Math ms sue please help again

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Jim has designed a rectangle with an area of 100 square feet and a perimeter of 401 feet.
What are the dimensions of Jim's rectangle?please show me how u got the answer.

  • Math ms sue please help again -

    A = LW

    Let's see which factors of 100 would add up to 401.

    10 * 10
    2 * 50
    4 * 25
    5 * 20

    P = 2L + 2W

    None of those come close to a perimeter of 401.

    Are you sure you copied the problem correctly?

  • Math ms sue please help again -

    Yes and there is only one way

  • Math ms sue please help again -

    lw = 100
    and 2l + 2w = 401
    from the 1st .... l = 100/w

    into the 2nd:
    2(100/w) + 2w = 401
    times w
    200 + 2w^2 = 401w

    2w^2 - 401w + 200 = 0
    by the formula:

    w = (401 ± √159201)/4
    = (401 ± 399)/4
    = 200 or .5

    or , it factors to
    (w-200)(2w - 1) = 0
    w = 200 or w = 1/2

    if w=200, the l = 100/200 = 1/2
    if w = 1/2, the l = 100/(1/2) = 200

    The rectangle is 200 feet by 1/2 ft

    check:
    area = (1/2)(200) = 100 , checks!
    perimeter = 200 + 200 + 1/2 + 1/2 = 401, checks!

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