Question

The perimeter of a rectangular playing field is 244 feet. If its length is 2 feet longer than twice its width, what are the dimensions of the field?

Answers

  1. Answer:
    L=82 and W=40
    Step-by-step explanation:
    The answer is W=40 and L=82
    If you’re wondering how to solve it, then let “P” represent perimeter “L” represent length and “W” represent width.
    P=2L+2W
    L=2W+2
    244=2(2W+2)+2W
    244=4W+4+2W
    244=6W+4
    6W=244
    244/6W=
    W=40
    Now plug the 40 in for w in: L=2W+2
    L=2(40)+2
    L=80+2
    L=82
    final answer: L=82 and W=40

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  2. The dimensions of the field are; L=82 and W=40

    What is the perimeter?

    It’s the sum of the length of the sides used to make the given figure.
    Given that perimeter of a rectangle playing field is 2the 44 feet. If its length is 2 feet longer than twice its width, then;
    let “P” represent perimeter “L” represent length and “W” represent width.
    Therefore, the  perimeter of a rectangular playing field,
    P=2L+2W
    Also, L = 2W+2
    Thus, P=2L+2W
    244=2(2W+2)+2W
    244=4W+4+2W
    244=6W+4
    6W=244
    244/6W
    W=40
    Now plug the 40 in for w;
    L=2W+2
    L=2(40)+2
    L=80+2
    L=82
    Hence, the dimensions of the field are; L=82 and W=40
    Learn more about perimeter here:
    #SPJ2

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