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2010 AMC 12A Problem 3

Problem 3 of 25EasierGeometryArithmetic

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}?

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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 s⋅AD.s\cdot AD. Because the overlap is 50%50\% of the rectangle, s⋅AD=12 AB⋅AD,s\cdot AD=\tfrac12\,AB\cdot AD, so AB=2s.AB=2s. Because it is 20%20\% of the square, s⋅AD=15s2,s\cdot AD=\tfrac15 s^2, so AD=s5.AD=\tfrac{s}{5}. Therefore ABAD=2ss5=10.\dfrac{AB}{AD}=\dfrac{2s}{\frac{s}{5}}=10. Thus, E is the correct answer.
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