17.152 In $\triangle ABC$, the median $AD=1; AB+AC=\frac{5}{2}, BC=2$. Then the area of $\triangle ABC$ is 
(A) $\frac{9}{8}$.
(B) $\frac{9}{4}$.
(C) 1.
(D) $\frac{9}{16}$.
(China Junior High School Mathematics Correspondence Competition, 1987)
[Solution] From the given information, the median $A D=1$, and since $B C=2$, we have $B C=2 A D$. Therefore, $\triangle A B C$ is a right triangle with $\angle B A C=90^{\circ}$.
From the given conditions and the Pythagorean theorem, we have
$$
\left\{\begin{array}{l}
A B+A C=\frac{5}{2}, \\
A B^{2}+A C^{2}=4 .
\end{array}\right.
$$

Thus,
$$
A B \cdot A C=\frac{9}{8} \text {. }
$$

Therefore,
$$
S_{\triangle A B C}=\frac{1}{2} A B \cdot A C=\frac{9}{16} \text {. }
$$

Hence, the answer is $(D)$.