5. As shown in Figure 2, given that the area of $\triangle A B C$ is $S$, $D$ is the trisection point of side $B C$, $E$ is the quadrisection point of side $A C$, and $F$ and $G$ are both the quintisection points of side $A B$. Then the area of quadrilateral $D E F G$ is $\qquad$
5. $\frac{29}{60} S$.

As shown in Figure 5, connect $C F$, $C G$, and $A D$. Then
$$
\begin{array}{l}
S_{\triangle M E F}=S \times \frac{1}{5} \times \frac{3}{4}=\frac{3}{20} S, \\
S_{\triangle B D C}=S \times \frac{3}{5} \times \frac{1}{3}=\frac{1}{5} S, \\
S_{\triangle C D E}=S \times \frac{2}{3} \times \frac{1}{4}=\frac{1}{6} S,
\end{array}
$$