Volume Of Cubes And Cuboid

Volume is the amount of space that is enclosed in a solid figure like a cube or a cuboid.

If we are given the dimensions of the cube/cuboid, we know how to calculate the volume by using the formula of

$\small​\mathsf{length \times breadth \times height} ​$
 

At P6, we are given the volume of a cube/cuboid, and we have to find:

1. find the length of the cube/cuboid.
2. find the base area of the cube/cuboid.
3. find the height of the cube/cuboid.

This article is written as per the course prescribed for Primary 6 Maths level.

This will provide you with a strong foundation and will help you build your concepts to be able to solve challenging questions related to this topic.

Volume Of Cube

A cube is a six-sided three-dimensional solid figure with square faces. All the edges of the cube are equal in length. 

We know, that the volume of a cube is  $\small​\mathsf{length \times breadth \times height} ​$.

Since, all the sides of the cube are the same, so 

$\small{\therefore \qquad \mathsf{length = breadth = height}}$.

$\small\textsf{Volume of a cube} = \mathsf{length \times length \times length}$.

$\small\textsf{Volume of a cube} = \mathsf{length^3}$

So, to find the length of one edge of the cube, we find the cube root of volume.

$\small\textsf{Length of one edge of a cube} = \sqrt [3]{\textsf{Volume}}$

Volume Of Cuboid

A cuboid is a six-sided 3-dimensional solid figure with rectangular and/or square faces. The opposite faces of the cuboid are the same.

$\small\textsf{Volume of a cuboid} = \mathsf{length \times breadth \times height}$

$\small\mathsf{V = L × B × H }$

Now, if the volume, breadth and height are given, how do we find the Length? 

We use the above triangle to help us with the formula.

$\small \mathsf{L = V \div ( B \times H )}$

Similarly, to find the breadth,

$\small \mathsf{B = V ÷ (L × H)}$

Lastly, to find the height,

$\small \mathsf{H = V \div (L \times B)}$

1. Finding the length of the edge of a cube/cuboid

Now that we have learned the formulae, let us try a few questions.

Question 1:

The volume of the cuboid below is $\small \mathsf{340 \,cm^3}$. Find the length of the cuboid.

Solution:

$\begin{align} \small \mathsf{Length (L)} &= \small \,? \\ \small \mathsf{Breadth (B)} &= \small \mathsf{4\;cm} \\ \small \mathsf{Height (H)} &= \small \mathsf{5\;cm} \\ \small \mathsf{Volume (V)} &= \small \mathsf{340\;cm^3} \\[2ex] \end{align}$

$\begin{align} \small \mathsf{Length} &= \small \mathsf{Volume \div ( Breadth \times Height )} \\   &= \small \mathsf{340 \,cm^3 \div (4 \,cm \times 5 \,cm)}\\   &= \small \mathsf{340 \,cm^3 \div 20 \,cm^2}\\   &= \small \mathsf{17 \,cm} \end{align}$

Answer:

$\small\textsf{17 cm}$

Question 2:

The volume of a cube is $\small\mathsf{1331 \,cm^3}$.  Find the length of the edge of the cube.

Solution:

$\begin{align} \small\textsf{Volume (V)} &= \small\mathsf{1331\,cm^3} \\ \small\textsf{Length (L)} &=\small\textsf{?} \end{align}$

$\begin{align} \small\textsf{Length of the edge of the cube} &= \small\mathsf{\sqrt [3]{1331} \,cm^3}\\ &=\small\mathsf{11 \,cm} \end{align}$

Answer:

$\small\textsf{11 cm}$

2. Finding the base area of a cube and a cuboid

$\small\textsf{Base area of a cuboid} ​​​​​​​= \mathsf{Length \times Breadth}$

We use the same triangle to help us with the formula.

$\small \mathsf{Base \;Area = V ÷ H}$
 

Lastly, to find the height,

$\small \mathsf{H = V ÷ Base \;Area}$
 

Question 1:

A solid has a volume of $\small \mathsf{1230 \;cm^3}$. If the height of the solid is $\small \mathsf{6 \;cm}$, find the base area of the solid. 

Solution:

$\begin{align}​ \small\textsf{Volume of solid} &= \small\mathsf{1230 \,cm^3} \\ \small\textsf{Height of solid} &= \small\mathsf{6 \,cm} \end{align}$

$\begin{align}​ \small\textsf{Volume of solid} &= \small\mathsf{Length \times Breadth \times Height}\\ &=\small\mathsf{Base Area \times Height} \end{align}$

$\begin{align}​ \small\textsf{Base area of solid} &= \small\mathsf{Volume \div Height} \\ &= \small\mathsf{1230 \,cm^3 \div 6\,cm} \\ &= \small\mathsf{205 \,cm^2} \end{align}$

Answer:

$\small\mathsf{205 \,cm^2}$

Question 2:

A solid has a volume of $\small\mathsf{1688 \,cm^3}$. If the height of the solid is $\small\mathsf{8\,cm}$, find the base area of the solid. 

