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The length of a rectangle is 3 inches less than twice its width if the area of the rectangle s 35 inches find the dimensions of the rectangle?
Wiki User
∙ 14y ago
L = 2W - 3 so W x (2W - 3) = 35, ie 2W^2 - 3W - 35 = 0
This factorises as (2W + 7)(w - 5) so the positive value for W is 5 in, making L = 7 in.
Wiki User
∙ 14y ago
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The length of a rectangle is 6 inches more than twice its width If the perimeter of the rectangle is 54 inches find the width of the rectangle?
Do-it-in-your-head method: Length + width is half the perimeter which is 27 inches; 27 minus 6 is 21 and one-third of 21 is 7 which is the width. (Length = 2 x 7 + 6 = 20)
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Let the length of the rectangle be L and the width W, then W = 2L - 6 The area of a rectangle = L x W : 108 = L x (2L - 6) = 2L2 - 6L. Expressing this as a standard quadratic equation gives : 2L2 - 6L - 108 = 0. This can be factored as (2L + 12)(L - 9) = 0. As we are only concerned with positive solutions then the negative solution can be ignored, thus, When L - 9 = 0 then L = 9. As L = 9 then W = 2L - 6 = 18 - 6 = 12. The dimensions of the rectangle are : Width 12 inches, Length 9 inches.
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