70 A regular quadrilateral frustum $A B C D-A_{1} B_{1} C_{1} D_{1}$ has a height of $6 \mathrm{~cm}$, with the side lengths of the upper and lower bases being $4 \mathrm{~cm}$ and $8 \mathrm{~cm}$, respectively. A section is made through the diagonal $B D_{1}$, parallel to the diagonal $A C$ of the base. Find the area of this section.
70 As shown in Figure 30, draw a line through $B$ parallel to $AC$, intersecting the extensions of $DA$ and $DC$ at $E$ and $F$ respectively. It is easy to see that $AE = DA = DC = CF = 8$.
Let $A A_{1}$, $B B_{1}$, $C C_{1}$, and $D D_{1}$ intersect at $S$.
Connect $D$ and $E$ to intersect $SA$ at $G$, and connect $D_{1}F$ to intersect $SC$ at $H$, then the section is $D_{1} G B H$.
Since $AE = 2 A_{1} D_{1}$, we have $D_{1} G = \frac{1}{3} DE$.
$$
S_{\triangle G E B} = \frac{2}{3} \times \frac{1}{2} S_{\triangle D_{1} E F} = \frac{1}{3} S_{\triangle D_{1} E F}.
$$

Similarly,
$$
S_{\triangle F H B} = \frac{1}{3} S_{\triangle D_{1} E F}.
$$

Since $A_{1} B_{1} = \frac{1}{2} AB$, we have $S D_{1} = \frac{1}{2} S D$. $B D_{1}$ is the median of $\triangle S B D$.
Let the center of the base $ABCD$ be $O$, then $SO = 2 \times 6 = 12$.
$$
\begin{array}{c}
S B^{2} = S D^{2} = 12^{2} + \left(8 \times \frac{\sqrt{2}}{2}\right)^{2} = 12^{2} + 32. \\
B D_{1}^{2} = \frac{1}{4} \left(S B^{2} + 2 B D^{2}\right) = \frac{1}{4} \left(12^{2} + 32 + 2 \times 8^{2} \times 2\right) = 108 \\
B D_{1} = \sqrt{108} = 6 \sqrt{3}.
\end{array}
$$

Since $AC \perp BD$, $AC \perp SO$, we have $AC \perp$ plane $SBD$, $BD_{1} \perp AC$, $BD_{1} \perp EF$.
$$
\begin{aligned}
S_{\text {section}} & = \left(1 - \frac{1}{3} - \frac{1}{5}\right) S_{\triangle D_{1} E F} = \frac{1}{3} \times \frac{1}{2} \times B D_{1} \times E F \\
& = \frac{1}{3} \times \frac{1}{2} \times 6 \sqrt{3} \times 2 \times 8 \sqrt{2} = 16 \sqrt{6}.
\end{aligned}
$$