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.

Continue reading at:

gogeometry.com/ArchBooLem04.htm

## Tuesday, December 16, 2008

### Archimedes' Book of Lemmas, Proposition #4, Arbelos

Labels:
arbelo,
Archimedes,
area,
book of lemmas,
circle,
diameter,
perpendicular

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AB dia of largest semicircle = 1

ReplyDeleteAD 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

AD^2 + DB^2 = AC^2 + BC^2 - 2 CD^2

ReplyDelete= 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

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

ReplyDeleteSumith Peiris

Moratuwa

Sri Lanka