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 $\frac{a}{b}?$

$1$

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

$\sqrt{3}$

$2$

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

Answer: B

Concepts:regular polygon equilateral triangle power scaling of length, area, and volume

Difficulty rating: 1420

Small Hint:

Compare the side length of the large equilateral triangle with the small equilateral triangles inside the hexagon

Big Hint:

Areas scale as the square of side length

Solution:

Let $s$ be the side length of the equilateral triangle and $A$ its area. Then $a=3s.$

A regular hexagon of side length $s$ consists of $6$ equilateral triangles of side length $s,$ so its area is $6A.$

Therefore, a regular hexagon of area $A$ has side length $\dfrac{s}{\sqrt 6} .$ Hence $b = 6\cdot \dfrac{s}{\sqrt 6} = s\sqrt 6 .$

It follows that $\dfrac ab = \dfrac{3s}{s\sqrt 6} = \dfrac{3 \sqrt 6}6 = \dfrac{\sqrt 6} 2.$

Thus, the correct answer is B.

2013 AMC 10B Problem 15

Answer: B

Difficulty rating: 1420

Small Hint:

Big Hint:

Solution:

Problem 15 in Other Years