Difference between revisions of "2013 AMC 10B Problems/Problem 15"
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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> | ||
Revision as of 20:05, 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 
?
