10.256. The lengths of two sides of an acute triangle are $\sqrt{13}$ and $\sqrt{10} \mathrm{~cm}$. Find the length of the third side, knowing that this side is equal to the height drawn to it.
Solution.

Let in $\triangle A B C$ (Fig. 10.61) $A B=\sqrt{10}$ cm, $B C=\sqrt{13}$ cm, height $B D=A C$. Since $\triangle A B C$ is an acute-angled triangle, point $D$ lies on the segment $A C$. Let $B D=A C=x$ cm. From $\triangle A D B\left(\angle A D B=90^{\circ}\right)$: $A D=\sqrt{A B^{2}-B D^{2}}=\sqrt{10-x^{2}} \cdot$ And from $\triangle C D B\left(\angle C D B=90^{\circ}\right): C D=\sqrt{B C^{2}-B D^{2}}=$ $=\sqrt{13-x^{2}} \cdot$ Therefore, $A D+D C=A C ; \sqrt{10-x^{2}}+\sqrt{13-x^{2}}=x$; $\sqrt{13-x^{2}}=x-\sqrt{10-x^{2}} ; 13-x^{2}=x^{2}-2 x \cdot \sqrt{10-x^{2}}+10-x^{2} ; 2 x \sqrt{10-x^{2}}=x^{2}-3$; $\left\{\begin{array}{l}4 x^{2}\left(10-x^{2}\right)=x^{2}-6 x^{2}+9 \\ x^{2} \geq 3 ;\end{array}\left\{\begin{array}{l}5 x^{4}-46 x^{2}+9=0, \\ x^{2} \geq 3 ;\end{array}\left\{\begin{array}{l}{\left[\begin{array}{l}x^{2}=9, \\ x^{2}=\frac{1}{5},\end{array} x^{2}=9\right.} \\ x^{2} \geq 3 ;\end{array}\right.\right.\right.$

From this, $A C=3 \text{ cm}$.

Answer: 3 cm.