The length of a rectangle is 3yd less than double the width, and the area of the rectangle is 27yd^2. Find the dimensions

Answer 1

Width = 4.5 yd

Length is 6 yd

A= 27 yd² = Length x width = l x w

l = 2w - 3

w(2w -3) = 27

2w² - 3w = 27

Subtract 27 from both sides of the equation to obtain a quadratic here

2w² - 3w - 27 = 0

Factoring gives

(2w - 9)(w + 3) = 0

w ≠ - 3

So solve for w below

2w - 9 = 0

Add 9 to both sides

2w = 9

Divide both sides by w

w = 9/2 or 4.5 yd

Substitute 4.5 into l = 2w - 3 to get the length.

l = 2(w) -3 = 2(4.5) -3 = 9 - 3 = 6 yd

Checking

Area = l x w

27yd² = 6 yd x 4.5 yd

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