The width of the rectangle is 12 cm and the length is 8 cm more. The area of the rectangle

The width of the rectangle is 12 cm and the length is 8 cm more. The area of the rectangle is calculated, which is equal to its perimeter.

By the condition of the problem, the length of the rectangle a is 8 cm greater than the width, and the width of the rectangle b is 12 cm, therefore, the length of the rectangle a is
a = b + 8 = 12 + 8 = 20 cm.
The area S of the rectangle is equal to the product of the length of the rectangle and its width, therefore:
S = a * b = 20 * 8 = 160 sq. Cm.
The perimeter p of the rectangle is equal to twice the sum of the length of the rectangle and its width, therefore:
p = (a + b) * 2 = (20 + 8) * 2 = 28 * 2 = 56 cm.
Answer: the area of this rectangle is 160 cm2, the perimeter of this rectangle is 56 cm.