Problem 2. If one side of the square is doubled, and the other is increased by $22 \mathrm{~mm}$, a rectangle is obtained whose perimeter is $2000 \mathrm{~mm}$ greater than the perimeter of the square. What is the length of the side of the given square?
Solution. If the side of the smaller square is denoted by $a$, then the sides of the rectangle are $2 a$ and $a+22$. The sum of the two sides of the rectangle with length $2 a$ is as much as the perimeter of the square. Therefore, the sum of the lengths of the other two sides, with length $a+22$, amounts to $2000 \mathrm{~mm}$. Thus,

$$
2 a+44=2000,2 a=1956, a=1956: 2=978 \mathrm{~mm}
$$