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let AB = a
BC = b
pythagoras thm
a^2 + b^2 = 36
also area 1/2ab = 10 => b=20/a put in above eqn
a^2 + 400/a^2 = 36
a^4 - 36a^2 +400 = 0
discriminant 36*36 - 4*400 = -304 negative which give imaginary values
ans:3.33cm
let AB=h BD=l DC=x AC=6cm
in Triangle ADC
DC is base BUt Height Would be AB
Hence area would be 1/2hx=10
hx=20
but as AD is internal angle bisector then using internal angle bisector theorem
AB/AC=BD/DC
AB.dC=BD.AC
hx=6l
but hx=20
hence l=10/3=3.33cm
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