2010 AMC 12A Problem 3

Attempt Problem 3 of the 2010 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 2010 AMC 12A solutions, or check the answer key.

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3.

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

44

55

66

88

1010

Answer: E
Concepts:area ratiopercentage
Difficulty rating: 1120
Solution:

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

Because the overlap is 50%50\% of the rectangle, sAD=12ABAD,s\cdot AD=\tfrac12\,AB\cdot AD, so AB=2s.AB=2s. Because it is 20%20\% of the square, sAD=15s2,s\cdot AD=\tfrac15 s^2, so AD=s5.AD=\tfrac{s}{5}.

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

Thus, E is the correct answer.

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