tag:blogger.com,1999:blog-6933544261975483399.post2516019788659372861..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 808: Parallelogram, Perpendicular, Right Triangle, Semicircle Area, DiameterAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-67582602441901241832012-09-29T18:54:29.376-07:002012-09-29T18:54:29.376-07:00http://imageshack.us/a/img141/4056/problem808.png
...http://imageshack.us/a/img141/4056/problem808.png<br />Let S3 is the red area ( see sketch)<br />In right triangle AED we have pi/4(AD^2)=pi/4(AE^2+ED^2)<br />So S+S3=S1+S2+S3 => S=S1+S2<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-12679948576502566582012-09-29T17:20:05.319-07:002012-09-29T17:20:05.319-07:00Let the area of segment FG be k.
From Pythagoras...Let the area of segment FG be k. <br /><br />From Pythagoras theorem, <br />AE^2 + DE^2 = AD^2 = BC^2<br /><br />Multiply both side by π/4, <br />(π/4)AE^2 + (π/4)DE^2 = (π/4)BC^2<br /><br />S1 + (S2 + k) = S + k<br />S1 + S2 = SJacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.com