8. Priscilla has been tasked with preparing the stage design for her school play. She needs a crescent moon, and she has a cardboard circle of radius $r$ to cut it from; she then places the compass on the edge of the circle, draws an arc of a circle of radius $r \sqrt{2}$, and cuts along the traced line. What is the area of the crescent moon she obtains?
(A) $r^{2}$
(B) $\frac{\sqrt{2}}{2} \pi r^{2}$
(C) $\frac{1}{3} \pi r^{2}$
(D) $\left(\frac{\pi}{4}+\frac{1}{2}\right) r^{2}$
(E) $\left(\frac{\pi}{4}+1\right) r^{2}$
8. The answer is $(\mathbf{A})$. Let $O$ be the center of the cardboard circle, $O^{\prime}$ the point where Priscilla sets the compass, $A$ and $B$ the intersections of the arc drawn by Priscilla with the edge of the circle. Then $A O^{\prime} B$ is a right isosceles triangle (we can justify this by noting that the triangles $O O^{\prime} A$ and $O O^{\prime} B$, since they have sides of length $r, r$ and $\sqrt{2} r$, are both right isosceles triangles). Let $A_{A O^{\prime} B}$ be the area of the sector $A O^{\prime} B$, $A_{C}$ the area of the cardboard circle, and $A_{T}$ the area of the triangle $A O^{\prime} B$. Then the area of the lune is $A_{C}-A_{A O^{\prime} B}-\frac{1}{2} A_{C}+A_{T}$, which is $\pi r^{2}-\frac{1}{4} \pi(\sqrt{2} r)^{2}-\frac{1}{2} \pi r^{2}+\frac{1}{2}(\sqrt{2} r)^{2}$, or exactly $r^{2}$.