6. As shown in Figure 2, in rectangle $A B C D$, $A B=5, B C=$ 8, points $E, F, G, H$ are on sides $A B, B C, C D, D A$ respectively, such that $A E=2, B F=5, D G=3, A H=3$. Point $O$ is on line segment $H F$ such that the area of quadrilateral $A E O H$ is 9. Then the area of quadrilateral $O F C G$ is
6.6.5.

As shown in Figure 5, connect $E F$, 
$F G$, $G H$, $H E$.
It is easy to see that quadrilateral $E F G H$ is a parallelogram, and
$$
\begin{array}{l}
S_{\text {OEFCH }} \\
=40-\left(S_{\triangle A E H}+S_{\triangle E B F}+S_{\triangle F C G}+S_{\triangle G D H}\right) \\
=19 .
\end{array}
$$

From $S_{\triangle E O H}+S_{\triangle F O G}=\frac{1}{2} S_{O E F C H}=9.5$, we get
$$
S_{\triangle F O G}=9.5-\left(9-S_{\triangle A B H}\right)=3.5 \text {. }
$$

Therefore, $S_{\text {quadrilateral } O F C G}=3.5+S_{\triangle F C G}=6.5$.