Tuesday, December 16, 2008

Archimedes' Book of Lemmas, Proposition #4, Arbelos

Problem 643
Exercise your brain. Archimedes wrote the "Book of Lemmas" more than 2200 years ago. Solve the proposition #4 (high school level) and lift up your geometry skills.


Archimedes' Book of Lemmas #4.
Continue reading at:
gogeometry.com/ArchBooLem04.htm
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3 comments:

  1. bjhopper bjhvash44@sbcglobal.netMarch 25, 2010 at 4:26 PM

    AB dia of largest semicircle = 1
    AD dia of smallest semicircle = F fraction of AB
    DB dia of intermediate semicircle = 1-F

    Arbelo area pi /8 - pi/8 . F^2 -pi/8.(1-F)^2
    This simplifies to pi/4(F-F^2)
    call CD a
    ACB is a right triangle and F/a= a/1-F
    a^2=F-F^2
    area of cirle dia a=pi/4 x( F-F^2) = to arbelo

    ReplyDelete
  2. PravinJuly 27, 2011 at 6:41 PM

    AD^2 + DB^2 = AC^2 + BC^2 - 2 CD^2
    = AB^2 - 2 CD^2
    Area of arbelos = Area of semicircle ACBA
    - Area of semicircle on AD as diameter
    - Area of semicircle on DB as diameter
    = (Pi/8) (AB^2 - AD^2 - DB^2)
    = (Pi/4) CD^2
    = Area of circle on CD as diameter

    ReplyDelete
  3. Sumith PeirisJanuary 10, 2016 at 10:17 AM

    Area of Arbelos = piAB^2/8- piAD^2/8 - piDB^2/8 = piAD.DB/4 = piCD^2/4 = Area of circle CD

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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