Sentence
stringlengths 50
2.16k
| video_title
stringlengths 16
104
|
|---|---|
And it's a 30-60-90 because the side opposite the 30 degrees is half the hypotenuse. And then the side opposite the 60 degrees is the square root of 3 times the other side that's not the hypotenuse. So with that said, this isn't supposed to be a review of 30-60-90 triangles, although I just did it. Let's actually find the trig ratios for the different angles. So if I were to ask you, or if anyone were to ask you, what is the sine of 30 degrees? And remember, 30 degrees is one of the angles in this triangle, but it would apply whenever you have a 30-degree angle and you're dealing with a right triangle. We'll have broader definitions in the future. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
Let's actually find the trig ratios for the different angles. So if I were to ask you, or if anyone were to ask you, what is the sine of 30 degrees? And remember, 30 degrees is one of the angles in this triangle, but it would apply whenever you have a 30-degree angle and you're dealing with a right triangle. We'll have broader definitions in the future. But if you say sine of 30 degrees, hey, this angle right over here is 30 degrees, so I can use this right triangle. And we just have to remember SOH CAH TOA. Let me rewrite it. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
We'll have broader definitions in the future. But if you say sine of 30 degrees, hey, this angle right over here is 30 degrees, so I can use this right triangle. And we just have to remember SOH CAH TOA. Let me rewrite it. SOH CAH TOA. SOH tells us what to do with sine. Sine is opposite over hypotenuse. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
Let me rewrite it. SOH CAH TOA. SOH tells us what to do with sine. Sine is opposite over hypotenuse. Sine of 30 degrees is the opposite side. That is the opposite side, which is 2, over the hypotenuse. The hypotenuse here is 4. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
Sine is opposite over hypotenuse. Sine of 30 degrees is the opposite side. That is the opposite side, which is 2, over the hypotenuse. The hypotenuse here is 4. It is 2 fourths, which is the same thing as 1 half. Sine of 30 degrees, you'll see, is always going to be equal to 1 half. Now, what is the cosine? | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
The hypotenuse here is 4. It is 2 fourths, which is the same thing as 1 half. Sine of 30 degrees, you'll see, is always going to be equal to 1 half. Now, what is the cosine? What is the cosine of 30 degrees? Once again, go back to SOH CAH TOA. The CAH tells us what to do with cosine. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
Now, what is the cosine? What is the cosine of 30 degrees? Once again, go back to SOH CAH TOA. The CAH tells us what to do with cosine. Cosine is adjacent over hypotenuse. So if we're looking at the 30-degree angle, it's the adjacent. This right over here is adjacent. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
The CAH tells us what to do with cosine. Cosine is adjacent over hypotenuse. So if we're looking at the 30-degree angle, it's the adjacent. This right over here is adjacent. It's right next to it. It's not the hypotenuse. It's the adjacent over the hypotenuse. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
This right over here is adjacent. It's right next to it. It's not the hypotenuse. It's the adjacent over the hypotenuse. So it's 2 square roots of 3 adjacent over the hypotenuse, over 4. Or if we simplify that, we divide the numerator and denominator by 2, it's the square root of 3 over 2. Finally, let's do the tangent. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
It's the adjacent over the hypotenuse. So it's 2 square roots of 3 adjacent over the hypotenuse, over 4. Or if we simplify that, we divide the numerator and denominator by 2, it's the square root of 3 over 2. Finally, let's do the tangent. The tangent of 30 degrees. We go back to SOH CAH TOA. TOA tells us what to do with tangent. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
Finally, let's do the tangent. The tangent of 30 degrees. We go back to SOH CAH TOA. TOA tells us what to do with tangent. It's opposite over adjacent. We go to the 30-degree angle, because that's what we care about. Tangent of 30. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
TOA tells us what to do with tangent. It's opposite over adjacent. We go to the 30-degree angle, because that's what we care about. Tangent of 30. Opposite is 2, and the adjacent is 2 square roots of 3. It's right next to it. It's adjacent to it. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
Tangent of 30. Opposite is 2, and the adjacent is 2 square roots of 3. It's right next to it. It's adjacent to it. Adjacent means next to. So 2 square roots of 3. So this is equal to, the 2's cancel out, 1 over the square root of 3. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
It's adjacent to it. Adjacent means next to. So 2 square roots of 3. So this is equal to, the 2's cancel out, 1 over the square root of 3. Or we can multiply the numerator and the denominator by the square root of 3. So we have square root of 3 over square root of 3. This is going to be equal to, the numerator is square root of 3, and then the denominator right over here is just going to be 3. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
So this is equal to, the 2's cancel out, 1 over the square root of 3. Or we can multiply the numerator and the denominator by the square root of 3. So we have square root of 3 over square root of 3. This is going to be equal to, the numerator is square root of 3, and then the denominator right over here is just going to be 3. So we've rationalized it. Square root of 3 over 3. Fair enough. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
This is going to be equal to, the numerator is square root of 3, and then the denominator right over here is just going to be 3. So we've rationalized it. Square root of 3 over 3. Fair enough. Now let's use the same triangle to figure out the trig ratios for the 60 degrees, since we've already drawn it. So what is the sine of 60 degrees? I think you're hopefully getting the hang of it now. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
