SOLUTION: If the length of a rectangle is four times it's width and the area is 256ft squared then what is the perimeter?

Question 316286: If the length of a rectangle is four times it's width and the area is 256ft squared then what is the perimeter?
Answer by Shogun(3) (Show Source): You can put this solution on YOUR website!
First things first: Let's jot down everything we know about the problem, to make it easier.
Area of a Rectange: Length * Width
Perimeter of a Rectangle: 2*Length + 2*Width
Now we know that the LENGTH of the rectangle is 4 TIMES its WIDTH.
So let's call the width "W". Then we can say that the length is 4W (since length is 4 times the width).
Now that we know both length and width, we can make a formula (from our area equation).
256 = 4W * W. <--- All we did here is substitute numbers back into the equation.
We can simplify that to get:
256 = 4W^2
64 = W^2
8 = W

So now we know that the width is 8 from our formula. That means we also know that the length is 32 (4 times the width, which is 8).
The final step is to find the perimeter, which we know is 2*Length + 2*Width.
2*32 + 2*8 = 64 + 16 = 80.

So we know that the perimeter is 80 ft.

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