Semicircle $\widehat{AB}$ has center $C$ and radius $1$. Point $D$ is on $\widehat{AB}$ and $\overline{CD}\perp\overline{AB}$. Extend $\overline{BD}$ and $\overline{AD}$ to $E$ and $F$, respectively, so that circular arcs $\widehat{AE}$ and $\widehat{BF}$ have $B$ and $A$ as their respective centers. Circular arc $\widehat{EF}$ has center $D$. The area of the shaded "smile" $AEFBDA$, is
$\text{(A) } (2-\sqrt{2})\pi\quad \text{(B) } 2\pi-\pi \sqrt{2}-1\quad \text{(C) } (1-\frac{\sqrt{2}}{2})\pi\quad\\ \text{(D) } \frac{5\pi}{2}-\pi\sqrt{2}-1\quad \text{(E) } (3-2\sqrt{2})\pi$
$\fbox{B}$ The area of the entire outer shape is the area of sector $ABE$, plus the area of sector $ABF$, minus the area of triangle $ABD$ (since it is part of both sectors), plus the area of sector $DEF$. We know $AC = CD = 1$, so the sector angles for $ABE$ and $ABF$ are $45$ degrees, and the radius of both of them is $2$. The radius of $DEF$ is $DE = BE - BD = 2 - BD$, and $BD$ can be found using Pythagoras in triangle $BCD$, giving $BD = \sqrt{2}$ and $DE = 2 - \sqrt{2}$, so after doing all the calculations, the area of the entire outer shape is $\pi(\frac{5}{2} - \sqrt{2}) - 1$. To get the area of the smile, we need to subtract the area of semicircle $ABD$, which is $\frac{1}{2} \pi 1^2 = \frac{\pi}{2}$, so the answer is $\pi(\frac{5}{2} - \frac{1}{2} - \sqrt{2}) - 1$ = $2\pi - \pi \sqrt{2} - 1$.