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# Using similar & congruent triangles

CCSS Math: HSG.SRT.B.5

## Video transcript

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 could 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. It has length 24. They tell us 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. And then finally, they tell us that FE is equal to 6. So this is FE, and then finally, they ask us, what is the area of CBHFD? 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. So we could figure out the area of the larger triangle, and then from that we could 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. And 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, then we'd say, hey, 1/2 base times height. We'd get the area of this triangle, and 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 could figure out this height, but we also don't quite yet know what this length right over here is. So let's just take it piece by piece. So 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, and they both have this right angle right over here. So they have two angles in common. They're going to be similar by angle-angle. And you can actually show that there's going to be a third angle in common, because these two are parallel lines. So we can write that triangle CGE is similar to triangle DFE, and 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. And so 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. So 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. And then, let's see. 6 over 18, this is just 1 over 3. 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 1/3. But we'll just do it this way. Divide both sides by 3. You get DF is equal to 8. So we found out that DF is equal to 8, that length right over there. And that's useful for us because we know that this length right over here is also equal to 8. And now what can we do? Well, it seems like we could make another similarity argument, because we have this angle right over here. It is congruent to that angle right over there, and we also have this angle, which is going to be 90 degrees. We have a 90-degree angle there. And actually, 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. And actually, we're going to show that these are actually congruent triangles that we're dealing with right over here. So we have two angles, and 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 are dealing with two congruent triangles. So we can write-- let me write it over here. I'll write it in pink. Triangle AHB is congruent to-- you want to get the corresponding vertices right-- is congruent to triangle EFD. And we know that, because we have angle-angle-side postulate for congruency. And 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. But that means if this side has length 6, then the corresponding side of this triangle is also going to have length 6. So we can write this length right over here is also going to be 6. I can imagine you can imagine where all 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. So how do we figure it out? We've almost figured out the entire base of this triangle, but we still haven't figured out the length of HG. Well, 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. 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. So 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 then C is the unlabeled angle. This is similar to triangle AGC. And what that does for us is now we can use the ratios to figure out what HG is equal to. So what could we say over here? Well, we could 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 a AG. 6 over AG, and I think you can see where this is going. You have 1/3. 1/3 is equal to 6 over AG, or we can cross-multiply here, and we can get AG is equal to 18. So 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. But now we have proven to ourselves that this base has length of-- well, we have 18 here, and then we have another 18 here. So it has a length of 36. So the entire base here is 36. So that is 36. And so now we can figure out the area of the entire isosceles triangle. So the area of ACE is going to be equal to 1/2 times the base, which is 36, times 24. And so this is going to be the same thing as 1/2 times 36 is 18. 18 times 24. I'll just do that over here on the top. So 18 times 24. 8 times 4 is 32. 1 times 4 is 4, plus 3 is 7. Then we put as 0 here, because we're now dealing not with 2, but 20. You have 2 times 8 is 16. 2 times 1 is 2, plus 1. So it's 360, and then you have a 2, 7 plus 6 is 13. 1 plus 3 is 4. So 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. So what is the area of each of these little wedges right over here? So it's going to be 1/2 times 8 times 6. So 1/2 times 8 is 4 times 6. So this is going to be 24 right over there, and this is going to be another 24 right over there. So this is going to be equal to 432 minus 24 minus 24, or minus 48, which is equal to-- and we could try to do this in our head. If we subtract 32, we're going to get to 400. And then we're going to have to subtract another 16. So if you subtract 10 from 400, you get to 390, so that 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. Yup, we're done.