[Solution] Let the areas of the two triangles be $S$ and $S+18$, and the corresponding side length of the larger triangle be $x$. By the given and the properties of similar figures, we have
$$
\frac{S+18}{S}=\frac{x^{2}}{3^{2}}=\left(\frac{x}{3}\right)^{2},
$$

Solving for $S$, we get
$$
S=\frac{18}{\left[\left(\frac{x}{3}\right)^{2}-1\right]} \text {. }
$$

Given that $\frac{x}{3}$ is an integer, then $x$ is a multiple of 3;
Also, since $S$ is an integer, $\left(\frac{x}{3}\right)^{2}-1$ must be a factor of 18. Therefore, $\left(\frac{x}{3}\right)^{2}$ can only be $2, 3, 4, 7, 10, 19$. Among these, only 4 is a perfect square, so $\left(\frac{x}{3}\right)^{2}=4$, which means $\frac{x}{3}=2$ (the negative value is discarded), hence $x=6$. Therefore, the answer is $(D)$.
$32 \cdot 65$ The difference in area between two similar triangles is 18 square feet, and the ratio of the area of the larger triangle to the smaller triangle is the square of an integer. The area of the smaller triangle (in square feet) is an integer, and one side is 3 feet, then the length of the corresponding side of the larger triangle (in feet) is
(A) 12.
(B) 9.
(C) $6 \sqrt{2}$.
(D) 6.
(E) $3 \sqrt{2}$.
(18th American High School Mathematics Examination, 1967)