# NCERT Solutions for Class 9 Maths Chapter 13

Download here the NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.1, 13.2, 13.3, 13.4, 13.5, 13.6, 13.7, 13.8 and 13.9 in English Medium as well as Hindi Medium updated for session 2021-2022. UP Board Students of Class 9 (High School) are using NCERT Book in session 2021-2022, so they can also using these textbooks solutions as UP Board Solutions for Class 9 Maths Chapter 13. Here, they can download Prashnavali 13.1, Prashnavali 13.2, Prashnavali 13.3, Prashnavali 13.4, Prashnavali 13.5, Prashnavali 13.6, Prashnavali 13.7, Prashnavali 13.8 and Prashnavali 13.9

9 in Hindi Medium to use offline for session 2021-22. NCERT Solutions for class 9 chapter 13 is applicable for CBSE Delhi Board, MP Board, UP Board – High School, UK Board, Gujrat Board and other board using NCERT Books 2021-22. NCERT Solutions 2021-2022 for other are also given in downloadable format.## NCERT Solutions for Class 9 Maths Chapter 13

### Class 9th Maths Exercise 13.1 Solutions

### Class 9th Maths Exercise 13.2 Solutions

### Class 9th Maths Exercise 13.3 Solutions

### Class 9th Maths Exercise 13.4 Solutions

### Class 9th Maths Exercise 13.5 Solutions

### Class 9th Maths Exercise 13.6 Solutions

### Class 9th Maths Exercise 13.7 Solutions

### Class 9th Maths Exercise 13.8 Solutions

### Class 9th Maths Exercise 13.9 Solutions

Class: 9 | Maths (English and Hindi Medium) |

Chapter 13: | Surface Areas and Volumes |

### 9th Maths Chapter 13 Solutions in English & Hindi Medium

Download free NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes all exercises which are given below in PDF format to download for new academic session 2021-22. NCERT Solutions as well as NCERT Solutions Offline Apps are updated for new academic session 2021-2022 based on latest CBSE Syllabus 2021-22. NCERT Books are now implemented in Uttar Pradesh also. So, students can download UP Board solutions for class 9 Maths chapter 13 all exercises from here.

#### Class 9 Maths Exercise 13.1 and 13.2 Solutions in Video

#### Class 9 Maths Exercise 13.3 and 13.4 Solutions in Video

#### Class 9 Maths Exercise 13.5 and 13.6 Solutions in Video

#### Class 9 Maths Exercise 13.7 and 13.8 Solutions in Video

#### Class 9 Maths Exercise 13.9 Solutions in Video

### Class 9 Maths Chapter 13 Practice Questions with Solution

##### What are the formulae for Cuboid?

Cuboid: A cuboid is a solid bounded by six rectangular plane surfaces for example match box, brick, books, etc. are cuboid.

1. Surface area (or total surface area) of cuboid = 2 (lb + bh + hl) square units

2. Lateral surface area of cuboid = 2(l + b)h square units

3. Diagonal of a cuboid = √[l² + b² + h²] units

4. Total length of a edges of a cuboid = 4 (l + b + h) units

5. Volume of cuboid = lbh cubic units

##### What are the formulae for Cube?

Cube: A cuboid whose length, breadth and height are equal, is called a cube.

1. Surface area (or total surface area) of cuboid = 6a² square units

2. Lateral surface area of cuboid = 4a² square units

3. Diagonal of a cuboid = √3a units

4. Total length of a edges of a cuboid = 12a units

5. Volume of cuboid = a³ cubic units

##### What is meant by a Cylinder? Write its formulae?

Cylinder: A solid generated by the revolution of a rectangle about one of its sides which is kept fixed is called right circular cylinder.

Curved Surface Area (CSA) = 2πrh square units

Total Surface Area (TSA) = 2πr(r + h) square units

Volume = πr²h cubic units

##### What do you understand by a Cone? Write its Formulae?

Cone: A right circular cone is solid generated by revolving a line segment which passes through a fixed point and which makes a constant angle with a fixed line.

Slant Height = √[r² + h²]

Curved Surface Area (CSA) = πrl square units

Total Surface Area (TSA) = πr(r + l) square units

Volume = 1/3 πr²h cubic units

##### What are the basic formulae of Sphare?

Sphere: A sphere is three dimensional figure which is made up of all points in the space, which lie at a constant distance, form a fixed point called the centre of the sphere and the constant distant is called its radius.

Curved Surface Area (CSA) = Total Surface Area (TSA) = 4πr² square units

Volume = 4/3 πr³ cubic units

#### Important Notes on 9th Maths Chapter 13

Surface Area:

Surface area of a solid body is the area of all of its surface together and it is always measured in square units. Surface area is also known as total surface area (TSA).

