11.051. Find the volume of a regular triangular pyramid, where the plane angle at the vertex is $90^{\circ}$, and the side of the base is 3 cm.
Solution.

$V=\frac{1}{3} S H \quad$ (Fig. 11.46); $S=\frac{9}{4} \sqrt{3} ; O C=\frac{a \sqrt{3}}{3}=\frac{3 \sqrt{3}}{3}=\sqrt{3} \quad$ (since $A B=B C=A C=a=3 \mathrm{~cm}) ; E O=\frac{a \sqrt{3}}{6}=\frac{\sqrt{3}}{2} . \mathrm{B} \triangle E S C \angle E S C=90^{\circ}, H^{2}=E O \times$ $\times O C=\frac{\sqrt{3}}{2} \cdot \sqrt{3}=\frac{3}{2} ; H=\sqrt{\frac{3}{2}} ; V=\frac{1}{3} \cdot \frac{9}{4} \sqrt{3} \cdot \sqrt{\frac{3}{2}}=\frac{9}{4 \sqrt{2}}=\frac{9 \sqrt{2}}{8}$.

Answer: $\frac{9 \sqrt{2}}{8} \mathrm{~cm}^{3}$.