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Difference between revisions of "2013 AMC 10B Problems/Problem 15"
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==Solution==
 
==Solution==
  
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We have that <math>\frac{(\frac{a}{3})^2 \sqrt3 }{4} =\frac{3 (\frac{b}{6})^2 \sqrt3){2}</math>.  Simplifying, we get that <math>\frac{a^2}{b^2}=\frac{3}{2}</math>. Taking the square root and rationalizing the denominator, we see that <math>\frac{a}{b}=\boxed{\textbf{(B) }\frac{\sqrt6}{2}}</math>.
+
Using the formulas for area of a regular triangle <math>(\frac{{s}^{2}\sqrt{3}}{4})</math> and regular hexagon <math>(\frac{3{s}^{2}\sqrt{3}}{2})</math> and plugging <math>\frac{a}{3}</math> and <math>\frac{b}{6}</math> into each equation, you find that <math>\frac{{a}^{2}\sqrt{3}}{36}=\frac{{b}^{2}\sqrt{3}}{24}</math>. Simplifying this, you get <math>\frac{a}{b}=\boxed{\textbf{(B)} \frac{\sqrt{6}}{2}}</math>

Revision as of 16:05, 22 February 2013

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

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 $\frac{a}{b}=\boxed{\textbf{(B)} \frac{\sqrt{6}}{2}}$

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