V-2 If one side of the rectangle is reduced by $3 \mathrm{~cm}$, and the other side is reduced by $2 \mathrm{~cm}$, we get a square whose area is $21 \mathrm{~cm}^{2}$ less than the area of the rectangle. Calculate the dimensions of the rectangle.

![](https://cdn.mathpix.com/cropped/2024_06_05_627a0068487f082c2c6cg-1.jpg?height=219&width=271&top_left_y=1438&top_left_x=1208)
Solution. If we denote the length of the side of the square by $a$, then the dimensions of the rectangle are $a+3$ and $a+2$. From the diagram, it is seen that the rectangle is divided into a square with area $a^{2}$ and three rectangles with areas $2a$, $3a$, and 6. Since the area of the square is $21 \mathrm{~cm}^{2}$ less than the area of the rectangle, it follows that $2a + 3a + 6 = 21$, or $5a = 15$. From this, we get $a = 3 \mathrm{~cm}$. Therefore, the dimensions of the rectangle are $6 \mathrm{~cm}$ and $5 \mathrm{~cm}$.

Note: For a diagram from which the equation $2a + 3a + 6 = 21$ is derived, 5 points are awarded.