10. As shown in the right 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 $\mathrm{MO}$ parallel to $\mathrm{AD}$ through point $G$, intersecting $\mathrm{CD}$ and $\mathrm{AB}$ at points $\mathrm{M}$ and $\mathrm{O}$, respectively. Draw $\mathrm{FN} \perp \mathrm{BC}$ at $\mathrm{N}$ through $\mathrm{F}$. Suppose the area of the square is "1", then the side length of the square is 1, and the area of each triangle in the figure is $\frac{1}{8}$. It is easy to see from the figure that $\mathrm{BI}=\mathrm{HA}=\frac{1}{4}$, quadrilateral $\mathrm{FGHI}$ is a parallelogram, $\mathrm{HI}=\mathrm{FG}=\frac{1}{2}$, $\mathrm{GO}=\mathrm{GM}=\frac{1}{2}$, $\mathrm{FN}=\frac{1}{8}$, so 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, $\mathrm{DG}=\sqrt{\frac{1}{2} \times \frac{1}{2}+\frac{3}{8} \times \frac{3}{8}}=\frac{5}{8}$
Therefore, the side length of the square is $5 \div \frac{5}{8}=8$, and the area of the square is $8 \times 8=64$.