12.378. A regular truncated quadrilateral pyramid is inscribed in a sphere of radius $R$, with the larger base passing through the center of the sphere, and the lateral edge forming an angle $\beta$ with the plane of the base. Find the volume of the truncated pyramid.
Solution.

The larger base $ABCD$ of the regular truncated pyramid

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1141.jpg?height=650&width=833&top_left_y=71&top_left_x=286)

Fig. 12.238

$ABCD A_{1}B_{1}C_{1}D_{1}$ (Fig. 12.238) passes through the center $O$ of the sphere and is inscribed in the great circle of this sphere, so $O$ is the center of the square $ABCD$, $OD = OD_{1} = R$.

$D_{1}E$ is the height of the truncated pyramid. $\angle D_{1}DE = \beta$.

$F$ is the midpoint of $DD_{1}$ and, since $OD = OD_{1}$, then $OF \perp DD_{1}$.

In $\triangle OFD (\angle OFD = 90^{\circ}): DF = OD \cos \angle D_{1}DE = R \cos \beta$.

$DD_{1} = 2DF = 2R \cos \beta$.

In $\triangle DED_{1} (\angle DED_{1} = 90^{\circ}):$

$D_{1}E = DD_{1} \sin \angle D_{1}DE = 2R \cos \beta \sin \beta = R \sin 2\beta$

$DE = DD_{1} \cos \angle D_{1}DE = 2R \cos^2 \beta$.

$OE = OD - DE = R - 2R \cos^2 \beta = -R \cos 2\beta \Rightarrow$

$\Rightarrow \cos 2\beta < 0, 90^{\circ} < 2\beta < 180^{\circ}, 45^{\circ} < \beta < 90^{\circ}$.

$B_{1}D_{1} = 2 \cdot OE = -2R \cos 2\beta$.

The area of the upper base $S_{1} = \frac{1}{2} B_{1}D_{1}^2 = 2R^2 \cos^2 2\beta$, the area of the lower base $S_{2} = \frac{1}{2} BD^2 = 2R^2$, the height of the pyramid $h = D_{1}E = R \sin 2\beta$, and its volume

\[
\begin{aligned}
& V = \frac{h}{3} (S_{1} + S_{2} + \sqrt{S_{1} S_{2}}) = \frac{1}{3} R \sin 2\beta \times \\
& \times (2R^2 \cos^2 2\beta + 2R^2 + 2R^2 \sqrt{\cos^2 2\beta}) = \frac{2}{3} R^3 \sin 2\beta (1 + \cos^2 2\beta - \cos 2\beta) \\
& \text{Answer: } \frac{2}{3} R^3 \sin 2\beta (1 + \cos^2 2\beta - \cos 2\beta)
\end{aligned}
\]