The area of the base of a frustum is $8 \mathrm{~cm}^{2}$, and the area of the top is $1 \mathrm{~cm}^{2}$. The pyramid is divided into two equal volumes by a plane parallel to the base. What is the area of the cross-section?
I. solution. Let's complete a frustum of a pyramid, whose base area is $A$, and top area is $B$ $(B<A)$, into a complete pyramid. The "added" and the resulting complete pyramids are centrally similar. If the height of the frustum is $m$, and the height of the complete pyramid is $m+x$, then from the similarity of the corresponding pyramids, we have $(m+x): x=\sqrt{A}: \sqrt{B}$, from which $x=m \frac{\sqrt{B}}{\sqrt{A}-\sqrt{B}}$. Therefore, the volume of the complete pyramid is $V=\frac{m+x}{3} A=\frac{m}{3} \frac{A \sqrt{A}}{\sqrt{A}-\sqrt{B}}$, and the volume of the frustum is $V\left(1-\left(\frac{x}{m+x}\right)^{3}\right)=V\left(1-\frac{B \sqrt{B}}{A \sqrt{A}}\right)$.

When the frustum is cut into two parts, the base area of the "lower" frustum is also $A$, and we denote the (unknown) area of its top by $C$. Since it can be completed to the same $V$ volume pyramid as the original frustum, its volume, according to the formula derived above, is $V\left(1-\frac{C \sqrt{C}}{A \sqrt{A}}\right)$. The condition of the problem is thus:

$$
1-\frac{C \sqrt{C}}{A \sqrt{A}}=\frac{1}{2}\left(1-\frac{B \sqrt{B}}{A \sqrt{A}}\right)
$$

Rearranging:

$$
\begin{aligned}
2 A \sqrt{A}-2 C \sqrt{C} & =A \sqrt{A}-B \sqrt{B} \\
\frac{A \sqrt{A}+B \sqrt{B}}{2} & =(\sqrt{C})^{3}
\end{aligned}
$$

The solution to this is

$$
C=\left(\frac{A \sqrt{A}+B \sqrt{B}}{2}\right)^{2 / 3}=\left(\frac{16 \sqrt{2}+1}{2}\right)^{2 / 3} \approx 5.19 \mathrm{~cm}^{2}
$$

II. solution. Let the volume of the frustum be $V$, the volumes of the parts be $V_{1}$ and $V_{2}$, the heights corresponding to them be $m_{1}$ and $m_{2}$, and the area of the plane section be $x$.

The volume of the complete pyramid is

$$
V=\frac{\left(m_{1}+m_{2}\right)}{3}(8+\sqrt{8}+1) ; \quad V_{1}=\frac{m_{1}}{3}(8+\sqrt{8 x}+x) ; \quad V_{2}=\frac{m_{2}}{3}(x+\sqrt{x}+1)
$$

From the equality of the volumes, and the fact that $V_{1}+V_{2}=V$, we can write for the triple of the volumes:

$$
\begin{aligned}
m_{1}(8+x+\sqrt{8 x}) & =m_{2}(x+\sqrt{x}+1) \\
m_{1}(8+x+\sqrt{8 x})+m_{2}(x+1+\sqrt{x}) & =\left(m_{1}+m_{2}\right)(9+\sqrt{8})
\end{aligned}
$$

From (1), solve for $m_{1}$:

$$
m_{1}=\frac{m_{2}(x+1+\sqrt{x})}{8+x+\sqrt{8 x}}
$$

Substituting this into (2):

$$
m_{2}(x+1+\sqrt{x})+m_{2}(x+1+\sqrt{x})=\left[\frac{m_{2}(x+1+\sqrt{x})}{8+x+\sqrt{8 x}}+m_{2}\right](9+\sqrt{8})
$$

Perform the indicated operations and simplify by $m_{2} \neq 0$, and introduce the new unknown $\sqrt{x}=a$. Then we arrive at the following quartic equation:

$$
2 a^{4}+(2+4 \sqrt{2}) a^{3}-(1+16 \sqrt{2}) a-(65+18 \sqrt{2})=0
$$

We can factor out $2 a^{3}$ from the first 2 terms, obtaining

$$
2 a^{3}(a+1+2 \sqrt{2})-(1+16 \sqrt{2}) a-(65+18 \sqrt{2})=0
$$

Let's check if $(a+1+2 \sqrt{2})$ is a divisor of the sum of the other two terms in the equation, i.e., if there is a number $c$ such that

$$
c(a+1+2 \sqrt{2})=(16 \sqrt{2}+1) a+65+18 \sqrt{2}
$$

The equality of the coefficients implies that

$$
c a=(16 \sqrt{2}+1) a
$$

This is satisfied by $c=16 \sqrt{2}+1$, and indeed

$$
(16 \sqrt{2}+1)(2 \sqrt{2}+1)=64+16 \sqrt{2}+2 \sqrt{2}+1=65+18 \sqrt{2}
$$

Thus, the polynomial can be factored as $\left(2 a^{3}-1-16 \sqrt{2}\right)(a+1+2 \sqrt{2})$. The second factor has only one negative root, which is not possible $(a=\sqrt{x})$. The first factor has only one real root $a=\sqrt[3]{\frac{1+16 \sqrt{2}}{2}} \approx 2.27752$, so the area of the plane section is $x=a^{2} \approx 5.187 \mathrm{~cm}^{2}$.

Remark. The problem, as can be seen from the solutions, was actually not necessarily about frustums of pyramids, but about generally "pyramid-like" bodies; only the area of the base and the top mattered for the final result, their shape was irrelevant. Therefore, we did not consider solutions that used a frustum of a cone instead of a frustum of a pyramid as incorrect.