2015 AMC 12B Problem 14

Question

A circle of radius

2

is centered at

A.

An equilateral triangle with side

4

has a vertex at

A.

What is the difference between the area of the region that lies inside the circle but outside the triangle and the area of the region that lies inside the triangle but outside the circle?

8 - \pi

\pi + 2

2\pi - \dfrac{\sqrt2}{2}

4(\pi - \sqrt3)

2\pi + \dfrac{\sqrt3}{2}

Answer: D

Concepts:area decomposition circle area equilateral triangle

Difficulty rating: 1740

Small hint:

Let

z

be the overlap area; the requested difference is

(\text{circle} - z) - (\text{triangle} - z)

Big hint:

That equals circle area minus triangle area; use

\pi r^2

and

\dfrac{\sqrt3}{4}s^2

Solution:

Let

z

be the area shared by the circle and triangle. The requested difference is

\begin{aligned} &(\text{circle} - z) - (\text{triangle} - z) \\ &= \text{circle} - \text{triangle}. \end{aligned}

The circle has area

\pi \cdot 2^2 = 4\pi,

and the equilateral triangle has area

\dfrac{\sqrt3}{4}\cdot 4^2 = 4\sqrt3.

The difference is

4\pi - 4\sqrt3 = 4(\pi - \sqrt3).

Thus, the correct answer is D.