hi please answer this question?

a rectangle is 6 times as long as it is wide, the perimater is 120 cm. find the dimension of rectangle found to the nearest tenth if necessary?

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  • 1 decade ago
    Favorite Answer

    Length = L

    Wide = W

    perimeter = (2L) + (2W)

    length = 6W

    120 cm = (2.6W) + (2W)

    120 cm = 12W + 2W

    120 cm = 14W

    W (wide) = 120/14 = 8,6

    L = 6W

    L= 6. (8.57)

    L= 51,4

  • 1 decade ago

    Width (x):

    2(x + 6x) = 120

    x + 6x = 60

    7x = 60

    x = 60/7 or 8 4/7

    Length:

    = 6(60/7)

    = 360/7 or 51 3/7

    Answer: Approximated to the nearest tenth dimension: 8.6 by 51.4 cm

    Proof (perimeter: 120 cm):

    = 2(8.6 cm) + 2(51.4 cm)

    = 17.2 cm + 102.8 cm

    = 120 cm

  • Anonymous
    1 decade ago

    Let's consider both the width and length to be 'x' and consider the perimeter of a rectangle, P = 2l + 2w, then:

    P = 2(6x + x)

    120 = 12x + 2x

    120 = 14x

    120/14 = 14x/14

    x = 120/14 or 8.6

  • 1 decade ago

    let breadth be x which means length is 6x

    perimeter, 2(6x+x)=120

    14x=120

    x=120/14

    x= 8.57=9(approx)

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  • Anonymous
    1 decade ago

    Perimeter = 2(length+Width)

    p =2(w+l)

    l=6w

    p=2(w+6w)

    120=2(7w)

    60= 7W

    W=8.571

    L= 51.426

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