10.260. The side of the triangle is 48 cm, and the height drawn to this side is 8.5 cm. Find the distance from the center of the circle inscribed in the triangle to the vertex opposite the given side, if the radius of the inscribed circle is 4 cm.
Solution.

In $\triangle ABC$ (Fig. 10.65) $AC=48$ cm, $BE$ is the altitude, $BE=8.5$ cm, $O$ is the center of the inscribed circle, $M, N, K$ are the points of tangency of this circle with the sides $AB, BC, AC$ of the triangle, respectively. Then $ON \perp BC$, $ON=4$ cm, $AM=AK$, $CK=CN$, $BM=BN$. The area of triangle $ABC$ $S=\frac{1}{2} AC \cdot BE=204 \, \text{cm}^2$. Then its semiperimeter $p=\frac{S}{ON}=51 \, \text{cm}$. But $p=\frac{AB+AC+BC}{2}=AK+KC+BN=AC+BN$. Therefore, $BN=p-AC=3 \, \text{cm}$. From $\triangle BNO\left(\angle BNO=90^{\circ}\right): OB=\sqrt{BN^2+ON^2}=5 \, \text{cm}$.

Answer: 5 cm.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0755.jpg?height=609&width=490&top_left_y=77&top_left_x=94)

Fig. 10.66