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Math Help - Perimeter of a rectangle

  1. #1
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    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?
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by GeinoD View Post
    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.
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  3. #3
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    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
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  4. #4
    Bar0n janvdl's Avatar
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    Quote Originally Posted by GeinoD View Post
    Ok, so I think I understand,

    Now by units, you mean in?
    Inches yes.

    Quote Originally Posted by GeinoD View Post
    One thing, when you did ...

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

    Remember W^2 = 36 and then i took the square root on both sides.
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  5. #5
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    Ahhh, ok I see.

    w^2=36

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

    Got it, thanks.
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