The length of a rectangle is 4 yd longer than its width. if the perimeter of the rectangle is 32 yd, find its area.

Hello:

let : x the length and : y the width

2 (x+y) the perimeter and; xy the area.

x = y + 4 ... (1)

2 (x+y) = 32 ... (2)

subsct x by (1) in (2):

2 (y+4 + y) = 32

2y + 8 + 2y = 32

4y = 24

y = 24/4 = 6 yd

x = 6+4=10 yd

the area is : 60 yd²