Answer :


Let AF be x


AE = x [Tangents from an external point are equal]


BC = 6 + 8 = 14 cm


AC = (x + 6) cm


AB = (x + 8) cm


OFA = OEA = ODB = 90° [Tangent to a circle is perpendicular to the radius at the point of contact]


s =


=


= = (x + 14)


Area of ΔACB =


=


= =


=


Area of ΔACB = ar (ΔAOC) + ar (ΔCOB) + ar (ΔBOA)


= [Area of right angled Δ = ]


= AC × 2 + CB × 2 + BA × 4


= 2 (x + 6 + 14 + 8 + x)


= 2 (2x + 28)


= 4 (x + 14)


Squaring both sides,


3x (x + 14) = (x + 14)2


(x + 14) [3x – (x + 14)] = 0


(x + 14) (2x – 14) = 0


x + 14 = 0 and 2x – 14 = 0


x = -14 and x = 7


x = 7 [ Sides cannot be negative]


AB = (x + 8) = 15 cm


AC = (x + 6) = 13 cm


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