11.069. The lateral edge of a regular triangular prism is equal to the height of the base, and the area of the section made through this lateral edge and the height of the base is $Q$. Determine the volume of the prism.
Solution.

The volume of the prism is found as $V=S \cdot H$ (Fig. 11.64). The area of the section $D D_{1} C_{1} C \quad Q=D C \cdot C C_{1}$. Since by condition $D C=C C_{1}$, then $D C=C C_{1}=\sqrt{Q}$. The area of $\triangle A B C: S=\frac{x^{2} \sqrt{3}}{4}$. From $\triangle C D B Q+\frac{x^{2}}{4}=x^{2}$. Hence $x^{2}=\frac{4 Q}{3}$. Then $S=\frac{Q \sqrt{3}}{3}=\frac{Q}{\sqrt{3}}$ and the required volume $V=\frac{Q \sqrt{Q}}{\sqrt{3}}$.

Answer: $Q \sqrt{Q / 3}$.