11.187. Calculate the surface area of a sphere inscribed in a triangular pyramid, all edges of which are equal to $a$.
Solution.

Consider the plane passing through the height of the pyramid and the apothem (Fig. 11.75). The radius of the circle in the obtained section is equal to the radius of the sphere. Since all edges of the pyramid are equal to $a$, then $S D=\frac{a \sqrt{3}}{2}$, $O D=\frac{a \sqrt{3}}{6}$. From $\triangle S O D$ we find $S O=\sqrt{S D^{2}-O D^{2}}=\frac{a \sqrt{6}}{3}$. Let the radius of the sphere be $r$, then $S O_{1}=S O-r$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0884.jpg?height=608&width=620&top_left_y=1222&top_left_x=639)

Fig. 11.75

In $\triangle S K O_{1}$ we have $O_{1} K^{2}=S O_{1}^{2}-S K^{2}$ or $r^{2}=(S O-r)^{2}-\frac{a^{2}}{3}$, hence $r^{2}=\frac{2 a^{2}}{3}-\frac{2 a r \sqrt{6}}{3}+r^{2}-\frac{a^{2}}{3}$, i.e., $r=\frac{a}{2 \sqrt{6}}$.

We obtain $S_{\text {sphere }}=4 \pi r^{2}=\frac{\pi a^{2}}{6}$.

Answer: $\quad \frac{\pi a^{2}}{6}$.