📑 Topics on this page (Jump to Topic):
📌 1. Formula Table
| Solid | Volume | Curved/lateral SA | Total SA |
|---|---|---|---|
| Cube | a³ | 4a² | 6a² |
| Cuboid | lbh | 2h(l + b) | 2(lb + bh + hl) |
| Cylinder | πr²h | 2πrh | 2πr(r + h) |
| Cone | ⅓πr²h | πrl | πr(r + l) |
| Sphere | (4/3)πr³ | 4πr² | 4πr² |
| Hemisphere | (2/3)πr³ | 2πr² | 3πr² |
- Cone: slant height l = √(r² + h²).
- Diagonal of cuboid = √(l² + b² + h²); of cube = a√3.
- In melting/recasting, volume is conserved: number = big volume / small volume.
⚠️ Exam Traps:
- Doubling the radius → sphere volume ×8, surface ×4.
- Same base and height: cone volume = ⅓ of cylinder.
📌 2. Prism and Pyramid
| Solid | Volume | Lateral surface (LSA) | Total surface (TSA) |
|---|---|---|---|
| Prism | Base area × height | Base perimeter × height | LSA + 2 × base area |
| Pyramid | ⅓ × base area × height | ½ × base perimeter × slant height | LSA + base area |
📘 Formulas:
- Square-based pyramid (side a, height h): slant height l = √(h² + (a/2)²)
- Base areas: equilateral triangle (√3/4)a²; regular hexagon (3√3/2)a²
- Regular tetrahedron (edge a): volume = a³/(6√2), TSA = √3a²
- Prism on an n-sided base: faces n + 2, vertices 2n, edges 3n
- Pyramid on an n-sided base: faces n + 1, vertices n + 1, edges 2n (Euler: F + V − E = 2)
⚠️ Exam Traps:
- A pyramid’s LSA uses the slant height; its volume uses the vertical height.
- Same base and height: pyramid volume = ⅓ of the prism (like cone = ⅓ of cylinder).
📌 3. Practice MCQs
📝 Practice MCQs (prepared by SF Prep — not PYQs)
Q1. The volume of a cylinder of radius 21 cm and height 6 cm is (π = 22/7): Practice
Correct answer: (B)
Solution: πr²h = 22/7 × 21² × 6 = 8316 cm³.
Solution: πr²h = 22/7 × 21² × 6 = 8316 cm³.
Q2. The curved surface area of a cone with radius 5 cm and height 12 cm is (π = 22/7): Practice
Correct answer: (C)
Solution: l = √(5² + 12²) = 13; πrl = 22/7 × 5 × 13 = 204.29 cm².
Solution: l = √(5² + 12²) = 13; πrl = 22/7 × 5 × 13 = 204.29 cm².
Q3. The volume of a sphere of radius 14 cm is (π = 22/7): Practice
Correct answer: (D)
Solution: (4/3)πr³ = (4/3) × 22/7 × 2744 = 11498.67 cm³.
Solution: (4/3)πr³ = (4/3) × 22/7 × 2744 = 11498.67 cm³.
Q4. The volume of a cube is 1000 cm³. Its diagonal is: Practice
Correct answer: (A)
Solution: Side = ∛1000 = 10; diagonal = a√3 = 10√3 cm.
Solution: Side = ∛1000 = 10; diagonal = a√3 = 10√3 cm.
Q5. The total surface area of a cuboid measuring 16 m × 7 m × 4 m is: Practice
Correct answer: (B)
Solution: 2(lb + bh + hl) = 2(112 + 28 + 64) = 408 m².
Solution: 2(lb + bh + hl) = 2(112 + 28 + 64) = 408 m².
Q6. A metal sphere of radius 9 cm is melted to make balls of radius 3 cm. How many balls can be made? Practice
Correct answer: (C)
Solution: Number = (R/r)³ = (9/3)³ = 27.
Solution: Number = (R/r)³ = (9/3)³ = 27.
Q7. If the radius of a sphere is made 3 times, its volume becomes how many times? Practice
Correct answer: (D)
Solution: V ∝ r³ → 3³ = 27 times.
Solution: V ∝ r³ → 3³ = 27 times.
Q8. The total surface area of a solid hemisphere of radius 7 cm is (π = 22/7): Practice
Correct answer: (A)
Solution: 3πr² = 3 × 22/7 × 7² = 462 cm².
Solution: 3πr² = 3 × 22/7 × 7² = 462 cm².
Q9. The volume of a cylinder of radius 3.5 cm and height 12 cm is (π = 22/7): Practice
Correct answer: (B)
Solution: πr²h = 22/7 × 3.5² × 12 = 462 cm³.
Solution: πr²h = 22/7 × 3.5² × 12 = 462 cm³.
