# 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:

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}

$\small\textsf{17 cm}$

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}

$\small\textsf{11 cm}$

## 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}

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

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}

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

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}}

$\small\textsf{8 cm}$

## 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}}

$\small\text{25 cm}$

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} }

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

### 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}}

$\small{\textsf{11 m​}}$

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}}

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

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}}

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

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}}

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

## 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.

Continue Learning
Algebra Distance, Speed and Time
Volume of Cubes and Cuboid Fundamentals Of Pie Chart
Finding Unknown Angles Number Patterns: Grouping & Common Difference
Fractions Of Remainder Fractions - Division
Ratio Repeated Identity: Ratio Strategies

## Test Yourself

QUESTION 1/3

The base of a fish tank measures 32 cm by 28 cm. Its volume is 10880 cm³ when it is 5/7 full of water. Find the height of the fish tank.

A.

17 cm

B.

19 cm

C.

20 cm

D.

21 cm

Explanation

Ans: (1) 17 cm

QUESTION 2/3

Peter wants to fill up his rectangular water tank measuring 150 cm by 30 cm by 40 cm to its brim. 5 litres of water flows out from Tap A per minute and 4 litres of water flows out from Tap B per minute. If both taps are turned on at the same time how long does it take to fill up the whole tank?

A.

9 minutes

B.

20 minutes

C.

36 minutes

D.

45 minutes

Explanation

Ans: (2) 20 min

QUESTION 3/3

Two cubes are glued together to form a solid as shown in the figure below. All the faces of the solid are painted. The total surface area painted is 90 cm². Find the volume of one cube.

A.

18 cm³

B.

27 cm³

C.

64 cm³

D.

81 cm³

Explanation

Ans: (2) 27 cm³

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