2012 AMC 10A Problem 18

Each arc has radius $1$ and length $\frac{2\pi}{3},$ so the corresponding circular sectors all have the same area. As the diagram shows, the sector pieces may be rearranged: everything added to the regular hexagon combines to exactly one unit circle. The regular hexagon is made of six equilateral triangles of side $2,$ so its area is $$6\left(\dfrac{\sqrt3}{4}\cdot2^2\right)=6\sqrt3.$$ Adding the unit circle gives a total enclosed area of $\pi+6\sqrt3.$ Thus, E is the correct answer.