A chord of a circle of radius $12\text{ cm}$ subtends an angle of $120^\circ$ at the centre. Find the area of the corresponding segment of the circle. (Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)
Step-by-Step Solution
Key Concept: $\text{Sector Area} = \dfrac{120}{360} \times 3.14 \times 144 = \dfrac{1}{3} \times 452.16 = 150.72\text{ cm}^2$.<br>Area of $\Delta OAB$ with $120^\circ = \dfrac{1}{2} r^2 \sin 120^\circ = \dfrac{1}{2} \times 144 \times \dfrac{\sqrt{3}}{2} = 36\sqrt{3} = 36 \times 1.73 = 62.28\text{ cm}^2$.<br>Segment Area $= 150.72 - 62.28 = 88.44\text{ cm}^2$.
$\text{Sector Area} = \dfrac{1}{3} \times 3.14 \times 144 = 150.72\text{ cm}^2$. [1.0 Mark]
$\text{Triangle Area} = \dfrac{1}{2} \times 144 \times \sin 120^\circ = 36\sqrt{3} = 36 \times 1.73 = 62.28\text{ cm}^2$. [1.0 Mark]
$\text{Segment Area} = 150.72 - 62.28 = 88.44\text{ cm}^2$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Finding sector area $= 150.72\text{ cm}^2$: 1.0 Mark
Finding triangle area with $120^\circ = 62.28\text{ cm}^2$: 1.0 Mark
Evaluating segment area $= 88.44\text{ cm}^2$: 1.0 Mark
Correct Answer: