Given: Radius $r = 21$ cm, $\theta = 60°$, $\pi = \dfrac{22}{7}$
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(i) Length of the arc:
$$\text{Length of arc} = \frac{\theta}{360} \times 2\pi r = \frac{60}{360} \times 2 \times \frac{22}{7} \times 21$$
$$= \frac{1}{6} \times \frac{44 \times 21}{7} = \frac{1}{6} \times 132 = 22 \text{ cm}$$
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(ii) Area of the minor segment:
$$\text{Area of sector} = \frac{\theta}{360} \times \pi r^2 = \frac{60}{360} \times \frac{22}{7} \times 21 \times 21 = \frac{1}{6} \times \frac{22 \times 441}{7} = 231 \text{ cm}^2$$
Area of triangle OAB: Since $\theta = 60°$ and OA = OB = 21 cm, $\triangle$OAB is equilateral.
$$\text{Area of } \triangle OAB = \frac{\sqrt{3}}{4} \times (21)^2 = \frac{\sqrt{3}}{4} \times 441 = \frac{441\sqrt{3}}{4} \text{ cm}^2$$
$$\text{Area of minor segment} = 231 - \frac{441\sqrt{3}}{4} = \frac{924 - 441\sqrt{3}}{4} = \frac{441(44 - 21\sqrt{3})}{84}$$
$$\boxed{= \left(231 - \frac{441\sqrt{3}}{4}\right) \text{ cm}^2 \approx 231 - 190.95 \approx 40.05 \text{ cm}^2}$$
Source: Chapter 11, Section 11.1 — Areas of Sector and Segment of a Circle
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