$32 \cdot 67$ In a right-angled $\triangle ABC$, the lengths of the three sides are positive integers, and the length of one of the legs is a root of the equation $2 x^{2}-23 x+11=0$. The area of this triangle is 
(A) $\frac{121}{2}$.
(B) 121 .
(C) 330 .
(D) 660 .
(Chinese Jiangsu Province Junior High School Mathematics Competition, 1986)
[Sol] Solving $2 x^{2}-23 x+11=0$ yields $x_{1}=\frac{1}{2}$ (discard), $x_{2}=11$. Let the other leg of $\triangle A B C$ be $a$, and the hypotenuse be $c$. By the Pythagorean theorem, we have $c^{2}-a^{2}=11^{2}$, i.e., $(c-a)(c+a)=121$.
$\because c-a \neq c+a$, and $c-a<c+a$, $\therefore S_{\triangle A B C}=\frac{1}{2} \cdot 60 \cdot 11=330$. Therefore, the answer is $(C)$.