12.432 The side of the lower base of a regular truncated quadrilateral pyramid is 5 times the side of the upper base. The lateral surface area of the pyramid is equal to the square of its height. Find the angle between the lateral edge of the pyramid and the plane of the base.
Solution.

Let $O_{1}$ and $O_{2}$ be the centers of the bases of the given regular truncated pyramid $A_{1} B_{1} C_{1} D_{1} A_{2} B_{2} C_{2} D_{2}$, with the sides of the bases being $2x$ and $10x$, and $K_{1}$ and $K_{2}$ be the midpoints of the edges $A_{1} D_{1}$ and $A_{2} D_{2}$. Then $K_{1} K_{2}$ is the height of the trapezoid $A_{1} D_{1} D_{2} A_{2}$ (Theorem 12). Let $h = O_{1} O_{2}$. By the problem statement,

$$
\begin{aligned}
& S_{A_{1} D_{1} D_{2} A_{2}} = \frac{h^{2}}{4} = 6x \cdot K_{1} K_{2} = \\
& = 6x \sqrt{O_{1} O_{2}^{2} + \left(K_{2} O_{2} - K_{1} O_{1}\right)^{2}} = 6x \sqrt{h^{2} + 16x^{2}} \Rightarrow
\end{aligned}
$$

$$
\Rightarrow \left(\frac{h}{x}\right)^{4} - 24^{2} \left(\frac{h}{x}\right)^{2} - 16 \cdot 24^{2} = 0 \Rightarrow \frac{h}{x} = 4 \sqrt{2(9 + 3 \sqrt{10})}
$$

But $\operatorname{tg} \angle D_{1} D_{2} O_{2} = \frac{h}{D_{2} O_{2} - D_{1} O_{1}} = \frac{h}{4x \sqrt{2}} \Rightarrow \angle D_{1} D_{2} O_{2} = \operatorname{arctg} \sqrt{9 + 3 \sqrt{10}}$.

Answer: $\operatorname{arctg} \sqrt{9 + 3 \sqrt{10}}$.