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8 October, 08:33

Find the dimensions of a rectangle whose length is a foot longer than twice its width and whose perimeter is 38 feet

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  1. 8 October, 09:23
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    P = 2 (L + W)

    P = 38

    L = 2W + 1

    38 = 2 (2W + 1 + W)

    38 = 2 (3W + 1)

    38 = 6W + 2

    38 - 2 = 6W

    36 = 6W

    36/6 = W

    6 = W < = = = width

    L = 2W + 1

    L = 2 (6) + 1

    L = 12 + 1

    L = 13 < = = = length
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