6. Given a tetrahedron $S-ABC$ with the base being an equilateral triangle of side length $4 \sqrt{2} \, \text{cm}$, and the edge $SC \perp$ the base $ABC$, with a length of $2 \, \text{cm}$. Points $D$ and $E$ are the midpoints of $AB$ and $BC$ respectively. If $P$ moves along $SE$, find the minimum area of $\triangle PCD$.
6. When $P$ moves on $SE$, the base $CD$ of $\triangle PCD$ remains unchanged, therefore, the minimum value of the area of $\triangle PCD$ is determined by the minimum distance from $P$ to $CD$. Draw $PF \perp BC$, with the foot of the perpendicular at $F$. Since $SC \perp$ the base $ABC$, the plane $SBC \perp$ the base $ABC$, and the intersection line is $BC$, so $PF \perp$ the base $ABC$. Draw $FG \perp CD$ at $G$, and connect $PG$, then $PG \perp CD$, thus the minimum value of $PG$ makes the area of $\triangle PCD$ the smallest.
Assume $PF=x, x \in (0,2)$, from $\triangle PEF \sim \triangle SEC$ we get
$$
\frac{x}{SC}=\frac{EF}{EC} \Rightarrow EF=\frac{EC}{SC} \cdot x \text{, }
$$

i.e., $EF=\sqrt{2} x$,
thus $CF=2\sqrt{2}-\sqrt{2} x$,
$$
GF=CF \sin 30^{\circ}=\sqrt{2}-\frac{\sqrt{2}}{2} x \text{. }
$$

From this, we get the functional relationship:
$$
PG^2=PF^2+GF^2=x^2+\left(\sqrt{2}-\frac{\sqrt{2}}{2} x\right)^2=\frac{3}{2}\left(x-\frac{2}{3}\right)^2+\frac{4}{3} \text{, }
$$

we know that when $x=\frac{2}{3}$, $PG_{\min}=\frac{2\sqrt{3}}{3}$.
Therefore, the minimum area of $\triangle PCD$ is $S_{\triangle PCD}=\frac{1}{2} CD \cdot PG=\frac{1}{2} \cdot \frac{\sqrt{3}}{2} \cdot 4\sqrt{2} \cdot \frac{2}{3} \sqrt{3}=2\sqrt{2}(\mathrm{~cm}^2)$.