Problem 3. A rectangle with an area of $99 \mathrm{~cm}^{2}$ has a width of $9 \mathrm{~cm}$. Calculate the area of the square whose perimeter is equal to the perimeter of the given rectangle.
Solution. If the area of the rectangle is $99 \mathrm{~cm}^{2}$, and the width is $9 \mathrm{~cm}$, then the length of the rectangle is $99: 9=11 \mathrm{~cm}$. The perimeter of the rectangle is $L=2 \cdot 9+2 \cdot 11=40 \mathrm{~cm}$. Then, the perimeter of the square is $40 \mathrm{~cm}$, and its side has a length of $40: 4=10 \mathrm{~cm}$. Finally, the area of the square is $P=10^{2}=100 \mathrm{~cm}^{2}$.