11.185. Find the surface area of a sphere inscribed in a pyramid, whose base is a triangle with sides 13, 14, and $15 \mathrm{~cm}$, if the vertex of the pyramid is 5 cm away from each side of the base.
Solution.

Let $MO$ be the height of the pyramid $MABC$ (Fig. 11.74), $AC=13$ cm, $BC=14$ cm, $AB=15$ cm, $MK \perp AC$, $MP \perp AB$, $MF \perp BC$, $MK=MP=MF=5$ cm.

Then $OK \perp AC$, $OP \perp AB$, $OF \perp BC$, $OK=OP=OF$, and therefore, point $O$ is the center of the circle inscribed in $\triangle ABC$, and $OK=r$ is the radius of this circle. The semiperimeter of triangle $ABC$ is $p=21$, and its area is calculated using Heron's formula $S_{\Delta}=84 \text{ cm}^2$, $r=\frac{S_{\Delta}}{p}=4 \text{ cm}$.

From $\triangle KOM (\angle KOM=90^{\circ}): MO=\sqrt{KM^2-KO^2}=3$ cm. $\angle MKO$, $\angle MPO$, $\angle MFO$ are the linear angles of the dihedral angles at the edges of the base, and since $\angle MKO=\angle MPO=\angle MFO$, the center $E$ of the sphere inscribed in the pyramid lies on the height $MO$, and $KE$ is the bisector of $\angle MKO$. Let the radius of the sphere $EO=R$. By the property of the angle bisector of a triangle from $\triangle MKO: \frac{EO}{ME}=\frac{OK}{MK}; \frac{R}{3-R}=\frac{4}{5}; R=\frac{4}{3}$. The surface area of the sphere $S=4 \pi R^2=\frac{64 \pi}{9} \text{ cm}^2$.

Answer: $\frac{64 \pi}{9} \text{ cm}^2$.