# Cl-5 (Math) | Ch-23 | Volume

## Pre-Requisites

• Basic understanding of figures

## Learning Objective

On successful completion of this lesson, students will be able to:

• demonstrate basic knowledge of solids, volume..
• analyze how to find volume of cube & cuboid.
• develop problem-solving skills.
• apply in real life.

## Dictionary

• Solid: Form of matter
• Volume: Occupied space
• Picture: Figure
• Graph: Grapgh or Sketch
• Percentage: %age
• Average: Mean
• Numeral: Number
• >: Greater than
• <: Lesser than
• =: Equal to
• Ascending: Increasing
• Descending: Decreasing
• Order: Sequence

### 23.1 Introduction

#### 23.1.1 Solid

=> Solid: An object having definite shape & size is called solid.

=> Solids are occur in different shapes & sizes.

=> Some of solids are:

• Cube
• Cuboid
• Cone
• Cylinder
• Sphere

#### 23.1.2 Volume

=> Volume: Solids occupy a certain amount of space called volume.

#### 23.1.3 Units of Volume

1. Cubic centimetre: The volume of a cube of edge 1 cm, is called 1 cubic centimetre.

=> It is denoted by 1 cu cm or 1cm³.

2. Cubic metre: The volume of a cube of edge 1 m, is called 1 cubic metre.

=> It is denoted by 1 cu m or 1 m³.

### 23.2 Cuboid

=> Cuboid: A solid bounded by six rectangular faces is called a cuboid.

e.g a brick , a match box, a paste box, a chalk box, a book etc.

#### Formula

V= (l x b x h) m³

Where l = length, b= breadth, h= height

### 23.3 Cube

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

e.g a dice, an ice cube, a sugar cube etc.

#### Formula

V = (a x a x a) m³

Where a = side

#### Solved Examples

Ex1. The length , breadth and height of a cuboid are 10 cm, 5 cm amd 2 cm respectively. Find its volume.

Sol. Given that:

l= 10 cm

b= 5 cm

h= 2 cm

V=l x b x h = 10 cm x 5 cm x 2 cm = 100 cm³

V= 100 cu cm.

Ex2. Find the volume of a cube whose each edge is 2 cm.

Sol. Given that:

a= 2 cm

V= a x a x a = 2x2x2 = 8 cu cm

Ex3. How many bricks each 24 cm long, 12.5 cm wideand 8 cm thick will be required to build a wall 18 m long, 2.5 m high and 40 cm thick?

Sol. Lenght of the wall = 18 m= 1800 cm

Height oof the wall = 2.5 m = 250 cm

Thickness oof the wall = 40 cm

Volume of the wall = l x b x h = (1800 x 250 x 40 ) cu cm

Now,

Length of a brick = 24 cm

Breadth of the brick = 12.5 cm

Thickness oof the brick = 8 cm

v = (24 x 12.5 x 8) cu cm

Therefore, The number of bricks required = V/v = (1800 x 250 x40) / (24 x 12.5 x 8)

= 7500 ans.

### 23.4 Practice Sets

1. Fill Type
2. Match Type
3. Order Type
4. Conversion Type
5. Compare Type
6. Error Type
7. Statement Type
8. True/False Type
9. Related Type
10. Day-to-Day Type
11. Miscellaneous Type

#### Set-1

1.Express each of the following as a fraction in its lowest terms.

2. Express each of the following as a decimal.

3. Convert each of the following fractions into a percentage.

4.Convert each of the following decimals into percentage.

5. Find the value of

6. Which of the following are meaningless?

7. Compare and put symbol >, < or =in the placeholder.

8. Fill in the blanks.

9. Cross (x) the Roman numerals which are not written correctly.

10. Write the following Roman numerals in ascending order.

11. Write the following Roman numerals in descending order.

12. Solve & write the answer in Roman numerals.

13. Complete the following table.

Ans: 1 ( ), 2( ), 3( ), 4( ), 5( ), 6( ), 7( ), 8( ), 9( ), 10( )

#### Set-2

Ans: 1 ( ), 2( ), 3( ), 4( ), 5( ), 6( ), 7( ), 8( ), 9( ), 10( )

#### Set-3

Ans: 1 ( ), 2( ), 3( ), 4( ), 5( ), 6( ), 7( ), 8( ), 9( ), 10( )

#### Set-4

Ans: 1 ( ), 2( ), 3( ), 4( ), 5( ), 6( ), 7( ), 8( ), 9( ), 10( )

#### Set-5

Ans: 1 ( ), 2( ), 3( ), 4( ), 5( ), 6( ), 7( ), 8( ), 9( ), 10( )

Ans:

1. ..
2. ..
3. ..
4. ..
5. ….
6. ….
7. ….

### 23.7 Exam

The exam will be conducted after the completion of the course.