# SOLUTION: the area of a rectangle is 56 square feet. its lenght is 10 feet longer than the width. find the lenght and width?

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 Click here to see ALL problems on Rectangles Question 418780: the area of a rectangle is 56 square feet. its lenght is 10 feet longer than the width. find the lenght and width?Found 2 solutions by praseenakos@yahoo.com, duckness73:Answer by praseenakos@yahoo.com(507)   (Show Source): You can put this solution on YOUR website!width = x feet length = x + 10 feet area = length x width 56 = x(x+18) ==> x + 14 = 0 or x-4 = 0 ==> x = -14 or x = 4 since length and width cannot be a negative number, you can ignore the negative value u got for x ==> width = 4 feet so length = 4 + 10 = 14 feet Answer by duckness73(43)   (Show Source): You can put this solution on YOUR website!Let x = width of the rectangle Let y = length of the rectangle Since the area of a rectangle is the width * length, and we know that this rectangle has an area of 56 square feet: x * y = 56 Since the length of the rectangle is 10 feet longer than the width: y = x + 10 Using the area equation from above, we have: x * (x + 10) = 56 which can be simplified as: x**2 + 10x = 56 x**2 + 10x - 56 = 0 which is a quadratic equation. Using factoring to solve this equation, we have: (x + 14)(x - 4) = 0 which means that x = -14 or x = 4. So the width of the rectangle is either -14 or 4 feet. So, which is it? Since it doesn't make any sense to have a rectangle with a negative width, the width of a rectangle cannot be -14 feet, so the width must be 4. Since the length (which is y) is 10 feet longer than the width, we have y = x + 10 = 4 + 10 = 14 So, the length (y) of the rectangle is 14 and the width (x) is 4.