2013 AMC 10B Problems/Problem 15

Problem

A wire is cut into two pieces, one of length $a$ and the other of length $b$ . The piece of length $a$ is bent to form an equilateral triangle, and the piece of length $b$ is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is $\frac{a}{b}$ ?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{\sqrt{6}}{2}\qquad\textbf{(C)}\ \sqrt{3} \qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{3\sqrt{2}}{2}$

Solution 1

Using the formulas for area of a regular triangle $(\frac{{s}^{2}\sqrt{3}}{4})$ and regular hexagon $(\frac{3{s}^{2}\sqrt{3}}{2})$ and plugging $\frac{a}{3}$ and $\frac{b}{6}$ into each equation, you find that $\frac{{a}^{2}\sqrt{3}}{36}=\frac{{b}^{2}\sqrt{3}}{24}$ . Simplifying this, you get $\frac{a}{b}=\boxed{\textbf{(B)} \frac{\sqrt{6}}{2}}$

Solution 2

The regular hexagon can be broken into 6 small equilateral triangles, each of which is similar to the big equilateral triangle. Each of the small equilateral triangle's area is $\frac{1}{6}$ th of the area of the big equilateral triangle. Therefore each side of the big triangle is $\sqrt{6}$ times the side of the small triangle. The desired ratio is $\frac{3\sqrt{6}}{6}=\frac{\sqrt{6}}{2}\Rightarrow\Box{B}.$

2013 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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2013 AMC 10B Problems/Problem 15

Contents

Problem

Solution 1

Solution 2

See also