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### Course: Geometry (all content) > Unit 12

Lesson 5: Solving problems with similar and congruent triangles# Using similar & congruent triangles

Sal uses the similarity of triangles and the congruence of others in this multi-step problem to find the area of a polygon. Created by Sal Khan.

## Want to join the conversation?

- Since GF + FE = 18, and GC = 24, can I use trigonometric functions to find the length of EC?(33 votes)
- Yes, you can! EC would equal sq.rt.(18^2+24^2) = 30(23 votes)

- how will you find the answer if it does not give you this information?(18 votes)
- You might not be able to solve the problem if not enough information is given.(13 votes)

- So... how do we know from the problem that line CG comes down to hit the base of triangle ACE at a right angle? Why can't it be slanted? Given that the fact that it creates a right angle is important for establishing similarity, I think it's a kind of important jump, and I don't see how you can deduce that from the problem.(12 votes)
- The little box at G on the original drawing is used to indicate a right angle. Since it was in the original drawing, it is considered a given premise of the the problem.(17 votes)

- Say I have a RIGHT ANGLED TRIANGLE. If I cut a PERPENDICULAR LINE with 90* angle at base, will the smaller triangle ALWAYS be similar to the entire triangle?(9 votes)
- I assume you mean like this:

http://moodle2.rockyview.ab.ca/pluginfile.php/63364/mod_book/chapter/23910/m10_3_m7/images/m7/m10_3_m7_010_opt.jpeg

If so, then yes. The smaller triangles will always be similar to the larger one. This is because they both share one angle, and they both have a 90 degree angle, and if two of their angles are equal then their last angle must be equal (because all angles add up to 180 degrees in a triangle). Since all their angles are equal, they are therefore similar.(13 votes)

- I got a dumb question: I know how to do this kind of problem but don't know how to do the practice problems, why is that? Maybe I used the wrong numbers to set up a proportion?(10 votes)
- Remember:the only DUMB question is the question not asked.

Answering your question, the practice may be set up differently so it may be a bit confusing.(6 votes)

- I can't do any of the problems correctly because I don't understand which sides to use during the problem. I know that the one you are searching for goes on the top of the first but I don't get which other side lengths to use. Can anyone help?(11 votes)
- i understand what your saying but i cant do any of the problems correctly(9 votes)
- At around seven minutes, Sal proves that triangle AHB is similar to triangle AGC. I had realized that since triangle AHB was congruent to triangle EFD, then triangle AGC is congruent to triangle EGC through AAS (<A is congruent to <C, <AGC and <EGC are both right angles, and CG is congruent to itself). Why did Sal take the extra step?(8 votes)
- I don't know; I was wondering that as well. Maybe he wanted to show us how we could go about doing a more complicated problem.(2 votes)

- How do you find corresponding sides in similar triangles?(6 votes)
- Corresponding sides? You do cross multiplication. In similar triangles, the fraction of the values of one from each triangle on one particular side is equal to another. Here, if two triangles are similar, and one has a side of 5, and the other has the exact same side (but with a different value) has a value of 8, then the fraction can be put as either 5/8 or 8/5, and the other side you want to find can be put in the same way. I put the value of the side on the triangle with 8 on the top, assuming I do 8/5, then put the other at the bottom. There's going to be an unknown value, so cross multiply. If it ends up as 8/5*6/x, then 8*x, 5*6, so 8x=30. And the answer to that is 15/4. Sorry about the long answer, but hope if helps!

(If I do turn out to be wrong, then I strongly apologize, just in case.)(6 votes)

- why dont the quizes ever give us this much information like i would love that but no they give one to two thing and tell us to figure the rest out like u never tell abt that its always something different from what u teach us(7 votes)

## 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.