11.083. Determine the volume of a sphere inscribed in a regular pyramid, where the height is equal to $h$, and the dihedral angle at the base is $60^{\circ}$.
Solution.

Consider the section of a regular pyramid through the apothem $AB$ and the height $BC = h$, on which lies the center of the sphere inscribed in the pyramid (Fig. 11.75). The sphere touches the lateral face at point $K$ on the apothem $AB$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-701.jpg?height=512&width=312&top_left_y=62&top_left_x=236)

Fig. 11.75

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-701.jpg?height=412&width=438&top_left_y=158&top_left_x=733)

Fig. 11.76

Then $\angle BAC = 60^\circ$, $\angle ABC = 30^\circ$. From $\triangle BKO$ we have $\sin 30^\circ = \frac{OK}{OB} = \frac{1}{2}$; $\frac{r}{h-r} = \frac{1}{2}$, $2r = h - r$, $r = \frac{1}{3}h$. Therefore, $V_{\mathrm{w}} = \frac{4}{3} \pi r^3 = \frac{4}{81} \pi h^3$.

Answer: $\frac{4}{81} \pi h^3$.