Q10. The curved surface area of a cone with radius 7 cm and height 24 cm is (π = 22/7): Practice
Correct answer: (C)
Solution: l = √(7² + 24²) = 25; πrl = 22/7 × 7 × 25 = 550 cm².
Solution: l = √(7² + 24²) = 25; πrl = 22/7 × 7 × 25 = 550 cm².
Q11. The volume of a sphere of radius 7 cm is (π = 22/7): Practice
Correct answer: (D)
Solution: (4/3)πr³ = (4/3) × 22/7 × 343 = 1437.33 cm³.
Solution: (4/3)πr³ = (4/3) × 22/7 × 343 = 1437.33 cm³.
Q12. The volume of a cube is 1728 cm³. Its diagonal is: Practice
Correct answer: (A)
Solution: Side = ∛1728 = 12; diagonal = a√3 = 12√3 cm.
Solution: Side = ∛1728 = 12; diagonal = a√3 = 12√3 cm.
Q13. The total surface area of a cuboid measuring 13 m × 8 m × 9 m is: Practice
Correct answer: (B)
Solution: 2(lb + bh + hl) = 2(104 + 72 + 117) = 586 m².
Solution: 2(lb + bh + hl) = 2(104 + 72 + 117) = 586 m².
Q14. If the radius of a sphere is made 2 times, its volume becomes how many times? Practice
Correct answer: (C)
Solution: V ∝ r³ → 2³ = 8 times.
Solution: V ∝ r³ → 2³ = 8 times.
Q15. The total surface area of a solid hemisphere of radius 3.5 cm is (π = 22/7): Practice
Correct answer: (D)
Solution: 3πr² = 3 × 22/7 × 3.5² = 231/2 cm².
Solution: 3πr² = 3 × 22/7 × 3.5² = 231/2 cm².
Q16. The volume of a cylinder of radius 7 cm and height 6 cm is (π = 22/7): Practice
Correct answer: (A)
Solution: πr²h = 22/7 × 7² × 6 = 924 cm³.
Solution: πr²h = 22/7 × 7² × 6 = 924 cm³.
Q17. The curved surface area of a cone with radius 14 cm and height 48 cm is (π = 22/7): Practice
Correct answer: (B)
Solution: l = √(14² + 48²) = 50; πrl = 22/7 × 14 × 50 = 2200 cm².
Solution: l = √(14² + 48²) = 50; πrl = 22/7 × 14 × 50 = 2200 cm².
Q18. The volume of a cube is 216 cm³. Its diagonal is: Practice
Correct answer: (C)
Solution: Side = ∛216 = 6; diagonal = a√3 = 6√3 cm.
Solution: Side = ∛216 = 6; diagonal = a√3 = 6√3 cm.
Q19. The total surface area of a cuboid measuring 18 m × 5 m × 9 m is: Practice
Correct answer: (D)
Solution: 2(lb + bh + hl) = 2(90 + 45 + 162) = 594 m².
Solution: 2(lb + bh + hl) = 2(90 + 45 + 162) = 594 m².
Q20. If the radius of a sphere is made 4 times, its volume becomes how many times? Practice
Correct answer: (A)
Solution: V ∝ r³ → 4³ = 64 times.
Solution: V ∝ r³ → 4³ = 64 times.
Q21. A triangular prism has a right-triangle base with legs 6 cm and 8 cm. The height of the prism is 10 cm. Find its volume. Practice
Correct answer: (C)
Solution: Base = ½ × 6 × 8 = 24 cm² → volume = 24 × 10 = 240 cm³.
Solution: Base = ½ × 6 × 8 = 24 cm² → volume = 24 × 10 = 240 cm³.
Q22. A pyramid has a square base of side 10 cm and height 12 cm. What is its lateral surface area? Practice
Correct answer: (A)
Solution: Slant height = √(12² + 5²) = 13 → LSA = ½ × 40 × 13 = 260 cm². (360 is the TSA.)
Solution: Slant height = √(12² + 5²) = 13 → LSA = ½ × 40 × 13 = 260 cm². (360 is the TSA.)
Q23. A pyramid has a base area of 45 cm² and a height of 8 cm. Its volume is: Practice
Correct answer: (D)
Solution: ⅓ × 45 × 8 = 120 cm³.
Solution: ⅓ × 45 × 8 = 120 cm³.
Q24. How many edges does a hexagonal prism have? Practice
Correct answer: (B)
Solution: n = 6 → edges = 3n = 18 (faces 8, vertices 12).
Solution: n = 6 → edges = 3n = 18 (faces 8, vertices 12).
Q25. A prism has an equilateral-triangle base of side 4 cm and height 10 cm. Its lateral surface area is: Practice
Correct answer: (D)
Solution: LSA = base perimeter × height = 12 × 10 = 120 cm². (120 + 8√3 is the TSA.)
Solution: LSA = base perimeter × height = 12 × 10 = 120 cm². (120 + 8√3 is the TSA.)