# Find the area of the rectangle whose length is 2 more that twice the width and the perimeter is 22 First, we have to solve for the length and the width.  We can do this by setting up the following equations.

l = 2w + 2

2l + 2w = 22

2 (2w +2) + 2w = 22

4w + 4 + 2w = 22

6w + 4 = 22

...

First, we have to solve for the length and the width.  We can do this by setting up the following equations.

l = 2w + 2

2l + 2w = 22

2 (2w +2) + 2w = 22

4w + 4 + 2w = 22

6w + 4 = 22

6w = 18

w = 3

If w = 3, then the length of the rectangle is 2 more than twice that.  That means that length = 2*3 + 2 or l = 6 + 2 or l = 8

So the width is 3 and the length is 8.  To get the area, we multiply these and we get 24.

So the area is 24.

Approved by eNotes Editorial Team Area of the rectangle is:

a = length * width = L*w

But we know that:

L = 2+ 2w

Also:

2L + 2W = 22

Now substitute with L value:

==> 2L + 2W = 22

==> 2(2+2w) + 2w = 22

==> 4 + 4w + 2w = 22

==> 6w = 18

==> w = 3

==> L = 2+ 2*3 = 8

==> a = L*w = 3*8 = 24

==> the area of the rectangle is 24 square unit.

Approved by eNotes Editorial Team