A sector of central angle $120^\circ$ is cut off from a circle of radius $21\text{ cm}$. Find:
(i) Length of the arc of the sector
(ii) Area of the sector
(iii) Area of the minor segment formed by the corresponding chord
Step-by-Step Solution
Key Concept: (i) $l = \dfrac{120}{360} \times 2 \times \dfrac{22}{7} \times 21 = 44\text{ cm}$.<br>(ii) Sector Area $= \dfrac{1}{3} \times \dfrac{22}{7} \times 441 = 462\text{ cm}^2$.<br>(iii) Triangle Area with $120^\circ = \dfrac{1}{2} \times 441 \times \sin 120^\circ = \dfrac{441\sqrt{3}}{4} \approx 190.95\text{ cm}^2$. Segment Area $= 462 - 190.95 = 271.05\text{ cm}^2$.
(i) $l = \dfrac{1}{3} \times 2 \times \dfrac{22}{7} \times 21 = 44\text{ cm}$. [1.5 Marks]
(ii) Sector Area $= \dfrac{1}{3} \times \dfrac{22}{7} \times 441 = 462\text{ cm}^2$. [1.5 Marks]
(iii) Triangle Area $= \dfrac{1}{2} \times 21 \times 21 \times \dfrac{\sqrt{3}}{2} = \dfrac{441\sqrt{3}}{4} \approx 190.95\text{ cm}^2$. [1.0 Mark]
Minor Segment Area $= 462 - 190.95 = 271.05\text{ cm}^2$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Part (i) Arc length $= 44\text{ cm}$: 1.5 Marks
Part (ii) Sector area $= 462\text{ cm}^2$: 1.5 Marks
Part (iii) Minor segment area $= 271.05\text{ cm}^2$: 2.0 Marks
Correct Answer: