△ABC circumscribe the circle with centre O
BD=8 cm
CD=6 cm
OD=4 cm
Since tangents drawn from extend point are equal
So, CE=CD=6 cm
BF=BD=8 cm
AE=AF=k cm
∴CB=CD+DB=6+8=14 cm =c
AB=AF+FC=(x+8) cm =b
AC=AE+EC=(x+6) cm =a
s=a+b+c=(x+6)+(x+8)+142=x+14 [∵ Heros formula =√s(s−a)(s−b)(s−c)]
Area of △ABC
=√(x+14)(x+14−(x+6))(x+14−(x+8))(x+14−14)
=√(x+14)(8)(6)x
=√46x2+672x
∴OD=OF=OE= radius =4 cm
And OD⊥BC,OF⊥AB & OE⊥AC (∵ tangent O ⊥ to radius)
ar(△ABC)=ar(△AOC)+ar(△AOB)+ar(△BOC)−−−−−(1)
ar(△AOC)=12×OE×AC
=12×4×(x+6)=2x+12
ar(△AOB)=12×OF×AB=12×4(x+8)=2x+16
ar(△BOC)=12×OD×BC=12×4×14=28
∴ from (1)
(√46x2+672x)2=(4x+56)2
∴48x2+672x=(4x)2+(56)2+2×4x×56
41x2−16x+672x−448x−3136=0
32(x2+7x−98)=0
(x−7)(x+14)=0, so x=7
as x correct be −19 (negative acceptable)
AC=x+6=7+6=13 cm
AB=x+8=7+8=15 cm