Fair enough. Now let's use the same triangle to figure out the trig ratios for the 60 degrees, since we've already drawn it. So what is the sine of 60 degrees? I think you're hopefully getting the hang of it now. Sine is opposite over adjacent. So, from the Sohcahtoa, for the 60 degree angle, what side is opposite? Well, it opens out into the 2 square roots of 3. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
I think you're hopefully getting the hang of it now. Sine is opposite over adjacent. So, from the Sohcahtoa, for the 60 degree angle, what side is opposite? Well, it opens out into the 2 square roots of 3. So the opposite side is 2 square roots of 3. And for the 60 degree angle, the adjacent, or sorry, it's opposite over hypotenuse. Don't want to confuse you. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
Well, it opens out into the 2 square roots of 3. So the opposite side is 2 square roots of 3. And for the 60 degree angle, the adjacent, or sorry, it's opposite over hypotenuse. Don't want to confuse you. So it's opposite over hypotenuse. So it's 2 square roots of 3 over 4. 4 is the hypotenuse. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
Don't want to confuse you. So it's opposite over hypotenuse. So it's 2 square roots of 3 over 4. 4 is the hypotenuse. So it is equal to, this simplifies to square root of 3 over 2. What is the cosine of 60 degrees? So remember, Sohcahtoa, cosine is adjacent over hypotenuse. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
4 is the hypotenuse. So it is equal to, this simplifies to square root of 3 over 2. What is the cosine of 60 degrees? So remember, Sohcahtoa, cosine is adjacent over hypotenuse. Adjacent is the 2 side. It's right next to the 60 degree angle. So it's 2 over the hypotenuse, which is 4. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
So remember, Sohcahtoa, cosine is adjacent over hypotenuse. Adjacent is the 2 side. It's right next to the 60 degree angle. So it's 2 over the hypotenuse, which is 4. So this is equal to 1 half. And then finally, what is the tangent? What is the tangent of 60 degrees? | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
So it's 2 over the hypotenuse, which is 4. So this is equal to 1 half. And then finally, what is the tangent? What is the tangent of 60 degrees? Well, tangent, Sohcahtoa, tangent is opposite over adjacent. Opposite the 60 degrees is 2 square roots of 3. And adjacent to that is 2. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
What is the tangent of 60 degrees? Well, tangent, Sohcahtoa, tangent is opposite over adjacent. Opposite the 60 degrees is 2 square roots of 3. And adjacent to that is 2. Adjacent to 60 degrees is 2. So it's opposite over adjacent. 2 square roots of 3 over 2, which is just equal to the square root of 3. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
And adjacent to that is 2. Adjacent to 60 degrees is 2. So it's opposite over adjacent. 2 square roots of 3 over 2, which is just equal to the square root of 3. And I just want to, you know, look how these are related. The sine of 30 degrees is the same thing as the cosine of 60 degrees. The cosine of 30 degrees is the same thing as the sine of 60 degrees. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
2 square roots of 3 over 2, which is just equal to the square root of 3. And I just want to, you know, look how these are related. The sine of 30 degrees is the same thing as the cosine of 60 degrees. The cosine of 30 degrees is the same thing as the sine of 60 degrees. And then these guys are the inverse of each other. I think if you think a little bit about this triangle, it will start to make sense why. We'll keep extending this and give you a lot more practice in the next few videos. | Basic trigonometry II Basic trigonometry Trigonometry Khan Academy.mp3 |
And I always do this before I have to convert between the two. If I do one revolution of a circle, how many radians is that going to be? Well, we know one revolution of a circle is 2 pi radians. And how many degrees is that? If I do one revolution around a circle, well, we know that that's 360. I can either write it with a little degree symbol right like that, or I could write it just like that. And this is really enough information for us to think about how to convert between radians and degrees. | Radian and degree conversion practice Trigonometry Khan Academy.mp3 |
And how many degrees is that? If I do one revolution around a circle, well, we know that that's 360. I can either write it with a little degree symbol right like that, or I could write it just like that. And this is really enough information for us to think about how to convert between radians and degrees. If we want to simplify this a little bit, we can divide both sides by 2, and you could have pi radians are equal to 180 degrees, or another way to think about it, going halfway around a circle in radians is pi radians, or the arc that subtends that angle is pi radiuses. And that's also 180 degrees. And if you want to really think about, well, how many degrees are there per radian, you can divide both sides of this by pi. | Radian and degree conversion practice Trigonometry Khan Academy.mp3 |
And this is really enough information for us to think about how to convert between radians and degrees. If we want to simplify this a little bit, we can divide both sides by 2, and you could have pi radians are equal to 180 degrees, or another way to think about it, going halfway around a circle in radians is pi radians, or the arc that subtends that angle is pi radiuses. And that's also 180 degrees. And if you want to really think about, well, how many degrees are there per radian, you can divide both sides of this by pi. So if you divide both sides of this by pi, you get one radian. I have to go from plural to singular. One radian is equal to 180 over pi degrees. | Radian and degree conversion practice Trigonometry Khan Academy.mp3 |
