Calculate the area of the designed region in a square of side $8\text{ cm}$ common between two quadrants of circles of radius $8\text{ cm}$ each.
Step-by-Step Solution
Key Concept: Designed Area $= 2 \times \text{Area of quadrant} - \text{Area of square} = 2\left(\dfrac{1}{4} \times \dfrac{22}{7} \times 64\right) - 64 = \dfrac{704}{7} - 64 = \dfrac{704 - 448}{7} = \dfrac{256}{7}\text{ cm}^2 = 36.57\text{ cm}^2$.
Area of two quadrants $= 2 \times \dfrac{1}{4} \times \dfrac{22}{7} \times 64 = \dfrac{704}{7}\text{ cm}^2$. [1.5 Marks]
Area of square $= 64\text{ cm}^2$. [0.5 Mark]
Designed Area $= \dfrac{704}{7} - 64 = \dfrac{256}{7}\text{ cm}^2 = 36.57\text{ cm}^2$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Calculating combined area of two quadrants $= 704/7\text{ cm}^2$: 1.5 Marks
Subtracting square area $= 64\text{ cm}^2$: 0.5 Mark
Evaluating designed region $= 256/7\text{ cm}^2$: 1.0 Mark
Correct Answer: