2020 AMC 12B Problem 11

The hexagon has area $\tfrac{3\sqrt3}{2}\cdot 2^2 = 6\sqrt3.$ Each semicircle has radius $1$ and area $\tfrac{\pi}{2},$ totaling $3\pi.$ Adjacent semicircle centers (side midpoints) are a distance $\sqrt3$ apart, so each adjacent pair overlaps in a lens of area $2\cos^{-1}\!\left(\tfrac{\sqrt3}{2}\right) - \tfrac{\sqrt3}{2} = \tfrac{\pi}{3} - \tfrac{\sqrt3}{2}.$ There are six such lenses. The union of the semicircles is $$3\pi - 6\left(\frac{\pi}{3} - \frac{\sqrt3}{2}\right) = \pi + 3\sqrt3.$$ Subtracting from the hexagon gives the shaded area $6\sqrt3 - (\pi + 3\sqrt3) = 3\sqrt3 - \pi.$ Thus, the correct answer is D.