2010 AMC 10B Problem 16

Drop the altitude from $O$ to $\overline{AB}$ at $X.$ Looking at the side lengths, we see that $\triangle OBX$ is a $30-60-90$ triangle. This means that the area of sector $AOB$ is $$\dfrac{1}{6} \cdot \pi \cdot \left(\dfrac{\sqrt3}{3}\right)^2 = \dfrac{\pi}{18}.$$ We also have that the area of $\triangle AOB$ is $$\dfrac{1}{2} \cdot 2 \cdot \dfrac{\sqrt3}{6} \cdot \dfrac{1}{2} = \dfrac{\sqrt3}{12}.$$ The area of the sector outside the square and inside the circle is then $$\dfrac{\pi}{18} - \dfrac{\sqrt3}{12}.$$ There are $4$ of these regions, which gives us a total area of $$4\left(\dfrac{\pi}{18} - \dfrac{\sqrt3}{12}\right) = \dfrac{2\pi}{9} - \dfrac{\sqrt3}{3}.$$ Thus, B is the correct answer.