And if you want to really think about, well, how many degrees are there per radian, you can divide both sides of this by pi. So if you divide both sides of this by pi, you get one radian. I have to go from plural to singular. One radian is equal to 180 over pi degrees. So all I did is I divided both sides by pi. And if you wanted to figure out how many radians are there per degree, you could divide both sides by 180. So you'd get pi over 180 radians is equal to 1 degree. | Radian and degree conversion practice Trigonometry Khan Academy.mp3 |
One radian is equal to 180 over pi degrees. So all I did is I divided both sides by pi. And if you wanted to figure out how many radians are there per degree, you could divide both sides by 180. So you'd get pi over 180 radians is equal to 1 degree. So now I think we are ready to start converting. So let's convert 30 degrees to radians. So let's think about it. | Radian and degree conversion practice Trigonometry Khan Academy.mp3 |
So you'd get pi over 180 radians is equal to 1 degree. So now I think we are ready to start converting. So let's convert 30 degrees to radians. So let's think about it. So I'm going to write it out. And actually, this might remind you of kind of unit analysis that you might do when you first did unit conversion, but it also works out here. So if I were to write 30 degrees, and this is how my brain likes to work with it, I like to write out the word degrees. | Radian and degree conversion practice Trigonometry Khan Academy.mp3 |
So let's think about it. So I'm going to write it out. And actually, this might remind you of kind of unit analysis that you might do when you first did unit conversion, but it also works out here. So if I were to write 30 degrees, and this is how my brain likes to work with it, I like to write out the word degrees. And then I say, well, I want to convert to radians. So I really want to figure out how many radians are there per degree. So let me write this down. | Radian and degree conversion practice Trigonometry Khan Academy.mp3 |
So if I were to write 30 degrees, and this is how my brain likes to work with it, I like to write out the word degrees. And then I say, well, I want to convert to radians. So I really want to figure out how many radians are there per degree. So let me write this down. I want to figure out how many radians do we have per degree. And I haven't filled out how many that is, but we see just the units will cancel out. If we have degrees times radians per degree, the degrees will cancel out and I'll be just left with radians. | Radian and degree conversion practice Trigonometry Khan Academy.mp3 |
So let me write this down. I want to figure out how many radians do we have per degree. And I haven't filled out how many that is, but we see just the units will cancel out. If we have degrees times radians per degree, the degrees will cancel out and I'll be just left with radians. If I multiply the number of degrees I have times the number of radians per degree, we're going to get radians. And hopefully that makes intuitive sense as well. And here we just have to think about, well, if I have pi radians, how many degrees is that? | Radian and degree conversion practice Trigonometry Khan Academy.mp3 |
If we have degrees times radians per degree, the degrees will cancel out and I'll be just left with radians. If I multiply the number of degrees I have times the number of radians per degree, we're going to get radians. And hopefully that makes intuitive sense as well. And here we just have to think about, well, if I have pi radians, how many degrees is that? Well, that's 180 degrees. It comes straight out of this right over here. Pi radians for every 180 degrees or pi over 180 radians per degree. | Radian and degree conversion practice Trigonometry Khan Academy.mp3 |
And here we just have to think about, well, if I have pi radians, how many degrees is that? Well, that's 180 degrees. It comes straight out of this right over here. Pi radians for every 180 degrees or pi over 180 radians per degree. And this is going to get us to 30 times pi over 180, which we'll simplify to 30 over 180 is 1 over 6. So this is equal to pi over 6. Actually, let me write the units out. | Radian and degree conversion practice Trigonometry Khan Academy.mp3 |
Pi radians for every 180 degrees or pi over 180 radians per degree. And this is going to get us to 30 times pi over 180, which we'll simplify to 30 over 180 is 1 over 6. So this is equal to pi over 6. Actually, let me write the units out. This is 30 radians, which is equal to pi over 6 radians. Now let's go the other way. Let's think about if we have pi over 3 radians, and I want to convert that to degrees. | Radian and degree conversion practice Trigonometry Khan Academy.mp3 |
Actually, let me write the units out. This is 30 radians, which is equal to pi over 6 radians. Now let's go the other way. Let's think about if we have pi over 3 radians, and I want to convert that to degrees. So what am I going to get if I convert that to degrees? Well, here we're going to want to figure out how many degrees are there per radian. And one way to think about it is, well, think about the pi and the 180. | Radian and degree conversion practice Trigonometry Khan Academy.mp3 |
Let's think about if we have pi over 3 radians, and I want to convert that to degrees. So what am I going to get if I convert that to degrees? Well, here we're going to want to figure out how many degrees are there per radian. And one way to think about it is, well, think about the pi and the 180. For every 180 degrees, you have pi radians. 180 degrees over pi radians, these are essentially the equivalent thing. Essentially, you're just multiplying this quantity by 1, but you're changing the units. | Radian and degree conversion practice Trigonometry Khan Academy.mp3 |
And one way to think about it is, well, think about the pi and the 180. For every 180 degrees, you have pi radians. 180 degrees over pi radians, these are essentially the equivalent thing. Essentially, you're just multiplying this quantity by 1, but you're changing the units. The radians cancel out, and then the pi's cancel out, and you're left with 180 over 3 degrees. 180 over 3 is 60, and we could either write out the word degrees, or you can write degrees just like that. Now let's think about 45 degrees. | Radian and degree conversion practice Trigonometry Khan Academy.mp3 |
