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### Course: Geometry (Eureka Math/EngageNY)>Unit 2

Lesson 4: Topic C: Lessons 18-20: Similarity applications

# 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?
• i understand what your saying but i cant do any of the problems correctly
• 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?