The length of the rectangle was increased by 5 dm, and the width was decreased by 3dm.

| July 27, 2021 | education

The length of the rectangle was increased by 5 dm, and the width was decreased by 3dm. How the perimeter has changed

Let’s denote the length of this rectangle by a, and the width of this rectangle by b.

The perimeter of any rectangle is equal to twice the sum of the length and width of this rectangle, therefore, the perimeter p1 of this rectangle is:

p1 = 2 * (a + b).

According to the condition of the problem, the length of the rectangle was increased by 5 dm, and the width was decreased by 3 dm, therefore, the perimeter p2 of the modified rectangle is:

p2 = 2 * (a + 5 + b – 3) = 2 * (a + b + 2) = 2 * (a + b) + 2 * 2 = p1 + 4.

Consequently, the perimeter of the original rectangle has increased by 4 dm.

Answer: the perimeter of the rectangle has increased by 4 dm.



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