times the length of XY, the corresponding sides. And I know that AB corresponds to XY because of the order in which I wrote this similarity statement. So some scaling factor times XY. We know that BC, the length of BC, needs to be that same scaling factor times the length of YZ. So that same scaling factor. And then we know the length of AC is going to be equal to that same scaling factor times XZ. So that's XZ, and this could be a scaling factor.
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Similar triangle basics Similarity Geometry Khan Academy.mp3
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We know that BC, the length of BC, needs to be that same scaling factor times the length of YZ. So that same scaling factor. And then we know the length of AC is going to be equal to that same scaling factor times XZ. So that's XZ, and this could be a scaling factor. So if ABC is larger than XYZ, then these Ks will be larger than 1. If they're the exact same size, if they're essentially congruent triangles, then these Ks will be 1. And if XYZ is bigger than ABC, then these scaling factors will be less than 1.
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Similar triangle basics Similarity Geometry Khan Academy.mp3
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So that's XZ, and this could be a scaling factor. So if ABC is larger than XYZ, then these Ks will be larger than 1. If they're the exact same size, if they're essentially congruent triangles, then these Ks will be 1. And if XYZ is bigger than ABC, then these scaling factors will be less than 1. But another way to write the same statement, notice all I'm saying is corresponding sides are scaled up versions of each other. This first statement right here, if you divide both sides by XY, you get AB over XY is equal to our scaling factor. And then the second statement right over here, if you divide both sides by YZ, you get B.
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Similar triangle basics Similarity Geometry Khan Academy.mp3
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And if XYZ is bigger than ABC, then these scaling factors will be less than 1. But another way to write the same statement, notice all I'm saying is corresponding sides are scaled up versions of each other. This first statement right here, if you divide both sides by XY, you get AB over XY is equal to our scaling factor. And then the second statement right over here, if you divide both sides by YZ, you get B. Let me do that same color. You get BC divided by YZ is equal to that scaling factor. Remember, in the example we just showed, that scaling factor was 3.
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Similar triangle basics Similarity Geometry Khan Academy.mp3
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And then the second statement right over here, if you divide both sides by YZ, you get B. Let me do that same color. You get BC divided by YZ is equal to that scaling factor. Remember, in the example we just showed, that scaling factor was 3. But now we're saying in the more general terms, similarity, as long as you have the same scaling factor. And then finally, if you divide both sides here by the length between X and XZ, or segment XZ's length, you get AC over XZ is equal to K as well. Or another way to think about it is the ratio between corresponding sides.
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Similar triangle basics Similarity Geometry Khan Academy.mp3
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Remember, in the example we just showed, that scaling factor was 3. But now we're saying in the more general terms, similarity, as long as you have the same scaling factor. And then finally, if you divide both sides here by the length between X and XZ, or segment XZ's length, you get AC over XZ is equal to K as well. Or another way to think about it is the ratio between corresponding sides. Notice, this is the ratio between AB and XY. The ratio between BC and YZ. The ratio between AC and XZ.
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Similar triangle basics Similarity Geometry Khan Academy.mp3
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Or another way to think about it is the ratio between corresponding sides. Notice, this is the ratio between AB and XY. The ratio between BC and YZ. The ratio between AC and XZ. That the ratio between corresponding sides all gives us the same constant. Or you could rewrite this as AB over XY is equal to BC over YZ is equal to AC over XZ, which would be equal to some scaling factor, which is equal to K. So if you have similar triangles, let me draw an arrow right over here. Similar triangles means that they're scaled up versions, and you can also flip and rotate and do all this stuff with congruency.
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Similar triangle basics Similarity Geometry Khan Academy.mp3
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So we're told that angle A is circumscribed about circle O. So this is angle A right over here. We're talking about this angle right over there. And when they say it's circumscribed about circle O, that means that the two sides of the angle, they're segments that would be part of tangent lines. So if we were to continue, so for example, that right over there, that line is tangent to the circle. And let me see if I could, and this line is also tangent to the circle. So you see that the sides of angle A are parts of those tangents.
