2012 AMC 10B Problems/Problem 14

Problem

Two equilateral triangles are contained in square whose side length is $2\sqrt 3$ . The bases of these triangles are the opposite side of the square, and their intersection is a rhombus. What is the area of the rhombus?

$\text{(A) } \frac{3}{2} \qquad \text{(B) } \sqrt 3 \qquad \text{(C) } 2\sqrt 2 - 1 \qquad \text{(D) } 8\sqrt 3 - 12 \qquad \text{(E)} \frac{4\sqrt 3}{3}$

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 $\boxed{D}$ is correct.

2012 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2012 AMC 10B Problems/Problem 14

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