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So let's just write that down, get that out of the way. We know that triangle ABE is similar to triangle CFE. And you want to make sure you get in the right order. F is where the 90 degree angle is. B is where the 90 degree angle is. And then E is where this orange angle is. So CFE, it's similar to triangle CFE.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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F is where the 90 degree angle is. B is where the 90 degree angle is. And then E is where this orange angle is. So CFE, it's similar to triangle CFE. Now let's see if we can figure out that same statement going the other way, looking at triangle DEB. So once again, you have a 90 degree angle here. If this is 90, then this is definitely going to be 90 as well.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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So CFE, it's similar to triangle CFE. Now let's see if we can figure out that same statement going the other way, looking at triangle DEB. So once again, you have a 90 degree angle here. If this is 90, then this is definitely going to be 90 as well. You have a 90 degree angle here at CFB. You have a 90 degree angle at DEF or DEB, however you want to call it. So they have one set of corresponding angles that are congruent.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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If this is 90, then this is definitely going to be 90 as well. You have a 90 degree angle here at CFB. You have a 90 degree angle at DEF or DEB, however you want to call it. So they have one set of corresponding angles that are congruent. And then you'll also see that they both share this angle right over here on the smaller triangle. So I'm now looking at this triangle right over here as opposed to the one on the right. So they both share this angle right over here, DBE.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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So they have one set of corresponding angles that are congruent. And then you'll also see that they both share this angle right over here on the smaller triangle. So I'm now looking at this triangle right over here as opposed to the one on the right. So they both share this angle right over here, DBE. Angle DBE is the same as angle CBF. So I've shown you already that we have this angle is congruent to this angle. And we have this angle as a part of both.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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So they both share this angle right over here, DBE. Angle DBE is the same as angle CBF. So I've shown you already that we have this angle is congruent to this angle. And we have this angle as a part of both. So it's obviously congruent to itself. So we have two corresponding angles that are congruent to each other. So we know that this larger triangle over here is similar to this smaller triangle over there.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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And we have this angle as a part of both. So it's obviously congruent to itself. So we have two corresponding angles that are congruent to each other. So we know that this larger triangle over here is similar to this smaller triangle over there. So let me write this down. So we also know that triangle DEB is similar to triangle CFB. Now what can we do from here?
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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So we know that this larger triangle over here is similar to this smaller triangle over there. So let me write this down. So we also know that triangle DEB is similar to triangle CFB. Now what can we do from here? Well, we know that the ratios of corresponding sides for each of those similar triangles are going to have to be the same. But we only have one side of one of the triangles. So in the case of AB and CFB, we've only been given one side.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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Now what can we do from here? Well, we know that the ratios of corresponding sides for each of those similar triangles are going to have to be the same. But we only have one side of one of the triangles. So in the case of AB and CFB, we've only been given one side. In the case of DEB and CFB, we've only been given one side right over here. So there doesn't seem to be a lot to work with. And this is why this is a slightly more challenging problem here.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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So in the case of AB and CFB, we've only been given one side. In the case of DEB and CFB, we've only been given one side right over here. So there doesn't seem to be a lot to work with. And this is why this is a slightly more challenging problem here. Let's just go ahead and see if we can assume one of the sides, or maybe a side that's shared by both of these larger triangles, and then maybe things will work out from there. So let's just assume that this length right over here, let's just assume that BE is equal to Y. So let me just write this down.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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And this is why this is a slightly more challenging problem here. Let's just go ahead and see if we can assume one of the sides, or maybe a side that's shared by both of these larger triangles, and then maybe things will work out from there. So let's just assume that this length right over here, let's just assume that BE is equal to Y. So let me just write this down. This whole length is going to be equal to Y, because this at least gives us something to work with. And Y is shared by both ABE and DEB, so that seems useful. And then we're going to have to think about the smaller triangles right over there.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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So let me just write this down. This whole length is going to be equal to Y, because this at least gives us something to work with. And Y is shared by both ABE and DEB, so that seems useful. And then we're going to have to think about the smaller triangles right over there. So maybe we'll call this length BFx. And then let's call FE. Well, if this is X, then this is Y minus X.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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And then we're going to have to think about the smaller triangles right over there. So maybe we'll call this length BFx. And then let's call FE. Well, if this is X, then this is Y minus X. So we've introduced a bunch of variables here, but maybe with all the proportionalities and things, just maybe things will work out, or at least we'll have a little bit more sense of where we can go with this actual problem. But now we can start dealing with the similar triangles. For example, so we want to figure out what CF is.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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Well, if this is X, then this is Y minus X. So we've introduced a bunch of variables here, but maybe with all the proportionalities and things, just maybe things will work out, or at least we'll have a little bit more sense of where we can go with this actual problem. But now we can start dealing with the similar triangles. For example, so we want to figure out what CF is. We now know that for these two triangles right here, the ratio of the corresponding sides are going to be constant. So for example, the ratio between CF and 9, their corresponding sides, the ratio between CF and 9 has got to be equal to the ratio between Y minus X, that's that side right there, Y minus X and the corresponding side of the larger triangle. Well, the corresponding side of the larger triangle is this entire length.