13. (12 points) As shown in the figure, the large square is divided into five equal-area pieces. If the length of $AB$ is 3.6 cm, then the area of the large square is $\qquad$ square centimeters.
【Solution】Solution: According to the analysis, let the side length of the square be one unit, as shown in the figure,
Since the square is divided into five equal areas, the area of each part is equal to $\frac{1}{5}$, so $AG=\frac{2}{5}$, the distance from $D$ to $FH$ = the distance from $C$ to $EF$ = $\frac{2}{5}$,
Since the distance from $A$ to the left $EG$ is equal to $\frac{2}{5}$ of the distance from $A$ to the top $EF$, the distance from $C$ to $EG$ is also equal to $\frac{2}{5}$ of the distance from $C$ to $EF$, which is $\frac{4}{25}$;
The distance from $C$ to $FH$ is $1-\frac{4}{25}=\frac{21}{25}$, similarly, the distance from $D$ to the right $FH$ is $\frac{2}{5}$, because the distance from $C$ to $EF$: the distance from $C$ to the right $FH$ = $\frac{2}{5}: \frac{21}{25}=10: 21$,
Therefore, the distance from $D$ to $EF$ is also equal to $\frac{10}{21}$ of the distance from $D$ to $FH$, which is: $\frac{10}{21} \times \frac{2}{5}=\frac{4}{21}$, so the distance from $D$ to $GH$ = $1-\frac{4}{21}=\frac{17}{21}$
The area of triangle $BDH$ = $\frac{1}{5}$, so $BH=\frac{1}{5} \times 2 \div \frac{17}{21}=\frac{42}{85}$, $AB=1-\frac{2}{5}-\frac{42}{85}=\frac{9}{85}$, which is 3.6 cm, so the side length of the square = $3.6 \div \frac{9}{85}=34$ (cm),
The area of the square = $34 \times 34=1156$ square cm.
Therefore, the answer is: 1156.