2020 AMC 10B Problem 14

By symmetry, the shaded region is made of six congruent pieces. One such piece is the union of two equilateral triangles with side length $1,$ minus a $60^\circ$ sector of a circle of radius $1.$ The two equilateral triangles have total area $$2\cdot\frac{\sqrt3}{4}=\frac{\sqrt3}{2}.$$ The sector has area $$\frac{60^\circ}{360^\circ}\cdot\pi(1)^2=\frac{\pi}{6}.$$ Thus one shaded piece has area $\frac{\sqrt3}{2}-\frac{\pi}{6},$ and the total shaded area is $$6\left(\frac{\sqrt3}{2}-\frac{\pi}{6}\right)=3\sqrt3-\pi.$$ Thus, D is the correct answer.