$21 \cdot 9$ Given that in the right triangle $\triangle ABC$, the hypotenuse $BC=2\pi$. A circle with center $A$ and radius $r$ is tangent to $BC$. Let the area of the remaining part of the right triangle $\triangle ABC$ after being covered by circle $A$ be $S$. Then the maximum value of $S$ is
(A) $\pi^{2}$.
(B) $\pi$.
(C) $\frac{\pi}{2}$.
(D) $\frac{\pi^{2}}{4}$.
(Anhui Province, Anqing County Junior High School Mathematics Competition, 1991)
[Solution] Let the two legs of $\triangle A B C$ be $x, y$, and the radius of circle $A$ be $r$. Then
$$
S=\frac{1}{2} x y-\frac{1}{4} \pi r^{2} .
$$

And, $S_{\triangle A B C}=\frac{1}{2} x y=\frac{1}{2} B C \cdot r=\pi r$, then $r=\frac{x y}{2 \pi}$, thus $S=\frac{1}{2} x y-\frac{1}{16 \pi}(x y)^{2}$, which can be regarded as a quadratic trinomial in terms of $x y$.
$$
\therefore \quad S_{\max }=\frac{0-\left(\frac{1}{2}\right)^{2}}{-4 \cdot \frac{1}{16 \pi}}=\pi .
$$

Or by $S=-\left(\frac{x y}{4 \sqrt{\pi}}-\sqrt{\pi}\right)^{2}+\pi$, we know $S_{\max }=\pi$. Therefore, the answer is $(B)$.