11.058. Determine the volume of a regular triangular pyramid if the height of the triangle serving as its base is $h$, and the apothem of the pyramid is $m$.
Solution.

$V=\frac{1}{3} S H, M C=h, S M=m$ (Fig. 11.53). $A B=B C=A C$, therefore

$C M=\frac{A B \sqrt{3}}{2} \Rightarrow A B=\frac{2 C M}{\sqrt{3}}=\frac{2 h}{\sqrt{3}}$ and $S=\frac{h^{2}}{\sqrt{3}} . M O=\frac{A B \sqrt{3}}{6}=\frac{h}{3}$. In $\triangle S O M \angle S O M=90^{\circ} \Rightarrow H=\sqrt{S M^{2}-M O^{2}}=\sqrt{m^{2}-\frac{h^{2}}{9}}=\sqrt{\frac{9 m^{2}-h^{2}}{3}}$.

Therefore, $V=\frac{1}{3} \cdot \frac{h^{2}}{\sqrt{3}} \cdot \frac{\sqrt{9 m^{2}-h^{2}}}{3}=\frac{\sqrt{3}}{27} h^{2} \sqrt{9 m^{2}-h^{2}}$.

Answer: $\frac{\sqrt{3}}{27} h^{2} \sqrt{9 m^{2}-h^{2}}$.