Difference between revisions of "ESC09-LAN"

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Draw a Perpendicular from "E" to "AB"Let it meet "AB" at "O"Consider the Triangles AEO and ABCThey are SIMILAR because angle "C" is 90 (Angle in a Semicircle!Angle "C"=Angle "O"=90Angle "A" is coommn)Therefor AE/AO = AB/ACThat is AE*AC=AB*AO.......(1)Similarly from the similar triangles BEO and BAD,BE/BO=AB/BDThat is BE*BD=AB*BO.....(2)(1)+(2)=> AB(AO+BO)=(AC*AE)+(BD*BE)That is AB*AB=(AC*AE)+(BD*BE)HENCE THE ANSWER!!!ASWATHY VARMACLASS XICOCHIN REFINERY SCHOOLAMBALAMUGAL.
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Latest revision as of 07:36, 8 June 2012

Just to follow up on the uptade of this topic on your website and wish to let you know just how much I appreciated the time you took to put together this valuable post. Inside the post, you spoke of how to definitely handle this challenge with all ease. It would be my own pleasure to accumulate some more suggestions from your website and come as much as offer other individuals what I learned from you. Many thanks for your usual great effort.

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