Q. 34

# In Figure 4, a tr

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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