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The hemispherical dome of a building needs to be painted (see Fig. 14.16). If the circumference of the base of the dome is 35.2 m, find the cost of painting it, given the cost of painting is ₹10 per 100 cm².
Given base circumference 2πr = 35.2 m, using π = 22/7 gives r = (35.2 × 7) / 44 = 5.6 m. The curved surface area of the dome is 2πr² = 2 × (22/7) × (5.6)² = 197.12 m². Converting to cm² gives 197.12 × 10,000 = 1,971,200 cm². At the rate of ₹10 per 100 cm², the cost is (1,971,200 / 100) × 10 = ₹1,97,Read more
Given base circumference 2πr = 35.2 m, using π = 22/7 gives r = (35.2 × 7) / 44 = 5.6 m. The curved surface area of the dome is 2πr² = 2 × (22/7) × (5.6)² = 197.12 m². Converting to cm² gives 197.12 × 10,000 = 1,971,200 cm². At the rate of ₹10 per 100 cm², the cost is (1,971,200 / 100) × 10 = ₹1,97,120.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 14 Math of Space: Surface Area and Volume Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-14/
See lessEstimate how many scoops of ice cream can be obtained from a cuboidal container of dimensions 10 cm × 15 cm × 20 cm.
The total volume of the cuboidal ice cream tub is 10 cm × 15 cm × 20 cm = 3000 cm³. Assuming an average spherical scoop has a diameter of 5 cm (radius r = 2.5 cm), each scoop has a volume of (4/3)πr³ = (4/3) × 3.14 × 2.5³ ≈ 65.45 cm³. Dividing the total volume by one scoop gives 3000 / 65.45 ≈ 46 scRead more
The total volume of the cuboidal ice cream tub is 10 cm × 15 cm × 20 cm = 3000 cm³. Assuming an average spherical scoop has a diameter of 5 cm (radius r = 2.5 cm), each scoop has a volume of (4/3)πr³ = (4/3) × 3.14 × 2.5³ ≈ 65.45 cm³. Dividing the total volume by one scoop gives 3000 / 65.45 ≈ 46 scoops.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 14 Math of Space: Surface Area and Volume Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-14/
See lessA cube of integer side length a is made of unit cubes. Write an expression giving the number of unit cubes to be added to make a cube of side length a + 1.
The initial cube of edge a has a volume of a³ unit cubes. The larger cube of side a + 1 requires a total volume of (a + 1)³ unit cubes. Expanding (a + 1)³ gives a³ + 3a² + 3a + 1. The number of unit cubes that must be added is the difference: (a + 1)³ − a³ = 3a² + 3a + 1. For more NCERT SolutRead more
The initial cube of edge a has a volume of a³ unit cubes. The larger cube of side a + 1 requires a total volume of (a + 1)³ unit cubes. Expanding (a + 1)³ gives a³ + 3a² + 3a + 1. The number of unit cubes that must be added is the difference: (a + 1)³ − a³ = 3a² + 3a + 1.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 14 Math of Space: Surface Area and Volume Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-14/
See lessSolve the following: (i) Could a person drink enough water in a lifetime to fill an entire room the size of your classroom? (ii) Estimate the number of bricks used to build the walls of your classroom.
(i) Drinking 2.5 litres daily for 80 years totals 2.5 × 365 × 80 = 73,000 litres = 73 m³. A typical classroom (8 m × 7 m × 3.5 m) has a volume of nearly 200 m³, so the answer is no. (ii) A standard classroom perimeter of 100 ft with height 10 ft gives 1,000 sq ft wall area. Accounting for doors, winRead more
(i) Drinking 2.5 litres daily for 80 years totals 2.5 × 365 × 80 = 73,000 litres = 73 m³. A typical classroom (8 m × 7 m × 3.5 m) has a volume of nearly 200 m³, so the answer is no. (ii) A standard classroom perimeter of 100 ft with height 10 ft gives 1,000 sq ft wall area. Accounting for doors, windows and mortar, approximately 8,000 to 10,000 bricks are needed.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 14 Math of Space: Surface Area and Volume Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-14/
See lessYou are given two options: (A) one chocolate cube of side 50 mm, (B) hundred chocolate cubes of side 10 mm. Which option gives you more chocolate?
For Option (A), the volume of the single chocolate cube of edge 50 mm is 50³ = 125,000 mm³. For Option (B), one small cube has a volume of 10³ = 1,000 mm³, making 100 such cubes equal to 100 × 1,000 = 100,000 mm³. Comparing the two volumes shows that Option (A) provides 25,000 mm³ more volume, givinRead more
For Option (A), the volume of the single chocolate cube of edge 50 mm is 50³ = 125,000 mm³. For Option (B), one small cube has a volume of 10³ = 1,000 mm³, making 100 such cubes equal to 100 × 1,000 = 100,000 mm³. Comparing the two volumes shows that Option (A) provides 25,000 mm³ more volume, giving you more chocolate.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 14 Math of Space: Surface Area and Volume Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-14/
See less