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
Let $X$ be the area of the overlap of the disk and the triangle. The two regions have areas $4\pi-X$ and $4\sqrt3-X$, so their difference is the area of the circle minus the area of the triangle: $4\pi - 4\sqrt3$.
A circle of radius $3$ is centered at $A$. An equilateral triangle with side $5$ 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?
