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(a)  ∠BOC = 2 *  ∠OAC = 2 * 2π/9 = 4π/9

Arc BC = (4π/9)* (2πr)/2π = 8π/3 cm (proved)

(b) Perimeter of ABC = ( π + 4π/9)* (2πr)/2π  + 2(6 cos(2π/9)) = 26π/3 + 9.19 = 36.14 cm

(c) Area of the minor sector that was cut off = Area of sector OAC – Area of triangle OA

= (π – 2 * 2π/9)πr²/2π – 9.19 * (6 * sin(2π/9))/2

= (5π/9)* 18 – 17.72

= 13.69 cm²

 

 

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