Length exceeds width by 2 feet. When each dimension is increased by 2 feet the area increases by 48 square feet. Find the dimensions.
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It is given that length exceeds width by 2 feet. And if each of the dimensions is increased by 2 feet, the area will increase by 48 square feet.
Let the initial width be W. The length is W + 2
The area now is width*length = W*(W + 2).
When the dimensions are increased by 2 feet, the width becomes W + 2 and the length becomes W + 2 + 2 = W + 4
The area is (W + 2)*(W + 4)
We know that (W + 2)*(W + 4) - W*(W + 2) = 48
=> W^2 + 6W + 8 - W^2 - 2W = 48
=> 4W + 8 = 48
=> 4W = 40
=> W = 10
Length = width + 2 = 12
The required dimensions are width = 10 feet and length = 12 feet.
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We'll put the width as x.
We'll put the length as x + 2.
The area of the rectangle is theproduct of length and width.
A = x(x+2) (1)
We'll increase dimensions by 2 feet:
The length will become x + 2 + 2 = x + 4
The width will become x + 2.
The area is increasing also by 48 sq. feet.
48 + A = (x+2)(x+4) (2)
We'll substitute A by (1):
48 + x(x+2) = (x+2)(x+4)
We'll move all terms in x to the left side:
x(x+2) - (x+2)(x+4) + 48 = 0
x^2 + 2x - x^2 - 6x - 8 + 48 = 0
We'll combine and eliminate like terms:
-4x + 40 = 0
-4x = -40
x = 10
The width is of 10 units and the length is of 12 units: l=12 units; w=10 units.
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