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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For example, so we want to figure out what CF is. We now know that for these two triangles right here, the ratio of the corresponding sides are going to be constant. So for example, the ratio between CF and 9, their corresponding sides, the ratio between CF and 9 has got to be equal to the ratio between Y minus X, that's that side right there, Y minus X and the corresponding side of the larger triangle. Well, the corresponding side of the larger triangle is this entire length. And that entire length right over there is Y. So it's equal to Y minus X over Y. So we could simplify this a little bit.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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Well, the corresponding side of the larger triangle is this entire length. And that entire length right over there is Y. So it's equal to Y minus X over Y. So we could simplify this a little bit. Well, I'll hold off for a second. Let's see if we can do something similar with this thing on the right. So once again, we have CF, its corresponding side on DEB.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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So we could simplify this a little bit. Well, I'll hold off for a second. Let's see if we can do something similar with this thing on the right. So once again, we have CF, its corresponding side on DEB. So now we're looking at the triangle CFB, not looking at triangle CFE anymore. So now when we're looking at this triangle, CF corresponds to DE. So we have CF over DE is going to be equal to, so CF over DE is going to be equal to X is going to be equal to, let me do that in a different color, it's going to be equal to, I'm using all my colors, it's going to be equal to X over this entire base right over here, so this entire B, which once again we know is Y, so over Y.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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So once again, we have CF, its corresponding side on DEB. So now we're looking at the triangle CFB, not looking at triangle CFE anymore. So now when we're looking at this triangle, CF corresponds to DE. So we have CF over DE is going to be equal to, so CF over DE is going to be equal to X is going to be equal to, let me do that in a different color, it's going to be equal to, I'm using all my colors, it's going to be equal to X over this entire base right over here, so this entire B, which once again we know is Y, so over Y. And now this looks interesting because it looks like we have three unknowns. We have CF, sorry, we know what DE is already. This is 12.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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So we have CF over DE is going to be equal to, so CF over DE is going to be equal to X is going to be equal to, let me do that in a different color, it's going to be equal to, I'm using all my colors, it's going to be equal to X over this entire base right over here, so this entire B, which once again we know is Y, so over Y. And now this looks interesting because it looks like we have three unknowns. We have CF, sorry, we know what DE is already. This is 12. I could have written CF over 12. The ratio between CF and 12 is going to be the ratio between X and Y. So we have three unknowns and only two equations, so it seems hard to solve it first because there's one unknown, another unknown, another unknown, another unknown, another unknown, and another unknown.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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This is 12. I could have written CF over 12. The ratio between CF and 12 is going to be the ratio between X and Y. So we have three unknowns and only two equations, so it seems hard to solve it first because there's one unknown, another unknown, another unknown, another unknown, another unknown, and another unknown. But it looks like I can write this right here, this expression, in terms of X over Y, and then we could do a substitution. So that's why this was a little tricky. So this one right here we can rewrite as CF, let me do it in that same green color, we can rewrite it as CF over 9 is equal to Y minus X over Y, the same thing as Y over Y minus X over Y, over, or 1 minus X over Y.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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So we have three unknowns and only two equations, so it seems hard to solve it first because there's one unknown, another unknown, another unknown, another unknown, another unknown, and another unknown. But it looks like I can write this right here, this expression, in terms of X over Y, and then we could do a substitution. So that's why this was a little tricky. So this one right here we can rewrite as CF, let me do it in that same green color, we can rewrite it as CF over 9 is equal to Y minus X over Y, the same thing as Y over Y minus X over Y, over, or 1 minus X over Y. All I did is I essentially, I guess you could say, distributed the 1 over Y times both of these terms. Y over Y minus X over Y, or 1 minus X minus Y. And this is useful because we already know what X minus Y is equal to.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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So this one right here we can rewrite as CF, let me do it in that same green color, we can rewrite it as CF over 9 is equal to Y minus X over Y, the same thing as Y over Y minus X over Y, over, or 1 minus X over Y. All I did is I essentially, I guess you could say, distributed the 1 over Y times both of these terms. Y over Y minus X over Y, or 1 minus X minus Y. And this is useful because we already know what X minus Y is equal to. Sorry, X over Y is equal to. We already know that X over Y is equal to CF over 12. So this right over here I can replace with this, CF over 12.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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And this is useful because we already know what X minus Y is equal to. Sorry, X over Y is equal to. We already know that X over Y is equal to CF over 12. So this right over here I can replace with this, CF over 12. So then we get, this is a home stretch here, CF, which is what we care about, CF over 9 is equal to 1 minus CF over 12. And now we have one equation with one unknown, and we should be able to solve this right over here. So we could add CF over 12 to both sides.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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So this right over here I can replace with this, CF over 12. So then we get, this is a home stretch here, CF, which is what we care about, CF over 9 is equal to 1 minus CF over 12. And now we have one equation with one unknown, and we should be able to solve this right over here. So we could add CF over 12 to both sides. So you have CF over 9 plus CF over 12 is equal to 1. We just have to find a common denominator here, and I think 36 will do the trick. So 9 times 4 is 36, so if you had to multiply 9 times 4, you have to multiply CF times 4.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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So we could add CF over 12 to both sides. So you have CF over 9 plus CF over 12 is equal to 1. We just have to find a common denominator here, and I think 36 will do the trick. So 9 times 4 is 36, so if you had to multiply 9 times 4, you have to multiply CF times 4. So you have 4 CF, 4 CF over 36 is the same thing as CF over 9, and then plus CF over 12 is the same thing as 3 CF over 36. And this is going to be equal to 1. And then we are left with 4 CF plus 3 CF is 7.