11.176. A regular quadrilateral pyramid is inscribed in a sphere of radius $R$. Determine the volume of this pyramid if the radius of the circle circumscribed around its base is $r$.
## Solution.

Let $EO$ be the height of the regular pyramid $EABCD$ inscribed in a sphere of radius $R$ (Fig. 11.69), $OC=r$. Points $A, E, C$ belong to the surface of the sphere. Therefore, $R$ is the radius of the circle circumscribed around $\triangle AEC$. Then $R=\frac{AE \cdot EC \cdot AC}{4 S_{\triangle AEC}}=\frac{EC^{2} \cdot AC}{4 \cdot \frac{1}{2} AC \cdot EO}=\frac{EC^{2}}{2 EO}$. Let

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0878.jpg?height=466&width=487&top_left_y=175&top_left_x=140)

Fig. 11.68

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0878.jpg?height=565&width=553&top_left_y=63&top_left_x=692)

Fig. 11.69

$EO=H$. Then from $\triangle EOC: EC^{2}=H^{2}+r^{2}$. Therefore, $\frac{H^{2}+r^{2}}{2 H}=R$; $H^{2}-2HR+r^{2}=0$; $H=R+\sqrt{R^{2}-r^{2}}$ or $H=R-\sqrt{R^{2}-r^{2}}$.

The area of the base of the pyramid $S=\frac{1}{2} AC^{2}=2r^{2}$, and its volume $V=\frac{1}{3} SH=\frac{2r^{2}(R \pm \sqrt{R^{2}-r^{2}})}{3}$.

Answer: $\frac{2r^{2}(R \pm \sqrt{R^{2}-r^{2}})}{3}$.