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### Course: Integrated math 2>Unit 8

Lesson 5: Solving problems with similar & 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?
• Yes, you can! EC would equal sq.rt.(18^2+24^2) = 30
• how will you find the answer if it does not give you this information?
• You might not be able to solve the problem if not enough information is given.
• 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.
• 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.
• 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?
• 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?
• 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.
• 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?
• 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?
• 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.
• i understand what your saying but i cant do any of the problems correctly
• How do you find corresponding sides in similar triangles?
• 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.)
• I am wondering, my solution arrived at the same answer though I used a different path, or algorithm, if I remember

If Triangle ACE = Isosceles triangle, then CEG = CAG
they are also similar, as are CEG ~ DFE, given this info
and the additional info of, GF + FE = GE * GC/2 = a of CGE
ge/gc : fe/df = (1) 18/24 :6/x (2) 18x = 144 (3)18x/18 = 144/18 = x = 8 = DF = BH

(&statement)because of the Isosceles triangle rule that CE = CA,
and angle = DEF & angle DFE (AA) as related to the above congruency (if =(&statement) then BA = DE, because of AA within an isosceles triangle

then we plug in all the values and get the same answer as your proof.

if this wasn't an isosceles triangle then your method would be required right?

but can't I assume these things? within the context of an isosceles triangle and two
alternate interior angles that must be congruent or equal triangles -this would seem like the faster way, though I am curious about another example with a scalene triangle?

addendum - I should have submitted the video link and time, will correct soon...

thanx,

j