10.327. Determine the area of a triangle if two of its sides are 1 and $\sqrt{15}$ cm, and the median to the third side is 2 cm.
Solution.

Let $B D$ be the median of $\triangle A B C$ (Fig. 10.115), $A B=1$ cm, $B C=\sqrt{15}$ cm, $B D=2$ cm. On the extension of segment $B D$ beyond point $D$, we lay off segment $D E=B D$. $A B C E$ is a parallelogram, so $A C^{2}+B E^{2}=2(A B+B C)^{2}$; $A C^{2}=2(1+15)-16 ; A C=4$ cm.

Then $A C^{2}=A B^{2}+B C^{2}$ and $\triangle A B C$ is a right triangle.

$S_{\triangle A B C}=\frac{1}{2} A B \cdot B C=\frac{\sqrt{15}}{2} \mathrm{~cm}^{2}$.

Answer: $\frac{\sqrt{15}}{2} \mathrm{~cm}^{2}$.