11.146. The base of the pyramid is a regular hexagon with a side length of $a$. One of the lateral edges is perpendicular to the base plane and equals the side of the base. Determine the total surface area of the pyramid.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0854.jpg?height=500&width=515&top_left_y=77&top_left_x=128)

Fig. 11.43 .1

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0854.jpg?height=412&width=440&top_left_y=117&top_left_x=771)

Fig. 11.43 .2
## Solution.

The correct hexagon $A B C D M N$ is the base of the pyramid KABCDMN (Fig. 11.43.1), the edge $K B$ is perpendicular to the plane of the base, $K B=A B=a$. Then $\triangle A B K=\triangle C B K$ (by two legs), $S_{\triangle A B K}=\frac{1}{2} A B \cdot B K=\frac{a^{2}}{2} \cdot \angle B A N=\angle A N M=120^{\circ}$ (Fig. 11.43.2). Then from $\triangle B A N A B=A N, B N=a \sqrt{3}, \angle A B N=\angle A N B=30^{\circ}$ and, consequently, $\angle B N M=90^{\circ} \cdot B N$ is the projection of $K N$ on the plane of the base of the pyramid. Therefore, $K N \perp N M$. Similarly, $K D \perp M D$ and since $B N=B D$, then $K N=K D, \Delta K N M=\Delta K D M, \Delta K A N=\Delta K C D$.

From $\triangle K B N\left(\angle K B N=90^{\circ}\right): K N=\sqrt{B N^{2}+K B^{2}}=2 a$.

Then $S_{\triangle K N M}=\frac{1}{2} K N \cdot M N=a^{2}$.

From $\triangle A B K\left(\angle A B K=90^{\circ}\right): A K=\sqrt{A B^{2}+B K^{2}}=a \sqrt{2}$. The semiperimeter of the triangle $\triangle A N K \quad p=\frac{a \sqrt{2}+2 a+a}{2}=\frac{3 a+a \sqrt{2}}{2}$, we will calculate its area using Heron's formula:

$$
S_{\triangle A N K}=\sqrt{\frac{3 a+a \sqrt{2}}{2}\left(\frac{3 a+a \sqrt{2}}{2}-a\right)} \times \sqrt{\left(\frac{3 a+a \sqrt{2}}{2}-a \sqrt{2}\right)\left(\frac{3 a+a \sqrt{2}}{2}-2 a\right)}=
$$

$=\sqrt{\frac{3 a+a \sqrt{2}}{4} \cdot(a \sqrt{2}+a)} \times \sqrt{\frac{3 a-a \sqrt{2}}{2} \cdot \frac{a \sqrt{2}-a}{2}}=$

$=\frac{a^{2}}{4} \sqrt{(3+\sqrt{2})(3-\sqrt{2})(\sqrt{2}+1)(\sqrt{2}-1)}=\frac{a^{2} \sqrt{2}}{4}$

The area of the regular hexagon $A B C D M N: S_{\text {base }}=\frac{3 a^{2} \sqrt{3}}{2}$. Therefore, the total surface area of the pyramid

$$
\begin{aligned}
& S=2\left(S_{\triangle A B K}+S_{\triangle K N M}+S_{\triangle A N K}\right)+S_{\mathrm{base}}=2\left(\frac{a^{2}}{2}+a^{2}+\frac{a^{2} \sqrt{2}}{4}\right)+\frac{3 a^{2} \sqrt{3}}{2}= \\
= & \frac{a^{2}(6+\sqrt{2}+3 \sqrt{3})}{2}
\end{aligned}
$$

Answer: $\frac{a^{2}(6+\sqrt{2}+3 \sqrt{3})}{2}$.