tag:blogger.com,1999:blog-6933544261975483399.post6906207516198197576..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Archimedes' Book of Lemmas, Proposition #4, ArbelosAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-83707793115940258512016-01-10T10:17:43.305-08:002016-01-10T10:17:43.305-08:00Area of Arbelos = piAB^2/8- piAD^2/8 - piDB^2/8 = ...Area of Arbelos = piAB^2/8- piAD^2/8 - piDB^2/8 = piAD.DB/4 = piCD^2/4 = Area of circle CD<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-78819048459017669362011-07-27T18:41:05.363-07:002011-07-27T18:41:05.363-07:00AD^2 + DB^2 = AC^2 + BC^2 - 2 CD^2
= AB^2 - 2 CD^2...AD^2 + DB^2 = AC^2 + BC^2 - 2 CD^2<br />= AB^2 - 2 CD^2 <br />Area of arbelos = Area of semicircle ACBA <br />- Area of semicircle on AD as diameter<br />- Area of semicircle on DB as diameter<br />= (Pi/8) (AB^2 - AD^2 - DB^2)<br />= (Pi/4) CD^2<br />= Area of circle on CD as diameterPravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-26388514993072917872010-03-25T16:26:35.386-07:002010-03-25T16:26:35.386-07:00AB dia of largest semicircle = 1
AD dia of smalles...AB dia of largest semicircle = 1<br />AD dia of smallest semicircle = F fraction of AB<br />DB dia of intermediate semicircle = 1-F<br /><br />Arbelo area pi /8 - pi/8 . F^2 -pi/8.(1-F)^2<br />This simplifies to pi/4(F-F^2)<br />call CD a<br />ACB is a right triangle and F/a= a/1-F<br />a^2=F-F^2<br />area of cirle dia a=pi/4 x( F-F^2) = to arbelobjhopper bjhvash44@sbcglobal.netnoreply@blogger.com