12.223. The base of the pyramid is a rhombus with an acute angle $\alpha$. All lateral faces form the same angle $\beta$ with the base plane.

The area of the section made through the larger diagonal of the base and the apex of the pyramid is $S$. Find the volume of the pyramid.
Solution.

Let the rhombus $ABCD$ be the base of the pyramid $LABCD$ (Fig. 12.90), $\angle BCD=\alpha, 0^{\circ}<\alpha<90^{\circ}, LO$-the height of the pyramid. All lateral faces form the same angle with the plane of the base, so $O$ is the center of the circle inscribed in the rhombus, which is the point of intersection

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0972.jpg?height=676&width=683&top_left_y=76&top_left_x=577)
of the diagonals of the rhombus.

$\angle BCD$ is acute, so

Fig. 12.90 $AC$ is the larger diagonal of the rhombus, $S_{\triangle ALC}=S$.

In the plane $DLC$, drop a perpendicular $LK$ to $CD$. Then $OK \perp CD$, $\angle LKO=\beta$.

The plane $ALC$ divides the given pyramid into two equal pyramids $DALC$ and $BALC$ with the common base $ALC$.

$DO \perp AC, DO \perp LO$. Therefore, $DO$ is perpendicular to the plane $ALC$ and is the height of the pyramid $DALC$.

The volume of the given pyramid $V=2 V_{DALC}=\frac{2}{3} S \cdot DO$.

Let $DO=a, \angle DOK=\angle DCO=\frac{1}{2} \angle BCD=\frac{\alpha}{2}$.

Then $OK=a \cos \frac{\alpha}{2}, OC=a \operatorname{ctg} \frac{\alpha}{2}, MO=a \cos \frac{\alpha}{2} \operatorname{tg} \beta$.

$$
\begin{aligned}
& S=\frac{1}{2} LO \cdot AC=LO \cdot OC \Rightarrow a \cos \frac{\alpha}{2} \operatorname{tg} \beta \cdot a \operatorname{ctg} \frac{\alpha}{2}=S ; a=\sqrt{\frac{S \operatorname{tg} \frac{\alpha}{2} \operatorname{ctg} \beta}{\cos \frac{\alpha}{2}}}= \\
& =\frac{\sqrt{S \sin \frac{\alpha}{2} \operatorname{ctg} \beta}}{\cos \frac{\alpha}{2}} \Rightarrow V=\frac{2 S \sqrt{S \sin \frac{\alpha}{2} \operatorname{ctg} \beta}}{3 \cos \frac{\alpha}{2}}
\end{aligned}
$$

Answer: $\frac{2 S \sqrt{S \sin \frac{\alpha}{2} \operatorname{ctg} \beta}}{3 \cos \frac{\alpha}{2}}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0973.jpg?height=708&width=632&top_left_y=312&top_left_x=96)

Fig. 12.91