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Thread: Circular park

  1. August 19th 2008, 10:53 PM #1
    devi
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    Circular park

    A circular park of 20m diameter has a circular path just inside the boundary with a width of 1m.The are of the path is
    15pi
    17pi
    19pi
    25pi
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  2. August 19th 2008, 11:00 PM #2
    Chris L T521
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    Quote Originally Posted by devi View Post
    A circular park of 20m diameter has a circular path just inside the boundary with a width of 1m.The are of the path is
    15pi
    17pi
    19pi
    25pi
    Find the area within the park, and then subtract from that the area within the path.

    So you should get 400\pi-361\pi=\dots

    However, I don't see the answer as one of the choices you gave...hmmm..I'm probably overlooking something pretty obvious...

    --Chris
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  3. August 19th 2008, 11:03 PM #3
    devi
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    Quote Originally Posted by Chris L T521 View Post
    Find the area within the park, and then subtract from that the area within the path.

    So you should get 400\pi-361\pi=\dots

    However, I don't see the answer as one of the choices you gave...hmmm..I'm probably overlooking something pretty obvious...

    --Chris
    The answer was given as 19Pi
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  4. August 19th 2008, 11:09 PM #4
    Chris L T521
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    Quote Originally Posted by devi View Post
    The answer was given as 19Pi
    Quote Originally Posted by Chris L T521 View Post
    Find the area within the park, and then subtract from that the area within the path.

    So you should get 400\pi-361\pi=\dots

    However, I don't see the answer as one of the choices you gave...hmmm..I'm probably overlooking something pretty obvious...

    --Chris
    I caught my error...It had a 20 m diameter...

    so the area of the park would be (10~m)^2\pi=100\pi~m^2

    The area of the park enclosed by the path would be (10~m-1~m)^2\pi=(9~m)^2\pi=81\pi~m^2

    So then we see that the difference between the two will give us the area of the path.

    100\pi~m^2-81\pi~m^2=\color{red}\boxed{19\pi~m^2}

    Does this make sense? Sorry about that; it was a mistake on my part...misread the problem.

    --Chris
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