In figure 4, a triangle ABC is drawn to circumscribe a circle of radius 4 cm, such that the segments BD and DC are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.

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