10.325. The base of the triangle is 20 cm, the medians of the lateral sides are 18 and 24 cm. Find the area of the triangle.
## Solution.

Let $A F$ and $C Q$ be the medians of $\triangle A B C$ (Fig. 10.113), $M$ the point of their intersection, $A C=20 \mathrm{~cm}, A F=18 \mathrm{~cm}, C Q=24 \mathrm{~cm}$. Then $A M=\frac{2}{3} A F=12 \mathrm{~cm}$, $C M=\frac{2}{3} C Q=16 \mathrm{~cm}$. Since $A M^{2}+M C^{2}=A C^{2}$, $\triangle A M C$ is a right triangle, and $S_{\triangle A M C}=\frac{1}{2} A M \cdot C M=96 \mathrm{~cm}^{2} ; S_{\triangle A B C}=3 S_{\triangle A M C}=288 \mathrm{~cm}^{2}$.

Answer: $288 \mathrm{~cm}^{2}$.