Problem 3. Two equal squares with an area of $100 \mathrm{~cm}^{2}$ are given. The side of one square is increased by $2 \mathrm{~cm}$, and the perimeter of the other square by $16 \mathrm{~cm}$. Which of the resulting squares will have a larger area and by how much?
Solution. If the area of the given squares is $100 \mathrm{~cm}^{2}$, then their sides are $10 \mathrm{~cm}$. The length of the side of one of the obtained squares is $12 \mathrm{~cm}$. If the perimeter of the other square is increased by $16 \mathrm{~cm}$, then its side has increased by $16: 4=4 \mathrm{~cm}$. Its side is $14 \mathrm{~cm}$.

The area of the first square is $P_{1}=12 \cdot 12=144 \mathrm{~cm}^{2}$, and the area of the second square is $P_{2}=14 \cdot 14 \mathrm{~cm}^{2}=196 \mathrm{~cm}^{2}$. The difference in their areas is

$$
P_{2}-P_{1}=196-144=52 \mathrm{~cm}^{2} \text {. }
$$