12.433 The base of a right prism is an isosceles trapezoid, the diagonals of which are perpendicular to the corresponding lateral sides. The angle between the diagonals of the trapezoid, opposite to its lateral side, is $\alpha$. The segment of a line connecting the vertex of the upper base with the center of the circle circumscribed around the lower base is $l$ and forms an angle $\boldsymbol{\beta}$ with the plane of the base. Find the volume of the prism.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-308.jpg?height=224&width=345&top_left_y=89&top_left_x=103)

Let the isosceles trapezoid \(ABCD\) be the base of the given prism, \(AB = CD\), \(O\) be the center of the circle circumscribed around \(ABCD\), and \(AC \perp CD\). Then \(AD\) is the diameter of the circle \(O\) (Theorem 6) and \(O\) is the midpoint of \(AD\).

Assume that \(K\) is the vertex of the upper base of the prism, which is connected to point \(O\) by a segment of length \(l\).

Since the prism is a right prism, the projection of \(K\) onto the lower base coincides with one of the vertices of the trapezoid \(ABCD\). Therefore, the projection of \(KO\) onto the lower base, equal to \(l \cdot \cos \beta\), coincides with the radius of the circle \(O\), i.e.,

\[ AD = 2l \cdot \cos \beta. \]

\(\angle DPC = \alpha \Rightarrow \angle PAD = \frac{\alpha}{2} \Rightarrow AC = AD \cdot \cos \frac{\alpha}{2} \Rightarrow\)

\(\Rightarrow S_{ABCD} = \frac{AC^2}{2} \cdot \sin \alpha = 2l^2 \cos^2 \beta \cdot \cos^2 \frac{\alpha}{2} \cdot \sin \alpha\).

Since the height of the prism is \(l \cdot \sin \beta\), the volume is \(2l^3 \cos^2 \beta \cdot \cos^2 \frac{\alpha}{2} \cdot \sin \alpha \cdot \sin \beta\).

Answer: \(l^3 \cos \beta \cdot \cos^2 \frac{\alpha}{2} \cdot \sin 2\beta \cdot \sin \alpha\).