# SOLUTION: The length of a rectangle is 1 cm longer than its width. If the diagonal of the rectangle is 4 cm what are the dimensions (length and width) of the rectangle. to start i get

Algebra ->  Algebra  -> Pythagorean-theorem -> SOLUTION: The length of a rectangle is 1 cm longer than its width. If the diagonal of the rectangle is 4 cm what are the dimensions (length and width) of the rectangle. to start i get       Log On

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 Click here to see ALL problems on Pythagorean-theorem Question 46608: The length of a rectangle is 1 cm longer than its width. If the diagonal of the rectangle is 4 cm what are the dimensions (length and width) of the rectangle. to start i get x^2 + (x+1)^2 = 4^2 x^2 + x^2 + 2x + 1 = 16 2x^2 + 2x - 15 = 0 x^2 + x - 7.5 = 0 This is where I can't go futher and I need to turn it in tonight. Any help please and thank you.Answer by Nate(3500)   (Show Source): You can put this solution on YOUR website!x^2 + x^2 + 2x + 1 = 16 2x^2 + 2x = 15 x^2 + x = 7.5 (x + 1/2)^2 = 30/4 + 1/4 = 31/4 x + 1/2 = +-/2 x = -1/2 +- /2 You can not have a negative measurement, so your answer: width = -1/2 + /2 length = 1/2 + /2