17・142 As shown in the figure, given $\angle C=90^{\circ}, A D=D B, D E \perp A B, A B=20$ and $A C=12$, then the area of quadrilateral $A D E C$ is
(A) 75.
(B) $58 \frac{1}{2}$.
(C) 48.
(D) $37 \frac{1}{2}$.
(E) Other.
[Solution] Given that $\triangle B D E \sim \triangle B C A$, then
$$
\frac{S_{\triangle B D E}}{S_{\triangle B C A}}=\frac{B D^{2}}{B C^{2}},
$$

where $\quad B C^{2}=A B^{2}-A C^{2}=20^{2}-12^{2}=16^{2}$,
and $B D=\frac{1}{2} A B=10$, we get

$$
\begin{aligned}
S_{\triangle B D E} & =\left(\frac{10}{16}\right)^{2} S_{\triangle B C A}=\left(\frac{5}{8}\right)^{2} S_{\triangle B C A}, \\
S_{\text {quadrilateral } A D E C} & =S_{\triangle A B C}-S_{\triangle B D E}=\left[1-\left(\frac{5}{8}\right)^{2}\right] S_{\triangle A B C} \\
& =\frac{13 \cdot 3}{8^{2}} \cdot \frac{A C \cdot B C}{2}=\frac{13 \cdot 3}{8^{2}} \cdot \frac{12 \cdot 16}{2}=58 \frac{1}{2} .
\end{aligned}
$$

Therefore, the answer is $(B)$.