Essentially, you're just multiplying this quantity by 1, but you're changing the units. The radians cancel out, and then the pi's cancel out, and you're left with 180 over 3 degrees. 180 over 3 is 60, and we could either write out the word degrees, or you can write degrees just like that. Now let's think about 45 degrees. So what about 45 degrees? And I'll write it like that just so you can figure it out as they're. Figure it out with that notation as well. | Radian and degree conversion practice Trigonometry Khan Academy.mp3 |
Now let's think about 45 degrees. So what about 45 degrees? And I'll write it like that just so you can figure it out as they're. Figure it out with that notation as well. How many radians will this be equal to? Well, once again, we're going to want to think about how many radians do we have per degree. So we're going to multiply this times, well, we know we have pi radians for every 180 degrees, or we could even write it this way, pi radians for every 180 degrees. | Radian and degree conversion practice Trigonometry Khan Academy.mp3 |
Figure it out with that notation as well. How many radians will this be equal to? Well, once again, we're going to want to think about how many radians do we have per degree. So we're going to multiply this times, well, we know we have pi radians for every 180 degrees, or we could even write it this way, pi radians for every 180 degrees. And here, this might be a little less intuitive, the degrees cancel out, and that's why I'd like to usually write out the word, and you're left with 45 pi over 180 radians. Actually, let me write this with the words written out. Maybe that's more intuitive when I'm thinking about it in terms of using the notation. | Radian and degree conversion practice Trigonometry Khan Academy.mp3 |
So we're going to multiply this times, well, we know we have pi radians for every 180 degrees, or we could even write it this way, pi radians for every 180 degrees. And here, this might be a little less intuitive, the degrees cancel out, and that's why I'd like to usually write out the word, and you're left with 45 pi over 180 radians. Actually, let me write this with the words written out. Maybe that's more intuitive when I'm thinking about it in terms of using the notation. So 45 degrees times, we have pi radians for every 180 degrees. So we are left with, when you multiply, 45 times pi over 180, the degrees have canceled out, and you're just left with radians, which is equal to what? 45 is half of 90, which is half of 180, so this is 1 4th. | Radian and degree conversion practice Trigonometry Khan Academy.mp3 |
Maybe that's more intuitive when I'm thinking about it in terms of using the notation. So 45 degrees times, we have pi radians for every 180 degrees. So we are left with, when you multiply, 45 times pi over 180, the degrees have canceled out, and you're just left with radians, which is equal to what? 45 is half of 90, which is half of 180, so this is 1 4th. This is equal to pi over 4 radians. Let's do one more over here. So let's say that we had negative pi over 2 radians. | Radian and degree conversion practice Trigonometry Khan Academy.mp3 |
45 is half of 90, which is half of 180, so this is 1 4th. This is equal to pi over 4 radians. Let's do one more over here. So let's say that we had negative pi over 2 radians. What's that going to be in degrees? Well, once again, we have to figure out how many degrees are each of these radians. We know that there are 180 degrees for every pi radians, so we're going to get the radians cancel out, the pi's cancel out, and so you have negative 180 over 2. | Radian and degree conversion practice Trigonometry Khan Academy.mp3 |
So in this problem here, we're told that the triangle ACE is isosceles. So that's this big triangle right here. It's isosceles, which means it has two equal sides. And we also know from isosceles triangles that the base angles must be equal. So these two base angles are going to be equal, and this side right over here is going to be equal in length to this side over here. We can say AC is going to be equal to CE. So we get all of that from this first statement right over there. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
And we also know from isosceles triangles that the base angles must be equal. So these two base angles are going to be equal, and this side right over here is going to be equal in length to this side over here. We can say AC is going to be equal to CE. So we get all of that from this first statement right over there. Then they give us some more clues or some more information. They say CG is equal to 24. So this is CG right over here. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
So we get all of that from this first statement right over there. Then they give us some more clues or some more information. They say CG is equal to 24. So this is CG right over here. It has length 24. They tell us BH is equal to DF. BH is equal to DF, so those two things are going to be congruent. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
So this is CG right over here. It has length 24. They tell us BH is equal to DF. BH is equal to DF, so those two things are going to be congruent. They're going to be the same length. Then they tell us that GF is equal to 12. So this is GF right over here. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
BH is equal to DF, so those two things are going to be congruent. They're going to be the same length. Then they tell us that GF is equal to 12. So this is GF right over here. So GF is equal to 12. That's that distance right over there. Then finally they tell us that FE is equal to 6. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
So this is GF right over here. So GF is equal to 12. That's that distance right over there. Then finally they tell us that FE is equal to 6. So this is FE. Then finally they ask us what is the area of CBHFD. So CBHFD. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
