2012 AMC 10B Problems/Problem 14

Solution

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 $2\sqrt{3} - 2$ . The formula for the area of an equilateral triangle of length s is $\frac{\sqrt{3}}{4}s^2$ . It follows that the area of the rhombus is:

$2\times\frac{\sqrt{3}}{4}(2\sqrt{3}-2)^2 = 8\sqrt{3} - 12$

Thus, answer choice D is correct. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2012 AMC 10B Problems/Problem 14

Solution