11.132. The base of a quadrilateral pyramid is a rectangle with an area of $S$; the lateral edges of the pyramid are equal and form an angle of $45^{\circ}$ with the base plane. The angle between the diagonals of the base is $60^{\circ}$. Find the volume of the pyramid.
Solution.

Given that $ABCD$ is a rectangle, draw $EO$ as the height of the pyramid $EABCD$ (Fig. 11.29). Then $\angle EAO = 45^\circ$, and since the lateral edges of the pyramid are equal, point $O$ is the center of the circle circumscribed around the rectangle $ABCD$ - the point of intersection of the diagonals. Let $EO = H$. Then $AO = EO = H$, $AC = 2H$, the area of the rectangle $S = \frac{1}{2} AC^2 \sin 60^\circ = 2H^2 \frac{\sqrt{3}}{2} ; H = \sqrt{\frac{S}{\sqrt{3}}} = \frac{\sqrt{S}}{\sqrt[4]{3}} = \frac{\sqrt{S} \cdot \sqrt[4]{27}}{3}$. The volume of the pyramid $V = \frac{1}{3} S H = \frac{S \sqrt{S} \cdot \sqrt[4]{27}}{9}$.

Answer: $\quad \frac{S \sqrt{S} \cdot \sqrt[4]{27}}{9}$.