Please solve q 105 ?

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1 Answer
CW
May 26, 2018

#(2) " " sqrt6-sqrt3#

Explanation:

enter image source here
Let #|ABC|# denote area of #ABC#
Given #AD=3, => |ABC|=1/2*3*BC=3/2BC#
let #|ABC|=3a, => a=1/2BC#,
#=> |AGH|=a=1/2BC#
#=> |AJK|=2a=BC#
As #DeltaABC, DeltaAJK and DeltaAGH# are similar,
#=> JK^2:BC^2=2a:3a=2:3#
#=> JK:BC=sqrt2:sqrt3, => color(red)(JK=sqrt2/sqrt3*BC)#
Similarly, #GH^2:BC^2=1a:3a=1:3#
#=> GH:BC=1:sqrt3, => color(red)(GH=1/sqrt3* BC)#
Now, #|AJK|=2a=BC=1/2*AF*JK#,
#=> AF=(2BC)/(JK)=2BC*sqrt3/(sqrt2BC)=sqrt6#
Similarly, #|AGH|=a=1/2BC=1/2*AE*GH#
#=> AE=(BC)/(GH)=BC*sqrt3/(BC)=sqrt3#
#=> EF=AF-AE=sqrt6-sqrt3# units

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