# determine the shape dimensionswhat are dimensions of rectangle if it's length is 2 more than width? when area of rectangle is increased by 48, then dimensions are increased by 2.

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Let the length of the rectangle be L. Its length is 2 more than the width. So width = L - 2

The area of the rectangle is L*(L - 2)

When the dimensions are increased by 2, the area is increased by 48 .

(L + 2)*L - L*( L - 2) = 48

=> L^2 + 2L - L^2 + 2L = 48

=> 4L = 48

=> L = 12

Width = 10

**The dimensions of the rectangle are 10 and 12.**

We'll put the width as x.

According to enunciation, we'll put the length = x + 2.

The area of the rectangle is the product 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.

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 to the left side:

x(x+2) - (x+2)(x+4) + 48 = 0

We'll remove the brackets:

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.**