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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So 9 times 4 is 36, so if you had to multiply 9 times 4, you have to multiply CF times 4. So you have 4 CF, 4 CF over 36 is the same thing as CF over 9, and then plus CF over 12 is the same thing as 3 CF over 36. And this is going to be equal to 1. And then we are left with 4 CF plus 3 CF is 7. CF over 36 is equal to 1. And then to solve for CF, we can multiply both sides times the reciprocal of 7 over 36. So 36 over 7, multiply both sides times that, 36 over 7.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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And then we are left with 4 CF plus 3 CF is 7. CF over 36 is equal to 1. And then to solve for CF, we can multiply both sides times the reciprocal of 7 over 36. So 36 over 7, multiply both sides times that, 36 over 7. This side things cancel out. And we are left with our final, we get our drum roll now. CF is equal to CF, so all of this stuff cancels out.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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So 36 over 7, multiply both sides times that, 36 over 7. This side things cancel out. And we are left with our final, we get our drum roll now. CF is equal to CF, so all of this stuff cancels out. CF is equal to 1 times 36 over 7, or it's just 36 over 7. And this was a pretty cool problem, because what it shows you is if you have two things, let's see, this thing is some type of a pole or a stick or maybe the wall of a building or who knows what it is. If this is 9 feet tall or 9 yards tall or 9 meters tall, and this over here, this other one is 12 meters tall or 12 yards or whatever units you want to use it, if you were to drape a string from either of them to the base of the other, from the top of one of them to the base of the other, regardless of how far apart these two things are going to be, we just said they're y apart, regardless of how far apart they are, the place where those two strings would intersect are going to be 36 sevenths high, or I guess 5 and 1 sevenths high, regardless of how far they are.
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Challenging similarity problem Similarity Geometry Khan Academy.mp3
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So try to figure out the measure of this angle. I encourage you to pause the video now and try it on your own. All right, now let's work through this together. And the key realization here is to think about this angle. Well, it is an inscribed angle. We see its vertex is sitting on the circle itself. And then think about the arc that it intercepts.
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Inscribed shapes find inscribed angle High School Math Khan Academy.mp3
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And the key realization here is to think about this angle. Well, it is an inscribed angle. We see its vertex is sitting on the circle itself. And then think about the arc that it intercepts. And we see that it intercepts. So let me draw these two sides of the angle. We see that it intercepts arc CD.
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Inscribed shapes find inscribed angle High School Math Khan Academy.mp3
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And then think about the arc that it intercepts. And we see that it intercepts. So let me draw these two sides of the angle. We see that it intercepts arc CD. It intercepts arc CD. And so the measure of this angle, since it's an inscribed angle, is going to be half the measure of arc CD. So if we could figure out the measure of arc CD, then we're going to be in good shape.
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Inscribed shapes find inscribed angle High School Math Khan Academy.mp3
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We see that it intercepts arc CD. It intercepts arc CD. And so the measure of this angle, since it's an inscribed angle, is going to be half the measure of arc CD. So if we could figure out the measure of arc CD, then we're going to be in good shape. So if we figure out the measure of arc CD, then we take half of that and we'll figure out what we care about. Well, what you might notice is that there is another inscribed angle that also intercepts arc CD. We have this angle right over here.
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Inscribed shapes find inscribed angle High School Math Khan Academy.mp3
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So if we could figure out the measure of arc CD, then we're going to be in good shape. So if we figure out the measure of arc CD, then we take half of that and we'll figure out what we care about. Well, what you might notice is that there is another inscribed angle that also intercepts arc CD. We have this angle right over here. It also intercepts arc CD. So you could call this angle C, whoops, you could call this angle CFD. This also intercepts the same arc.
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Inscribed shapes find inscribed angle High School Math Khan Academy.mp3
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We have this angle right over here. It also intercepts arc CD. So you could call this angle C, whoops, you could call this angle CFD. This also intercepts the same arc. So there's two ways you could think about it. Two inscribed angles that intercept the same arc are going to have the same angle measure. So just off of that, you could say that this is going to be, that these two angles are going to have the same measure.
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Inscribed shapes find inscribed angle High School Math Khan Academy.mp3
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This also intercepts the same arc. So there's two ways you could think about it. Two inscribed angles that intercept the same arc are going to have the same angle measure. So just off of that, you could say that this is going to be, that these two angles are going to have the same measure. So you could say this is going to be 50 degrees. Or you could actually solve what the measure of arc CD is. It's going to be twice the measure of the inscribed angle that intercepts it.
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Inscribed shapes find inscribed angle High School Math Khan Academy.mp3
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So just off of that, you could say that this is going to be, that these two angles are going to have the same measure. So you could say this is going to be 50 degrees. Or you could actually solve what the measure of arc CD is. It's going to be twice the measure of the inscribed angle that intercepts it. So the measure of arc CD is going to be 100 degrees, twice the 50 degrees. And then you use that and you say, well, if the measure of that arc is 100 degrees, then an inscribed angle that intercepts it is going to have half its measure. It's going to be 50 degrees.
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Inscribed shapes find inscribed angle High School Math Khan Academy.mp3
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Now what I'm going to do is start cutting up this left cylinder here and shifting things around. So if I just cut it in two and take that bottom cylinder, that bottom half, and shift it a bit, have I changed its volume? Well, clearly I have not changed its volume. I still have the same volume, the combined volume of both of these half cylinders, I could say, are equal to the original cylinder. Now what if I were to cut it up even more? So let me cut it up now into three. Well, once again, I still haven't changed my original volume.