Then finally they tell us that FE is equal to 6. So this is FE. Then finally they ask us what is the area of CBHFD. So CBHFD. So they're asking us for the area of this part right over here. That part and that part right over there. That is CBHFD. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
So CBHFD. So they're asking us for the area of this part right over here. That part and that part right over there. That is CBHFD. So let's think about how we can do this. We can figure out the area of the larger triangle. Then from that we can subtract the areas of these little pieces at the end. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
That is CBHFD. So let's think about how we can do this. We can figure out the area of the larger triangle. Then from that we can subtract the areas of these little pieces at the end. Then we'll be able to figure out this middle area, this area that I've shaded. We don't have all the information yet to solve that. We know what the height or the altitude of this triangle is, but we don't know its base. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
Then from that we can subtract the areas of these little pieces at the end. Then we'll be able to figure out this middle area, this area that I've shaded. We don't have all the information yet to solve that. We know what the height or the altitude of this triangle is, but we don't know its base. If we knew its base, we'd say 1 half base times height, we'd get the area of this triangle. Then we'd have to subtract out these areas. We don't have full information there either. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
We know what the height or the altitude of this triangle is, but we don't know its base. If we knew its base, we'd say 1 half base times height, we'd get the area of this triangle. Then we'd have to subtract out these areas. We don't have full information there either. We don't know this height. Once we know that height, then we can figure out this height, but we also don't quite yet know what this length right over here is. Let's just take it piece by piece. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
We don't have full information there either. We don't know this height. Once we know that height, then we can figure out this height, but we also don't quite yet know what this length right over here is. Let's just take it piece by piece. The first thing we might want to do, and you might guess because we've been talking a lot about similarity, is making some type of argument about similarity here because there's a bunch of similar triangles. For example, triangle CGE shares this angle with triangle DFE. They both share this orange angle right here. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
Let's just take it piece by piece. The first thing we might want to do, and you might guess because we've been talking a lot about similarity, is making some type of argument about similarity here because there's a bunch of similar triangles. For example, triangle CGE shares this angle with triangle DFE. They both share this orange angle right here. They both have this right angle right over here. They have two angles in common. They are going to be similar by angle-angle. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
They both share this orange angle right here. They both have this right angle right over here. They have two angles in common. They are going to be similar by angle-angle. You can actually show that there's going to be a third angle in common because these two are parallel lines. We can write that triangle CGE is similar to triangle DFE. We know that by angle-angle. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
They are going to be similar by angle-angle. You can actually show that there's going to be a third angle in common because these two are parallel lines. We can write that triangle CGE is similar to triangle DFE. We know that by angle-angle. We have one set of corresponding angles congruent, and then this angle is in both triangles, so it is a set of corresponding congruent angles right over there. Then once we know that they are similar, we can set up the ratio between sides because we have some information about some of the sides. We know that the ratio between DF and this side right over here, which is a corresponding side, the ratio between DF and CG, which is 24, is going to be the same thing as the ratio between FE, which is 6, and GE, which is not 12. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
We know that by angle-angle. We have one set of corresponding angles congruent, and then this angle is in both triangles, so it is a set of corresponding congruent angles right over there. Then once we know that they are similar, we can set up the ratio between sides because we have some information about some of the sides. We know that the ratio between DF and this side right over here, which is a corresponding side, the ratio between DF and CG, which is 24, is going to be the same thing as the ratio between FE, which is 6, and GE, which is not 12. It's 12 plus 6. It is 18. Then let's see. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
We know that the ratio between DF and this side right over here, which is a corresponding side, the ratio between DF and CG, which is 24, is going to be the same thing as the ratio between FE, which is 6, and GE, which is not 12. It's 12 plus 6. It is 18. Then let's see. 6 over 18, this is just 1 over 3. You get 3DF. You get 3DF is equal to 24. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
Then let's see. 6 over 18, this is just 1 over 3. You get 3DF. You get 3DF is equal to 24. I just cross-multiplied, or you could multiply both sides by 24, multiply both sides by 3. You would get this. Actually, you could just multiply both sides times 24, and you'll get 24 times 1 third, but we'll just do it this way. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
You get 3DF is equal to 24. I just cross-multiplied, or you could multiply both sides by 24, multiply both sides by 3. You would get this. Actually, you could just multiply both sides times 24, and you'll get 24 times 1 third, but we'll just do it this way. Divide both sides by 3. You get DF is equal to 8. We found out that DF is equal to 8, that length right over there. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
