The perimeter of the rectangle is 30 dm. If the length of one side of it is increased by 20%, and the other by 40%

The perimeter of the rectangle is 30 dm. If the length of one side of it is increased by 20%, and the other by 40%, then the perimeter of the rectangle will become equal to 40 dm. Find the sides of the resulting rectangle.

1. Let the length of the original rectangle be x dm, and its width y dm. Then their perimeter is 2 (x + y) and is 30 dm.

2. If the length of the rectangle is increased by 20%, then it will be equal to:

x + 20% x = x + 0.2x = 1.2x dm;

3. If the width of the rectangle is increased by 40%, then it will be equal to:

y + 40% y = y + 0.4y = 1.4y dm;

4. Then the perimeter of the resulting rectangle is 2 (1.2x + 1.4y) and is 40 dm. Let’s compose and solve the system of equations:

{2 (x + y) = 30;

{2 (1.2x + 1.4y) = 40;

{x + y = 15;

{1.2x + 1.4y = 20;

Let’s use the substitution method:

x = 15 – y;

1.2 (15 – y) + 1.4y = 20;

18 – 1.2y + 1.4y = 20;

0.2y = 2;

y = 10;

This means that the width of the original rectangle is 10 dm, and its length is 15 – 10 = 5 dm;

5. Find the sides of the resulting rectangle:

length 1.2 * 5 = 6 dm;

width 1.4 * 10 = 14 dm;

Answer: the length of the resulting rectangle is 6 dm, and the width is 14 dm.