7. (15 points) As shown in the figure, an equilateral triangle is constructed outward from each side of the square $A B C D$ ( $\triangle A B F$, $\triangle A D E$, $\triangle$ $C D H$, $\triangle B C G$). Given that the side length of the square $A B C D$ is 10, the area of the shaded part in the figure is $\qquad$.


【Solution】Solution: According to the analysis, $\triangle F C D$ is an isosceles triangle, $\angle F B C=150^{\circ}$, then $\angle B F C=15^{\circ}, \angle B M F$ $=90^{\circ}$,
$\triangle B M F$ and $\triangle B M C$ have equal areas, $\triangle A B C$ and $\triangle B M C$ have equal areas, then the area of $\triangle A B N$ is equal to the area of $\triangle F C B$,

The area of the shaded part in the middle is equivalent to the area of the square $A B C D$ minus the areas of $\triangle F C B$ and $\triangle A D H$, the area of $\triangle F C B$ is $10 \times 5 \div 2=25$, so the area of the shaded part $=100-25 \times 2=50$. Therefore, the answer is: 50.