Mathematics

Question

Question 2 The length of a rectangle is 5 yd more than twice the width, and the area of the rectangle is 63 yd2 . Find the dimensions of the rectangle.

2 Answer

  • L = 2W + 5
    A = LW = 63
    L = 63/W
    W=63/L

    L = 2W + 5
    63/W = 2W + 5
    63 = 2WW + 5W
    2WW + 5W - 63 = 0

    2WW + 5W - 63 = 0 
    is already in quadratic equation form: ax^2 + bx + c = 0

    use the quadratic formula to solve for two solutions:
    x = ( -b +- sqrt( bb - 4ac ) ) / 2a

    the two solutions are:
    x = 9/2
    x = -7
    you cant have a negative solution for length so use the positive one

    solve for length:
    L=5+2(9/2)
    L=5+18/2
    L=5+9
    L=14


    W = 9/2
    L = 14

  • The length and the width of the rectangle is 14 yards and 4.5 yards respectively.

    Let the width of the rectangle be represented by w

    Therefore, the length of the rectangle will be represented by 2w + 5.

    Therefore, the area of the rectangle will be: = Length × Width

    w(2w + 5) = 63

    2w² + 5w = 63

    2w² + 5w - 63 = 0

    2w² + 14w - 9w - 63 = 0

    2w(w + 7) - 9(w + 7) = 0

    Therefore, 2w - 9 = 0

    2w = 9

    w = 9/2

    w = 4.5

    Length = 2w + 5 = 2(4.5) + 5 = 14

    The length and the width of the rectangle is 14 yards and 4.5 yards respectively.

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