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Math Help - Jigsaw puzzle involving perimeter of semi-circles

  1. #1
    Senior Member Mukilab's Avatar
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    Jigsaw puzzle involving perimeter of semi-circles

    Intermediate Mathematical Challenge 2010

    Puzzle number 9

    I'm not satisfied with the explanation because I did 26+4\pi - 22+2\pi to get 4+2\pi

    With the 26+4 pi being what I got from them being separate pieces (3+3+3+3+3+3+1+1+1+1+1+1+4(\frac{2\pi}{2}))

    And the second part being the same way without using two 1s and four of 2pi over 2

    Could anyone please help me?
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  2. #2
    A Plied Mathematician
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    I don't think you were adding properly. I get, for the perimeter of the un-fitted pieces,

    30+6\pi, and for the fitted pieces, I get

    22+2\pi.

    The difference is then 8+4\pi, or answer C.
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  3. #3
    Senior Member Mukilab's Avatar
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    How can you get 6 pi if there are only 4 sides with semi circle sticking out/indented?

    If anything now I seem to be getting 8 pi for the first perimeter.

    Mind explaining how you got that answer?
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  4. #4
    A Plied Mathematician
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    I don't mind at all! There are three pieces. Each piece has two semi-circular regions of perimeter pi. Ergo, three pieces times two regions times pi gives you six pi.
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  5. #5
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    Quote Originally Posted by Mukilab View Post
    Intermediate Mathematical Challenge 2010

    Puzzle number 9

    I'm not satisfied with the explanation because I did 26+4\pi - 22+2\pi to get 4+2\pi

    With the 26+4 pi being what I got from them being separate pieces (3+3+3+3+3+3+1+1+1+1+1+1+4(\frac{2\pi}{2}))

    And the second part being the same way without using two 1s and four of 2pi over 2

    Could anyone please help me?
    The perimeters of all pieces summed are

    (1) 1+1+1+1+3+3+{\pi}+{\pi} for the piece on the left

    (2) same for the centre piece

    (3) same for the piece on the right.

    You correctly calculated the perimeter for the assembled jigsaw.
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  6. #6
    Senior Member Mukilab's Avatar
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    Quote Originally Posted by Ackbeet View Post
    I don't mind at all! There are three pieces. Each piece has two semi-circular regions of perimeter pi. Ergo, three pieces times two regions times pi gives you six pi.
    sorry it seems I ignored the middle one

    Thanks for pointing this out
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  7. #7
    A Plied Mathematician
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    All good now, then?
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