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

We have that $\frac{(\frac{a}{3})^2 \sqrt3 }{4} =\frac{3 (\frac{b}{6})^2 \sqrt3){2}$ (Error compiling LaTeX. Unknown error_msg). Simplifying, we get that $\frac{a^2}{b^2}=\frac{3}{2}$ . Taking the square root and rationalizing the denominator, we see that $\frac{a}{b}=\boxed{\textbf{(B) }\frac{\sqrt6}{2}}$ .

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2013 AMC 10B Problems/Problem 15

Problem

Solution