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What are the dimensions of a rectangle with an area of fifty square feet and a perimeter of 201 feet?
Wiki User
∙ 14y ago
Area of a rectangle = length(L) x width(W)
Perimeter of a rectangle = 2L + 2W.
50 = LW : therefore W = 50/L
201 = 2L + 2 x (50/L) = 2L + 100/L : Multiply by L
201L = 2L2 + 100 : Therefore 2L2 - 201L + 100 = 0
This quadratic equation factorises (2L - 1)(L -100) = 0
When 2L - 1 = 0 then L = 1/2
When L - 100 = 0 then L = 100
The dimensions of the rectangle are therefore length = 100 ft, width = 1/2 foot
Wiki User
∙ 14y ago
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