In Figure 4, a tr

Let Δ ABC be drawn to circumscribe a circle with centre O and radius 4 cm and circle touches the sides BC, CA and AB at D, E and F respectively.

Given CD = 6 cm and BD = 8 cm

∴ BF = BD = 8 cm and CE = CD = 6 cm

[Since length of two tangents drawn from an external point of circle are equal]

Now, let AF = AE = x cm

Then, AB = c = (x + 8) cm, BC = a = 14 cm, CA = b = (x + 6) cm

We know that 2s = a + b + c

⇒ 2s = (x + 8) + 14 + (x + 6)

⇒ 2s = 2x + 28

⇒ s = x + 14

⇒ s – a = (x + 14) – 14 = x

⇒ s – b = (x + 14) – (x + 6) = 8

⇒ s – c = (x + 14) – (x + 8) = 6

We know that area of triangle

⇒ Area of Δ ABC

⇒ Area (Δ ABC) = area (Δ OBC) + area (Δ OCA) + area (Δ OAB)

= 1/2 (BC) (OD) + 1/2 (CA) (OE) + 1/2 (AB) (OF)

= 1/2 (14) (4) + 1/2 (x + 6) (4) + 1/2 (x + 8) (4)

= 28 + 2x + 12 + 2x + 16

= 4x + 56

⇒

⇒

Squaring on both sides,

⇒ 48x (x + 14) = 16 (x + 14)²

⇒ 48x (x + 14) – 16 (x + 14)² = 0

⇒ 16 (x + 14) [3x – (x + 14)] = 0

⇒ 16 (x + 14) (2x – 14) = 0

⇒ 16 (x + 14) = 0 or 2x – 14 = 0

⇒ x = -14 or 2x = 14

⇒ x = -14 or x = 7

But x cannot be negative so x ≠ -14

∴ x = 7 cm

∴ The sides AB = x + 8 = 7 + 8 = 15 cm

And AC = x + 6 = 7 + 6 = 13 cm