In order to solve problems which require application of the

area and

perimeter for rectangles, it is necessary to

A typical problem involving the

area and

perimeter of a

rectangle gives us the area,

perimeter and/or

length and width of the rectangle. We may also be given a relationship between the

area and

perimeter or between the

length and width of the rectangle. We need to calculate some of these quantities given information about the others.

To get started, relate the

length and width. We know that the

length is twice the width so

*l = 2w*

We know the

area is 32, so we use the

area formula for a rectangle:

*lw* = *2w(w)* = * 2w*^{2} = 32

Solving for *w:*

*
*

* 2w*^{2} = 32

* w*^{2} = 16

*w* = 4

* l = 2w*

l = 8

and the dimensions are 8 x 4.

The

perimeter is the sum of the lengths of all four sides or

*2w + 2l* = 2(4) + 2(8) = 24