2010 AMC 12A Problem 3

Below is the professionally curated solution for Problem 3 of the 2010 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 12A solutions, or check the answer key.

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Concepts:area ratio percentage

Difficulty rating: 1120

Rectangle $ABCD,$ pictured below, shares $50\%$ of its area with square $EFGH.$ Square $EFGH$ shares $20\%$ of its area with rectangle $ABCD.$ What is $\dfrac{AB}{AD}?$

$4$

$5$

$6$

$8$

$10$

Solution:

Let $s$ be the side length of square $EFGH.$ The shaded overlap has width $s$ and height $AD,$ so its area is $s\cdot AD.$

Because the overlap is $50\%$ of the rectangle, $s\cdot AD=\tfrac12\,AB\cdot AD,$ so $AB=2s.$ Because it is $20\%$ of the square, $s\cdot AD=\tfrac15 s^2,$ so $AD=\tfrac{s}{5}.$

Therefore $\dfrac{AB}{AD}=\dfrac{2s}{s/5}=10.$

Thus, E is the correct answer.

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