The length of a rectangle is 44 inches than twice the width. If the diagonal is 2 inches more than the​ length, find the dimensions of the rectangle.

Answer 1

L= 44+2W

2W = L-44

W = L/2 - 22

D= 2+L

D^2 = L^2 + W^2

(2+L)^2 = L^2 + (L/2 -22)^2

L^2 + 4L + 4 = L^2 + L^2/4 - 22L + 484

L^2/4 - 26L + 480 = 0

L^2 - 104L + 1920 = 0

L = 104/2 + or - (1/2)sqr(104^2 -4(1920))

L = 52 + or - .5sqr(10816 - 7680)

L = 52 + or - .5sqr3136

L = 52 + or - 56/2

L = 52+ or - 28

L = 80 or 24

L= 44+2W

W = (L-44)/2 = (80-44)/2 = 36/2 = 18

Length = 80 inches, Width = 18 inches