6. As shown in the figure, given that the area of the equilateral $\triangle A B C$ is $S, D$ is the midpoint of $A B, D E \perp B C, E F \perp A C, F G \perp$ $A B$. Then the area of quadrilateral $D E F G$ is ${ }^{\circ}$ . $\qquad$
6. $\frac{51}{128} S$

As shown in the figure, draw the heights $A M, B N$, and $C D$, and let $S_{1}, S_{2}, S_{3}$ be the areas of $\triangle D B E, \triangle E C F, \triangle F A G$ respectively. From $\triangle D B M \backsim \triangle A B M$, we have $\frac{S_{1}}{S}=\frac{1}{8}$. From $\triangle E C F \backsim \triangle B C N$, we have $\frac{S_{2}}{S}=\frac{9}{32}$. From $\triangle F A G \backsim \triangle C A D$, we have $\frac{S_{3}}{S}=\frac{5}{128}$. Therefore, $S_{1}+S_{2}+S_{3}=\frac{77}{128} S$.