Difference between revisions of "2013 AMC 10B Problems/Problem 15"
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==Solution== | ==Solution== | ||
| − | 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>. | + | 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 20:10, 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 
.