Solution:

$\begin{align} \small\textsf{Volume of solid} &= \small\mathsf{1688 \,cm^3} \\ \small\textsf{Height of solid} &= \small\mathsf{8 \,cm} \end{align}$

$\begin{align} \small\textsf{Volume of solid} &= \small\mathsf{Length \times Breadth \times Height}\\ &=\small\mathsf{Base Area \times Height} \end{align}$

$\begin{align} \small\textsf{Base area of solid} &= \small\mathsf{Volume \div Height} \\ &= \small\mathsf{1688 \,cm^3 \div 8\,cm} \\ &= \small\mathsf{211 \,cm^2} \end{align}$

Answer:

$\small\mathsf{211 \,cm^2}$

Question 3: 

The volume of the cuboid is $\small\mathsf{1048 \,cm^3}$ and the area of the shaded face is $\small\mathsf{131 \,cm^2}$. Find the length of the unknown edge of the cuboid.

Solution:

$\small{ \begin{align}​​​ \textsf{Volume} &= \mathsf{1048\,cm^3} \\ \textsf{Shaded area} &= \mathsf{131\,cm^2} \end{align}}$

$\small\bbox[8px,border:2px solid red] { \textsf{Volume} =  \textsf{Shaded Area × Length} }$

$\small{ \begin{align}​​​ \textsf{Length of solid}  &= \mathsf{Volume \div Shaded Area} \\     &= \mathsf{1048 \,cm^3 \div 131 \,cm^2} \\     &= \mathsf{8 \,cm} \end{align}}$

Answer:

$\small\textsf{8 cm}$

3. Finding the height or water level of the container

Using the concepts learned thus far, let us try to solve the following questions.

Question 1:

A rectangular tank contains $\small\mathsf{12.5 \,\ell}$ of water. If the base area of the tank is $\small\mathsf{500 \;cm^2}$, what is the height of the water level of the tank? $\small\mathsf{(1 \,\ell = 1000 \,cm^3)}$

Solution:

$\small{\begin{align}​​ \mathsf{1\,\ell} &= \mathsf{1000 \,cm^3} \\ \mathsf{12.5 \,\ell}  &= \mathsf{12.5 \times 1000 \,cm^3}\\ &= \mathsf{12 \,500 \,cm^3} \end{align}}$

$\small{\begin{align}​​ \textsf{Volume of water in the tank} &= \mathsf{12\,500 \,cm^3} \\ \textsf{Base area of the tank} &= \mathsf{500 \,cm^2} \end{align}}$

$\small{ \bbox[8px, border:2px solid red]{ \textbf{Volume} = \textbf{Base Area × Height} }}$

$\small{\begin{align}​​ \textsf{Height of water in the tank} &= \mathsf{12 \,500 \,cm^3 \div 500 \,cm^2} \\ &= \mathsf{25 \,cm} \end{align}}$

Answer:

$\small\text{25 cm}$

Question 2:

A rectangular container had a base area of $\small\displaystyle\mathsf{750 \,cm^2}$. Sally poured some mango syrup into the container till it was $\small\displaystyle\mathsf{\frac {3}{8}}$ full. She then poured $\small\displaystyle\mathsf{11\frac {1}{4}}$ litres of water into the container until it was completely full. What was the height of the rectangular container?

Solution:

$\small{ \begin{align}​​ \mathsf{1 \,litre} &= \mathsf{1000 \,cm^3} \\ \mathsf{11\frac{1}{4} \textsf{ litres of water}} &= \mathsf{11.25 × 1000 \,cm^3} \\ &= \mathsf{11 \,250 \,cm^3​} \end{align} }$a

$\small{ \begin{align}​​ \textsf{Volume of the full container} &= \textsf{Volume of mango syrup} \\ & \qquad\quad + \\ & \quad\, \textsf{Volume of water} \end{align} }$
 

$\small{ \begin{align}​​ \textsf{Volume of mango syrup} &= \mathsf{\frac{3}{8}} \textsf{ of total Volume}\\ \textsf{Volume of water} &= \mathsf{1-\frac{3}{8}}\\ &=\mathsf{\frac{5}{8}} \textsf{ of total volume}​ \end{align} }$

$\small{ \begin{align}​​ \mathsf{\frac {5}{8} \textsf{ of total volume}} &= \mathsf{11\,250 \,cm^3} \\ \mathsf{\frac {1}{8} \textsf{ of total volume}} &= \mathsf{11\,250 \,cm^3 \div 5}\\ &= \mathsf{2250 \,cm^3} \end{align} }$

$\small{ \begin{align}​​ \mathsf{\frac {3}{8} \textsf{ of total volume}} &= \mathsf{2250 \,cm^3 \times 8}\\ &= \mathsf{18\,000 \,cm^3} \end{align} }$

$\small{ \begin{align}​​ \textsf{Height of the container} &= \mathsf{Volume \div Base Area} \\ &= \mathsf{18\,000 \,cm^3 ÷ 750 \,cm^2} \\ &= \mathsf{24 \,cm} \end{align} }$

Answer:

$\small{\textsf{24 cm}}$

Practice Questions

Question 1:

The shaded face of the cuboid is a square. The length of the cuboid is $\small{\textsf{12 m}}$ and its volume is $\small{\mathsf{1452 \,m^3}}$. Find the length of one side of the square face.

