$32 \cdot 61$ In a right-angled triangle, one of the legs has a length of 11, and the lengths of the other two sides are also natural numbers. Therefore, its perimeter is 
(A) 132.
(B) 121.
(C) 120.
(D) None of the above
(China Shanghai Junior High School Mathematics Competition, 1984)
[Solution] Let the two legs of the right triangle be 11 and $b$, and the hypotenuse be $c$. By the Pythagorean theorem, we have
$11^{2}+b^{2}=c^{2}$, which means $(c+b)(c-b)=121$.
$\because c-b<c+b$, and 121 has only two factorizations $1 \cdot 121$ and $11 \cdot 11$.
$\therefore\left\{\begin{array}{l}c+b=121, \\ c-b=1,\end{array} \quad\right.$ solving gives $\left\{\left\{\begin{array}{l}c=61, \\ b=60 .\end{array}\right.\right.$
Thus, the perimeter of the triangle is $11+60+61=132$.
Therefore, the answer is $(A)$.