tag:blogger.com,1999:blog-6933544261975483399.post2199383505400568355..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1057: Triangle, Area, Circle with Perpendicular Diameters, Tangent, SecantAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-60380164119951524812014-11-23T09:51:17.402-08:002014-11-23T09:51:17.402-08:00<CBF=<ECF from tangent angle theorem, <FA...<CBF=<ECF from tangent angle theorem, <FAB=<FCB from inscribed angles, and <FAB=<CEF from parallel property. Therefore <FCB=<FEC, making triangle CFB similar to triangle ECF. FB/CF=CF/FE, which results in CF^2=25=FB*FE=2*[FEB]. Ivan Bazarovnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-19046455597371983362014-11-07T20:34:55.831-08:002014-11-07T20:34:55.831-08:00Tr. BFE = Tr. BEA - Tr. BFA = Tr. BCA - Tr. BHA....Tr. BFE = Tr. BEA - Tr. BFA = Tr. BCA - Tr. BHA. Now draw a perpendicular HX to AB.It can be proven easily that HX =(2R²-25)/2R if R be the radius of the circle. <br />Hence Tr. BFE = (R*R) - ((2R²-25)/2R)*(2R/2) = 25/2 = 12.5 sq. units.Ajithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com