11.230 The height of a regular triangular pyramid is equal to $\boldsymbol{H}$. Find its total surface area if a plane, passing through the vertex of the base of the pyramid perpendicular to the height of the opposite lateral face, forms an angle of $30^{\circ}$ with the plane of the axis.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-279.jpg?height=342&width=383&top_left_y=187&top_left_x=189)

Let in the given regular pyramid $F A B C$, the height $F L$ of the lateral face $F A C$ be perpendicular to the section $B M N$. Since $M N \| A C$ and $M N \perp F L$, the line $l$ of intersection of the planes $B M N$ and $A B C$ is parallel to $M N$ and $A C$ (Theorem 14) $\Rightarrow B L \perp l$. Furthermore, $B M = B N (\triangle F B N = \Delta F B M) \Rightarrow B K \perp M N (K$ is the intersection point of $M N$ and $F L) \Rightarrow B K \perp l \Rightarrow$

$\Rightarrow \angle K B L$ is the linear angle of the dihedral angle formed by $B M N$ and $A B C$.

By the condition, $\angle K B L = 30^{\circ}, \angle F L B = 60^{\circ}$.

The height $H = F O$ of the given pyramid lies in the plane $\triangle F B L \Rightarrow$

$\Rightarrow F L = \frac{H}{\sin 60^{\circ}}, L O = H \cdot \operatorname{ctg} 60^{\circ}, A C = 2 \sqrt{3} \cdot L O \quad(O$ is the center of the regular $\triangle A B C$ )

$\Rightarrow 3 S_{F A C} + S_{A B C} = \frac{3}{2} F L \cdot A C + \frac{\sqrt{3}}{4} A C^{2} = 3 \sqrt{3} H^{2}$.

Answer: $3 \sqrt{3} H^{2}$.