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 area formulas for an equilateral triangle and regular hexagon , with side length and plugging and into each equation, we find that . Simplifying this, we 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
See also
2013 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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