A chord of a circle of radius $10\text{ cm}$ subtends a right angle at the centre. Find the area of the corresponding minor segment. (Use $\pi = 3.14$)
Step-by-Step Solution
Key Concept: Minor sector area $= \dfrac{90}{360} \pi (10)^2 = 78.5\text{ cm}^2$. $\text{Area}(\Delta OAB) = \dfrac{1}{2} \times 10 \times 10 = 50\text{ cm}^2$.
Sector area $= \dfrac{1}{4} \times 3.14 \times 100 = 78.5\text{ cm}^2$. [1.0 Mark]
Triangle area $= \dfrac{1}{2} \times 10 \times 10 = 50\text{ cm}^2$.
Minor segment area $= 78.5 - 50 = 28.5\text{ cm}^2$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Calculating sector area $= 78.5\text{ cm}^2$: 1.0 Mark
Subtracting triangle area 50 to get $28.5\text{ cm}^2$: 1.0 Mark
Correct Answer: