2012 AMC 10B Problems/Problem 14

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Solution

2012 AMC-10B-14.jpg

Observe that the rhombus is made up of two congruent equilateral triangles with side length equal to GF. Since AE has length sqrt(3) and triangle AEF is a 30-60-90 triangle, it follows that EF has length 1. By symmetry, HG also has length 1. Thus GF has length 2sqrt(3) - 2. The formula for the area of an equilateral triangle of length s is (s^2)sqrt(3)/4. It follows that the area of the rhombus is:

2((2sqrt(3) - 2)^2)sqrt(3)/4 = (sqrt(3)/2)(16 - 8sqrt(3)) = 8sqrt(3)-12. Thus, answer choice D is correct.