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let ABC be a triangle right angled at B.Let AD be the angle bisector of A.suppose AC=6cm and the area of the triangle ADC is 10cm^2, wat is length of BD?
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
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
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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