Find the area of the minor segment of a circle of radius $14\text{ cm}$, when the angle of the corresponding sector is $60^\circ$.
Step-by-Step Solution
Key Concept: $\text{Area of sector} = \dfrac{60}{360} \times \dfrac{22}{7} \times 196 = \dfrac{308}{3}\text{ cm}^2$. $\text{Area of equilateral } \Delta = \dfrac{\sqrt{3}}{4} \times 196 = 49\sqrt{3}\text{ cm}^2$. $\text{Segment Area} = \dfrac{308}{3} - 49\sqrt{3}\text{ cm}^2 = 102.67 - 84.87 = 17.80\text{ cm}^2$.
$\text{Sector Area} = \dfrac{1}{6} \times \dfrac{22}{7} \times 196 = \dfrac{308}{3} \approx 102.67\text{ cm}^2$. [1.0 Mark]
$\text{Triangle Area} = \dfrac{\sqrt{3}}{4} \times 196 = 49\sqrt{3} \approx 84.87\text{ cm}^2$. $\text{Segment Area} = 102.67 - 84.87 = 17.80\text{ cm}^2$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Finding sector area $= 308/3\text{ cm}^2$: 1.0 Mark
Subtracting triangle area to get $17.80\text{ cm}^2$: 1.0 Mark
Correct Answer: