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Math Help - Area of Semi-Circle region from Chord

  1. #1
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    Question Area of Semi-Circle region from Chord



    Hello,
    I would like to know how I am supposed to approach this problem.
    A small semi-circle is inserted into a bigger one.
    The only measurement that I have been given is that chord WX = 36mm, and is parallel to diameter YZ.
    I need to find the area of the grey area, and I am completely stuck for any idea on how to tackle this with only the one measurement.

    p.s. The diameter of the smaller semi-circle is not necessarily half the diameter of the larger one.

    Thanks if anyone can help.
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  2. #2
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    This shape is called the Arbelos (or Shoemaker's Knife) it has been discussed in the Book of Lemmas by Archimedes Proposition 14.

    NOTE: You need to find a link called "Book of Lemmas", it is not possible to post this as a redirect.
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  3. #3
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    Thanks, but I am still none the wiser as to how to solve the question at hand.

    I found proposition 14 in the Book of Lemmas, but I couldn't figure out how to utilise any of the information.

    Also, it seems that all of the Arbelos' have two semi-circles inscribed and not just one as in my problem.

    Thanks in advance for any further help!
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  4. #4
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    Hello, Chris Rickets!



    A small semi-circle is inserted into a bigger one.
    The only measurement is that chord WX = 36mm, and is parallel to diameter YZ.
    Find the area of the grey area.

    Let r = radius of small circle.
    Let R = radius of large circle.

    Let O be the center of the large circle.
    Then: . OW \,=\, R

    From O erect a perpendicular, cutting WX at P.
    \. . Note that: OP \,= \,r

    In right triangle WPO, we have: . OP^2 + WP^2 \:=\:OW^2
    . . So: . r^2 + WP^2 \:=\:R^2\quad\Rightarrow\quad WP \:=\:\sqrt{R^2-r^2}

    Since WZ = 36,\:WP = 18\!:\;\sqrt{R^2-r^2} \:=\:18\quad\Rightarrow\quad R^2-r^2\:=\:324 .[1]


    The area of the large semicircle is: . \frac{1}{2}\pi R^2
    The area of the small semicircle is: . \frac{1}{2}\pi r^2

    Then the grey area is: . A\:=\:\frac{1}{2}\pi R^2 - \frac{1}{2}\pi r^2 \:=\:\frac{1}{2}\pi\left(R^2-r^2\right) .[2]


    Substitute [1] into [2]: . A \:=\:\frac{1}{2}\pi(324) \:=\:\boxed{162\pi\text{ mm}^2}

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  5. #5
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    Thank you so much!
    I just couldn't get my head around it!
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