Volume:

Space occupied by an object (solid body) is called the volume of the object. Volume is always measured in cubic units.

### Important Questions on 9th Maths Chapter 13

##### The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm³ of wood has a mass of 0.6 g.

Inner radius of cylindrical pipe r = 24/2 = 12 cm,

outer radius R = 28/2 = 14 cm and

length h = 35 m

Volume of cylindrical wooden pipe

= π(R²-r²)h

= 22/7 × (14²-12²)×35

= 22 × (196 – 144) × 5

= 22 × 52 × 5

= 5720 cm³

Mass of cylindrical wooden pipe

= 5720 × 0.6 g

= 3432 g

= 3.432 kg [∵1 cm³ of wood has a mass of 0.6 g]

Hence, the volume of cylindrical wooden pipe is 3.432 kg.

##### A soft drink is available in two packs – (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

Length of tin can l = 5 cm, breadth b = 4 cm and height h = 15 cm

Volume of tin can = lbh = 5 × 4 × 15 = 300 cm³

Radius of plastic cylinder r = 7/2 = 3.5 cm and height H = 10 cm

Volume of plastic cylinder

= πr²H = 22/7 × 3.5 × 3.5 × 10

= 22 × 0.5 × 3.5 × 10

= 385 cm³

Difference between capacities of two packs = 385 – 300 = 85 cm³

Hence, the capacity of plastic cylindrical pack is greater than tin can by 85 cm³.

##### If the lateral surface of a cylinder is 94.2 cm² and its height is 5 cm, then find radius of its base.

Lateral surface area of cylinder C = 94.2 cm² and height h = 5 cm.

Let, the radius of cylinder = r cm

Lateral surface area of cylinder C = 2πrh

⇒ 94.2 = 2 × 3.14 × r × 5

⇒ r = 94.2/(3.14×10) = 3 cm

Hence, the radius of base is 3 cm.

##### It costs ₹ 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of ₹ 20 per m², find inner curved surface area of the vessel.

Cost of painting the inner curved surface of cylindrical vessel = ₹ 2200 and

height h = 10 m.

Let, the inner radius of cylindrical vessel = r m

The inner curved surface area of cylindrical vessel = 2πrh

The cost of painting is at the rate of ₹ 20 per m² = ₹ 20 × 2πrh

According to question,

₹ 20 × 2πrh = ₹ 2200

⇒ 2πrh = 2200/20 = 110

Hence, the inner curved surface area is 110 m².

##### A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?

Radius of cylindrical bowl r = 7/2 = 3.5 and

height of soup inside the cylindrical bowl h = 4 cm

Volume of cylindrical bowl

= πr²h = 22/7 × (3.5)² × 4

= 22/7× 3.5 × 3.5 × 4

= 22 × 0.5 × 3.5 × 4 = 154 cm³

Therefore, the volume of soup per day for 250 patient

= 250 × 154

= 38500 cm³

Hence, hospital has to prepare 38500 cm³ soup daily to serve 250 patients.

##### The curved surface area of a right circular cylinder of height 14 cm is 88 cm². Find the diameter of the base of the cylinder.

Curved surface area of cylinder 88 cm² and height h = 14 cm

Let, the radius of base of cylinder = r cm

Curved surface area of cylinder = 2πrh

⇒ 88 = 2 × 22/7 × r × 14

⇒ 88 = 88r

⇒ r = 1 cm

Hence, the diameter of base of cylinder

= 2r = 2 × 1 = 2 cm

##### The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m².

Radius of roller r = 84/2 = 42 cm = 0.42 m and

length h = 120 cm = 1.2 m

Outer curved surface area of roller

= 2πrh = 2 × 22/7 × 0.42 × 1.2 = 2 × 22 × 0.06 × 1.2 = 3.168 m²

Area of ground levelled in on revolution = 3.168 m²

Therefore, area of ground levelled in 500 revolutions

= 500 × 3.168 = 1584 m²

Hence, the area of playground is 1584 m².

##### Curved surface area of a right circular cylinder is 4.4 m². If the radius of the base of the cylinder is 0.7 m, find its height.

Curved surface area of cylinder 4.4 m² and radius r = 0.7 m

Let, the height of cylinder = h m

Curved surface area of cylinder = 2πrh

⇒ 4.4 = 2 × 22/7 × 0.7 × h

⇒ 4.4 = 4.4h

⇒ h = 1 m

Hence, the height of the cylinder is 1 m.

##### Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)

Radius of hemisphere r = 10 cm

Surface area of hemisphere = 3πr²

= 3 × 3.14 × 10 × 10

= 942 cm²

Hence, the total surface area of hemisphere is 942 cm².