tag:blogger.com,1999:blog-6933544261975483399.post7433926191192678404..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 825: Circle, Semicircle, Arc, Chord, Midpoint, Sector, Triangle, AreaAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-77000785390679517412012-11-27T22:23:34.535-08:002012-11-27T22:23:34.535-08:00Since ∠DOC=∠OCB=67.5°, so OD//BC.
Hence,
S = ar...Since ∠DOC=∠OCB=67.5°, so OD//BC. <br /><br />Hence, <br />S = area of sector BOC (with angle 45°)<br />= 1/2×6^2×π/4<br />= 9π/2Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-67026420407447731442012-11-27T22:21:27.076-08:002012-11-27T22:21:27.076-08:00See attached sketch
http://img706.imageshack.us/im...See attached sketch<br />http://img706.imageshack.us/img706/5949/problem825.png<br />Let Area(XYZ)= area of triangle XYZ<br />Let Sector(XYZ)= area of sector XYZ<br />Let R=OA=6<br />We have Area of white area= Area(ADB)+2.S1<br />But S1= Sector(AOD)-Area(AOD)<br />And Area(ADB)=2.Area(AOD)<br />So white area=2.Sector(AOD) => S= sector(AOB)-white area=(pi/2-3/8.pi)*R^2= 9/2.pi<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-63707914014226989582012-11-27T22:12:58.988-08:002012-11-27T22:12:58.988-08:00∠DOA = 135/2 = 67.5 = ∠CBO
So OD // BC.
By sharing...∠DOA = 135/2 = 67.5 = ∠CBO<br />So OD // BC.<br />By sharing the same base BC and same height, ΔBCD has the same area as ΔOCB.<br />S = sector(OBC) = π *(36)*(1/8) = (9/2) π<br />W Fungnoreply@blogger.com