$17 \cdot 135$ In an isosceles triangle, the altitude to the base is 8, and the perimeter is 32. Then the area of the triangle is
(A) 56.
(B) 48.
(C) 40.
(D) 32.
(E) 24.
(9th American High School Mathematics Examination, 1958)
[Solution] Let the base of the isosceles triangle be $b$, then the legs are $\frac{1}{2}(32-b)$. By the Pythagorean theorem, we have
$$
\left[\frac{1}{2}(32-b)\right]^{2}=\left(\frac{b}{2}\right)^{2}+8^{2} \text {, }
$$

Therefore, the correct choice is (B).
$$
b=12, \quad S_{\triangle}=\frac{1}{2} \cdot 12 \cdot 8=48 \text {. }
$$