2013 AMC 12A Problem 9

Below is the professionally curated solution for Problem 9 of the 2013 AMC 12A, 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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Difficulty rating: 1460

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

$48$

$52$

$56$

$60$

$72$

Solution:

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

Half the perimeter of parallelogram $ADEF$ is $AF + FE = AF + FC = AC = 28.$ The entire perimeter is $56.$

Thus, the correct answer is C.

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