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Cavalieri's principle in 3D Solid geometry High school geometry Khan Academy.mp3
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I still have the same volume, the combined volume of both of these half cylinders, I could say, are equal to the original cylinder. Now what if I were to cut it up even more? So let me cut it up now into three. Well, once again, I still haven't changed my original volume. It's still the same volume as the original, and I'll just cut it up in 2 3rds, and if I shift them around a little bit, I'm not changing the volume, and I could keep doing that. I could cut it up into a bunch of them. Notice, this still has the same original volume.
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Cavalieri's principle in 3D Solid geometry High school geometry Khan Academy.mp3
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Well, once again, I still haven't changed my original volume. It's still the same volume as the original, and I'll just cut it up in 2 3rds, and if I shift them around a little bit, I'm not changing the volume, and I could keep doing that. I could cut it up into a bunch of them. Notice, this still has the same original volume. I've just cut it up into a bunch of sections, a bunch, I've cut it horizontally, and now I'm just shifting things around, but that doesn't change the volume. And I can do it a bunch of times. This looks like some type of poker chips or gambling chips, where I could have my original cylinder, and now I've cut it horizontally into a bunch of these, I guess you could say chips, but clearly it has the same combined volume.
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Cavalieri's principle in 3D Solid geometry High school geometry Khan Academy.mp3
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Notice, this still has the same original volume. I've just cut it up into a bunch of sections, a bunch, I've cut it horizontally, and now I'm just shifting things around, but that doesn't change the volume. And I can do it a bunch of times. This looks like some type of poker chips or gambling chips, where I could have my original cylinder, and now I've cut it horizontally into a bunch of these, I guess you could say chips, but clearly it has the same combined volume. I can shift it around a bit, but it has the same volume. And this leads us to an interesting question, and it's actually a principle known as Cavalieri's Principle, which is if I have two figures that have the same height, and at any point along that height, the cross-sectional area is the same, then the two figures have the same volume. Now, how does what I just say apply to what's going on here?
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Cavalieri's principle in 3D Solid geometry High school geometry Khan Academy.mp3
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This looks like some type of poker chips or gambling chips, where I could have my original cylinder, and now I've cut it horizontally into a bunch of these, I guess you could say chips, but clearly it has the same combined volume. I can shift it around a bit, but it has the same volume. And this leads us to an interesting question, and it's actually a principle known as Cavalieri's Principle, which is if I have two figures that have the same height, and at any point along that height, the cross-sectional area is the same, then the two figures have the same volume. Now, how does what I just say apply to what's going on here? Well, clearly both of these figures have the same height, and then at any point here, wherever I did the cuts, at this point, at the same point on this original cylinder, well, my cross-sectional area is going to be the same, because it's going to be the same area as the base in the case of this cylinder, and so it meets Cavalieri's Principle. But Cavalieri's Principle's nothing exotic. It comes straight out of common sense.
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Cavalieri's principle in 3D Solid geometry High school geometry Khan Academy.mp3
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Now, how does what I just say apply to what's going on here? Well, clearly both of these figures have the same height, and then at any point here, wherever I did the cuts, at this point, at the same point on this original cylinder, well, my cross-sectional area is going to be the same, because it's going to be the same area as the base in the case of this cylinder, and so it meets Cavalieri's Principle. But Cavalieri's Principle's nothing exotic. It comes straight out of common sense. I can just do more cuts like this, and you can see that I have, you could say a more continuous-looking skewed cylinder, but this will have the same volume as our original cylinder. When I shift it around like this, it's not changing the volume, and that's not just true for cylinders. I could do the exact same argument with some form of a prism.
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Cavalieri's principle in 3D Solid geometry High school geometry Khan Academy.mp3
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It comes straight out of common sense. I can just do more cuts like this, and you can see that I have, you could say a more continuous-looking skewed cylinder, but this will have the same volume as our original cylinder. When I shift it around like this, it's not changing the volume, and that's not just true for cylinders. I could do the exact same argument with some form of a prism. Once again, they have the same volume. I could cut the left one in half and shift it around. It doesn't change its volume.
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Cavalieri's principle in 3D Solid geometry High school geometry Khan Academy.mp3
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I could do the exact same argument with some form of a prism. Once again, they have the same volume. I could cut the left one in half and shift it around. It doesn't change its volume. I could cut it more and shift those around. It still doesn't change the volume. So Cavalieri's Principle seems to make a lot of intuitive sense here.
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Cavalieri's principle in 3D Solid geometry High school geometry Khan Academy.mp3
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It doesn't change its volume. I could cut it more and shift those around. It still doesn't change the volume. So Cavalieri's Principle seems to make a lot of intuitive sense here. If I have two figures that have the same height, and at any point along that height, the cross-sectional area is the same, then the figures have the same volume. So these figures also have the same volume. And I could do it with interesting things like, say, a pyramid.
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Cavalieri's principle in 3D Solid geometry High school geometry Khan Academy.mp3
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So Cavalieri's Principle seems to make a lot of intuitive sense here. If I have two figures that have the same height, and at any point along that height, the cross-sectional area is the same, then the figures have the same volume. So these figures also have the same volume. And I could do it with interesting things like, say, a pyramid. These two pyramids have the same volume, and if I were to cut the left pyramid halfway along its height and shift the bottom like this, that doesn't change its volume. And I can keep doing that with more and more cuts. And, because at any point here, these figures have the same height, and at any point on that height, the cross-sectional area is the same, and so they have the same volumes.
