The perimeter of the square and the rectangle are equal. The area of the square is 81 cm².

The perimeter of the square and the rectangle are equal. The area of the square is 81 cm². The length of the rectangle is 2 times its width. Find the area of the rectangle.

Find the perimeter of the square.
Let us denote by a the side of the square.
According to the condition of the problem, the area of ​​the square is 81 cm², therefore, the following relationship takes place:
a² = 81 = 9².
Since the side of the square is positive, then a = 9.
The perimeter of the square is:
4 * a = 4 * 9 = 36.
Let’s denote by x the length of the rectangle, and by y – its width.
According to the condition of the problem, the perimeters of the square and rectangle are equal, therefore, the following relation holds:
2 * (x + y) = 36.
It is also known that the length of a rectangle is 2 times its width, therefore, the following relationship holds:
x = 2 * y.
We solve the resulting system of equations. Substituting into the first equation the value x = 2 * y from the second equation, we get:
2 * (2 * y + y) = 36.
We solve the resulting equation:
2 * (3 * y) = 36;
6 * y = 36;
y = 36/6:
y = 6.
Knowing y, we find x:
x = 2 * y = 2 * 6 = 12.
Find the area S of the rectangle:
S = x * y = 12 * 6 = 72 cm².

Answer: the area of ​​the rectangle is 72 cm².