SECTION-C12Solve the following : (4 marks each)13. What will be the co

1.	SECTION-C12Solve the following : (4 marks each)13. What will be the cost of covering thewhite portion in figure with a silver foilif the rate is Rs. 100 per m2 ? (Ď-3.14)121212
Answer» Side of a square = 12 m Radius of each sector = 12 m θ = 90° White part is divided into two parts of equal area. Area of part 1 = Area of square ABCD - Area of sector BAPC Area of part 1= (Side)² - (πr²θ)/360° = (12)² - (3.14 × 12 × 12 × 90°/360°) = 144 - (3.14 × 3 × 12) = 144 - 3.14 × 36 = 144 - 113.04 Area of part 1 = 30.96 m² Area of Part 1= Area of part 2 = 30.96 m² Total Area of white part = 2× 30.96 = 61.92 m² Cost of covering 1m² area by silver foil = ₹100 Cost of covering 61.92 m² area by silver foil = ₹(61.92 × 100) = ₹ 6192 Hence , the Cost of covering by silver foil = ₹ 6192 .

SECTION-C12Solve the following : (4 marks each)13. What will be the cost of covering thewhite portion in figure with a silver foilif the rate is Rs. 100 per m2 ? (Ď-3.14)121212

Answer»

Side of a square = 12 m

Radius of each sector = 12 m

θ = 90°

White part is divided into two parts of equal area.

Area of part 1 = Area of square ABCD - Area of sector BAPC

Area of part 1= (Side)² - (πr²θ)/360°

= (12)² - (3.14 × 12 × 12 × 90°/360°)

= 144 - (3.14 × 3 × 12)

= 144 - 3.14 × 36

= 144 - 113.04

Area of part 1 = 30.96 m²

Area of Part 1= Area of part 2 = 30.96 m²

Total Area of white part = 2× 30.96 = 61.92 m²

Cost of covering 1m² area by silver foil = ₹100

Cost of covering 61.92 m² area by silver foil = ₹(61.92 × 100) = ₹ 6192

Hence , the Cost of covering by silver foil = ₹ 6192 .