A [semicircle](https://artofproblemsolving.com/wiki/index.php/Semicircle) of [diameter](https://artofproblemsolving.com/wiki/index.php/Diameter) $1$ sits at the top of a semicircle of diameter $2$, as shown. The shaded [area](https://artofproblemsolving.com/wiki/index.php/Area) inside the smaller semicircle and outside the larger semicircle is called a lune. Determine the area of this lune. 

$\mathrm{(A) \ } \frac{1}{6}\pi-\frac{\sqrt{3}}{4}\qquad \mathrm{(B) \ } \frac{\sqrt{3}}{4}-\frac{1}{12}\pi\qquad \mathrm{(C) \ } \frac{\sqrt{3}}{4}-\frac{1}{24}\pi\qquad \mathrm{(D) \ } \frac{\sqrt{3}}{4}+\frac{1}{24}\pi\qquad \mathrm{(E) \ } \frac{\sqrt{3}}{4}+\frac{1}{12}\pi$
The shaded area $[A]$ is equal to the area of the smaller semicircle $[A+B]$ minus the area of a [sector](https://artofproblemsolving.com/wiki/index.php/Sector) of the larger circle $[B+C]$ plus the area of a [triangle](https://artofproblemsolving.com/wiki/index.php/Triangle) formed by two [ radii](https://artofproblemsolving.com/wiki/index.php/Radius) 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 $[A] = [A+B]-[B+C]+[C] = \left(\frac{1}{8}\pi\right)-\left(\frac{1}{6}\pi\right)+\left(\frac{\sqrt{3}}{4}\right)=\boxed{\mathrm{(C)}\ \frac{\sqrt{3}}{4}-\frac{1}{24}\pi}$. We have thus solved the problem.