10.321. The medians of a triangle are 3, 4, and 5 cm. Find the area of the triangle.
## Solution.

Let $A N, B M, C P$ be the medians of $\triangle A B C$ (Fig. 10.110), $O$-the point of their intersection, $A N=3 \text{ cm}, B M=4 \text{ cm}, C P=5 \text{ cm}$. On the extension of segment $B M$ beyond point $M$, lay off segment $D M=O M$. Then $D O=B O=$ $=\frac{2}{3} B M=\frac{8}{3}$ cm. Drop a perpendicular $A K$ to $B M . S_{\triangle A D O}=$ $=\frac{1}{2} D O \cdot A K=\frac{1}{2} B O \cdot A K=S_{\triangle A B O}=\frac{1}{3} S_{\triangle A B C}$. Since $A O C D$ is a parallelogram, then $A D=O C=\frac{10}{3} \text{ cm} . A O=\frac{2}{3} A N=2 \text{ cm}$. Using Heron's formula, we find $S_{\triangle A D O}=\frac{8}{3} \text{ cm}^{2}$. Then $S_{\triangle A B C}=3 S_{\triangle 1 D O}=8 \text{ cm}^{2}$.

Answer: $8 \text{ cm}^{2}$.