RD Sharma Solutions for Class 9 Maths Chapter 20 Surface Area and Volume of A Right Circular Cone

Here are the RD Sharma Solutions Class 9 Maths Chapter 20 Surface Area and Volume of A Right Circular Cone Solutions, designed to help students prepare effectively for their exams. By referring to these solutions and practicing the problems, students can boost their confidence and improve their scores.

RD Sharma Solution for Class 9 Chapter 20 - Question with Answers

Basic Conceptual Questions

Q1. What is the formula for the curved surface area (CSA) of a right circular cone?
A1. CSA = π × r × l

Q2. What is the total surface area (TSA) of a cone?
A2. TSA = π × r × (l + r)

Q3. What is the volume of a cone?
A3. Volume = (1/3) × π × r² × h

Q4. Define 'slant height' of a cone.
A4. Slant height is the distance from the tip of the cone to a point on the edge of the circular base.

Q5. How is slant height (l) calculated?
A5. l = √(r² + h²)

Direct Formula-Based Questions

Q6. Find the CSA of a cone with radius 7 cm and slant height 25 cm.
A6. CSA = π × 7 × 25 = 550 cm² (approx.)

Q7. Find the TSA of a cone with radius 5 cm and slant height 13 cm.
A7. TSA = π × 5 × (13 + 5) = π × 5 × 18 = 282.74 cm² (approx.)

Q8. Volume of a cone with radius 6 cm and height 9 cm?
A8. Volume = (1/3) × π × 6² × 9 = 339.29 cm³ (approx.)

Q9. If the slant height is 10 cm and height is 8 cm, find the radius.
A9. r = √(l² - h²) = √(100 - 64) = √36 = 6 cm

Q10. Find height if radius is 9 cm and slant height is 15 cm.
A10. h = √(l² - r²) = √(225 - 81) = √144 = 12 cm

Application-Based Questions

Q11. A cone has a radius of 4 cm and height of 3 cm. Find its volume.
A11. Volume = (1/3) × π × 4² × 3 = 50.27 cm³ (approx.)

Q12. A cone has a CSA of 314 cm² and slant height 10 cm. Find the radius.
A12. r = CSA / (π × l) = 314 / (3.14 × 10) = 10 cm

Q13. A cone has TSA of 471.24 cm², radius 7 cm. Find slant height.
A13. TSA = πr(l + r) ⇒ 471.24 = 3.14 × 7 × (l + 7)
Solve: (l + 7) = 471.24 / 21.98 = 21.43 ⇒ l = 14.43 cm

Q14. Height and slant height of a cone are 9 cm and 15 cm. Find CSA.
A14. r = √(l² - h²) = √(225 - 81) = √144 = 12 cm
CSA = π × 12 × 15 = 565.2 cm² (approx.)

Q15. A cone of volume 261.8 cm³ has height 6 cm. Find the radius.
A15. Use V = (1/3)πr²h ⇒ r² = (3 × 261.8) / (π × 6) = 41.7
r = √41.7 ≈ 6.46 

Mixed Practice Problems

Q16. Radius of base = 5 cm, height = 12 cm. Find slant height.
A16. l = √(25 + 144) = √169 = 13 cm

Q17. Find CSA of cone: r = 3.5 cm, l = 10 cm
A17. CSA = π × 3.5 × 10 = 110 cm² (approx.)

Q18. Volume of cone: r = 10 cm, h = 21 cm
A18. Volume = (1/3) × π × 100 × 21 = 2200 cm³ (approx.)

Q19. TSA of cone: r = 4 cm, l = 5 cm
A19. TSA = π × 4 × (5 + 4) = 113.04 cm² (approx.)

Q20. Slant height = 17 cm, radius = 8 cm. Find CSA.
A20. CSA = π × 8 × 17 = 427.26 cm² (approx.)

Word Problems

Q21. A cone-shaped ice cream has radius 2.5 cm, height 4 cm. Find volume.
A21. V = (1/3)π × 2.5² × 4 = 26.18 cm³

Q22. A conical tent has a radius of 14 m and height 24 m. Find its CSA.
A22. l = √(14² + 24²) = √(196 + 576) = √772 ≈ 27.8 m
CSA = π × 14 × 27.8 = 1223.46 m² (approx.)

Q23. Water is poured into a conical vessel of radius 3 cm, height 12 cm. How much water can it hold?
A23. Volume = (1/3)π × 3² × 12 = 113.1 cm³

Q24. A metallic cone is melted to make small spheres of radius 1 cm. How many spheres are made from a cone of r = 3 cm, h = 12 cm?
A24. Volume of cone = 113.1 cm³
Volume of 1 sphere = (4/3)π × 1³ = 4.19 cm³
Number = 113.1 / 4.19 ≈ 27 spheres

Q25. A cone’s TSA is 1256 cm² and radius is 10 cm. Find slant height.
A25. TSA = πr(l + r) ⇒ 1256 = π × 10 × (l + 10)
(l + 10) = 1256 / 31.4 = 40 ⇒ l = 30 cm

True or False

Q26. CSA of cone = π × r².
A26. False (Correct: CSA = π × r × l)

Q27. TSA of cone is always more than CSA.
A27.  True (TSA = CSA + base area)

Q28. Volume of cone is less than volume of cylinder with same base and height.
A28.  True (Cone’s volume is 1/3 of cylinder)

Fill in the Blanks

Q29. The slant height of a cone is the ______ of a right triangle formed with height and radius.
A29. Hypotenuse

Q30. The shape of the base of a cone is a ______.