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