# RD‌ ‌Sharma‌ ‌Chapter‌ ‌21 ‌Class‌ ‌9‌ ‌Maths‌ Exercise‌ ‌21.1 ‌Solutions‌

RD‌ ‌Sharma‌ ‌Chapter‌ ‌21 ‌Class‌ ‌9‌ ‌Maths‌ Exercise‌ ‌21.1 ‌Solutions‌‌ have been provided here to understand the concepts. Specifically, it can be useful for the students when they go for doing their own homework while preparing for the tests through studying. Moreover, the solutions are provided here for Class 9 is in a comprehensive fashion, in a stepwise fashion. It can help learn the concepts well, at the same time obtaining the specific answers.

Moreover, we have attached the RD‌ ‌Sharma‌ ‌Chapter‌ ‌21 ‌Class‌ ‌9‌ ‌Maths‌ Exercise‌ ‌21.1 ‌Solutions‌ PDF from which students can practice with various types of questions. The level of the questions is prepared as per the Class 9 students. By practicing with these several types of questions students understand the concept and they can score well in the exam.

Learn about RD Sharma Class 9 Chapter 21 (Surface Area And Volume of Sphere)

## Download RD‌ ‌Sharma‌ ‌Chapter‌ ‌21 ‌Class‌ ‌9‌ ‌Maths‌ Exercise‌ ‌21.1 ‌Solutions‌ PDF

Solutions for Class 9 Maths Chapter 21 Surface Area and Volume of A Sphere Exercise 21.1

## What topics are covered in RD‌ ‌Sharma‌ ‌Chapter‌ ‌21 ‌Class‌ ‌9‌ ‌Maths‌ Exercise‌ ‌21.1 ‌Solutions‌?

The Class 9 Maths Chapter 21 Surface Area and Volume of A Sphere Exercise 21.1 come from the following topics.

1. It covers the topic called Section of a sphere by a plane.
2. The question comes from the topics called Hemisphere and Spherical shell
3. There are many questions one can find from the topic called the Surface Area of a sphere as well.
4. There is a huge range of questions one can find within the exercise from the topic called Surface Area of a hemisphere.

Apart from these, one can explore the questions from the topic called Surface Area of a spherical shell as part of the RD‌ ‌Sharma‌ ‌Chapter‌ ‌21 ‌Class‌ ‌9‌ ‌Maths‌ Exercise‌ ‌21.1 ‌Solutions‌.

## Important Definitions RD‌ ‌Sharma‌ ‌Chapter‌ ‌21 ‌Class‌ ‌9‌ ‌Maths‌ Exercise‌ ‌21.1 ‌Solutions‌

This exercise of Chapter 21 Surface Area And Volume of Sphere is based on the following topics. Also, the definitions of each topic are given below with an example.

1. Section of a sphere by a plane
2. Hemisphere
3. Spherical shell
4. Surface Area of a sphere
5. Surface Area of a hemisphere and
6. Surface Area of a spherical shell

### Section of a sphere by a plane

The section of a sphere by a plane or the plane section of the sphere is often referred to as the spherical sections that can be bigger circles for planes that go through the center of the sphere or smaller circles for the rest.

### Hemisphere

Any plane having the center of a sphere dissects it in two equivalent hemispheres.

### Spherical shell

In simplest terms, a spherical shell can be understood as the sold wrapped amidst a couple of concentric spheres.

### Surface Area of a sphere

The surface area of a sphere can be calculated through the formula A=4 πr2, where ‘r’ is the radius of the sphere.

### Surface Area of a hemisphere

The surface area of a hemisphere can be calculated through the formula SA= 3πr2 square units

Here π is a constant that is equal to 3.14 approximately, and “r” is the radius of the hemisphere.

### Examples of RD‌ ‌Sharma‌ ‌Chapter‌ ‌21 ‌Class‌ ‌9‌ ‌Maths‌ Exercise‌ ‌21.1 ‌Solutions‌

Ques: Find the surface area of a sphere of radius:

(a) 10.5 cm (b) 5.6 cm (c) 14 cm

Solution:

Surface area of the sphere = 4πr2

Here, r = radius of a sphere

Surface area = 4x 22/ 7 x(10.5) x 2

= 1386

Surface area is 1386 cm2

Surface area = 4 × 22/ 7 × (5.6) x 2

= 394.24

Surface area = 394.24 cm2

Surface area = 4 × 22/ 7 × (14) x 2

= 2464

Surface area = 2464 cm2

Ques- Find the surface area of a sphere of diameter:

(a) 14 cm (b) 21 cm (c) 3.5 cm

Solution:

Surface area of a sphere = 4πr2

here, r = radius of a sphere

(a) Diameter = 14 cm

As, Radius = Diameter/ 2 = 14 / 2 cm = 7 cm

Surface area = 4 × 22/ 7 × (7) x 2

= 616

Surface area = 616 cm2

(ii) Diameter = 21cm

So, Radius = Diameter/ 2 = 21/ 2 cm = 10.5 cm

Surface area= 4 × 22 /7 × (10.5) x 2

= 1386

Surface area = 1386 cm2

(iii) Diameter= 3.5cm

So, Radius = Diameter/ 2 = 3.5 / 2 cm = 1.75 cm

Surface area = 4 × 22/ 7 × (1.75) x 2

= 38.5

Surface area = 38.5 cm2

Ques: Find the total surface area of a hemisphere and a solid hemisphere each of radius 10 cm. (π = 3.14)

Solution:

Radius of a hemisphere = Radius of the solid hemisphere = 10 cm

Surface area of hemisphere = 2πr2

= 2 × 3.14 × (10) x 2 cm2

= 628 cm2

And, surface area of solid hemisphere = 3πr2

= 3 × 3.14 × (10) x 2 cm2

= 942 cm2.