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2005 AMC 10A Problem 8

Problem 8 of 25EasierGeometry

In the figure, the length of side ABAB of square ABCDABCD is 50,\sqrt{50}, EE is between BB and H,H, and BE=1.BE = 1. What is the area of the inner square EFGH?EFGH?

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Solution

The triangles ABH,ABH, BCE,BCE, CDF,CDF, and DAGDAG are congruent right triangles. In △BCE\triangle BCE the hypotenuse is BC=50BC = \sqrt{50} and BE=1,BE = 1, so CE=50−1=7.CE = \sqrt{50 - 1} = 7. Since BH=CE=7BH = CE = 7 and EE lies on BHBH with BE=1,BE = 1, the inner square’s side is EH=7−1=6,EH = 7 - 1 = 6, giving area 62=36.6^2 = 36. Thus, the correct answer is C.
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Tagged: square (geometry) · Pythagorean Theorem · congruence (geometry)

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