Actually, you could just multiply both sides times 24, and you'll get 24 times 1 third, but we'll just do it this way. Divide both sides by 3. You get DF is equal to 8. We found out that DF is equal to 8, that length right over there. That's useful for us because we know that this length right over here is also equal to 8. Now what can we do? It seems like we can make another similarity argument because we have this angle right over here. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
We found out that DF is equal to 8, that length right over there. That's useful for us because we know that this length right over here is also equal to 8. Now what can we do? It seems like we can make another similarity argument because we have this angle right over here. It is congruent to that angle right over there. We also have this angle, which is going to be 90 degrees. We have a 90-degree angle there. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
It seems like we can make another similarity argument because we have this angle right over here. It is congruent to that angle right over there. We also have this angle, which is going to be 90 degrees. We have a 90-degree angle there. That by itself is actually enough to say that we have two similar triangles. We don't even have to show that they have a congruent side here. Actually, we're going to show that these are actually congruent triangles that we're dealing with right over here. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
We have a 90-degree angle there. That by itself is actually enough to say that we have two similar triangles. We don't even have to show that they have a congruent side here. Actually, we're going to show that these are actually congruent triangles that we're dealing with right over here. We have two angles. Actually, we could just go straight to that because when we talk about congruency, if you have an angle that's congruent to another angle, another angle that's congruent to another angle, and then a side that's congruent to another side, you're dealing with two congruent triangles. We can write triangle A, H, B is congruent to triangle E, F, D. We know that because we have angle, angle, side. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
Actually, we're going to show that these are actually congruent triangles that we're dealing with right over here. We have two angles. Actually, we could just go straight to that because when we talk about congruency, if you have an angle that's congruent to another angle, another angle that's congruent to another angle, and then a side that's congruent to another side, you're dealing with two congruent triangles. We can write triangle A, H, B is congruent to triangle E, F, D. We know that because we have angle, angle, side. Angle, angle, side postulate for congruency. If the two triangles are congruent, that makes things convenient. That means if this side is 8, that side is 8. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
We can write triangle A, H, B is congruent to triangle E, F, D. We know that because we have angle, angle, side. Angle, angle, side postulate for congruency. If the two triangles are congruent, that makes things convenient. That means if this side is 8, that side is 8. We already knew that. That's how we established our congruency. That means if this side has length 6, then the corresponding side on this triangle is also going to have length 6. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
That means if this side is 8, that side is 8. We already knew that. That's how we established our congruency. That means if this side has length 6, then the corresponding side on this triangle is also going to have length 6. We can write this length right over here is also going to be 6. I can imagine you can imagine where all of this is going to go, but we want to prove to ourselves. We want to know for sure what the area is. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
That means if this side has length 6, then the corresponding side on this triangle is also going to have length 6. We can write this length right over here is also going to be 6. I can imagine you can imagine where all of this is going to go, but we want to prove to ourselves. We want to know for sure what the area is. We don't want to say, hey, maybe this is the same thing as that. Let's just actually prove it to ourselves. How do we figure out? | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
We want to know for sure what the area is. We don't want to say, hey, maybe this is the same thing as that. Let's just actually prove it to ourselves. How do we figure out? We've almost figured out the entire base of this triangle, but we still haven't figured out the length of HG. Now we can use a similarity argument again because we can see that triangle ABH is actually similar to triangle ACG. They both have this angle here, and then they both have a right angle. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
How do we figure out? We've almost figured out the entire base of this triangle, but we still haven't figured out the length of HG. Now we can use a similarity argument again because we can see that triangle ABH is actually similar to triangle ACG. They both have this angle here, and then they both have a right angle. They have one. ABH has a right angle there. ACG has a right angle right over there. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
They both have this angle here, and then they both have a right angle. They have one. ABH has a right angle there. ACG has a right angle right over there. So you have two angles, two corresponding angles are equal to each other. You're now dealing with similar triangles. We know that triangle ABH, I'll just write it as AHB since I already wrote it this way, AHB is similar to triangle AGC. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
