10.221. Two sides of a triangle are 6 and $8 \mathrm{~cm}$. The medians drawn to these sides are perpendicular to each other. Find the third side of the triangle.
## Solution.

In $\triangle A B C$ (Fig. 10.31) $A C=6$ cm, $B C=8$ cm, $B K$ and $A N$ are medians. $O$ is the point of their intersection, $A N \perp B K$. Let $O N=x$ cm, $O K=y$ cm. Then $A O=2 x$ cm, $B O=2 y$ cm. From $\triangle B O N\left(\angle B O N=90^{\circ}\right): B O^{2}+O N^{2}=B N^{2} ;$ $4 y^{2}+x^{2}=16(1)$. From $\triangle A O K\left(\angle A O K=90^{\circ}\right): A O^{2}+O K^{2}=A K^{2} ; 4 x^{2}+y^{2}=9(2)$. Adding (1) and (2), we have $5 x^{2}+5 y^{2}=25, x^{2}+y^{2}=5$. From $\triangle A O B$ $\left(\angle A O B=90^{\circ}\right): A B=\sqrt{A O^{2}+B O^{2}}=\sqrt{4 x^{2}+4 y^{2}}=2 \sqrt{x^{2}+y^{2}}=2 \sqrt{5}(\mathrm{~cm})$.

Answer: $2 \sqrt{5} \mathrm{~cm}$.