11.011. The base of a regular pyramid is a polygon, the sum of the interior angles of which is $720^{\circ}$. Determine the volume of the pyramid if its lateral edge, equal to $l$, forms an angle of $30^{\circ}$ with the height of the pyramid.
## Solution.

From the formula for the sum of the interior angles of a polygon, we determine the type of the polygon: $720^{\circ}=180^{\circ}(n-2)$. Hence, $n=6$, i.e., we have a hexagon; since the angle between the lateral edge and the height is $30^{\circ}$, the height $h=l \cos 30^{\circ}=\frac{\sqrt{3}}{2} l$. The radius of the circumscribed circle around the hexagon $R=l \sin 30^{\circ}=\frac{l}{2}$. It is equal to the side of the hexagon $a=\frac{l}{2}$. The radius of the circle inscribed in the hexagon is $r=\frac{a}{2 \operatorname{tg} \frac{180^{\circ}}{6}}=\frac{a \sqrt{3}}{2}=\frac{l \sqrt{3}}{4}$. Therefore, the area of the base $S_{\text {base }}=\frac{6 a r}{2}=\frac{6 \sqrt{3} l^{2}}{16}$, and the volume $V=\frac{1}{3} S_{\text {base }} \cdot h=\frac{3 l^{3}}{16}$. Answer: $\frac{3 l^{3}}{16}$.