11.019. The lateral edge of a regular triangular pyramid is equal to $l$, and the height is equal to $h$. Determine the volume of the pyramid.
## Solution.

Let $A S=l, S H=h$ (see Fig. 11.5). Then

$A H=\sqrt{A S^{2}-S H^{2}}=\sqrt{l^{2}-h^{2}}$. In a regular pyramid, $A H$ is the radius of the circumscribed circle around the equilateral $\triangle A B C$, i.e., $A H=R=\frac{a \sqrt{3}}{3}$ or $\sqrt{l^{2}-h^{2}}=\frac{a \sqrt{3}}{3}$, from which $a=\frac{3 \sqrt{l^{2}-h^{2}}}{\sqrt{3}}=\sqrt{3}\left(l^{2}-h^{2}\right)$. The area of the base $S_{\text {base }}=\frac{a^{2} \sqrt{3}}{4}=\frac{3 \sqrt{3}\left(l^{2}-h^{2}\right)}{4}$. Then the volume $V=\frac{1}{3} S_{\text {base }} \cdot h=\frac{1}{3} h \frac{3 \sqrt{3}\left(l^{2}-h^{2}\right)}{4}=\frac{\sqrt{3} h\left(l^{2}-h^{2}\right)}{4}$.

Answer: $\frac{\sqrt{3} h\left(l^{2}-h^{2}\right)}{4}$.