2013 AMC 10A Problem 12

Attempt Problem 12 of the 2013 AMC 10A 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 10A solutions, or check the answer key.

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12.

In ABC,\triangle ABC, AB=AC=28AB=AC=28 and BC=20.BC=20. Points D,E,D,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:similarityparallelogramperimeter
Difficulty rating: 1420
Solution:

Note that DBEABC\triangle DBE \sim \triangle ABC and FECABC\triangle FEC \sim \triangle ABC due to the parallel lines.

This tells us that DB=DEDB = DE and FE=FC.FE = FC. We have that the perimeter of ADEFADEF is AD+DE+EF+AF AD + DE + EF + AF =AD+DB+FC+AF= AD + DB + FC + AF =AB+AC = AB + AC =56.= 56.

Thus, C is the correct answer.

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