The perimeter of the rectangle is 40 dm. Find the area of a rectangle if its length is 3 times its width.

Let’s denote by x the length of the rectangle, and by y the width of the rectangle.
According to the condition of the problem, the perimeter of this rectangle is 40 dm, therefore, the following relationship holds:
2 * (x + y) = 40.
It is also known that the length of this rectangle is 3 times the width of the rectangle, therefore, the following relationship holds:
x = 3 * y.
We solve the resulting system of equations.
Substituting into the first equation the value x = 3 * y from the second equation, we get:
2 * (3 * y + y) = 40.
We solve the resulting equation:
2 * 4 * y = 40;
8 * y = 40;
y = 40/8;
y = 5 dm.
Knowing y, we find x:
x = 3 * y = 3 * 5 = 15 dm.
Find the area S of this rectangle:
S = x * y = 15 * 5 = 75 dm².
Answer: the area of ​​the rectangle is 75 dm².