The area of a rectangle is 66 yd2 , and the length of the rectangle is 1 yd more than twice the width. Find the dimensions of the rectangle.

Question

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Answer 1

 

Let:

  L = length

  W = width

 

LW = 66      {area = length*width;    eq1]

L = 1+2W       [eq2; “length of the rectangle is 1 yd more than twice the width”]

 

(1+2W)(W)=66      [substitute eq2 for L in eq1]

W + 2W²= 66       [distribute]

2W²+ W – 66 = 0     [put all term on left side]

(2W-11)(W+6) = 0      [factor or use Quadratic Formula]

W = 5.5                  [W=-6 is impossible]

 

Substitute for W in either formula (let’s use eq2}:

 L = 1 + 2(5.5)

 L = 12