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Tangents of circles problem (example 1) Mathematics II High School Math Khan Academy.mp3
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And when they say it's circumscribed about circle O, that means that the two sides of the angle, they're segments that would be part of tangent lines. So if we were to continue, so for example, that right over there, that line is tangent to the circle. And let me see if I could, and this line is also tangent to the circle. So you see that the sides of angle A are parts of those tangents. And point B and point C are where those tangents actually sit on the circle. So given all of that, they're asking us, what is the measure of angle A? So we're trying to figure this out right here.
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Tangents of circles problem (example 1) Mathematics II High School Math Khan Academy.mp3
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So you see that the sides of angle A are parts of those tangents. And point B and point C are where those tangents actually sit on the circle. So given all of that, they're asking us, what is the measure of angle A? So we're trying to figure this out right here. And I encourage you to pause the video and figure it out on your own. So the key insight here, and there's multiple ways that you could approach this, is to realize that a radius is going to be perpendicular to a tangent, or a radius that intersects a tangent is going to be perpendicular to it. So let me label this.
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Tangents of circles problem (example 1) Mathematics II High School Math Khan Academy.mp3
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So we're trying to figure this out right here. And I encourage you to pause the video and figure it out on your own. So the key insight here, and there's multiple ways that you could approach this, is to realize that a radius is going to be perpendicular to a tangent, or a radius that intersects a tangent is going to be perpendicular to it. So let me label this. So this is going to be a right angle, and this is going to be a right angle. And any quadrilateral, the sum of the angles are going to add up to 360 degrees. And if you wonder where that comes from, well you could divide a quadrilateral into two triangles, where the sum of all the interior angles of a triangle are 180, and since you have two of them, 360 degrees.
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Tangents of circles problem (example 1) Mathematics II High School Math Khan Academy.mp3
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So let me label this. So this is going to be a right angle, and this is going to be a right angle. And any quadrilateral, the sum of the angles are going to add up to 360 degrees. And if you wonder where that comes from, well you could divide a quadrilateral into two triangles, where the sum of all the interior angles of a triangle are 180, and since you have two of them, 360 degrees. So this 92 plus 90 plus 90 plus question mark is going to be equal to 360 degrees. So let me write that down. 92 plus 90 plus 90, so plus 180, plus, let's just call this x, x degrees, plus x.
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Tangents of circles problem (example 1) Mathematics II High School Math Khan Academy.mp3
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And if you wonder where that comes from, well you could divide a quadrilateral into two triangles, where the sum of all the interior angles of a triangle are 180, and since you have two of them, 360 degrees. So this 92 plus 90 plus 90 plus question mark is going to be equal to 360 degrees. So let me write that down. 92 plus 90 plus 90, so plus 180, plus, let's just call this x, x degrees, plus x. They all have to add up to be 360 degrees. So let's see, we could subtract 180 from both sides, and so if we do that we would have 92 plus x is equal to 180. And if we subtract 92 from both sides, we get x is equal to, let's see, 180 minus 90 would be 90, and then we're going to subtract two more, so it's going to be x is equal to 88.
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Tangents of circles problem (example 1) Mathematics II High School Math Khan Academy.mp3
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92 plus 90 plus 90, so plus 180, plus, let's just call this x, x degrees, plus x. They all have to add up to be 360 degrees. So let's see, we could subtract 180 from both sides, and so if we do that we would have 92 plus x is equal to 180. And if we subtract 92 from both sides, we get x is equal to, let's see, 180 minus 90 would be 90, and then we're going to subtract two more, so it's going to be x is equal to 88. So the measure of angle A is 88 degrees. Now let's do one more of these. These are surprisingly fun.
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Tangents of circles problem (example 1) Mathematics II High School Math Khan Academy.mp3
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And if we subtract 92 from both sides, we get x is equal to, let's see, 180 minus 90 would be 90, and then we're going to subtract two more, so it's going to be x is equal to 88. So the measure of angle A is 88 degrees. Now let's do one more of these. These are surprisingly fun. All right, so it says angle A is circumscribed about circle O. We've seen that before in the last question, and they said what is the measure of angle D? So we want to find, let me make sure I'm on the right layer, we want to find the measure of that angle, and let's call that x again.
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Tangents of circles problem (example 1) Mathematics II High School Math Khan Academy.mp3
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These are surprisingly fun. All right, so it says angle A is circumscribed about circle O. We've seen that before in the last question, and they said what is the measure of angle D? So we want to find, let me make sure I'm on the right layer, we want to find the measure of that angle, and let's call that x again. So what can we figure out? Well, just like in the last question, we have a quadrilateral here. Quadrilateral A, B, O, C, and we know two of the angles.
