11.126. The side of the base of a regular hexagonal pyramid is equal to $a$. All diagonal sections of it are equal. Find the volume and lateral surface area of the pyramid.
## Solution.

Let's draw SO - the height of the regular pyramid $S A B C D E F$ (Fig. 11.22). Let $S O=H$. Then $S_{\triangle A S D}=\frac{A D \cdot S O}{2}=\frac{2 a H}{2}=a H$.

$K$ is the intersection point of the diagonals $B O$ and $A C$ of the rhombus $A B C O$.

$$
O K=\frac{1}{2} B O=\frac{a}{2}, A K=\frac{a \sqrt{3}}{2}, A C=
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0838.jpg?height=634&width=523&top_left_y=111&top_left_x=740)

Fig. 11.22

$=2 A K=a \sqrt{3}, O K \perp A C . O K-$ is the projection of $S K$ on the base plane.

Then $S K \perp A C$ and $S_{\triangle A S C}=\frac{A C \cdot S K}{2}=\frac{a \sqrt{3}}{2} S K$.

By the condition $S_{\triangle A S D}=S_{\triangle A S C}$, then $a H=\frac{a \sqrt{3}}{2} S K ; \frac{H}{S K}=\frac{\sqrt{3}}{2}$. From $\triangle S O K\left(\angle S O K=90^{\circ}\right): \sin \angle S K O=\frac{H}{S K}=\frac{\sqrt{3}}{2} ; \angle S K O=60^{\circ} ; H=S O=$ $=K O \operatorname{tg} \angle S K O=\frac{a \sqrt{3}}{2}$. The area of the base $S_{\text {base }}=6 S_{\triangle A O B}=6 \frac{a^{2} \sqrt{3}}{4}=$ $=\frac{3 a^{2} \sqrt{3}}{2}$. The volume of the pyramid $V=\frac{1}{3} S_{\text {base }} H=\frac{3 a^{3}}{4}$. Let $T-$ be the midpoint of $C D$, then $T O=\frac{a \sqrt{3}}{2}$. Since $S O=\frac{a \sqrt{3}}{2}$, then $\triangle S O T$ is a right isosceles triangle, $\angle S T O=45^{\circ}$ and the lateral surface area of the pyramid $S_{6}=\frac{S_{\text {base }}}{\cos \angle S T O}=\frac{3 a^{2} \sqrt{3}}{2} \sqrt{2}=\frac{3 a^{2} \sqrt{6}}{2}$.

Answer: $\frac{3 a^{3}}{4} ; \frac{3 a^{2} \sqrt{6}}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0839.jpg?height=428&width=473&top_left_y=194&top_left_x=139)

Fig. 11.23

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0839.jpg?height=537&width=490&top_left_y=79&top_left_x=720)

Fig. 11.24