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2012 AMC 8 Problems/Problem 23

Revision as of 17:34, 17 March 2015 by SirCalcsALot (talk | contribs) (→‎Solution 2)

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

Solution 2

Let the side length of the equilateral triangle be $s$ and the side length of the hexagon be $y$. Since the perimeters are equal, we must have $3s=6y$ which reduces to $s=2y$. Substitute this value in to the area of an equilateral triangle to yield $\dfrac{(2y)^2\sqrt{3}}{4}=\dfrac{4y^2\sqrt{3}}{4}$.

Setting this equal to $4$ gives us $\dfrac{4y^2\sqrt{3}}{4}=4\implies 4y^2\sqrt{3}=16\implies y^2\sqrt{3}=4$.

Substitue $y^2\sqrt{3}$ into the area of a regular hexagon to yield $\dfrac{3(4)}{2}=6$.

Therefore, our answer is $\boxed{\textbf{(C)}\ 6}$.

Notes

The area of an equilateral triangle with side length $s$ is $\dfrac{s^2\sqrt{3}}{4}$.


The area of a regular hexagon with side length $s$ is $\dfrac{3s^2\sqrt{3}}{2}$.

See Also

2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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