RD Sharma Solutions for Class 9 Maths Chapter 20 Surface Area and Volume of A Right Circular Cone
The RD Sharma Solutions for Class 9 Chapter 20 include questions and answers about the surface area and volume of a right circular cone. These solutions explain the important ideas related to the cone and its measurements in an easy and clear way. Expert teachers at IL have prepared these answers keeping in mind the understanding level of Class 9 students.
Each answer is explained step by step so that students can understand the concepts better. This helps them not only score well in exams but also build a strong understanding of the subject.
Using RD Sharma Solutions for Class 9 Maths helps students learn the concepts clearly and gain confidence. Solving textbook problems with the help of these solutions removes confusion and improves overall knowledge. Students can also download the solutions in PDF format for free using the links given below.
RD Sharma Solution for Class 9 Chapter 20 - Download PDF
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
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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 ______.
RD Sharma Solutions for Class 9 Maths Chapter 20
What does Chapter 20 in Class 9 RD Sharma cover?
Chapter 20 covers the surface area and volume of a right circular cone, including curved surface area, total surface area, and volume formulas and applications.
How can I prepare Chapter 20 Surface Areas and Volumes effectively?
Start by understanding the formulas and concepts, then practice solved examples from RD Sharma. Follow it up with exercises and previous year questions.
Are RD Sharma solutions for Chapter 20 helpful for board exams?
Yes, the RD Sharma solutions provide detailed steps and are aligned with CBSE Class 9 syllabus, making them ideal for concept clarity and exam preparation.
Where can I find the RD Sharma Class 9 Chapter 20 solutions PDF?
You can download the Chapter 20 solutions in PDF format from educational websites like Infinity Learn, Vedantu, or by referring to the official RD Sharma solution books.
What are the key formulas in Chapter 20 Surface Areas and Volumes?
Important formulas include:
Surface Area of Cube: 6a²
Surface Area of Sphere: 4πr²
Volume of Cylinder: πr²h
Volume of Cone: (1/3)πr²h*
Is Chapter 20 difficult in RD Sharma Class 9?
The chapter is moderately easy if you understand the concepts. Regular practice of formulas and application-based questions makes it manageable.
Can this chapter help in competitive exams like Olympiads?
Yes, concepts from Surface Areas and Volumes are frequently asked in math Olympiads and other school-level competitive exams.