$20 \cdot 72$ In a circle $O$ with radius $r$, a chord $AB$ is drawn, whose length is $r$. A perpendicular is drawn from $O$ to $AB$, with the foot of the perpendicular being $M$. A perpendicular is drawn from $M$ to $OA$, with the foot of the perpendicular being $D$. Then the area of $\triangle MDA$ is
(A) $\frac{3 r^{2}}{16}$.
(B) $\frac{\pi r^{2}}{16}$.
(C) $\frac{\pi r^{2} \sqrt{2}}{8}$.
(D) $\frac{\sqrt{3} r^{2}}{32}$.
(E) $\frac{\sqrt{6} r^{2}}{48}$.
(20th American High School Mathematics Examination, 1969)
[Solution] Given that $A M=\frac{1}{2} A B=\frac{r}{2}$.
Also, in the right triangle $\triangle A D M$,
$$
\begin{aligned}
\angle \mathrm{AMD} & =90^{\circ}-\angle M A D \\
& =90^{\circ}-60^{\circ}=30^{\circ},
\end{aligned}
$$

then $A D=\frac{1}{2} A M=\frac{r}{4}$.
Similarly, in the right triangle $\triangle O M D$, $M D=\frac{1}{2} O M=\frac{1}{2} \cdot \frac{\sqrt{3}}{2} r=\frac{\sqrt{3}}{4} r$.
$$
\therefore \quad S_{\triangle M D A}=\frac{1}{2} A D \cdot M D=\frac{\sqrt{3}}{32} r^{2} \text {. }
$$

Therefore, the answer is $(D)$.