Answer :


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


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