10.244. A circle with a radius of 3 cm is inscribed in a triangle. Calculate the lengths of the sides of the triangle if one of them is divided by the point of tangency into segments of 4 and $3 \mathrm{~cm}$.
Solution.

Let $O$ be the center of the circle with radius 3 inscribed in $\triangle ABC$ (Fig. 10.51). $M, N, K$ are the points of tangency of this circle with the sides $AB, BC, AC$ respectively. $BN=3 \text{ cm}, NC=4 \text{ cm}$. Then $ON \perp BC, OM \perp AB$, $ON=OM=3 \text{ cm}$. $BM=BN=3 \text{ cm}$. Therefore, $BMON$ is a rhombus, and since it has two right angles, $BMON$ is a square. Consequently, $\triangle ABC$ is a right triangle. Let $AM=x \text{ cm}$. Then $AK=AM=x \text{ cm}$. $CK=CN=4 \text{ cm}$. Therefore, $AB=(x+3) \text{ cm}$, $BC=7 \text{ cm}$, $AC=(x+4) \text{ cm}$, and $AC^2=AB^2+BC^2$. Then $(x+4)^2=49+(3+x)^2; x=21$. Thus, $AC=25 \text{ cm}$, $AB=24 \text{ cm}$.

Answer: 24 cm; 25 cm; 7 cm.