7. (15 points) As shown in the figure, it is known that the area of square $A B C D$ is $2520$; $E$, $F$, $G$, $H$ are the points on the sides closer to the vertices of the square, dividing each side into four equal parts. Connect $A G$, $E C$, $H B$, $D F$. Then the area of the “$X$” part in the figure is $\qquad$ .
【Solution】Solution: As shown in the figure:
The lengths of the two diagonals of the middle rhombus are $A E$ and $\frac{1}{3} \mathrm{AD}$,
$$
\begin{array}{l}
A E=\frac{1}{4} A D \\
\frac{1}{3} \times \frac{1}{4} \div 2=\frac{1}{24}
\end{array}
$$

Therefore, the overlapping area is $\frac{1}{24}$ of the square's area,
The areas of the two parallelograms are both $\frac{1}{4}$ of the square's area,
$$
\begin{array}{l}
\frac{1}{4}+\frac{1}{4}-\frac{1}{24}=\frac{11}{24} \\
2520 \times \frac{11}{24}=1155
\end{array}
$$

Answer: The area of the " $X$ " part in the figure is 1155.
The answer is: 1155.