smc

The shaded area $[A]$ is equal to the area of the smaller semicircle $[A+B]$ minus the area of a sector of the larger circle $[B+C]$ plus the area of a triangle formed by two radii of the larger semicircle and the diameter of the smaller semicircle $[C]$. The area of the smaller semicircle is $[A+B]=\frac{1}{2}\pi \cdot {\left(\frac{1}{2}\right)}^2=\frac{1}{8}\pi$. Since the radius of the larger semicircle is equal to the diameter of the smaller half circle, the triangle is an equilateral triangle and the sector measures $60^{\circ }$. The area of the $60^{\circ }$ sector of the larger semicircle is $[B+C]=\frac{60}{360}\pi \cdot {\left(\frac{2}{2}\right)}^2=\frac{1}{6}\pi$. The area of the triangle is $[C]=\frac{1^2\sqrt{3}}{4}=\frac{\sqrt{3}}{4}$. So the shaded area is $[A]=[A+B]-[B+C]+[C]=\left(\frac{1}{8}\pi \right)-\left(\frac{1}{6}\pi \right)+\left(\frac{\sqrt{3}}{4}\right)=\boxed{\frac{\sqrt{3}}{4}-\frac{1}{24}\pi }$. We have thus solved the problem.