Q. 305.0( 1 Vote )

# In figure 4, a tr

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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