10.324. A circle is inscribed in a triangle. The lines connecting the center of the circle with the vertices divide the area of the triangle into parts with areas 4, 13, and $15 \mathrm{~cm}^{2}$. Find the sides of the triangle.
## Solution.

Let the sides of the triangle be denoted by $a, b, c$. Then the areas of the parts of the triangle are $\frac{1}{2} a r, \frac{1}{2} b r, \frac{1}{2} c r$, i.e., $a r=8, b r=26, c r=30$, from which we get $a=\frac{8}{r}, b=\frac{26}{r}, c=\frac{30}{r}$. Using Heron's formula, we find
$S=\sqrt{\frac{32}{r} \cdot \frac{24}{r} \cdot \frac{6}{r} \cdot \frac{2}{r}}=\frac{96}{r^{2}}$. But $S=4+13+15=32 \text{cm}^{2} ;$ therefore, $\frac{96}{r^{2}}=32$, $r=\sqrt{3}$ cm. Thus, $a=\frac{8}{\sqrt{3}} \text{cm}, b=\frac{26}{\sqrt{3}}$ cm, $c=\frac{30}{\sqrt{3}}$ cm.

Answer: $\frac{8}{\sqrt{3}}, \frac{26}{\sqrt{3}}, \frac{30}{\sqrt{3}}$ cm.