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2012 AMC 10B Problems/Problem 14: Difference between revisions

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Thus, answer choice D is correct.
Thus, answer choice D is correct.
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Revision as of 12:15, 4 July 2013

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. These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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