【Question 7】As shown in the figure, on a square piece of paper with a side length of 15 cm, cut along a $45^{\circ}$ angle from 4 cm away from each vertex, forming a small square in the middle. The area of this small square is $\qquad$ (square centimeters).
As shown in the figure, extend one side $AB$ of the smaller square to intersect with one side of the larger square at point $C$, and connect $CD$.
$\therefore \triangle C E D$ is an isosceles right triangle with legs of $4 \mathrm{~cm}$
$$
\therefore C D^{2}=C E^{2}+D E^{2}=4^{2}+4^{2}=32\left(\mathrm{~cm}^{2}\right)
$$

And $CD$ is equal to the side length of the smaller square
$\therefore$ The area of the shaded square is $32 \mathrm{~cm}^{2}$