# A square with a side of 8 cm was cut into two rectangles. The perimeter of one of these rectangles is 22 cm.

**A square with a side of 8 cm was cut into two rectangles. The perimeter of one of these rectangles is 22 cm. How much is the area of one of the rectangles more than the area of the other rectangle?**

Since the rectangle is obtained from a square with a side of 8 cm, then the two sides of this rectangle will be equal to 8 cm.

Let the other two sides of the rectangle be x cm, then, according to the problem statement, we can compose the following equation:

2 * (x + 8) = 22,

x + 8 = 11,

x = 11 – 8,

x = 3 (cm).

The area of this rectangle will be equal to:

S = 3 * 8 = 24 (cm²).

Therefore, the second rectangle has sides of 8 cm and

8 – 3 = 5 cm. So the area of this rectangle is equal to:

S = 8 * 5 = 40 (cm²).

Thus, we find that the area of the second rectangle is greater than the area of the first by

40 – 24 = 16 cm².