5. Given a square $A B C D$ with side length $4, E$ and $F$ are points on sides $A B$ and $B C$ respectively, such that $A E=3, B F=2$. Line segments $A F$ and $D E$ intersect at point $G$. Then the area of quadrilateral $D G F C$ is $\qquad$.
5. $\frac{84}{11}$.

As shown in Figure 3, extend $C B$ and the extension of $D E$ to intersect at point $K$. Then
$$
\begin{array}{l}
\frac{K B}{K B+4}=\frac{B E}{C D}=\frac{1}{4} \\
\Rightarrow K B=\frac{4}{3} \text {. } \\
\text { Therefore, } \frac{K G}{G D}=\frac{K F}{A D} \\
=\frac{\frac{4}{3}+2}{4}=\frac{5}{6} \text {, } \\
\frac{K G}{K D}=\frac{5}{11} \text {. } \\
\text { Also, } \frac{K F}{K C}=\frac{\frac{4}{3}+2}{\frac{4}{3}+4}=\frac{5}{8} \text {, } \\
S_{\triangle K C D}=\frac{1}{2} \times 4\left(\frac{4}{3}+4\right)=\frac{32}{3} \text {. } \\
\text { Therefore, } S_{\triangle K G F}=\frac{5}{11} \times \frac{5}{8} S_{\triangle X C D}=\frac{100}{33} \text {, } \\
S_{\text {quadrilateral DCFC }}=S_{\triangle X C D}-S_{\triangle K C F}=\frac{32}{3}-\frac{100}{33}=\frac{84}{11} \text {. } \\
\end{array}
$$