11.106. The side of the base of a regular hexagonal pyramid is equal to $a$. Calculate the volume of the pyramid if it is known that its lateral surface area is 10 times greater than the area of the base.
Solution.

$S O$ is the height of the regular pyramid SABCDEF (Fig. 11.3), $A B=B C=F A=a$. Then $C O=O D=C D=a$, and $S_{A B C D E F}=6 S_{\triangle C O D}=$ $=6 \frac{a^{2} \sqrt{3}}{4}=3 \frac{a^{2} \sqrt{3}}{2}=S_{1} . S M-$ is the apothem of the lateral face CSD. Then $O M \perp C D, O M=\frac{a \sqrt{3}}{2}, \angle S M O-$ is the angle between the plane of the lateral face and the plane of the base.

Let $\angle S M O=\varphi, 0^{\circ}<\varphi<90^{\circ}, S_{2}$ be the area of the lateral surface of the pyramid. Then $\frac{S_{1}}{S_{2}}=\cos \varphi$. Since by the condition $S_{2}=10 S_{1}$, then $\frac{1}{\cos \varphi}=10$ and $\operatorname{tg} \varphi=\sqrt{\frac{1}{\cos \varphi}-1}=\sqrt{100-1}=3 \sqrt{11}$.

From $\triangle S O M\left(\angle S O M=90^{\circ}\right): S O=O M \operatorname{tg} \varphi=\frac{a \sqrt{3}}{2} \cdot 3 \sqrt{11}=\frac{3 a \sqrt{33}}{2}$.

The volume of the pyramid $V=\frac{1}{3} S_{1} \cdot S O=\frac{1}{3} \cdot \frac{3 a^{2} \sqrt{3}}{2} \cdot \frac{3 a \sqrt{33}}{2}=\frac{9 a^{3} \sqrt{11}}{4}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0822.jpg?height=678&width=581&top_left_y=77&top_left_x=81)

Fig. 11.3

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0822.jpg?height=757&width=535&top_left_y=78&top_left_x=742)

Fig. 11.4

Answer: $\frac{9 a^{3} \sqrt{11}}{4}$.