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Cavalieri's principle in 3D Solid geometry High school geometry Khan Academy.mp3
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And I could do it with interesting things like, say, a pyramid. These two pyramids have the same volume, and if I were to cut the left pyramid halfway along its height and shift the bottom like this, that doesn't change its volume. And I can keep doing that with more and more cuts. And, because at any point here, these figures have the same height, and at any point on that height, the cross-sectional area is the same, and so they have the same volumes. But once again, it is intuitive. And it goes all the way to the case where you have, you could view it as a continuous pyramid right over here that has been skewed. So no matter how much you skew it, it's going to have the same volume as our original pyramid because they have the same height, and the cross-sectional area at any point in the height is going to be the same.
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Cavalieri's principle in 3D Solid geometry High school geometry Khan Academy.mp3
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And, because at any point here, these figures have the same height, and at any point on that height, the cross-sectional area is the same, and so they have the same volumes. But once again, it is intuitive. And it goes all the way to the case where you have, you could view it as a continuous pyramid right over here that has been skewed. So no matter how much you skew it, it's going to have the same volume as our original pyramid because they have the same height, and the cross-sectional area at any point in the height is going to be the same. We could actually do this with any figure. So these spheres have the same volume. I could cut the left one in half, halfway along its height, and shift it like this.
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Cavalieri's principle in 3D Solid geometry High school geometry Khan Academy.mp3
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So no matter how much you skew it, it's going to have the same volume as our original pyramid because they have the same height, and the cross-sectional area at any point in the height is going to be the same. We could actually do this with any figure. So these spheres have the same volume. I could cut the left one in half, halfway along its height, and shift it like this. Clearly, I'm not changing the volume. And I could make more cuts like that. And clearly, it has still the same volume.
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Cavalieri's principle in 3D Solid geometry High school geometry Khan Academy.mp3
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I could cut the left one in half, halfway along its height, and shift it like this. Clearly, I'm not changing the volume. And I could make more cuts like that. And clearly, it has still the same volume. And this meets Cavalieri's principle because they have the same height, and the cross-section at any point along that height is going to be the same. So even though I can cut that one up and I can shift it, it looks like a different type of object, a different type of thing, but they have the same height, and the cross-sections at any point are the same area. So we have the same volume, which is a useful thing to know, not just to know the principle, but hopefully this video helps you gain some of the intuition for why it makes intuitive sense.
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Cavalieri's principle in 3D Solid geometry High school geometry Khan Academy.mp3
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Looks pretty good. And now let me move this center so it sits on our original circle. So they now sit on each other, or their centers now sit on each other. So I can make it. That looks pretty good. And now let's think about something. If I were to draw this segment right over here, this of course has the length of the radius.
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Constructing equilateral triangle inscribed in circle Geometry Khan Academy.mp3
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So I can make it. That looks pretty good. And now let's think about something. If I were to draw this segment right over here, this of course has the length of the radius. Now let's do another one. And that's either of their radii, because they have the same length. Now let's center this at our new circle.
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Constructing equilateral triangle inscribed in circle Geometry Khan Academy.mp3
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If I were to draw this segment right over here, this of course has the length of the radius. Now let's do another one. And that's either of their radii, because they have the same length. Now let's center this at our new circle. And take it out here. Now this is equal to the radius of the new circle, which is the same as the radius of the old circle. It's going to be the same as this length here.
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Constructing equilateral triangle inscribed in circle Geometry Khan Academy.mp3
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Now let's center this at our new circle. And take it out here. Now this is equal to the radius of the new circle, which is the same as the radius of the old circle. It's going to be the same as this length here. So these two segments have the same length. Now if I were to connect that point to that point, this is a radius of our original circle. And so it's going to have the same length as these two.
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Constructing equilateral triangle inscribed in circle Geometry Khan Academy.mp3
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It's going to be the same as this length here. So these two segments have the same length. Now if I were to connect that point to that point, this is a radius of our original circle. And so it's going to have the same length as these two. So this right over here, I have constructed an equilateral triangle. Now why is this at all useful? Well, we know that the angles in an equilateral triangle are 60 degrees.
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Constructing equilateral triangle inscribed in circle Geometry Khan Academy.mp3
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And so it's going to have the same length as these two. So this right over here, I have constructed an equilateral triangle. Now why is this at all useful? Well, we know that the angles in an equilateral triangle are 60 degrees. So we know that this angle right over here is 60 degrees. Now why is this being 60 degrees interesting? Well, imagine if we constructed another triangle out here, just symmetrically, but kind of flipped down, just like this.
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Constructing equilateral triangle inscribed in circle Geometry Khan Academy.mp3
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Well, we know that the angles in an equilateral triangle are 60 degrees. So we know that this angle right over here is 60 degrees. Now why is this being 60 degrees interesting? Well, imagine if we constructed another triangle out here, just symmetrically, but kind of flipped down, just like this. Well, the same argument, this angle right over here between these two edges, this is also going to be 60 degrees. So this entire interior angle, if we add those two up, are going to be 120 degrees. Now why is that interesting?