Solution:

$\small{ \begin{align}​​​​ \textsf{Volume of cuboid} &= \textsf{Length × Breadth × Height} \\ &= \textsf{Area of the shaded face × Height} \end{align}}$

$\small{ \begin{align}​​​​ \textsf{Area of the square face} &= \textsf{Volume ÷ Height}\\ &= \mathsf{1452 \,m^3 ÷ 12 \,m}\\ &= \mathsf{121 \,m^2} \end{align}}$

$\small{ \begin{align}​​​​ &\textsf{Since, the shaded face of the cuboid is a square, then }\\ &\textsf{Length} = \textsf{Breadth} \end{align}}$

$\small{ \begin{align}​​​​ \textsf{Area of the square} &= \textsf{Length × Length} \\ \textsf{Length} &= \mathsf{\sqrt{121} m^2} \\ &= \mathsf{11 \,m​} \end{align}}$

Answer:

$\small{\textsf{11 m​}}$

Question 2:

The shaded face of a cuboid is a square. The length of a cuboid is $\small{\textsf{28 cm}}$ and its volume is $\small{\mathsf{1008 \,cm^3}}$. Find the length of one side of the square face.

Solution:

$\small{\begin{align}​​​​​ \textsf{Volume of a cuboid} &=  \mathsf{Length \times Breadth \times Height}\\ &=\mathsf{Length \times \textsf{Area of the shaded square}} \end{align}}$

$\small{\begin{align}​​​​​ \textsf{Area of the square base} &= \mathsf{Volume \div Length} \\ &= \mathsf{1008 \,cm^3 \div 28 \,cm} \\ &= \mathsf{36 \,cm^2} \end{align}}$

$\small{\begin{align}​​​​​ \textsf{Side of the square} &= \mathsf{\sqrt{36} \,cm^2} \\ &= \mathsf{6 \,cm​} \end{align}}$

Answer:

$\small{\textsf{6 cm}}$

Question 3:

Sam filled a rectangular tank with a square base partially filled with water, as shown in the figure below. The volume of the water in the tank is $\small{\mathsf{972 \,cm^3}}$. Find the length of the rectangular tank.

Solution:

$\small{\begin{align}​​​ \textsf{Volume of water} &= \mathsf{972 \,cm^3} \end{align}}$

$\small\bbox[8px,border:2px solid red] { \textbf{Volume of the water} = \textbf{Base area × Height} }$

$\small{\begin{align}​​​ \textsf{Base area of the tank} &= \textsf{Volume ÷ Height} \\ &= \mathsf{972 \,cm^3 \div 12 \,cm} \\ &= \mathsf{81 \,cm^2} \end{align}}$

$\small{\begin{align}​​​ \textsf{Side of the square base} &= \mathsf{\sqrt{81} \,cm^2} \\ &= \mathsf{9 \,cm} \end{align}}$

$\small{\begin{align}​​​ \textsf{Length of the rectangular tank} &= \mathsf{9 \,cm} ​ \end{align}}$

Answer:

$\small{\textsf{9 cm}}$

Question 4:

The area of one of the faces of a cube is $\small{\mathsf{144 \,cm^2}}$. What is the volume of four such cubes?

Solution:

$\small{\begin{align}​​​ \textsf{Side of the cube} &= \mathsf{\sqrt{144} \,cm} \\ &= \mathsf{12 \,cm} \end{align}}$

$\small{\begin{align}​​​ \textsf{Volume of each cube} &= \textsf{Length × Breadth × Height}\\ &= \mathsf{12 \,cm \times 12 \,cm \times 12 \,cm}\\ &= \mathsf{1728 \,cm^3} \end{align}}$

$\small{\begin{align}​​​ \textsf{Volume of four cubes} &= \mathsf{4 \times 1728 \,cm^3} \\   &= \mathsf{6912 \,cm^3} ​​ \end{align}}$

Answer:

$\small{\mathsf{6912 \,cm^3}}$

Summary

In Primary 6 Maths Volume, we need to know the following:

• How to calculate the volume of a cube/cuboid?
• Given the volume of a cube or cuboid, how can we find out the unknown side?
• Given the volume of a cube or cuboid, how to find the area of one of the faces of the cube or cuboid? 
• Given the volume of the cube and the area of one face, how to find the unknown side?
• Given the volume of water in a rectangular/cubical tank, how to find the unknown side?

Remember, practice is the key to perfection.

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