A rectangle is 7 times as long as it is wide. The perimeter is 64 feet. Find the dimensions. The length is ???feet where the width is

Question

A rectangle is 7 times as long as it is wide. The perimeter is
64 feet. Find the dimensions. The length is ???feet where the width is ???feet.

Please show a step by step on how to do this.

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Kaylee 2 weeks 2021-09-09T10:09:36+00:00 2 Answers 0

Answers ( )

    0
    2021-09-09T10:11:02+00:00

    Answer: Length = 28 feet

    Width = 4 feet

    Step-by-step explanation:

    Let L represent the length of the rectangle.

    Let W represent the width of the rectangle.

    The formula for determining the perimeter of a rectangle is expressed as

    Perimeter = 2(L + W)

    The perimeter of the rectangle is 64 feet. This means that

    2(L + W) = 64

    Dividing through by 2, it becomes

    L + W = 32 – – – – – – – – – – – -1

    The rectangle is 7 times as long as it is wide. This means that

    L = 7W

    Substituting L = 7W into equation 1, it becomes

    7W + W = 32

    8W = 32

    W = 32/8

    W = 4

    L = 7W = 7 × 4

    L = 28

    0
    2021-09-09T10:11:13+00:00

    Answer:

    The length is 28ft and the width is 4ft.

    Step-by-step explanation:

    Let x equals to the width. Since the length is 7 times as long as the width, we know that it has to be 7x.

    The perimeter equation of a rectangle would be P=2(l+w)

    Here are the givens:

    l= 7x

    w= x

    P=32 ft

    P=2(l+w)

    All you need to do is substitute x and 7x into width and length in the perimeter equation

    P=2(7x+x)

    P=2(8x)

    Substitute 64ft into “P”

    64=16x

    x=4

    After finding x=2, we would substitute x into the length and the width equations

    w=4

    l=4(7)=28

    so the length is 28ft and the width is 4ft.

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