2015 AMC 12B Problems/Problem 14

Problem

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?

$\textbf{(A)}\; 8-\pi \qquad\textbf{(B)}\; \pi+2 \qquad\textbf{(C)}\; 2\pi-\dfrac{\sqrt{2}}{2} \qquad\textbf{(D)}\; 4(\pi-\sqrt{3}) \qquad\textbf{(E)}\; 2\pi-\dfrac{\sqrt{3}}{2}$

Solution

The area of the circle is $\pi \cdot 2^2 = 4\pi$ , and the area of the triangle is $\frac{4^2 \cdot\sqrt{3}}{4} = 4\sqrt{3}$ . 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 is the same as the difference between the area of the circle and the area of the triangle (because from both pieces we are subtracting the area of the two shapes' intersection), so the answer is $4\pi-4\sqrt{3} = \boxed{\textbf{(D)}\; 4(\pi-\sqrt{3})}$ .

2015 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2015 AMC 12B Problems/Problem 14

Problem

Solution

See Also