2011 AMC 10A Problem 18

The area of this region is the area of circle $C$ minus the area of the overlapping regions with $A$ and $B.$ From the diagram, we can find the area of half of one of the overlapping regions by finding the area of the sector and subtracting the area of the triangle. This area is then $$\dfrac{1}{4} \pi \cdot 1^2 - \dfrac{1}{2} \cdot 1 \cdot 1 = \dfrac{\pi}{4} - \dfrac{1}{2}.$$ There are four of these that we must subtract, which leaves us with a final answer of $$\pi \cdot 1^2 - 4(\dfrac{\pi}{4} - \dfrac{1}{2}) = 2.$$ Thus, C is the correct answer.