Difference between revisions of "2012 AMC 8 Problems/Problem 23"

Revision as of 14:50, 17 October 2016

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 $\boxed{\textbf{(C)}\ 6}$ .

Solution 3

Let the side length of the triangle be $s$ and the side length of the hexagon be $t$ . As explained in Solution 1, $s=2t$ , or $t=\frac{s}{2}$ . The area of the triangle is $\frac{s^2\sqrt3}{4}=4$ and the area of the hexagon is $\frac{t^2\sqrt3}{4} \cdot 6=\frac{3t^2\sqrt3}{2}$ . Substituting $\frac{s}{2}$ in for $t$ , we get $\frac{\frac{3s^2\sqrt3}{4}}{2}=\frac{3s^2\sqrt3}{8}.$ $\frac{s^2\sqrt3}{4}=4 \implies \frac{s^2\sqrt3}{8}=2 \implies \frac{3s^2\sqrt3}{8}=\boxed{\textbf{(C)}\ 6}$ .

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 $\dfrac{3s^2\sqrt{3}}{2}$ .

@@ Line 18: / Line 18: @@
 Therefore, our answer is <math>\boxed{\textbf{(C)}\ 6}</math>.
+==Solution 3==
+Let the side length of the triangle be <math>s</math> and the side length of the hexagon be <math>t</math>. As explained in Solution 1, <math>s=2t</math>, or <math>t=\frac{s}{2}</math>. The area of the triangle is <math>\frac{s^2\sqrt3}{4}=4</math> and the area of the hexagon is <math>\frac{t^2\sqrt3}{4} \cdot 6=\frac{3t^2\sqrt3}{2}</math>. Substituting <math>\frac{s}{2}</math> in for <math>t</math>, we get
+<cmath>\frac{\frac{3s^2\sqrt3}{4}}{2}=\frac{3s^2\sqrt3}{8}.</cmath>
+<math>\frac{s^2\sqrt3}{4}=4 \implies \frac{s^2\sqrt3}{8}=2 \implies \frac{3s^2\sqrt3}{8}=\boxed{\textbf{(C)}\ 6}</math>.
 == Notes ==

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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