A chord subtends an angle of 120^o at the centre of a circle of radius 3.5cm. Find the perimeter of the minor sector containing the chord, [Take \(\pi = \frac{22}{7}\)]

14\(\frac{1}{3}\)cm

12\(\frac{5}{6}\)cm

8\(\frac{1}{7}\)cm

7\(\frac{1}{3}\)cm

Correct answer is A

perimeter of minor sector

2r + \(\frac{\theta}{360} \times 2 \pi r\)

= 2 x 3.5 + \(\frac{120^o}{360^o} \times 2 \frac{22}{7} \times 3.5\)

= 7 + \(\frac{154}{21}\)

= 7 + 7.33

= 14.33

= 14\(\frac{1}{3}\)cm

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