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17 January, 01:16

The length of a rectangle is three more than twice the width. Determine the dimensions that will give a total area of 27m^ {2}. What is the minimum area that this rectangle can have.

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Answers (2)
  1. 17 January, 02:15
    0
    3m and 9m

    Step-by-step explanation:

    length of the rectangle = 2w+3

    width of rectangle = w

    area = L*W

    if area = 27m²

    then (2w+3) w=27

    =2w²+3w-27=0

    using factorization w = 3 or - 4.5

    w=3m

    L=9m
  2. 17 January, 03:43
    0
    Answer: The length is 10metres and the width is 3 1/2metres or 3.5metres

    Step-by-step explanation:

    If the length of the rectangle is 3m more than * 2 the width, the dimensions (length & width) that will give an area of 27m^2 = ?

    Firstly, the area of a rectangle = (L+W) * 2

    Since the length is 3 more than * 2 the width,

    The length, L us therefore = (2*w) + 3 = 2w+3

    Recall that the area is 27m^2

    Now, substitute L for 2w+3

    [ (2w+3) + w]*2 = 27

    (2w+3+w) * 2 = 27

    (3w+3) * 2 = 27

    6w+6=27

    6w = 27-6

    6w = 21

    w = 21/6

    w = 3 1/2m

    If the width = 3 1/2, and area = (L+w) * 2, we substitute "w" for 3 1/2 in the equation below:

    (L + 3 1/2) * 2 = 27

    2L + 7 = 27

    2L = 27-7

    2L = 20

    L = 10m

    Therefore, the length = 10m and the width

    = 3 1/2m
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