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. | |
Revision as of 00:22, 6 June 2012
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.
