RD Sharma Chapter 20 Class 9 Maths Exercise 20.1 Solutions can be significant for addressing questions around right circular cones. To be specific, it can be significant for solving questions around surface area and volume of right circular cones. As the questions of the problems have been provided by the experts, students can stay assured of its accuracy. These exercises can be absolutely relevant from an examination point of view.
The solutions provided here are thoroughly analyzed as per the questions. Hence, they can be significant for understanding the concepts of surface area and volume of right circular cones and the exercises based on the same. RD Sharma Chapter 20 Class 9 Maths Exercise 20.1 Solutions focuses on two specific topics called the Right Circular Cone Introduction and The Surface Area of Right Circular Cone. Those having strong fundamentals about the concepts of right circular cones at the CBSE level can find it easier to attain.
Learn about Class 9 Chapter 20 (Surface Area And Volume Of Right Circular Cone)
Download RD Sharma Chapter 20 Class 9 Maths Exercise 20.1 Solutions PDF
Important Definitions RD Sharma Chapter 20 Class 9 Maths Exercise 20.1 Solutions
Right Circular Cone Introduction
The right circular cone is primarily a cone having its base as a circle. Its axis remains perpendicular to the base. It is also termed the cone of revolution.
Surface Area of a Right Circular Cone
The surface area of the Right Circular Cone can be calculated through the below formula.
A= πr(r+ √h2+r2)
Here ‘r’ is the radius of the base circle, and ‘h’ is the height of the cone.
Examples of RD Sharma Chapter 20 Class 9 Maths Exercise 20.1 Solutions
Ques- Find the rounded surface space of a cone if its slant height is 60cm and the radius of its base is 21cm.
Solution-
Slant height of cone (a) = 60 cm
The radius of the base of the cone (b) = 21 cm
Now,
Rounded surface space of the right circular cone = πba = 22/7 x 21 x 60 = 3960 cm2
Hence the rounded surface space of the right circular cone is 3960 cm2
Ques- The radius of a cone is 5cm, and the vertical height is 12cm. Find the area of the rounded surface.
Solution-
Radius of cone (c) = 5 cm
Height of cone (b) = 12 cm
Find Slant Height of cone (a):
We know, a2 = b2 + c2
a2 = 52 +122
a2 = 25 + 144 = 169
Or a = 13 cm
Now,
C.S.A = πcb =3.14 x 5 x 13 = 204.28
Hence, the rounded surface space of the cone is 204.28 cm2
Ques- Find the entire surface space of a right circular cone with a radius of 6 cm and a height of 8 cm.
Solution-
Radius of cone (b) = 6 cm
Height of cone (a) = 8 cm
The total Surface area of a cone (T.S.A)=?
Find slant height of cone:
We know, c2 = b2 + a2
=62+82
= 36 + 64
= 100
or c = 10 cm
Now,
The entire Surface space of the cone (T.S.A) = Rounded surface space of a cone
+ Area of the circular base
= πbc + πb2
= (22/7 x 6 x 10) + (22/7 x 6 x 6)
= 1320/7 + 792/7
= 301.71
Hence, the area of the base is 301.71cm2.
Frequently Asked Questions of RD Sharma Chapter 20 Class 9 Maths Exercise 20.1 Solutions
Ques 1- Can it be useful for class 9 examination preparation?
Ans- Yes, a student can expect absolutely similar questions in the examinations.
Ques 2- Are the questions only based on the surface area?
Ans- Simultaneously, revolving around surface area calculation of right circular cone, it also covers the questions around the volume of right circular cones. One can find the whole range of questions here.