ACG has a right angle right over there. So you have two angles, two corresponding angles are equal to each other. You're now dealing with similar triangles. We know that triangle ABH, I'll just write it as AHB since I already wrote it this way, AHB is similar to triangle AGC. You want to make sure you get the vertices in the right order. A is the orange angle, G is the right angle, and C is the unlabeled angle. This is similar to triangle AGC. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
We know that triangle ABH, I'll just write it as AHB since I already wrote it this way, AHB is similar to triangle AGC. You want to make sure you get the vertices in the right order. A is the orange angle, G is the right angle, and C is the unlabeled angle. This is similar to triangle AGC. What that does for us is now we can use the ratios to figure out what HG is equal to. What can we say over here? We can say that 8 over 24, BH over its corresponding side of the larger triangle, so we say 8 over 24 is equal to 6 over not HG but over AG. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
This is similar to triangle AGC. What that does for us is now we can use the ratios to figure out what HG is equal to. What can we say over here? We can say that 8 over 24, BH over its corresponding side of the larger triangle, so we say 8 over 24 is equal to 6 over not HG but over AG. I think you can see where this is going. You have 1 third is equal to 6 over AG, or we can cross multiply here and we can get AG is equal to 18. This entire length right over here is 18. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
We can say that 8 over 24, BH over its corresponding side of the larger triangle, so we say 8 over 24 is equal to 6 over not HG but over AG. I think you can see where this is going. You have 1 third is equal to 6 over AG, or we can cross multiply here and we can get AG is equal to 18. This entire length right over here is 18. If AG is 18 and AH is 6, then HG is 12. This is what you might have guessed if you were just trying to guess the answer right over here. Now we have proven to ourselves that this base has length of 18 here and then we have another 18 here. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
This entire length right over here is 18. If AG is 18 and AH is 6, then HG is 12. This is what you might have guessed if you were just trying to guess the answer right over here. Now we have proven to ourselves that this base has length of 18 here and then we have another 18 here. It has a length of 36. The entire base here is 36. Now we can figure out the area of this larger, of the entire isosceles triangle. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
Now we have proven to ourselves that this base has length of 18 here and then we have another 18 here. It has a length of 36. The entire base here is 36. Now we can figure out the area of this larger, of the entire isosceles triangle. The area of ACE is going to be equal to 1 half times the base, which is 36, times 24. This is going to be the same thing as 1 half times 36 is 18 times 24. I'll just do that over here on the top. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
Now we can figure out the area of this larger, of the entire isosceles triangle. The area of ACE is going to be equal to 1 half times the base, which is 36, times 24. This is going to be the same thing as 1 half times 36 is 18 times 24. I'll just do that over here on the top. 18 times 24, 8 times 4 is 32, 1 times 4 is 4 plus 3 is 7. We put a zero here because we're not dealing with 2 but 20. You have 2 times 8 is 16, 2 times 1 is 2 plus 1, so it's 360. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
I'll just do that over here on the top. 18 times 24, 8 times 4 is 32, 1 times 4 is 4 plus 3 is 7. We put a zero here because we're not dealing with 2 but 20. You have 2 times 8 is 16, 2 times 1 is 2 plus 1, so it's 360. Then you have the 2, 7 plus 6 is 13, 1 plus 3 is 4. The area of ACE is equal to 432, but we're not done yet. This area that we care about is the area of the entire triangle minus this area and minus this area right over here. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
You have 2 times 8 is 16, 2 times 1 is 2 plus 1, so it's 360. Then you have the 2, 7 plus 6 is 13, 1 plus 3 is 4. The area of ACE is equal to 432, but we're not done yet. This area that we care about is the area of the entire triangle minus this area and minus this area right over here. What is the area of each of these little wedges right over here? It's going to be 1 half times 8 times 6. 1 half times 8 is 4 times 6, so this is going to be 24 right over there. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
This area that we care about is the area of the entire triangle minus this area and minus this area right over here. What is the area of each of these little wedges right over here? It's going to be 1 half times 8 times 6. 1 half times 8 is 4 times 6, so this is going to be 24 right over there. This is going to be another 24 right over there. This is going to be equal to 432 minus 24 minus 24 or minus 48, which is equal to, and we could try this to do this in our head, if we subtract 32, we're going to get to 400. Then we're going to have to subtract another 16. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
1 half times 8 is 4 times 6, so this is going to be 24 right over there. This is going to be another 24 right over there. This is going to be equal to 432 minus 24 minus 24 or minus 48, which is equal to, and we could try this to do this in our head, if we subtract 32, we're going to get to 400. Then we're going to have to subtract another 16. If you subtract 10 from 400, you get 390, so you get to 384, whatever the units were. If these were in meters, then this would be meters squared. If this was centimeters, this would be centimeters squared. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
Then we're going to have to subtract another 16. If you subtract 10 from 400, you get 390, so you get to 384, whatever the units were. If these were in meters, then this would be meters squared. If this was centimeters, this would be centimeters squared. Did I do that right? Let me go the other way. If I add 40 to this, 24 plus another 8 gets me to 432. | Finding area using similarity and congruence Similarity Geometry Khan Academy (2).mp3 |
