$20 \cdot 121$ As shown in the figure, the center of the semicircle $\overparen{A B}$ is $C$, with a radius of 1, and point $D$ is on $\overparen{A B}$, such that $C D \perp A B$. Extend $B D$ and $A D$ to points $E$ and $F$ respectively, so that $\overparen{A E}$ and $\overparen{B F}$ are arcs of circles with centers at $B$ and $A$ respectively, and the diameter of the semicircle as the radius. $E F$ is an arc with center at $D$. Then the area of the shaded region in the figure is
(A) $(2-\sqrt{2}) \pi$.
(B) $2 \pi-\pi \sqrt{2}-1$.
(C) $\left(1-\frac{\sqrt{2}}{2}\right) \pi$.
(D) $\frac{5 \pi}{2}-\pi \sqrt{2}-1$.
(E) $(3-2 \sqrt{2}) \pi$.
(43rd American High School Mathematics Examination, 1992)
[Solution] Let I, II, III represent the corresponding parts in the figure, then we have
$$
\begin{array}{c}
S_{1+\mathbb{I I}}=2 S_{\text {trapezoid } B A F}-S_{\text {semicircle } A B}-S_{\triangle A D B} \\
=\pi-\frac{\pi}{2}-1=\frac{\pi}{2}-1 . \\
S_{\mathbb{I I}}=\frac{1}{4} \pi r^{2}=\frac{1}{4} \pi(2-\sqrt{2})^{2} \\
=\frac{3}{2} \pi-\sqrt{2} \pi .
\end{array}
$$

Therefore, $\quad S_{\text {shaded }}=S_{\mathrm{I}+\mathbb{I I}}+S_{\mathbb{I I}}=2 \pi-\sqrt{2} \pi-1$.
Hence, the answer is (B).