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


2.56. To find the radius of the sphere, we will pass a plane through the height of the pyramid and the apothem (Fig. 2.46). The radius of the circle in the obtained section is equal to the radius of the sphere. Since all the edges of the pyramid are equal to $a$, then

$$
S D=\frac{a \sqrt{3}}{2}, O D=\frac{a \sqrt{3}}{6} \cdot U_{3} \triangle S O D
$$

$$
\text { we find } S O=\sqrt{S D^{2}-O D^{2}}=\frac{a \sqrt{6}}{3} \text {. }
$$

Let the radius of the sphere be $r$; then $S O_{1}=S O-r$. In $\triangle S K O_{1}$ we have $O_{1} K^{2}=S O_{1}^{2}-S K^{2}$, or $r^{2}=(S O-r)^{2}-\frac{a^{2}}{3}$, from which

$$
r^{2}=\frac{2 a^{2}}{3}-\frac{2 a r \sqrt{6}}{3}+r^{2}-\frac{a^{2}}{3}, \text { i.e. } r=\frac{a}{2 \sqrt{6}}
$$

Finally, we find $S_{\text {tara }}=4 \pi r^{2}=\frac{\pi a^{2}}{6}$.

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