In particular, we're gonna think about rotations and reflections in this video. And both of those are rigid transformations, which means that the length between corresponding points do not change. So for example, let's say we take this circle A, it's centered at point A, and we were to rotate it around point P. Point P is the center of rotation. And just say, for the sake of argument, we rotate it clockwise a certain angle. So let's say we end up right over, so we're gonna rotate that way. And let's say our center ends up right over here. So our new circle, the image after the rotation, might look something like this. | Properties perserved after rigid transformations.mp3 |
And just say, for the sake of argument, we rotate it clockwise a certain angle. So let's say we end up right over, so we're gonna rotate that way. And let's say our center ends up right over here. So our new circle, the image after the rotation, might look something like this. And I'm hand drawing it, so you gotta forgive that it's not that well hand drawn of a circle. But the circle might look something like this. And so the clear things that are preserved, or maybe it's not so clear, and we're gonna hopefully make them clear right now. | Properties perserved after rigid transformations.mp3 |
So our new circle, the image after the rotation, might look something like this. And I'm hand drawing it, so you gotta forgive that it's not that well hand drawn of a circle. But the circle might look something like this. And so the clear things that are preserved, or maybe it's not so clear, and we're gonna hopefully make them clear right now. Things that are preserved under a rigid transformation like this rotation right over here, this is clearly a rotation. Things that are preserved, well, you have things like the radius of the circle, the radius length, I could say, to be more particular. The radius here is two. | Properties perserved after rigid transformations.mp3 |
And so the clear things that are preserved, or maybe it's not so clear, and we're gonna hopefully make them clear right now. Things that are preserved under a rigid transformation like this rotation right over here, this is clearly a rotation. Things that are preserved, well, you have things like the radius of the circle, the radius length, I could say, to be more particular. The radius here is two. The radius here is also two, right over there. You have things like the perimeter. Well, if the radius is preserved, the perimeter of a circle, which we call a circumference, well, that's just a function of the radius. | Properties perserved after rigid transformations.mp3 |
The radius here is two. The radius here is also two, right over there. You have things like the perimeter. Well, if the radius is preserved, the perimeter of a circle, which we call a circumference, well, that's just a function of the radius. We're talking about two times pi times the radius. So the perimeter, of course, is going to be preserved. In fact, that follows from the fact that the length of the radius is preserved. | Properties perserved after rigid transformations.mp3 |
Well, if the radius is preserved, the perimeter of a circle, which we call a circumference, well, that's just a function of the radius. We're talking about two times pi times the radius. So the perimeter, of course, is going to be preserved. In fact, that follows from the fact that the length of the radius is preserved. And of course, if the radius is preserved, then the area is also going to be preserved. The area is just pi times the radius squared. So if they have the same radius, they're gonna have all of these in common. | Properties perserved after rigid transformations.mp3 |
In fact, that follows from the fact that the length of the radius is preserved. And of course, if the radius is preserved, then the area is also going to be preserved. The area is just pi times the radius squared. So if they have the same radius, they're gonna have all of these in common. And you can also, that feels intuitively right. So what is not preserved? Not preserved. | Properties perserved after rigid transformations.mp3 |
So if they have the same radius, they're gonna have all of these in common. And you can also, that feels intuitively right. So what is not preserved? Not preserved. And this is in general true of rigid transformations, is that they will preserve the distance between corresponding points. If we're transforming a shape, they'll preserve things like perimeter and area. In this case, instead of perimeter, I could say circumference, circumference. | Properties perserved after rigid transformations.mp3 |
Not preserved. And this is in general true of rigid transformations, is that they will preserve the distance between corresponding points. If we're transforming a shape, they'll preserve things like perimeter and area. In this case, instead of perimeter, I could say circumference, circumference. So they'll preserve things like that. They'll preserve angles. We don't have clear angles in this picture, but they'll preserve things like angles. | Properties perserved after rigid transformations.mp3 |
In this case, instead of perimeter, I could say circumference, circumference. So they'll preserve things like that. They'll preserve angles. We don't have clear angles in this picture, but they'll preserve things like angles. But what they won't preserve is the coordinates, coordinates of corresponding points. They might sometimes, but not always. So for example, the coordinate of the center here is for sure going to change. | Properties perserved after rigid transformations.mp3 |
We don't have clear angles in this picture, but they'll preserve things like angles. But what they won't preserve is the coordinates, coordinates of corresponding points. They might sometimes, but not always. So for example, the coordinate of the center here is for sure going to change. We go from the coordinate negative three comma zero to here we went to the coordinate negative one comma two. So the coordinates are not preserved, coordinates of the center. Let's do another example with a non-circular shape, and we'll do a different type of transformation. | Properties perserved after rigid transformations.mp3 |