### 11.206 The side of the base of a regular triangular pyramid is equal to

 $a$; the area of its section, which has the shape of a square, is equal to $m^{2}$. Find the ratio of the lateral surface area of the pyramid to the area of the base.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-263.jpg?height=299&width=324&top_left_y=1043&top_left_x=157)

Let $K L M N$ be the section of the given regular pyramid $F A B C$, which is a square. $M N \| L K \Rightarrow$ $M N\|F B C \Rightarrow M N\| F C$ (Theorem 14). For the same reason, $M L \| A B$. From the similarity of triangles, we have:

$$
\frac{A M}{A C}=\frac{M N}{F C} \Rightarrow F C=\frac{a m}{a-m}(A M=a-M C=a-m, M N=m)
$$

Therefore, the apothem $h$ of the lateral face is $\sqrt{F C^{2}-\frac{a^{2}}{4}}=\frac{a \sqrt{3 m^{2}-a^{2}+2 a m}}{2(a-m)}$, the area of the lateral surface is $3 \frac{a h}{2}$, and the area of the base is $\frac{a^{2} \sqrt{3}}{4}$. Therefore, their ratio is $\frac{\sqrt{9 m^{2}-3 a^{2}+6 a m}}{a-m}$.

Answer: $\frac{\sqrt{9 m^{2}-3 a^{2}+6 a m}}{a-m}$.