7. (10 points) As shown in the figure, the length $AB$ of rectangle $ABCD$ is 20 cm, and the width $BC$ is 16 cm. Inside the rectangle, there are two overlapping squares $DEFG$ and $BHIJ$. It is known that the perimeters of the three shaded rectangles are equal. Therefore, the area of rectangle INFM is $\qquad$ square centimeters.
【Solution】Solution: The total perimeter of the shaded part is: $(20+16) \times 2=72$, the perimeter of quadrilateral IMFN is $72 \div 3=24$, so $M F+F N=12$ (1),

Since the sides of the square are equal: $M F+M G=F N+E N$, then $M F-F N=E N-G M$,

Therefore, $E N-G M=E N+B J-(G M+B H)=A B-B C=4$, then $M F-F N=4$ (2),

According to equations (1) and (2): $(12+4) \div 2=8,(12-4) \div 2=4$, the area of rectangle IMFN is $4 \times 8=32$. Therefore, the answer is: 32.