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15 June, 23:15

Find the length and width of a rectangle whose perimeter is 40 feet and whose area is 96 square feet.

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  1. 15 June, 23:37
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    Let the length and width be l and w, respectively. Then, perimeter = 2 (l + w) = 40, and area = lw = 96. In other words:

    l + w = 20

    lw = 96

    From here, we could guess and check to get that the length and width are 12 and 8, but let's do this rigorously:

    By the first equation, l = 20 - w

    Substituting into the second equation:

    (20 - w) * w = 96

    w^2 - 20w + 96 = 0

    w^2 - 20w + 100 = 4

    (w - 10) ^2 = 4

    w - 10 = 2 or - 2

    w = 12 or 8

    The rest is trivial.
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