Class 9 Volume of a Cuboid

Topic covered

• Volume of a Cuboid

Volume of a Cuboid

You have already learnt about volumes of certain figures (objects) in earlier classes. Recall that solid objects occupy space.

The measure of this occupied space is called the Volume of the object.

`color{green}("Note :")` If an object is solid, then the space occupied by such an object is measured, and is termed the Volume of the object.

On the other hand, if the object is hollow, then interior is empty, and can be filled with air, or some liquid that will take the shape of its container. In this case, the volume of the substance that can fill the interior is called the capacity of the container.

In short, the volume of an object is the measure of the space it occupies, and the capacity of an object is the volume of substance its interior can accommodate. Hence, the unit of measurement of either of the two is cubic unit.

So, if we were to talk of the volume of a cuboid, we would be considering the measure of the space occupied by the cuboid.

Further, the area or the volume is measured as the magnitude of a region. So, correctly speaking, we should be finding the area of a circular region, or volume of a cuboidal region, or volume of a spherical region, etc.

But for the sake of simplicity, we say, find the area of a circle, volume of a cuboid or a sphere even though these mean only their boundaries.

Observe Fig. 13.23. Suppose we say that the area of each rectangle is A, the height up to which the rectangles are stacked is h and the volume of the cuboid is V. Can you tell what would be the relationship between V, A and h?

The area of the plane region occupied by each rectangle × height
= Measure of the space occupied by the cuboid
So, we get `A × h = V`

That is,

`color{green}("Volume of a Cuboid = base area × height = length × breadth × height")`

or `l × b × h,` where l, b and h are respectively the length, breadth and height of the cuboid.

`color{orange}("Note :")` When we measure the magnitude of the region of a space, that is, the space occupied by a solid, we do so by counting the number of cubes of edge of unit length that can fit into it exactly.
Therefore, the unit of measurement of volume is cubic unit.


`color{red}("Volume of a Cube = edge × edge × edge = a^3")`

where a is the edge of the cube (see Fig. 13.24).
So, if a cube has edge of 12 cm, then volume of the cube `color{blue}(= 12 × 12 × 12 cm^3)` `color{navy}(= 1728 cm^3.)`

Recall that you have learnt these formulae in earlier classes. Now let us take some examples to illustrate the use of these formulae:

Q 3210078819

A wall of length 10 m was to be built across an open ground. The height of the wall is 4 m and thickness of the wall is 24 cm. If this wall is to be built up with bricks whose dimensions are 24 cm × 12 cm × 8 cm, how many bricks would be required?
Class 9 Chapter 13 Example 11

Since the wall with all its bricks makes up the space occupied by it, we need to find the volume of the wall, which is nothing but a cuboid.

Here, `\ \ \ \ \ \"Length" = 10 m = 1000 cm`
`\ \ \ \ \ \ \ \ \ "Thickness" = 24 cm`
Therefore, `\ \ \ "Volume of the wall = length × thickness × height"`
`\ \ \ \ \ \ \ \ \= 1000 × 24 × 400 cm^3`
Now, each brick is a cuboid with length = 24 cm, breadth = 12 cm and height = 8 cm So, volume of each brick = length × breadth × height `= 24 × 12 × 8 cm^3`

`"So, number of bricks required = text(volume of the wall)/text(volume of each brick)`

`= (1000 × 24 × 400)/(24 ×12 × 8)`
`= 4166.6`
So, the wall requires 4167 bricks.
Q 3210178910

A child playing with building blocks, which are of the shape of cubes, has built a structure as shown in Fig. 13.25. If the edge of each cube is 3 cm, find the volume of the structure built by the child.
Class 9 Chapter 13 Example 12

Volume of each cube `= "edge" × "edge "× "edge" \ \ 3 × 3 × 3 cm^3 = 27 cm^3`
Number of cubes in the structure `= 15`
Therefore, volume of the structure `= 27 × 15 cm^3`
`= 405 cm^3`