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18 January, 15:38

The length of a rectangle is 3 times the width. If the perimeter is to be greater than 56 meter. What are the lossible values for the width? (use w as the width)

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Answers (2)
  1. 18 January, 17:19
    0
    w > 7m

    Step-by-step explanation:

    let l is the length of rectangle and w is the width of rectangle,

    Perimeter of rectangle is 2 (l+w).

    From the question statement, we observe that

    l=3w

    2 (l+w) >56

    Put l=3w in above inequality,

    2 (3w+w) >56

    2 (4w) >56

    8w>56

    Multiplying by 1/8 on both sides of equation we get

    (1/8) 8w> (1/8) 56

    w>7m

    We get the conclusion that width must be greater than 7m.
  2. 18 January, 18:07
    0
    w > 7

    Step-by-step explanation:

    L = 3w

    2L + 2w > 56

    sub for L

    2 (3w) + 2w > 56

    6w + 2w > 56

    8w > 56

    w > 7

    Check: L = 3 (8)

    L = 24

    2 (24) + 2 (8) > 56

    48 + 16 > 56

    64 > 56
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