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.
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.