A chord of a circle of radius $15\text{ cm}$ subtends an angle of $60^\circ$ at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)
Step-by-Step Solution
Key Concept: $\text{Sector Area} = \dfrac{60}{360} \times 3.14 \times 225 = 117.75\text{ cm}^2$. $\text{Triangle Area} = \dfrac{1.73}{4} \times 225 = 97.3125\text{ cm}^2$. Minor Segment Area $= 117.75 - 97.3125 = 20.4375\text{ cm}^2$. Total Circle Area $= 3.14 \times 225 = 706.5\text{ cm}^2$. Major Segment Area $= 706.5 - 20.4375 = 686.0625\text{ cm}^2$.
$\text{Sector Area} = \dfrac{1}{6} \times 3.14 \times 225 = 117.75\text{ cm}^2$. [1.0 Mark]
$\text{Minor Segment Area} = 117.75 - \dfrac{1.73}{4} \times 225 = 117.75 - 97.3125 = 20.4375\text{ cm}^2$. [1.0 Mark]
$\text{Major Segment Area} = 3.14 \times 225 - 20.4375 = 706.5 - 20.4375 = 686.0625\text{ cm}^2$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Finding minor sector area $= 117.75\text{ cm}^2$: 1.0 Mark
Evaluating minor segment area $= 20.4375\text{ cm}^2$: 1.0 Mark
Evaluating major segment area $= 686.0625\text{ cm}^2$: 1.0 Mark
Correct Answer: