$32 \cdot 66$ In a right-angled triangle, one of the legs is 7, and the other two sides are integers. Then its area could be
(A) 165.
(B) 84.
(C) 108.
(D) Cannot be determined.
("Wisdom Cup" Junior High School Mathematics Competition, 1991)
[Solution] If the lengths of the two legs of a right triangle are $a, b$, and the length of the hypotenuse is $c$, then its area is
$$
S=\frac{1}{2} a b,
$$
$\because a=7$, and among $(A),(B),(C)$, the only one divisible by 7 is 84.
If we let $a=7$, then from $84=\frac{1}{2} a b=\frac{1}{2} \cdot 7 \cdot b$, we get $b=24$.
Also, $a^{2}+b^{2}=7^{2}+24^{2}=625=25^{2}, \quad \therefore c=25$.
Therefore, the answer is $(B)$.