2013 AMC 12A Problem 9

Attempt Problem 9 of the 2013 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 12A solutions, or check the answer key.

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9.

In ABC,\triangle ABC, AB=AC=28AB = AC = 28 and BC=20.BC = 20. Points D,D, E,E, and FF are on sides AB,\overline{AB}, BC,\overline{BC}, and AC,\overline{AC}, respectively, such that DE\overline{DE} and EF\overline{EF} are parallel to AC\overline{AC} and AB,\overline{AB}, respectively. What is the perimeter of parallelogram ADEF?ADEF?

4848

5252

5656

6060

7272

Answer: C
Concepts:similarityisosceles triangleparallelogram
Difficulty rating: 1460
Solution:

Because EFAB,EF \parallel AB, triangle FECFEC is similar to triangle ABC,ABC, which is isosceles, so FE=FC.FE = FC.

Half the perimeter of parallelogram ADEFADEF is AF+FEAF + FE =AF+FC= AF + FC =AC=28.= AC = 28. The entire perimeter is 56.56.

Thus, the correct answer is C.

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