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RD Sharma Class 9 Solutions Chapter 20 - Exercise 20.1

Last updated date: 17th May 2024
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RD Sharma Class 9 Solutions Chapter 20 - Surface Area and Volume of Right Circular Cone (Ex 20.1) Exercise 20.1 - Free PDF

Free PDF download of RD Sharma Class 9 Solutions Chapter 20 - Surface Area and Volume of Right Circular Cone Exercise 20.1 solved by Expert Mathematics Teachers on All Chapter 20 - Surface Area and Volume of Right Circular Cone Ex 20.1 Questions with Solutions for RD Sharma Class 9 Maths to help you to revise the complete syllabus and score higher marks in the exams. Register for online coaching for IIT JEE (Mains & Advanced) and other engineering entrance exams. Students can download NCERT Solutions for Class 9 Science created by the best teachers at Vedantu for free. You can also download NCERT Solution PDF for all subjects to prepare for their forthcoming exams.

There are solutions available here for Class 9 Chapter 20 - Surface Area and Volume of a Right Circular Cone Exercise 20.1. Our experts have prepared these solutions with accurate answers so that students are prepared to answer questions in their tests. In this chapter of RD Sharma Class 9, students find several questions pertaining to the surface area of a right circular cone.

Maths Exercise 20.1 Solutions from RD Sharma Chapter 20 can be helpful in answering questions related to right circular cones. For example, it can be helpful in determining the surface area and volume of the right circular cones. Since the experts have provided the questions for the problems, students can be sure that they will be accurate. Students can use these exercises to prepare for their examinations.

Each solution is carefully analyzed in accordance with the questions. Consequently, they provide excellent support for understanding the surface area and volume of right circular cones, as well as exercises that use them. In RD Sharma's Chapter 20 - Class 9 Maths Exercise 20.1 Solutions, two topics are covered: Introduction to Right Circular Cones, and Surface Area of Right Circular Cones. The CBSE level of exams is easier to solve for those who have solid understandings of the right circular cones.

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Competitive Exams after 12th Science

Circular Right Cone Properties

Let us go through the properties of the right cone as discussed in RD Sharma Class 9 Chapter 20:

  • The circular base has an axis, which starts at the vertex and extends to the base, showing the axis of each cone.

  • It is the distance measured from the vertex to the outer line of the circular base of this cone that is known as its slant height. This is indicated by the letter "l.".

  • Generally, the altitude of a right cone is defined using the perpendicular line between the basal vertex and the vertex. A cone's axis is represented by the letter 'h', which coincides with it.

  • Consider the perpendicular as the axis of rotation when a right triangle is rotated about its perpendicular. Here, the solid constructed is the required cone. Hypotenuses generate lateral surface area. This is the area generated by the hypotenuse of the triangle.

  • Right circular cones formed by sections parallel to the axis of the cone form circles.

  • As the vertex and two points of the base of a right circular cone are contained in an isosceles triangle, it is a right circular cone section.

FAQs on RD Sharma Class 9 Solutions Chapter 20 - Exercise 20.1

1. What are the benefits of following RD Sharma solutions for Class 9 Maths Chapter 20?

These solutions provide complete and quality information about different concepts in Maths related to Chapter 20 of Chapter 9 of the RD Sharma Solutions for Class 9 Maths textbook. Answers for the questions have been given in an easy-to-remember format to further assist students in understanding and remembering the answers. Therefore, RD Sharma Textbooks for Class 9 can certainly be considered as a reliable source of reading material for scoring high in exams. This collection of Class 9 Maths solutions can help you score good marks to a great extent.

2. How do RD Sharma Solutions for Class 9 Maths Chapter 20 help students prepare for exams?

Wherever necessary our solution module includes examples and diagrams to explain the questions. RD Sharma Solutions for Class 9 are a must for CBSE board students aiming for excellent grades. Students can benefit from RD Sharma Solutions for Class 9 Maths Chapter 20 by having a better understanding of the topics covered. The students will score high marks if they solve the questions from each exercise.

3. In RD Sharma Solutions for Class 9 Maths Chapter 20, what are the topics covered?

Chapter 20 of RD Sharma's Solutions for Class 9 Maths consists of the following topics:

1. Circular cones with right axes are ones with axes perpendicular to the plane of their bases. A right circular cone with a circular base of radius r and an axis at right angles to the base. Cone height is the line connecting the vertex of the cone to the centre of the base.

2. In a right circular cone, the surface area equals the radius of the base plus the height of the slant: *r2+*Lr, where r is the radius of the base and L is the height of the slant. Similarly, the lateral area is defined as the curvature of the surface.

4. Following concepts from RD Sharma Class 9 Chapter 20 solve: The diameter of a cone is 21 cm. If the slant height is 28 cm, find the area of the base.

The cone height formula is used to determine the height of the cone. Using cone height formulas, cone height is calculated by h = 3V/*r2, where V = Volume of the cone, r = Radius of the cone, and l = Slant Height of the cone.= 21 cm

Slant height (l) = 28 cm

∴ l2 = r2 + h2

⇒ r2 = l2– h2 = (28)2 – (21 )2

⇒ 784 – 441 = 343 …(i)

Now area of base = πr2= 227 x 343 [From (i)]

= 22 x 49 = 1078 cm2

5. Following concepts from RD Sharma Class 9 Chapter 20, solve: Calculate the surface area of a right circular cone that has a radius of 6 cm and a height of 8 cm.

The radius of the cone (r) is 6 cm

8 cm is the height of the cone (h)

What is the total surface area of the cone (T.S.A.)?

To find the slant height of a cone, follow these steps:

We know, l2 = r2 + h2


= 36 + 64

= 100

or l = 10 cm


The total Surface area of the cone (T.S.A) = Curved surface area of cone + Area of circular base

= πrl + πr2

= (22/7 x 6 x 10) + (22/7 x 6 x 6)

The product of 1320/7 and 792/7 is 1320/7

The value is 301.71

This means that the base is 301.71cm2.

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