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Constructing equilateral triangle inscribed in circle Geometry Khan Academy.mp3
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Well, imagine if we constructed another triangle out here, just symmetrically, but kind of flipped down, just like this. Well, the same argument, this angle right over here between these two edges, this is also going to be 60 degrees. So this entire interior angle, if we add those two up, are going to be 120 degrees. Now why is that interesting? Well, if this interior angle is 120 degrees, then that means that this arc right over here is 120 degrees, or it's a third of the way around the triangle. This right over here is a third of the way around the triangle. Since that's a third of the way around the triangle, if I were to connect these two dots, that is going to be, this right over here, is going to be a side of our equilateral triangle.
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Constructing equilateral triangle inscribed in circle Geometry Khan Academy.mp3
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Now why is that interesting? Well, if this interior angle is 120 degrees, then that means that this arc right over here is 120 degrees, or it's a third of the way around the triangle. This right over here is a third of the way around the triangle. Since that's a third of the way around the triangle, if I were to connect these two dots, that is going to be, this right over here, is going to be a side of our equilateral triangle. This right over here, it's secant to an arc that is 1 third of the entire circle. And now I can keep doing this. Let's move, I'll reuse these, let's move our circle around.
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Constructing equilateral triangle inscribed in circle Geometry Khan Academy.mp3
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Since that's a third of the way around the triangle, if I were to connect these two dots, that is going to be, this right over here, is going to be a side of our equilateral triangle. This right over here, it's secant to an arc that is 1 third of the entire circle. And now I can keep doing this. Let's move, I'll reuse these, let's move our circle around. And so now I'm going to move my circle along the circle. And what I want, again, I just want to intersect these two points. And so now, let's see, I could take one of these, take it there, take it there, same exact argument.
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Constructing equilateral triangle inscribed in circle Geometry Khan Academy.mp3
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Let's move, I'll reuse these, let's move our circle around. And so now I'm going to move my circle along the circle. And what I want, again, I just want to intersect these two points. And so now, let's see, I could take one of these, take it there, take it there, same exact argument. This angle that I haven't fully drawn, or this arc, you could say, is 120 degrees. So this is going to be one side of our equilateral triangle. It's secant to an arc of 120 degrees.
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Constructing equilateral triangle inscribed in circle Geometry Khan Academy.mp3
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And so now, let's see, I could take one of these, take it there, take it there, same exact argument. This angle that I haven't fully drawn, or this arc, you could say, is 120 degrees. So this is going to be one side of our equilateral triangle. It's secant to an arc of 120 degrees. So let's move this around again. Actually, we don't even have to move this around anymore. We could just connect those last two dots.
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Constructing equilateral triangle inscribed in circle Geometry Khan Academy.mp3
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It's secant to an arc of 120 degrees. So let's move this around again. Actually, we don't even have to move this around anymore. We could just connect those last two dots. So we could just connect this one. Actually, I just want to connect that one to that one. And just like that, and we're done.
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Constructing equilateral triangle inscribed in circle Geometry Khan Academy.mp3
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And I want to think about which of these figures are going to be unchanged if I were to rotate it 180 degrees. So let's do two examples of that. So I have two copies of this square. If I were to take one of these copies and rotate it 180 degrees, so let me show you what that looks like. And we're going to rotate around its center 180 degrees. We're going to rotate around the center. So this is it.
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Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3
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If I were to take one of these copies and rotate it 180 degrees, so let me show you what that looks like. And we're going to rotate around its center 180 degrees. We're going to rotate around the center. So this is it. So we're rotating it. That's rotated 90 degrees. And then we've rotated 180 degrees.
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Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3
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So this is it. So we're rotating it. That's rotated 90 degrees. And then we've rotated 180 degrees. And notice, the figure looks exactly the same. This one, the square, is unchanged by 180 degree rotation. Now what about this trapezoid right over here?
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Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3
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And then we've rotated 180 degrees. And notice, the figure looks exactly the same. This one, the square, is unchanged by 180 degree rotation. Now what about this trapezoid right over here? Let's think about what happens when it's rotated by 180 degrees. So that is 90 degrees and 180 degrees. So this has now been changed.
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Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3
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Now what about this trapezoid right over here? Let's think about what happens when it's rotated by 180 degrees. So that is 90 degrees and 180 degrees. So this has now been changed. Now I have the short side, or I have my base is short and my top is long before my base was long and my top was short. So when I rotated 180 degrees, I didn't get to the exact same figure. I have essentially an upside down version of it.
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Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3
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So this has now been changed. Now I have the short side, or I have my base is short and my top is long before my base was long and my top was short. So when I rotated 180 degrees, I didn't get to the exact same figure. I have essentially an upside down version of it. So what I want you to do for the rest of these is pause the video and think about which of these will be unchanged and which of them will be changed when you rotate by 180 degrees. So let's look at this star thing. And one way that my brain visualizes is imagine the center.
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Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3
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I have essentially an upside down version of it. So what I want you to do for the rest of these is pause the video and think about which of these will be unchanged and which of them will be changed when you rotate by 180 degrees. So let's look at this star thing. And one way that my brain visualizes is imagine the center. That's what we're rotating around. And then if you rotate 180 degrees, imagine any point, say this point, relative to the center. If you were to rotate it 90 degrees, you would get over here.
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Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3
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And one way that my brain visualizes is imagine the center. That's what we're rotating around. And then if you rotate 180 degrees, imagine any point, say this point, relative to the center. If you were to rotate it 90 degrees, you would get over here. And then if you rotate 180 degrees, you go over here. You go on the opposite side of the center from where it is. So from that point to the center, you keep going that same distance, you'll end up right over there.
