A rectangle has a length that is 4 more than twice the width. If the perimeter is 128 yards, find the rectangle's dimensions.

The length of rectangle is 44 yards and width of rectangle is 20 yards

Solution:

Let the length of rectangle be "a"

Let the width of rectangle be "b"

Rectangle has a length that is 4 more than twice the width

Length = 4 + 2 (width)

Length = 4 + 2b

The perimeter is 128 yards

The perimeter of rectangle is given by:

Perimeter = 2 (length + width)

128 = 2 (4 + 2b + b)

128 = 2 (4 + 3b)

Divide both sides by 2

64 = 4 + 3b

3b = 60

b = 20

Substitute b = 20 in length

Length = 4 + 2 (20)

Length = 4 + 40 = 44

Thus length of rectangle is 44 yards and width of rectangle is 20 yards