Question

A wire is cut into two pieces, one of length $a$ and the other of length $b.$ The piece of length $a$ is bent to form an equilateral triangle, and the piece of length $b$ is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is $\frac{a}{b}?$

$1$

$\dfrac{\sqrt{6}}{2}$

$\sqrt{3}$

$2$

$\dfrac{3\sqrt{2}}{2}$

Solution:

Let the side length of the equilateral triangle be $s.$ Then, let its area be $A.$ This would make $a=3s.$

As such, a hexagon with side $s$ would have $6$ equilateral triangles, with side length $s,$ making its area $6A.$

Therefore, the side length of the hexagon with area $A$ is equal to $\dfrac{s}{\sqrt 6} .$ As such, $b = 6\cdot \dfrac{s}{\sqrt 6} = s\sqrt 6 .$

This makes $\dfrac ab = \dfrac{3s}{s\sqrt 6} = \dfrac{3 \sqrt 6}6 = \dfrac{\sqrt 6} 2.$

Thus, the correct answer is B .

2013 AMC 10B Problem 15

Solution:

Problem 15 in Other Years