12.101. The side of the base of a regular quadrilateral prism is equal to $a$. The angle between the intersecting diagonals of two adjacent lateral faces is $\boldsymbol{\alpha}$. Find the volume of the prism.
Solution.

Let $A B C D A_{1} B_{1} C_{1} D_{1}$ be a regular quadrilateral prism, $A B=a, \angle A_{1} D C_{1}=\alpha$ (Fig. 12.101). We have $A_{1} C_{1}=a \sqrt{2}$. Let $A_{1} D=x ;$

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-789.jpg?height=408&width=420&top_left_y=216&top_left_x=174)

Fig. 12.101

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-789.jpg?height=546&width=406&top_left_y=75&top_left_x=749)

Fig. 12.102

then $D C_{1}=x$. From $\Delta A_{1} D C_{1}$, using the cosine theorem, we find

$$
\begin{aligned}
& 2 a^{2}=2 x^{2}-2 x^{2} \cos \alpha=4 x^{2} \sin ^{2} \frac{\alpha}{2} \text {, hence } x=\frac{a \sqrt{2}}{2 \sin \frac{\alpha}{2}} \cdot \text { From } \triangle D D_{1} A_{1} \text { we have } \\
& D D_{1}=\sqrt{\frac{2 a^{2}}{4 \sin ^{2} \frac{\alpha}{2}}-a^{2}}=\frac{a \sqrt{2\left(1-2 \sin ^{2} \frac{\alpha}{2}\right)}}{2 \sin \frac{\alpha}{2}}=\frac{a \sqrt{2 \cos \alpha}}{2 \sin \frac{\alpha}{2}} \cdot \text { Therefore } \\
& V_{\text {pr }}=\frac{a^{3} \sqrt{2 \cos \alpha}}{2 \sin \frac{\alpha}{2}}
\end{aligned}
$$

Answer: $\frac{a^{3} \sqrt{2 \cos \alpha}}{2 \sin \frac{\alpha}{2}}$.