2012 AMC 12A Problem 14

Each arc has length $\dfrac{2\pi}{3}$ on a unit circle, so it is a $120^\circ$ sector. The nine equal sectors can be reassembled so that the enclosed region equals the regular hexagon of side $2$ plus one full circle of radius $1.$ A regular hexagon of side $2$ splits into $6$ equilateral triangles of side $2,$ so its area is $6 \cdot \dfrac{\sqrt3}{4} \cdot 2^2 = 6\sqrt3.$ Adding the unit circle’s area $\pi$ gives $\pi + 6\sqrt3.$ Thus, the correct answer is E.