## RD Sharma Class 9 Solutions Chapter 20 - Surface Area and Volume of Right Circular Cone (Ex 20.1) Exercise 20.1 - Free PDF

## 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

∴ l^{2 }= r^{2 }+ h^{2}

⇒ r^{2 }= l^{2}– h^{2} = (28)^{2} – (21 ^{)2}

⇒ 784 – 441 = 343 …(i)

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

= 22 x 49 = 1078 cm^{2}

**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

=62+82

= 36 + 64

= 100

or l = 10 cm

Now,

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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