11.172. The angle between the generatrix of the cone and the plane of the base is $30^{\circ}$. The lateral surface area of the cone is $3 \pi \sqrt{3}$ square units. Determine the volume of a regular hexagonal pyramid inscribed in the cone.
Solution.

A regular hexagon $ABCDEF$ is the base of a regular pyramid inscribed in a cone with height $MO$ (Fig. 11.66). Then $\angle MEO$ is the angle between the generatrix of the cone and the plane of the base, $\angle MEO = 30^\circ$. Let $EO = R, ME = l$.

Then from $\triangle MOE \left(\angle MOE = 90^\circ\right)$, $l = ME = \frac{OE}{\cos \angle MEO} = \frac{R}{\cos 30^\circ} = \frac{2R}{\sqrt{3}}$; $OM = OE \tan \angle MEO = \frac{R \sqrt{3}}{3}$.

The lateral surface area of the cone $S_5 = \pi R l = 3 \pi \sqrt{3}$.

Then $R \cdot \frac{2R}{\sqrt{3}} = 3 \sqrt{3}$; $R^2 = \frac{9}{2}$; $R = \frac{3}{\sqrt{2}}$. The area of the base of the pyramid $S = 6 S_{\triangle EOF} = 6 \frac{R^2 \sqrt{3}}{4} = \frac{3 R^2 \sqrt{3}}{2} = \frac{3 \sqrt{3}}{2} \cdot \frac{9}{2} = \frac{27 \sqrt{3}}{4}$ (sq. units); $OM = \frac{R \sqrt{3}}{3} = \frac{\sqrt{3}}{\sqrt{2}}$. The volume of the pyramid $V = \frac{1}{3} S \cdot OM = \frac{1}{3} \cdot \frac{27 \sqrt{3}}{4} \cdot \frac{\sqrt{3}}{\sqrt{2}} = \frac{27}{4 \sqrt{2}} = \frac{27 \sqrt{2}}{8}$ (cubic units). Answer: $\frac{27 \sqrt{2}}{8}$ cubic units.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0876.jpg?height=571&width=505&top_left_y=106&top_left_x=141)

Fig. 11.66

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0876.jpg?height=617&width=402&top_left_y=65&top_left_x=792)

Fig. 11.67