2012 AMC 8 Problems/Problem 23

Problem

An equilateral triangle and a regular hexagon have equal perimeters. If the triangle's area 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 1

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 side length of the large one. Thus, the area of one of the small equilateral triangles is $1$ . The area of the hexagon is then $1 \times 6 = \boxed{\textbf{(C)}\ 6}$ .

Solution 2 (Hard)

Let the side length of the small triangle be $2x$ , so the side length of the hexagon is $x$ . By the Pythagorean theorem, the height of the triangle is $x\sqrt{3}$ , so the area of the triangle is $\frac{x^2}{\sqrt{3}}$ We are given that the area is $4$ , so x must be $\frac{2}{\sqrt[4]{3}}$ . Therefore, by the Pythagorean theorem, the height of one of the triangles in the hexagon (made by subdividing the hexagon into $6$ congruent triangles), is $\frac{\sqrt{3}}{\sqrt[4]{3}}$ , so the area is $\frac{\frac{2}{\sqrt[4]{3}}\times\frac{\sqrt{3}}{\sqrt[4]{3}}}{2}$ , which equals $1$ . Since there are $6$ such triangles in the hexagon, the area of the hexagon is $\boxed{\textbf{(C)}\ 6}$ .

Video Solution

https://youtu.be/SctoIY1cbss ~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/j3QSD5eDpzU?t=2101

~ pi_is_3.14

2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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All AJHSME/AMC 8 Problems and Solutions

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