20.66 The three vertices of an inscribed triangle in a circle divide the circumference into three arcs of lengths $3, 4, 5$. The area of this triangle is
(A) 6.
(B) $\frac{18}{\pi^{2}}$.
(C) $\frac{9}{\pi^{2}}(\sqrt{3}-1)$.
(D) $\frac{9}{\pi^{2}}(\sqrt{3}+1)$.
(E) $\frac{9}{\pi^{2}}(\sqrt{3}+3)$.
(40th American High School Mathematics Examination, 1989)
[Solution] Given that the circumference of the circle is $3+4+5=12$, let the radius of the circle be $R$, then $2 \pi R=12$, so $R=\frac{6}{\pi}$. If the center of the circle is $O$, then $\angle A O B=90^{\circ}$, $\angle B O C=150^{\circ}, \angle C O A=120^{\circ}$.
Thus, $S_{\triangle A B C}=S_{\triangle O A B}+S_{\triangle O B C}+S_{\triangle U C A}$
$$
\begin{aligned}
= & \frac{1}{2} R^{2}(\sin \angle A O B+\sin \angle B O C+ \\
& \sin \angle C O A) \\
= & \frac{1}{2} \cdot \frac{36}{\pi^{2}} \cdot\left(1+\frac{1}{2}+\frac{\sqrt{3}}{2}\right) \\
= & \frac{9}{\pi^{2}}(3+\sqrt{3}) .
\end{aligned}
$$

Therefore, the answer is $(E)$.