$20 \cdot 120$ As shown in the figure, on the diameter $AB$ of a semicircle, a point $C$ is taken, and semicircles are constructed with $AC$ and $CB$ as diameters, respectively. A line $CD$ perpendicular to $AB$ is drawn through $C$, intersecting the circumference at $D$, with the length of $CD$ being $h$. Then the area of the shaded part is
(A) $\frac{1}{3} \pi h^{2}$.
(B) $\frac{1}{4} \pi h^{2}$.
(C) $\frac{1}{5} \pi h^{2}$.
(D) $\frac{1}{6} \pi h^{2}$.
[Solution] Let the radius of the large circle be $r$, and $CO=x$. According to the problem, we have
$$
\begin{aligned}
S_{\text {shaded }} & =\frac{\pi r^{2}}{2}-\frac{\pi AC^{2}}{8}-\frac{\pi CB^{2}}{8} \\
& =\frac{\pi r^{2}}{2}-\frac{\pi(r-x)^{2}}{8}-\frac{\pi(r+x)^{2}}{8} . \\
& =\frac{\pi r^{2}}{2}-\frac{\pi\left(r^{2}+x^{2}\right)}{4}=\frac{\pi}{4}\left(r^{2}-x^{2}\right),
\end{aligned}
$$

In the right triangle $\triangle COD$, $OD^{2}-CO^{2}=CD^{2}$, which means $r^{2}-x^{2}=h^{2}$, thus
$$
S_{\text {shaded }}=\frac{\pi}{4} h^{2} \text {. }
$$

Therefore, the answer is $(B)$.