$20 \cdot 33$ The area between two equal parallel chords that are 8 inches apart in a circle with a radius of 8 inches is equal to
(A) $21 \frac{1}{3} \pi-32 \sqrt{3}$.
(B) $32 \sqrt{3}+21 \frac{1}{3} \pi$.
(C) $32 \sqrt{3}+42 \frac{2}{3} \pi$.
(D) $16 \sqrt{3}+42 \frac{2}{3} \pi$.
(E) $42 \frac{2}{3} \pi$.
(3rd American High School Mathematics Examination, 1952)
[Solution] As shown in the figure, connect $O B$, and draw the diameter $E D / / A B$ of $\odot O$. Then $S_{\triangle C O B}=\frac{1}{2} \cdot 4 \cdot 4 \sqrt{3}=8 \sqrt{3}$,
and $S_{\text {sector } B O D}=\frac{30^{\circ}}{360^{\circ}} \cdot \pi \cdot 8^{2}=\frac{16}{3} \pi$. Therefore, the area between the two parallel chords of the circle is
$$
4\left(8 \sqrt{3}+\frac{16}{3} \pi\right)=32 \sqrt{3}+21 \frac{1}{3} \pi \text {. }
$$

Hence, the correct choice is $(B)$.