20.34 A circular lawn with a diameter of 12 feet is intersected by a straight gravel path 3 feet wide, one side of which passes through the center of the lawn. The number of square feet in the remaining lawn area is
(A) $36 \pi-34$.
(B) $30 \pi-15$.
(C) $36 \pi-33$.
(D) $35 \pi-9 \sqrt{3}$.
(E) $30 \pi-9 \sqrt{3}$.
(24th American High School Mathematics Examination, 1973)
[Solution] As shown in the figure, let $O$ be the center of the circular lawn, and $M$ be the midpoint of the other side of the road. If $AB$ represents the arc cut by the road, then the area of the half-road $OMBA$ includes a $30^{\circ}$ sector $OAB$ and a $30^{\circ}-60^{\circ}$ right triangle $\triangle OMB$.
Thus,
$$
\begin{aligned}
S= & \pi \cdot 6^{2}-2 S_{\text {sector }}-2 S_{\triangle O M B} \\
= & 36 \pi-2 \cdot \frac{1}{2} \cdot \frac{\pi}{6} \cdot 6 \cdot 6-2 \cdot \frac{1}{2} \\
& \cdot 3 \cdot 3 \sqrt{3} \\
= & 30 \pi-9 \sqrt{3} .
\end{aligned}
$$

Therefore, the answer is $(E)$.