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Suppose you have a cup in the shape of a right circular cone. Fill it with water to half the depth of the cone. What fraction of the volume of the cup is occupied by the water?
Let the conical cup have height H and radius R. When filled to half depth, water height is h = H/2. By similar triangles, the water surface radius is r = R/2. Water volume is Vwater = (1/3)π(R/2)²(H/2) = (1/8) × (1/3)πR²H = (1/8)Vcup. Hence, the water occupies 1/8 of the total volume of the cup.Read more
Let the conical cup have height H and radius R. When filled to half depth, water height is h = H/2. By similar triangles, the water surface radius is r = R/2. Water volume is Vwater = (1/3)π(R/2)²(H/2) = (1/8) × (1/3)πR²H = (1/8)Vcup. Hence, the water occupies 1/8 of the total volume of the cup.
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 right triangle with sides 6 cm, 8 cm and 10 cm is rotated through 360° about the side of 8 cm. Find the volume and the curved surface area of the solid so formed.
Rotating the right triangle about the 8 cm side generates a right circular cone having height h = 8 cm, base radius r = 6 cm and slant height l = 10 cm. Volume is V = (1/3)πr²h = (1/3)π × 6² × 8 = 96π ≈ 301.71 cm³. Curved surface area is CSA = πrl = π × 6 × 10 = 60π ≈ 188.57 cm². For more NCERead more
Rotating the right triangle about the 8 cm side generates a right circular cone having height h = 8 cm, base radius r = 6 cm and slant height l = 10 cm.
Volume is V = (1/3)πr²h = (1/3)π × 6² × 8 = 96π ≈ 301.71 cm³.
Curved surface area is CSA = πrl = π × 6 × 10 = 60π ≈ 188.57 cm².
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 lessWhat length of tarpaulin 3 m wide is required to make a conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that is required for stitching margins and wastage in cutting is 20 cm.
Given height h = 8 m and radius r = 6 m, slant height is l = √(8² + 6²) = 10 m. Curved surface area of the tent is πrl = 3.14 × 6 × 10 = 188.4 m². Required tarpaulin length before wastage is 188.4 / 3 = 62.8 m. Adding extra length of 20 cm = 0.2 m for stitching and wastage yields a total required leRead more
Given height h = 8 m and radius r = 6 m, slant height is l = √(8² + 6²) = 10 m. Curved surface area of the tent is πrl = 3.14 × 6 × 10 = 188.4 m². Required tarpaulin length before wastage is 188.4 / 3 = 62.8 m. Adding extra length of 20 cm = 0.2 m for stitching and wastage yields a total required length of 62.8 + 0.2 = 63 m.
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 joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
For each conical cap, base radius r = 7 cm and height h = 24 cm. The slant height is l = √(r² + h²) = √(7² + 24²) = √625 = 25 cm. Since caps are open at the base, sheet area per cap equals CSA = πrl = (22/7) × 7 × 25 = 550 cm². Therefore, 10 caps require 10 × 550 = 5500 cm² of sheet. For moreRead more
For each conical cap, base radius r = 7 cm and height h = 24 cm. The slant height is l = √(r² + h²) = √(7² + 24²) = √625 = 25 cm. Since caps are open at the base, sheet area per cap equals CSA = πrl = (22/7) × 7 × 25 = 550 cm². Therefore, 10 caps require 10 × 550 = 5500 cm² of sheet.
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 lessThe curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find (i) the radius of the base, (ii) the total surface area of the cone.
Given CSA = 308 cm² and slant height l = 14 cm. For (i), curved surface area is πrl. Substituting gives (22/7) × r × 14 = 308, so 44r = 308, yielding base radius r = 7 cm. For (ii), total surface area is CSA + πr² = 308 + (22/7) × 7² = 308 + 154 = 462 cm². For more NCERT Solutions of Class 9Read more
Given CSA = 308 cm² and slant height l = 14 cm.
For (i), curved surface area is πrl. Substituting gives (22/7) × r × 14 = 308, so 44r = 308, yielding base radius r = 7 cm.
For (ii), total surface area is CSA + πr² = 308 + (22/7) × 7² = 308 + 154 = 462 cm².
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