11.142. In a regular triangular prism, a plane is drawn through the side of the lower base and the opposite vertex of the upper base, forming an angle of $45^{\circ}$ with the plane of the lower base. The area of the section is $S$. Find the volume of the prism.
## Solution.

$\triangle A C_{1} B$ is the section of the regular prism $A B C A_{1} B_{1} C_{1}$ (Fig. 11.39) mentioned in the problem, and $K$ is the midpoint of $A B$. Then $C K \perp A B$, $C_{1} K \perp A B$, $\angle C_{1} K C$ is the angle of inclination of the plane $A C_{1} B$ to the plane of the base, $\angle C_{1} K C=45^{\circ} \cdot S_{\triangle A B C}=S_{\triangle A C_{1} B} \cdot \cos \angle C_{1} K C=\frac{S \sqrt{2}}{2}$. Let $C C_{1}=H$.

Then $C K=C C_{1}=H, B K=\frac{H \sqrt{3}}{3}, S_{\triangle A B C}=\frac{1}{2} A B \cdot C K=B K \cdot C K=$ $=\frac{H^{2} \sqrt{3}}{3}$. Thus, $\frac{H^{2} \sqrt{3}}{3}=\frac{S \sqrt{2}}{2} ; H=\sqrt{\frac{S \sqrt{6}}{2}}$.

The volume of the prism is $V=S_{\triangle A B C} \cdot H=\frac{S \sqrt{2}}{2} \cdot \sqrt{\frac{S}{2} \sqrt{6}}=\frac{S \sqrt{S} \cdot \sqrt[4]{6}}{2}$. Answer: $\frac{S \sqrt{S} \cdot \sqrt[4]{6}}{2}$.