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14 August, 16:21

The area of a rectangle is 66 ft^2 and the length of the rectangle is 7 feet less than three times the width find the dimensions of the rectangle

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Answers (2)
  1. 14 August, 16:41
    0
    Step-by-step explanation:

    Length * width = Area

    3 (x-7) * x=66

    x^2-7x-22=0

    Solve for 'x' using quadratic formula

    Now width=9.35234995ft

    Length=7.05704985ft
  2. 14 August, 16:47
    0
    Length = 11Width = 6

    The area of a rectangle is 66 ft^2:

    L * W = 66

    Length of the rectangle is 7 feet less than three times the width:

    L = 3W-7

    Substitute L in terms of W:

    (3W-7) * W = 66

    Factorise the equation:

    3W^2 - 7W = 66

    3W^2 - 7W - 66 = 0

    Factors of 66 =

    1 66, 2 33, 3 22, 6 11

    (3W + 11) (W - 6) = 0

    Solve for W:

    3W + 11 = 0

    3W = 11

    W = 3/11

    W - 6 = 0

    W = 0 + 6

    W = 6

    Using the original equation, find L:

    L = 3W-7

    L = 3 (6) - 7

    L = 18-7

    L = 11

    L * W = 66

    11 * 6 = 66
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