2012 AMC 8 Problems/Problem 23

Problem

An equilateral triangle and a regular hexagon have equal perimeters. If the area of the triangle is 4, what is the area of the hexagon?

$\textbf{(A)}\hspace{.05in}4\qquad\textbf{(B)}\hspace{.05in}5\qquad\textbf{(C)}\hspace{.05in}6\qquad\textbf{(D)}\hspace{.05in}4\sqrt3\qquad\textbf{(E)}\hspace{.05in}6\sqrt3$

Solution

Let the perimeter of the equilateral triangle be $3s$ . The side length of the equilateral triangle would then be $s$ and the sidelength of the hexagon would be $\frac{s}{2}$ .

A hexagon contains six equilateral triangles. One of these triangles would be similar to the large equilateral triangle in the ratio $1 : 4$ , since the sidelength of the small equilateral triangle is half the sidelength of the large one. Thus, the area of one of the small equilateral triangles is $1$ . The area of the hexagon is then $1 \times 6 = \boxed{\textbf{(C)}\ 6}$ .

2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2012 AMC 8 Problems/Problem 23

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