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Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3
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If you were to rotate it 90 degrees, you would get over here. And then if you rotate 180 degrees, you go over here. You go on the opposite side of the center from where it is. So from that point to the center, you keep going that same distance, you'll end up right over there. So this one looks like it won't be changed, but let's verify it. So we're gonna rotate 90 degrees, and then we have 180 degrees. It is unchanged.
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Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3
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So from that point to the center, you keep going that same distance, you'll end up right over there. So this one looks like it won't be changed, but let's verify it. So we're gonna rotate 90 degrees, and then we have 180 degrees. It is unchanged. Now let's look at this parallelogram right over here. So its center, if we think about its center where my cursor is right now, think about this point. The distance between that point and the center, if we were to keep going that same distance again, you would get to that point.
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Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3
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It is unchanged. Now let's look at this parallelogram right over here. So its center, if we think about its center where my cursor is right now, think about this point. The distance between that point and the center, if we were to keep going that same distance again, you would get to that point. Likewise, the distance between this point and the center, if we were to go that same distance again, you would get to that point. So it seems like that point would end up there, that point would end up there, and vice versa. So I think this one will be unchanged by rotation.
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Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3
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The distance between that point and the center, if we were to keep going that same distance again, you would get to that point. Likewise, the distance between this point and the center, if we were to go that same distance again, you would get to that point. So it seems like that point would end up there, that point would end up there, and vice versa. So I think this one will be unchanged by rotation. So let's verify it. So you go 90 degrees, and then you go 180 degrees. Or I should say, it will be unchanged by rotation of 180 degrees around its center.
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Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3
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So I think this one will be unchanged by rotation. So let's verify it. So you go 90 degrees, and then you go 180 degrees. Or I should say, it will be unchanged by rotation of 180 degrees around its center. We got the same figure. Now let's think about this triangle. So if you think about the center of the figure, let's say the center of the figure is right around here.
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Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3
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Or I should say, it will be unchanged by rotation of 180 degrees around its center. We got the same figure. Now let's think about this triangle. So if you think about the center of the figure, let's say the center of the figure is right around here. If you take this point, go to the center of the figure, and then go that distance again, you end up in a place where there's no point right now. So that point is going to end up there, this point is going to end up there, this point is gonna end up here. So you're not going to have the same figure anymore.
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Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3
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So if you think about the center of the figure, let's say the center of the figure is right around here. If you take this point, go to the center of the figure, and then go that distance again, you end up in a place where there's no point right now. So that point is going to end up there, this point is going to end up there, this point is gonna end up here. So you're not going to have the same figure anymore. And so we can rotate it to verify. So that's rotated 90 degrees, and then that's rotated 180 degrees. So we've kind of turned this thing on its side.
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Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3
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So you're not going to have the same figure anymore. And so we can rotate it to verify. So that's rotated 90 degrees, and then that's rotated 180 degrees. So we've kind of turned this thing on its side. It is not the same thing. Now let's think about this figure right over here. Well, this figure, if you rotate it 180 degrees, this point is now going to be down here, and this point is going to be up here.
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Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3
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So we've kind of turned this thing on its side. It is not the same thing. Now let's think about this figure right over here. Well, this figure, if you rotate it 180 degrees, this point is now going to be down here, and this point is going to be up here. So you're gonna make, essentially it's going to be an upside down version of the same kite. And we can view that, we can visualize that now. So it's going to be different, but let's just show it.
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Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3
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Well, this figure, if you rotate it 180 degrees, this point is now going to be down here, and this point is going to be up here. So you're gonna make, essentially it's going to be an upside down version of the same kite. And we can view that, we can visualize that now. So it's going to be different, but let's just show it. So that is 90 degrees, and now this is 100 and 180 degrees. If it was actually symmetric, if it actually was symmetric about the horizontal axis, then we would have a different scenario. We would have a different scenario with this thing right over here.
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Rotating polygons 180 degrees about their center Transformations Geometry Khan Academy.mp3
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We're told that point P was rotated about the origin zero comma zero by 60 degrees. Which point is the image of P? Pause this video and see if you can figure that out. Alright, now let's think about it. It's being, this is point P. It's being rotated around the origin, zero comma zero by 60 degrees. So if originally point P is right over here, and we're rotating by positive 60 degrees, that means we go counterclockwise by 60 degrees. So this looks like about 60 degrees right over here.
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Positive and negative rotaion of points example.mp3
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Alright, now let's think about it. It's being, this is point P. It's being rotated around the origin, zero comma zero by 60 degrees. So if originally point P is right over here, and we're rotating by positive 60 degrees, that means we go counterclockwise by 60 degrees. So this looks like about 60 degrees right over here. One way to think about 60 degrees is that that's 1 3rd of 180 degrees. So does this look like 1 3rd of 180 degrees? Remember, 180 degrees would be almost a full line.
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Positive and negative rotaion of points example.mp3
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So this looks like about 60 degrees right over here. One way to think about 60 degrees is that that's 1 3rd of 180 degrees. So does this look like 1 3rd of 180 degrees? Remember, 180 degrees would be almost a full line. So that indeed does look like 1 3rd of 180 degrees, 60 degrees. It gets us to point C. Point C, and it looks like it's the same distance from the origin we have just rotated by 60 degrees. Point D looks like it's more than 60 degree rotation, so I won't go with that one.
