$17 \cdot 136$ In a triangle, one side is 2, the median on this side is 1, and the sum of the other two sides is $1+\sqrt{3}$. Then the area of this triangle is 
(A) 1.
(B) $\frac{1}{2} \sqrt{3}$.
(C) $\sqrt{3}$.
(D) Cannot be determined.
(Tianjin, China Junior High School Mathematics Competition, 1983)
[Solution] Given one side of a triangle is 2, and the length of the median to this side is 1, it is known that the triangle is a right triangle with the hypotenuse being 2.

The sum of the two legs is $1+\sqrt{3}$. By the Pythagorean theorem, their lengths can be found to be 1 and $\sqrt{3}$.
Thus, $S_{\triangle}=\frac{1}{2} \cdot 1 \cdot \sqrt{3}=\frac{\sqrt{3}}{2}$.
Or, given the sum of the two sides is $1+\sqrt{3}$, let the two sides be $a, b$,

i.e., $\square$
Then
Therefore, the answer is $(B)$.
$$
\begin{array}{c}
(a+b)^{2}=a^{2}+2 a b+b^{2}=c^{2}+4 S_{\triangle}, \\
S_{\triangle}=\frac{1}{4}\left[(a+b)^{2}-c^{2}\right]=\frac{1}{4} \cdot 2 \sqrt{3}=\frac{\sqrt{3}}{2} .
\end{array}
$$