SOLUTION: The length of a rectangle is 44 inches greater than twice the width. If the diagonal is 2 inches more than the length, find the dimensions Length 2w+44 Diagonal 2w+46

Algebra ->  Equations -> SOLUTION: The length of a rectangle is 44 inches greater than twice the width. If the diagonal is 2 inches more than the length, find the dimensions Length 2w+44 Diagonal 2w+46      Log On


   

Question 1067641: The length of a rectangle is
44
inches
greater
than twice the width. If the diagonal is 2 inches more than the length, find the dimensions
Length 2w+44
Diagonal 2w+46
Answer by LinnW(1048) About Me  (Show Source):
You can put this solution on YOUR website!
The length and with are the legs of a triangle,
and the diagonal is the hypotenuse.
So w%5E2+%2B+%282w%2B44%29%5E2+=+%282w%2B46%29%5E2
subtract %282w%2B46%29%5E2 from each side.
w%5E2+%2B+%282w%2B44%29%5E2+-+%282w%2B46%29%5E2=+0
w%5E2+%2B+4w%5E2+%2B+176w+%2B+1936+-+%28+4w%5E2+%2B+184w+%2B+2116+%29+=+0
w%5E2+%2B+-8w+-180+=+0
factoring,
%28w%2B10%29%28w-18%29+=+0
So w = -10 or w = 18
Since we need a positive width, w = 18
The length is 2w+44 = 2(18) + 44 = 36 + 44 = 80
The diagonal is 2w + 46 = 2(18) + 46 = 36 + 46 = 82
Notice that 18^2 + 80^2 = 82^2