$17 \cdot 146$ In $\triangle A B C$, the medians $C E$ and $B D$ of sides $A B$ and $A C$ are perpendicular to each other, and $B D=8, C E=12$. Then $S_{\triangle A B C}$ is equal to
(A) 24.
(B) 32.
(C) 48.
(D) 64.
(E) 96.
(48th American High School Mathematics Examination, 1997)
[Solution] Connect $E D$. Given $B D \perp C E$, then $S_{\text {quadrilateral } B(C) E}=\frac{1}{2} \cdot B D \cdot C E=48$. Since $B D$ and $C E$ are medians of the triangle, $S_{\triangle A D E}=\frac{1}{4} S_{\triangle A B C}$, thus $S_{\text {quadrilateral BCDE }}=$
$$
\frac{3}{4} S_{\triangle A B C} \text {, }
$$

Therefore, $S_{\triangle A B C}=48 \cdot \frac{4}{3}=64$.
So the answer is $(D)$.
Note: When $B D=4, C E=6$, the problem is from the 1998 Chinese Junior High School Mathematics League.