# NCERT Solutions for Exercise 13.4 Class 9 Maths Chapter 13 - Surface Area and Volumes

NCERT Solutions for Class 9 Maths exercise 13.4 deals with the concept of the sphere, hemisphere, and it's surface area . A three dimensional object which is round in shape is known as a sphere. Radius is the distance between surface and centre of the sphere and diameter is the distance from one point to another point on the surface of the sphere, passing through the centre. The diameter of the sphere is given by 2r where r is the radius of the sphere. In exercise 13.4 Class 9 Maths, The total area covered by the surface of a sphere in a three-dimensional space is known as the surface area of the sphere. The amount of space occupied by the sphere is known as the volume of the sphere . The nine questions in NCERT solutions for Class 9 Maths chapter 13 exercise 13.4 are based on the notion of surface areas and volumes of spheres. Class 9 Maths chapter 13 exercise 13.4 thoroughly explains the concepts of surface area and volume. The following activities are included along with Class 9 Maths chapter 13 exercise 13.4 .

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• Surface Area and Volumes Exercise 13.1
• Surface Area and Volumes Exercise 13.2
• Surface Area and Volumes Exercise 13.3
• Surface Area and Volumes Exercise 13.5
• Surface Area and Volumes Exercise 13.6
• Surface Area and Volumes Exercise 13.7
• Surface Area and Volumes Exercise 13.8
• Surface Area and Volumes Exercise 13.9

## Surface Area and Volumes Class 9 Chapter 13 Exercise: 13.4

Q1 (i) Find the surface area of a sphere of radius: .

We know,

The surface area of a sphere of radius =

Required surface area =

Q1 (ii) Find the surface area of a sphere of radius:

We know,

The surface area of a sphere of radius =

Required surface area =

Q1 (iii) Find the surface area of a sphere of radius:

We know,

The surface area of a sphere of radius =

Required surface area =

Q2 (i) Find the surface area of a sphere of diameter: 14 cm

Given,

The diameter of the sphere =

We know,

The surface area of a sphere of radius =

Required surface area =

Q2 (ii) Find the surface area of a sphere of diameter: 21 cm

Given,

The diameter of the sphere =

We know,

The surface area of a sphere of radius =

Required surface area =

Q2 (iii) Find the surface area of a sphere of diameter:

Given,

The diameter of the sphere =

We know,

The surface area of a sphere of radius =

Required surface area =

Q3 Find the total surface area of a hemisphere of radius 10 cm. (Use )

We know,

The total surface area of a hemisphere = Curved surface area of hemisphere + Area of the circular end

The required total surface area of the hemisphere =

Q4 The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Given,

We know,

The surface area of a sphere of radius =

The ratio of surface areas of the ball in the two cases =

Therefore, the required ratio is

Q5 A hemispherical bowl made of brass has inner diameter . Find the cost of tin-plating it on the inside at the rate of Rs 16 per .

Given,

The inner radius of the hemispherical bowl =

We know,

The curved surface area of a hemisphere =

The surface area of the hemispherical bowl =

Now,

Cost of tin-plating = Rs 16

Cost of tin-plating =

Therefore, the cost of tin-plating it on the inside is

Q6 Find the radius of a sphere whose surface area is .

Given,

The surface area of the sphere =

We know,

The surface area of a sphere of radius =

Therefore, the radius of the sphere is

Q7 The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

Let diameter of Moon be and diameter of Earth be

We know,

The surface area of a sphere of radius =

The ratio of their surface areas =

Therefore, the ratio of the surface areas of the moon and earth is

Q8 A hemispherical bowl is made of steel, thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Given,

The inner radius of the bowl =

The thickness of the bowl =

Outer radius of the bowl = (Inner radius + thickness) =

We know, Curved surface area of a hemisphere of radius =

The outer curved surface area of the bowl =

Therefore, the outer curved surface area of the bowl is

Q9 (i) A right circular cylinder just encloses a sphere of radius (see Fig. ). Find surface area of the sphere,

Given,

The radius of the sphere =

Surface area of the sphere =

Q9 (ii) A right circular cylinder just encloses a sphere of radius (see Fig. ). Find curved surface area of the cylinder,

Given,

The radius of the sphere =

The surface area of the sphere =

According to the question, the cylinder encloses the sphere.

Hence, the diameter of the sphere is the diameter of the cylinder.

Also, the height of the cylinder is equal to the diameter of the sphere.

We know, the curved surface area of a cylinder =

Therefore, the curved surface area of the cylinder is

Q9 (iii) A right circular cylinder just encloses a sphere of radius (see Fig. ). Find ratio of the areas obtained in (i) and (ii).

The surface area of the sphere =

And, Surface area of the cylinder =

So, the ratio of the areas =

## More About NCERT Solutions for Class 9 Maths Exercise 13.4

The NCERT solutions for Class 9 Maths exercise 13.4 is mainly focused on the surface area and the volume of the sphere. The surface area of the sphere is calculated by the product of four times the area of the circle. A=4πr2. The volume of the sphere is equal to 4Πr3/3. An exact half of a sphere is known as the hemisphere. When a sphere is cut at the exact centre along its diameter which leaves two equal hemispheres. There are two types of surface area. They are

• Lateral Surface Area

• Total Surface Area

The total surface area of hemisphere is equal to 3πr2 whereas the Lateral surface area of hemisphere is equal to 2πr2. In NCERT solutions for Class 9 Maths exercise 13.4, the formulas for computing surface areas and volume for the sphere and hemisphere are thoroughly explored.

Also Read| Surface Areas And Volumes Class 9 Notes

## Benefits of NCERT Solutions for Class 9 Maths Exercise 13.4 :

• NCERT solutions for Class 9 Maths exercise 13.4, helps in finding the radius and diameter of a sphere and hemisphere by using the formula of total surface area of sphere and hemisphere respectively.

• NCERT book Exercise 13.4 Class 9 Maths, the questions are explained clearly with proper geometric figures and explanations in a step-by-step procedure for our good understanding that will help us to secure more marks.

• By answering the NCERT syllabus Class 9 Maths chapter 13 exercise 13.4 exercises, we may improve our grades in the first and second terms, and the formulae for calculating surface area and volume can aid us in solving competitive problems, therefore we should remember them.

Also See:

• NCERT Solutions for Class 9 Maths Chapter 13 – Surface Area and Volumes

• NCERT Example Solutions Class 9 Maths Chapter 13 – Surface Area and Volumes

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