# 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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### 2 Answers

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.**

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.**