# Thread: Perimeter of a rectangle

1. ## Perimeter of a rectangle

I am trying to find the perimeter of a Rectangle,
However I think I am doing something wrong.

The length of a rectangle is four times its width. If the area of the rectangle is 144in^2, find its perimeter.

Ok, so first things first, I find that L=4W

2L+2w=144in^2

144*144=20736 in or 1728 ft

2(4w)+2w=1728ft
8W+2W=1728ft
10w=1728ft
/10 /10
w=172.8ft

Did I do this correctly?

2. Originally Posted by GeinoD
I am trying to find the perimeter of a Rectangle,
However I think I am doing something wrong.

The length of a rectangle is four times its width. If the area of the rectangle is 144in^2, find its perimeter.
P = 2L + 2W

L = 4W

Thus P = 8W + 2W = 10W

We also know that L x W = 144

But L = 4W

4W^2 = 144

W^2 = 36

W = 6

And P = 10W

So the perimeter = 10 x 6 = 600 units.

3. Ok, so I think I understand,

Now by units, you mean in?

So P=600in?

One thing, when you did
4w^2=144
/4 /4

w^2=36

How did you get the 6?
w^2=6

4. Originally Posted by GeinoD
Ok, so I think I understand,

Now by units, you mean in?
Inches yes.

Originally Posted by GeinoD
One thing, when you did ...

...How did you get the 6?
w^2=6
$\displaystyle W^2 \neq 6$

Remember $\displaystyle W^2 = 36$ and then i took the square root on both sides.

5. Ahhh, ok I see.

w^2=36

Sq root of w^2=36
is w=6

Got it, thanks.