$17 \cdot 140$ An isosceles triangle with base $\sqrt{2}$, the medians to the two equal sides are perpendicular to each other, then the area of this triangle is
(A) 1.5 .
(B) 2 .
(C) 2.5 .
(D) 3.5 .
(E) 4 .
(10th American High School Mathematics Examination, 1959)
[Solution] Given that $B E$ and $C F$ are the medians of the legs of an isosceles triangle, their intersection point $G$ is the centroid of the triangle, and $\triangle B G C$ is an isosceles triangle.

Furthermore, since the medians intersect at a right angle, $\triangle B G C$ is an isosceles right triangle. Therefore, $S_{\triangle B G C}=\frac{1}{2}$.
Also, $S_{\triangle A B C}=3 S_{\triangle B G C}$ (noting that $G$ is the centroid),

Thus,
$$
S_{\triangle A B C}=3 \cdot \frac{1}{2}=\frac{3}{2} .
$$

Therefore, the answer is $(A)$.