$17 \cdot 134$ In an isosceles $\triangle A B C$, the length of the altitude from one of the equal sides is 1. This altitude forms a $45^{\circ}$ angle with the base. Then the area of $\triangle A B C$ is
(A) 1 .
(B) 0.5 .
(C) 0.25 .
(D) $\sqrt{3}$.
(E) None of the above answers is correct.
(China Beijing Junior High School Mathematics Competition, 1983)
[Solution] The altitude $BD$ from one waist to the base forms an angle $\angle DBC=45^{\circ}$ with the base, so the base angle $\angle C=45^{\circ}$. Therefore, $\triangle ABC$ is an isosceles right triangle (with $D$ coinciding with $A$, as shown in the figure).
Thus, $AB=AC=1$,

So $S_{\triangle ABC}=\frac{1}{2} AB \cdot AC=\frac{1}{2}=0.5$.
Therefore, the answer is $(B)$.