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Thread: Harriet's Rectangular Yard

  1. #1
    Super Member harpazo's Avatar
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    Harriet's Rectangular Yard

    Harriet wants to put up fencing around 3 sides of her rectangular yard and leave a side of 20 feet unfenced. If the yard has an area of 680 square feet, how many feet of fencing does she need?

    My Effort:

    A = L*W

    A = 680 feet^2

    L = 20 feet

    680 = 20W

    680/20 = W

    34 = W

    One side of the rectangular yard is 20 feet less.

    So, P = L + 2W.

    P = 20 feet + 2(34 feet)

    P = 20 feet + 68 feet

    P = 88 feet

    Correct?
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Harriet's Rectangular Yard

    Yes, if F is the length of fence needed, and W and L and the width and length respectively of the rectangular yard, and we choose W = 20 then:

    A = LW or L = A/W

    F = 2L + W = 2A/W + W = 2(680)/20 + 20 = 68 + 20 = 88
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  3. #3
    Super Member harpazo's Avatar
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    Re: Harriet's Rectangular Yard

    Quote Originally Posted by MarkFL View Post
    Yes, if F is the length of fence needed, and W and L and the width and length respectively of the rectangular yard, and we choose W = 20 then:

    A = LW or L = A/W

    F = 2L + W = 2A/W + W = 2(680)/20 + 20 = 68 + 20 = 88
    Cool.
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