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 $\qquad$
Answer: 64.
The problem only provides one specific data point, DG=5, and requires finding the area of the square. We can use an unknown variable $\mathrm{x}$ as a bridge to find the relationship between DG and the side length of the square. Once we have the relationship, we can determine the side length and then calculate the area. Let the side length be $8x$.
$S_{\triangle \mathrm{ADH}}=S_{\triangle \mathrm{BCI}}$ so $\mathrm{BI}=\mathrm{AH}$. If the entire area of the square is 8 parts, the area of $\triangle \mathrm{ADH}$ is 1 part, and $A H$ is $\frac{1}{4}$ of $A B$, i.e., $A H=B I=2 x, H I=4 x$. If we connect $C H$, then the area of $\triangle C H I$ is twice that of $\triangle C B I$, which is 2 parts, and the area of $\triangle F H I$ is 1 part, so $CF=FI$. However, we need to relate this to DG, so we draw an auxiliary line: extend FG to intersect DH at N and DA at M (as shown in the right figure). According to the swallowtail model, DN=NH. In trapezoid CDHI, $\mathrm{CF}=\mathrm{FI}, \mathrm{DN}=\mathrm{NH}$, so $\mathrm{FN}$ is parallel to $\mathrm{CD}$ and $\mathrm{HI}$ (parallel lines divide segments proportionally), so $D M=M A$.
After establishing the parallelism, the distance between $\mathrm{FG}$ and $\mathrm{HI}$ is equal, and since $S_{\triangle \mathrm{GFH}}=S_{\triangle \mathrm{IHF}}$, $\mathrm{FG}=\mathrm{HI}=4 \mathrm{x}$. Also, because the area of $\triangle F D G$ is twice that of $\triangle D G N$, $G N=2 x$. By symmetry, $N M=(8 x-6 x) \div 2=x$, so $G M=3 x$. In the right triangle $\triangle G D M$, $D G=5 x$. Since $D G=5$, $x=1$. Finally, the area of the square is $8 \times 8=64$.