2013 AMC 10B Problems/Problem 15
Contents
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 1
Using the formulas for area of a regular triangle and regular hexagon and plugging and into each equation, you find that . Simplifying this, you get
Solution 2
The regular hexagon can be broken into 6 small equilateral triangles, each of which is similar to the big equilateral triangle. The big triangle's area is 6 times the area of one of the little triangles. Therefore each side of the big triangle is times the side of the small triangle. The desired ratio is $\frac{3\sqrt{6}}{6}=\frac{\sqrt{6}}{2}\Rightarrow\boxed{\textbf{(B)}$ (Error compiling LaTeX. Unknown error_msg)
See also
2013 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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