10. (10 points) As shown in the figure, square $A B C D$ is divided into 8 triangles of equal area. If $D G=5$, then the area of square $A B C D$ is $\qquad$ .
【Analysis】As shown in the figure, draw $MO$ parallel to $AD$ through point $G$, intersecting $CD$ and $AB$ at points $M$ and $O$ respectively, and draw $FN \perp BC$ at $N$ through point $F$. If the area of the square is "1", then the side length of the square is 1. According to the fact that the square $ABCD$ is divided into 8 triangles of equal area, we can find the lengths of $BI$ and $HA$, and then, based on the fact that quadrilateral $FGHI$ is a parallelogram, we can find the lengths of $FG$, $GO$, and $GM$. In the right triangle $GMD$, we can use the Pythagorean theorem to find the length of $DG$. Finally, if $DG=5$, we can find the side length of the square and then calculate its area.
【Solution】As shown in the figure, draw $MO$ parallel to $AD$ through point $G$, intersecting $CD$ and $AB$ at points $M$ and $O$ respectively, and draw $FN \perp BC$ at $N$ through point $F$.

Assume the area of the square is "1", then the side length of the square is 1. Each small triangle in the figure has an area of $\frac{1}{8}$, so $BI=HA=\frac{1}{4}$.
Since quadrilateral $FGHI$ is a parallelogram, $HI=FG=\frac{1}{2}$, $GO=GM=\frac{1}{2}$, and $FN=\frac{1}{8}$. In the right triangle $GMD$, $MG=\frac{1}{2}$, $MD=1-\frac{1}{8}-\frac{1}{2}=\frac{3}{8}$. According to the Pythagorean theorem, $DG=\sqrt{\frac{1}{2} \times \frac{1}{2}+\frac{3}{8} \times \frac{3}{8}}=\frac{5}{8}$. Therefore, the side length of the square is $5 \div \frac{5}{8}=8$, and the area of the square $=8 \times 8=64$. Hence, the answer is: 64.