The perimeter of the first rectangle, 26cm, of the second, is 18cm longer, the lengths of these figures are the same

The perimeter of the first rectangle, 26cm, of the second, is 18cm longer, the lengths of these figures are the same. Find the area of the rectangles if the width of the second is 10cm. How much the area of one rectangle is larger than the area of the other.

By the condition of the problem, the perimeter of the second rectangle is 18 cm larger than the perimeter, which means it will be equal to:

26 + 18 = 44 (cm).

Since its width is 10 cm, we can find its length:

2 * (x + 10) = 44,

x + 10 = 22,

x = 22 – 10,

x = 12 (cm).

The first rectangle, according to the problem, has the same length, which means the width is:

2 * (12 + y) = 26,

12 + y = 13,

y = 13 – 12,

y = 1 (cm).

The area of the first rectangle is:

S = 12 * 1 = 12 (cm²).

The area of the second rectangle is:

S = 12 * 10 = 120 (cm²).

Thus, the area of the second rectangle is larger than the area of the first by:

120 – 12 = 108 (cm²).