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Positive and negative rotaion of points example.mp3
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Remember, 180 degrees would be almost a full line. So that indeed does look like 1 3rd of 180 degrees, 60 degrees. It gets us to point C. Point C, and it looks like it's the same distance from the origin we have just rotated by 60 degrees. Point D looks like it's more than 60 degree rotation, so I won't go with that one. Alright, let's do one more of these. So we're told point P was rotated by negative 90 degrees. The center of rotation is indicated.
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Positive and negative rotaion of points example.mp3
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Point D looks like it's more than 60 degree rotation, so I won't go with that one. Alright, let's do one more of these. So we're told point P was rotated by negative 90 degrees. The center of rotation is indicated. Which point is the image of P? So once again, pause this video and try to think about it. Alright, so we have our center of rotation, this is our point P, and we're rotating by negative 90 degrees.
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Positive and negative rotaion of points example.mp3
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The center of rotation is indicated. Which point is the image of P? So once again, pause this video and try to think about it. Alright, so we have our center of rotation, this is our point P, and we're rotating by negative 90 degrees. So this means we are going clockwise. So we're going in that direction, and 90 degrees is easy to spot, it's a right angle. And so it would look like that, and it looks like it is getting us right to point A.
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Positive and negative rotaion of points example.mp3
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And I want to imagine what type of a shape I would get if I were to make a vertical cut. And just to refresh ourselves or give us a sense of what a vertical cut is, imagine if this was made out of Jell-O or something kind of fairly soft, but it's still a three-dimensional solid, and I were to make a cut. Let's say I were to make a cut right over here. So let's say I have this big, sharp metal thing. Let me draw it like that. So you have this big, sharp metal thing. Let me draw it a little bit neater.
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Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3
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So let's say I have this big, sharp metal thing. Let me draw it like that. So you have this big, sharp metal thing. Let me draw it a little bit neater. So you have this big, sharp metal thing that I'm going to cut right over here. So this is the thing that I'm going to make the cut. And I'm going to go straight down.
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Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3
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Let me draw it a little bit neater. So you have this big, sharp metal thing that I'm going to cut right over here. So this is the thing that I'm going to make the cut. And I'm going to go straight down. This is a vertical cut that we're talking about. So this is the thing that I'm going to cut with. Let me make it big enough so that it can capture the shape that will result.
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Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3
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And I'm going to go straight down. This is a vertical cut that we're talking about. So this is the thing that I'm going to cut with. Let me make it big enough so that it can capture the shape that will result. So this thing right over here, it's right in front. And I'm going to make it go straight down and cut through this Jell-O or whatever you want to call it, this rectangular pyramid of Jell-O. And what would be the resulting shape of the intersection between the Jell-O and this thing that I'm using to cut it?
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Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3
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Let me make it big enough so that it can capture the shape that will result. So this thing right over here, it's right in front. And I'm going to make it go straight down and cut through this Jell-O or whatever you want to call it, this rectangular pyramid of Jell-O. And what would be the resulting shape of the intersection between the Jell-O and this thing that I'm using to cut it? And now I encourage you to pause your video and think about what the resulting shape would be. And the shape would be in two dimensions, right? This purple surface is a two-dimensional.
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Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3
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And what would be the resulting shape of the intersection between the Jell-O and this thing that I'm using to cut it? And now I encourage you to pause your video and think about what the resulting shape would be. And the shape would be in two dimensions, right? This purple surface is a two-dimensional. You could view it as part of a plane. And so where this intersects when you cut down this rectangular pyramid is the shape we're looking for. So I encourage you to pause the video and think about it or try to come up with it on your own.
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Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3
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This purple surface is a two-dimensional. You could view it as part of a plane. And so where this intersects when you cut down this rectangular pyramid is the shape we're looking for. So I encourage you to pause the video and think about it or try to come up with it on your own. So let's think about it. And let me draw the rectangular pyramid again. So that's the same one.
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Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3
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So I encourage you to pause the video and think about it or try to come up with it on your own. So let's think about it. And let me draw the rectangular pyramid again. So that's the same one. And now let me see what it would look like once I've done my cut, once I've brought this thing down. So then this is where I cut. So I cut it right over here.
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Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3
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So that's the same one. And now let me see what it would look like once I've done my cut, once I've brought this thing down. So then this is where I cut. So I cut it right over here. And then it'll exit the bottom. It'll cut along this side like that, cut along that side like that. And then it'll exit the bottom right over there.
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Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3
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So I cut it right over here. And then it'll exit the bottom. It'll cut along this side like that, cut along that side like that. And then it'll exit the bottom right over there. And so let me draw my whole thing. And so once I slice it down, it will look like this. My best shot at drawing it.
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Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3
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And then it'll exit the bottom right over there. And so let me draw my whole thing. And so once I slice it down, it will look like this. My best shot at drawing it. It will look like this. This is a rectangular. This is a vertical cut.
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Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3
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My best shot at drawing it. It will look like this. This is a rectangular. This is a vertical cut. So I brought this thing down. And now the intersection between the thing that I'm cutting with and this pyramid is going to be this shape right over here. It cut into the top right over there.
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Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3
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This is a vertical cut. So I brought this thing down. And now the intersection between the thing that I'm cutting with and this pyramid is going to be this shape right over here. It cut into the top right over there. It would get all the way to the bottom right over there. And along this side, it would cut right there. And along that side, it would cut right over there.
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Slice a rectangular pyramid Perimeter, area, and volume Geometry Khan Academy.mp3
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