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Math Help - Distance

  1. #1
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    Distance

    A boxing ring is in the shape of a square, 33 feet on each side. How far apart are the fighters when one is in the center and the other is in the corner.
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  2. #2
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    Quote Originally Posted by pashah
    A boxing ring is in the shape of a square, 33 feet on each side. How far apart are the fighters when one is in the center and the other is in the corner.
    The diagnol of a square is the line from one corner to the opposite corner (it's the hypotenuse of a right triangle with the same sides as the square)

    therefore we find the distance from one corner to the opposite corner:

    \sqrt{a^2+b^2}=c

    \sqrt{33^2+33^2}=c

    \sqrt{1089+1089}=c

    \sqrt{2178}=c

    \sqrt{2178}=c

    but we only want to go half that distance, so we divide that answer by 2:

    \frac{\sqrt{2178}}{2}=\frac{c}{2}

    \frac{\sqrt{2178}}{\sqrt4}=\frac{c}{2}

    \sqrt{\frac{2178}{4}}=\frac{c}{2}

    \sqrt{544.5}=\frac{c}{2}

    23.3345\approx\frac{c}{2}

    so the distance from a corner to the center is about 23 feet
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  3. #3
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    Quote Originally Posted by pashah
    A boxing ring is in the shape of a square, 33 feet on each side. How far apart are the fighters when one is in the center and the other is in the corner.
    The length of the square diagnol is,
    s\sqrt{2}
    Thus,
    33\sqrt{2}\approx 46.53
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  4. #4
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    here's a diagram of what I'm talking about (note: x and y are boxers)(also note: it's supposed to be a square):

    Code:
    	 
    -------x
    |    / |
    |   /  | a
    |  y   |
    | /    |
    |/     |
    -------
      b
    To Hacker: one of the boxers is in the CENTER of the square
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  5. #5
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    \sqrt{\bigg({\frac{33}{2}\bigg)^2}+{\bigg(\frac{33  }{2}\bigg)^2}}} \approx 23.33
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