Four, Given the pyramid $S-ABC$ with height $SO=3$, the base edge length is 6. A perpendicular is drawn from point $A$ to the opposite side $SBC$, with the foot of the perpendicular being $O'$. On $AO'$, take a point $P$ such that $\frac{AP}{PO'}=8$. Find the area of the section passing through point $P$ and parallel to the base.
$\sqrt{3}$
IV. 【Analysis and Solution】As shown in the figure, since $S-ABC$ is a regular triangular pyramid, $O$ is the centroid of $\triangle ABC$. Connecting $AO$ and extending it to intersect $BC$ at $D$, because $D$ is the midpoint of $BC$, $BC \perp$ plane $SAD$, and $AO' \perp BC$, so $AO'$ lies on plane $SAD$, thus $O'$ must be on $DS$, hence
$$
\begin{array}{l}
AD=6 \times \frac{\sqrt{3}}{2}=3 \sqrt{3}, \\
OD=\frac{1}{3} AD=\sqrt{3},
\end{array}
$$
$$
\begin{array}{l}
SD=\sqrt{3^{2}+(\sqrt{3})^{2}}=\sqrt{12} . \\
\text { and } \frac{O' D}{A D}=\frac{O D}{S D},
\end{array}
$$

Then $O' D=\frac{O D}{S D} \cdot A D=\frac{3}{2} \sqrt{3}$.
Let the intersection point of the section passing through point $P$ and parallel to the base with $SD$ be $O''$, then $\frac{O'' D}{O' D}=\frac{A P}{O' A}=\frac{8}{9}$, i.e., $O'' D=\frac{4}{3} \sqrt{3}, SO''$ $=SD-O'' D=\frac{2}{3} \sqrt{3}, \frac{SO''^2}{SD^2}=\frac{1}{9}$,
thus the area of the required section is $\frac{1}{9} S_{\triangle ABC}=\sqrt{3}$.