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