Difference between revisions of "2013 AMC 10B Problems/Problem 15"
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==Problem== | ==Problem== | ||
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A wire is cut into two pieces, one of length <math>a</math> and the other of length <math>b</math>. The piece of length <math>a</math> is bent to form an equilateral triangle, and the piece of length <math>b</math> is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is <math>\frac{a}{b}</math>? | A wire is cut into two pieces, one of length <math>a</math> and the other of length <math>b</math>. The piece of length <math>a</math> is bent to form an equilateral triangle, and the piece of length <math>b</math> is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is <math>\frac{a}{b}</math>? | ||
<math> \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} </math> | <math> \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} </math> | ||
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| + | ==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>. | ||
Revision as of 20:08, 21 February 2013
Problem
A wire is cut into two pieces, one of length 
and the other of length 
. The piece of length is bent to form an equilateral triangle, and the piece of length is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is 
?
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 
. Taking the square root and rationalizing the denominator, we see that 
.