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Tangents of circles problem (example 1) Mathematics II High School Math Khan Academy.mp3
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So we want to find, let me make sure I'm on the right layer, we want to find the measure of that angle, and let's call that x again. So what can we figure out? Well, just like in the last question, we have a quadrilateral here. Quadrilateral A, B, O, C, and we know two of the angles. We know this is going to be a right angle. We have a radius intersecting with a tangent, or part of a tangent, I guess you could say, and then this would be a right angle. And so by the same logic as we saw in the last question, this angle plus this angle plus this angle plus the central angle are going to add up to 360 degrees.
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Tangents of circles problem (example 1) Mathematics II High School Math Khan Academy.mp3
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Quadrilateral A, B, O, C, and we know two of the angles. We know this is going to be a right angle. We have a radius intersecting with a tangent, or part of a tangent, I guess you could say, and then this would be a right angle. And so by the same logic as we saw in the last question, this angle plus this angle plus this angle plus the central angle are going to add up to 360 degrees. So let's call the measure of the central angle, let's call that y over there. So we have y plus 80 degrees, or we'll just assume everything's in degrees, so y plus 80 plus 90 plus 90, so I could say plus another 180, is going to be equal to 360 degrees, sum of the interior angles of a quadrilateral. And so let's see, you have y plus 80 is equal to 180.
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Tangents of circles problem (example 1) Mathematics II High School Math Khan Academy.mp3
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And so by the same logic as we saw in the last question, this angle plus this angle plus this angle plus the central angle are going to add up to 360 degrees. So let's call the measure of the central angle, let's call that y over there. So we have y plus 80 degrees, or we'll just assume everything's in degrees, so y plus 80 plus 90 plus 90, so I could say plus another 180, is going to be equal to 360 degrees, sum of the interior angles of a quadrilateral. And so let's see, you have y plus 80 is equal to 180. If I just subtract 180 from both sides, I'm going to subtract 80 from both sides, and we get y is equal to 100, or the measure of this is 100, the measure of this interior angle right over here is 100 degrees, which also tells us that the measure of this arc, because that interior angle intercepts arc CB right over here, that tells us that the measure of arc CB is also 100 degrees. And so if we're trying to find this angle, the measure of angle D, is the inscribed angle that intercepts the same arc. And we've seen in previous videos that that inscribed angle that intercepts that arc is going to have half the arc's measure.
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Tangents of circles problem (example 1) Mathematics II High School Math Khan Academy.mp3
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And so let's see, you have y plus 80 is equal to 180. If I just subtract 180 from both sides, I'm going to subtract 80 from both sides, and we get y is equal to 100, or the measure of this is 100, the measure of this interior angle right over here is 100 degrees, which also tells us that the measure of this arc, because that interior angle intercepts arc CB right over here, that tells us that the measure of arc CB is also 100 degrees. And so if we're trying to find this angle, the measure of angle D, is the inscribed angle that intercepts the same arc. And we've seen in previous videos that that inscribed angle that intercepts that arc is going to have half the arc's measure. So this is 100 degree measured arc, then the measure of this angle right over here is going to be 50 degrees. So the measure of angle D is 50 degrees. And we are done.
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Tangents of circles problem (example 1) Mathematics II High School Math Khan Academy.mp3
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We've done some examples on this playlist where if you had an angle like that, you might call that a 30-degree angle. If you have an angle like this, you could call that a 90-degree angle, and we'd often use this symbol just like that. If you were to go 180 degrees, you'd essentially form a straight line. Let me make these proper angles. If you go 360 degrees, you've essentially done one full rotation. If you watch figure skating on the Olympics and someone does a rotation, they'll say, oh, they did a 360, or especially in some skateboarding competitions and things like that. But the one thing to realize, and it might not be obvious right from the get-go, is this whole notion of degrees, this is a human-constructed system.
