12.256. The base of the pyramid is a rectangle, where the angle between the diagonals is $\alpha$. One of the lateral edges is perpendicular to the base plane, and the largest edge forms an angle $\beta$ with the base plane. The radius of the sphere circumscribed around the pyramid is $R$. Find the volume of the pyramid.
## Solution.

Let the rectangle $ABCD$ be the base of the pyramid $EABCD$ (Fig. 12.122), the angle between its diagonals is $\alpha$, the lateral edge $EB$ is perpendicular to the plane of the base of the pyramid and is its height, $AB$, $CB$, $DB$ are the projections of the lateral edges $AE$, $EC$, $DE$ on the plane of the base, $DB > AB$, $DB > CB \Rightarrow DE$ is the largest lateral edge and $\angle EDB = \beta$.

Points $O_1$ - the midpoint of $ED$, point $O$ - the intersection of the diagonals of the rectangle - the midpoint of $BD$. Thus, $OO_1$ is the midline of $\triangle EDB$ and $OO_1 \parallel BE$. Therefore, the line $OO_1$ is perpendicular to the plane of the base and its points are equidistant from the vertices of the base. Since $O$ is removed from the vertex of the pyramid by a distance $O_1E = O_1D$, then $O_1$ is equidistant from all vertices of the pyramid and is the center of the circumscribed sphere, $O_1D = O_1E = R$.

In $\triangle EBD\left(\angle EBD = 90^\circ\right):$

\[
\begin{aligned}
& EB = ED \sin \angle EDB = 2R \sin \beta \\
& BD = ED \cos \angle EDB = 2R \cos \beta
\end{aligned}
\]

The area of the base of the pyramid $S = \frac{1}{2} BD^2 \sin \alpha = 2R^2 \cos^2 \beta \sin \alpha$.

The volume of the pyramid:

$V = \frac{1}{3} S \cdot EB = \frac{2}{3} R^2 \cos^2 \beta \sin \alpha \cdot 2R \sin \beta = \frac{2}{3} R^3 \sin 2\beta \cos \beta \sin \alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1009.jpg?height=644&width=503&top_left_y=1192&top_left_x=96)

Fig. 12.123
Answer: $\frac{2}{3} R^3 \sin 2\beta \cos \beta \sin \alpha$.