Problema Solution
I need the process of the following problem:
The area of each of two rectangles is 72 square cm. The lenghts and widths of each rectangles are integers. The lenght of rectangle 2 is 5 cm greater than the length of rectangle 1; rectangle 2's width is 15 cm less than rectangle 1. Find the positive difference in the perimeters of rectangle 1 and rectangle 2.
Answer provided by our tutors
Rectangle A
length = l
width = b
perimeter = 2l + 2b
Rectangle 2
length = l+5
width = b- 15
perimeter = 2l + 10 +2b -30 = 2l +2b -20
So positive difference = 20 cm