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Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3
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Let me make these proper angles. If you go 360 degrees, you've essentially done one full rotation. If you watch figure skating on the Olympics and someone does a rotation, they'll say, oh, they did a 360, or especially in some skateboarding competitions and things like that. But the one thing to realize, and it might not be obvious right from the get-go, is this whole notion of degrees, this is a human-constructed system. This is not the only way that you can measure angles. If you think about it, you say, well, why do we call a full rotation 360 degrees? There are some possible theories, and I encourage you to think about them.
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Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3
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But the one thing to realize, and it might not be obvious right from the get-go, is this whole notion of degrees, this is a human-constructed system. This is not the only way that you can measure angles. If you think about it, you say, well, why do we call a full rotation 360 degrees? There are some possible theories, and I encourage you to think about them. Why does 360 degrees show up in our culture as a full rotation? Well, there's a couple of theories there. One is ancient calendars, and even our calendar is close to this, but ancient calendars were based on 360 days in a year.
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Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3
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There are some possible theories, and I encourage you to think about them. Why does 360 degrees show up in our culture as a full rotation? Well, there's a couple of theories there. One is ancient calendars, and even our calendar is close to this, but ancient calendars were based on 360 days in a year. Some ancient astronomers observed that things seemed to move 1 360th of the sky per day. Another theory is the ancient Babylonians liked equilateral triangles a lot, and they had a base 60 number system, so they had 60 symbols. We only have 10.
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Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3
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One is ancient calendars, and even our calendar is close to this, but ancient calendars were based on 360 days in a year. Some ancient astronomers observed that things seemed to move 1 360th of the sky per day. Another theory is the ancient Babylonians liked equilateral triangles a lot, and they had a base 60 number system, so they had 60 symbols. We only have 10. We have a base 10. They had 60. In our system, we like to divide things into 10.
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Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3
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We only have 10. We have a base 10. They had 60. In our system, we like to divide things into 10. They probably like to divide things into 60. If you had a circle and you divided it into 6 equilateral triangles, and each of those equilateral triangles you divided into 60 sections because you have a base 60 number system, then you might end up with 360 degrees. What I want to think about in this video is an alternate way of measuring angles.
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Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3
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In our system, we like to divide things into 10. They probably like to divide things into 60. If you had a circle and you divided it into 6 equilateral triangles, and each of those equilateral triangles you divided into 60 sections because you have a base 60 number system, then you might end up with 360 degrees. What I want to think about in this video is an alternate way of measuring angles. That alternate way, even though it might not seem as intuitive to you from the get-go, in some ways is much more mathematically pure than degrees. It's not based on these cultural artifacts of base 60 number systems or astronomical patterns. To some degree, an alien on another planet would not use degrees, especially if the degrees are motivated by these astronomical phenomena, but they might use what we're going to define as a radian.
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Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3
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What I want to think about in this video is an alternate way of measuring angles. That alternate way, even though it might not seem as intuitive to you from the get-go, in some ways is much more mathematically pure than degrees. It's not based on these cultural artifacts of base 60 number systems or astronomical patterns. To some degree, an alien on another planet would not use degrees, especially if the degrees are motivated by these astronomical phenomena, but they might use what we're going to define as a radian. There's a certain degree of purity here. Radians. Let's just cut to the chase and define what a radian is.
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Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3
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To some degree, an alien on another planet would not use degrees, especially if the degrees are motivated by these astronomical phenomena, but they might use what we're going to define as a radian. There's a certain degree of purity here. Radians. Let's just cut to the chase and define what a radian is. Let me draw a circle here. My best attempt at drawing a circle. Not bad.
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Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3
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Let's just cut to the chase and define what a radian is. Let me draw a circle here. My best attempt at drawing a circle. Not bad. Let me draw the center of the circle. Now let me draw this radius. You might already notice the word radius is very close to the word radians, and that's not a coincidence.
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Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3
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Not bad. Let me draw the center of the circle. Now let me draw this radius. You might already notice the word radius is very close to the word radians, and that's not a coincidence. Let's say that this circle has a radius of length r. Let's construct an angle. I'll call that angle theta. Let's construct an angle theta.
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Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3
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You might already notice the word radius is very close to the word radians, and that's not a coincidence. Let's say that this circle has a radius of length r. Let's construct an angle. I'll call that angle theta. Let's construct an angle theta. Let's call this angle right over here theta. Let's just say, for the sake of argument, that this angle is just the exact right measure. If you look at the arc that subtends this angle, that seems like a very fancy word.
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Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3
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