[Solution] Let the three vertices of the triangle be $A, B, C$, and assume the length of side $BC$ is 21 (as shown in the figure), and let the three side lengths be
$$
21, x, 48-21-x=27-x \text{. }
$$

Draw the altitude $AD$ from $A$ to side $BC$, and let $BD=y$, then
$$
AD^{2}=x^{2}-y^{2}=(27-x)^{2}-(21-y)^{2},
$$

Simplifying, we get $48-9x+7y=0$, so $x=5+\frac{3+7y}{9}$.
The smallest integer $y$ that makes $x$ an integer is $y=6$, and the corresponding $x=10$.
Furthermore, when $y=15$, $x=17$, and $AC=27-x=10$.
For $y>15$ and values of $y$ that make $x$ an integer, they do not meet the requirements (the corresponding values cannot form the three sides of a triangle).
Therefore, the answer is $(B)$.
$32 \cdot 64$ A triangle has integer side lengths and area, one side length is 21, and the perimeter is 48, then the length of the shortest side is
(A) 8.
(B) 10.
(C) 12.
(D) 14.
(E) 16.
(14th American High School Mathematics Examination, 1963)