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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