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

 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 Math 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 A 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  length \(\times\) breadth \(\times\) height. 

Since, all the sides of the cube are the same, so length \(=\) breadth \(=\) height.

Volume Of A Cube \(=\)  length \(\times\) breadth \(\times\) height. 

Volume Of A Cube \(=\) length³

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

Length Of One Edge Of A Cube \(=\sqrt [3]{Volume}\)

Volume Of A 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.

Volume of a cuboid  \(=\)  length \(\times\) breadth \(\times\) height

\(V = L × B × H \)

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.

\(L = V \div ( B \times H )\)

Similarly, to find the breadth,

\(B = V ÷ (L × H)\)

Lastly, to find the height,

\(H = V ÷ (L × B)\)

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 \(340 \;cm^3\). Find the length of the cuboid.

Solution:

\(\begin{align}​​ ​ \text{Length (L)} &=\;? \\ \text{Breadth (B)} &= 4\;cm \\ \text{Height (H)} & = 5\;cm \\ \text{Volume (V)} &= 340\;cm^3 \\ \\ ​ \text{Length} &= \text{Volume} \div (\text{ Breadth} \times \text{Height }) \\   &= 340 \;cm^3 \div (4 \;cm \times 5 \;cm)\\   &= 340 \;cm^3 \div 20 \;cm^2\\   &= 17 \;cm \end{align}\)

Answer:

\(17 \;cm\)

Question 2:

The volume of a cube is \(1331 \;cm^3\).  Find the length of the edge of the cube.

Solution:

\(\begin{align} \text{Volume (V)} &= 1331\;cm^3 \\ \text{Length (L)} &=\;? \\ \\ \text{Length of the edge of the cube} &= \sqrt [3]{1331} \;cm^3\\ &=11 \;cm \end{align} \)

Answer:

\(11 \;cm\)

Finding The Base Area Of A Cube And A Cuboid

Base Area Of A Cuboid \(=\) Length \(\times\) Breadth 

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

\(Base \;Area = V ÷ H\)

Lastly, to find the height,

\(H = V ÷ Base \;Area\)

Question 1:

A solid has a volume of \(1230 \;cm^3\). If the height of the solid is \(6 \;cm\), find the base area of the solid. 

Solution:

\(\begin{align}​ \text{Volume of solid} &= 1230 \;cm^3 \\ \text{Height of solid} &= 6 \;cm \\ \\ \text{Volume Of Solid} &= \text{Length} × \text{Breadth} × \text{Height}\\ &=\text{Base Area}× \text{Height} \\\\ \text{Base Area Of Solid} &= \text{Volume} ÷ \text{Height} \\ &= 1230 \;cm^3 \div 6\;cm \\ &= 205 \;cm^2 \end{align}\)

Answer:

\(205 \;cm^2\)

Question 2:

A solid has a volume of \(1688 \;cm^3\). If the height of the solid is \(8\;cm\), find the base area of the solid. 

Solution:

\(\begin{align} \text{Volume of solid} &= 1688 \;cm^3 \\ \text{Height of solid} &= 8 \;cm \\ \\ \text{Volume Of Solid} &= \text{Length} × \text{Breadth} × \text{Height}\\ &=\text{Base Area}× \text{Height} \\\\ \text{Base Area Of Solid} &= \text{Volume} ÷ \text{Height} \\ &= 1688 \;cm^3 \div 8\;cm \\ &= 211 \;cm^2 \end{align}\)

Answer:

\(211 \;cm^2\)

Question 3: 

The volume of the cuboid is \(1048 \;cm^3\) and the area of the shaded face is \(131 \;cm^2\). Find the length of the unknown edge of the cuboid.

Solution:

\(\begin{align}​​​ \text{Volume} &= 1048\;cm^3 \\ \text{Shaded area} &= 131\;cm^2 \\ \\ \textbf{Volume} &=  \textbf{Shaded Area} \times \textbf{Length} \\ \\ ​ \text{Length Of Solid}  &= \text{Volume} \div \text{Shaded Area} \\    &= 1048 \;cm^3 \div 131 \;cm^2 \\    &= 8 \;cm \end{align}\)

Answer:

\(8\;cm\)

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 \(12.5 \;l\) of water. If the base area of the tank is \(500 \;cm^2\), what is the height of the water level of the tank ? \((1 \;l = 1000 \;cm^3)\)

Solution:

\(\begin{align}​​ 1\;l &= 1000 \;cm^3 \\ 12.5 \;l  &= 12.5 \times 1000 \;cm^3\\ &= 12 \,500 \;cm^3 \\ \\ ​ \text{Volume of water in the tank} &= 12\,500 \;cm^3 \\ \text{Base area of the tank} &= 500 \;cm^2 \\ \\ \textbf{Volume} &= \textbf{Base Area} \times \textbf{Height} \\ \\ \text{Height of water in the tank} &= 12 \,500 \;cm^3 \div 500 \;cm^2 \\ &= 25 \;cm ​ \end{align}\)

Answer:

\(25 \;cm\)

Question 2:

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

Solution:

\(\begin{align}​​ ​1 \;litre &= 1000 \;cm^3 \\ 11\frac{1}{4} \;litres \text{ of water} &= 11.25 × 1000 \;cm^3 \\ &= 11 \,250 \;cm^3​ \\ \\ \text{Volume of the full container} &= \text{Volume of mango syrup} + \text{Volume of water}\\ \text{Volume of mango syrup} &= \frac{3}{8} \text{ of total Volume}\\ \text{Volume of water} &= 1-\frac{3}{8}\\ &=\frac{5}{8} \text{ of total volume}​ \\ \\ \frac {5}{8} \;\text{ of total volume } &= 11\,250 \;cm^3 \\ \frac {1}{8} \;\text{ of total volume } &= 11\,250 \;cm^3 \div 5\\ &= 2250 \;cm^3 \\ \\ \frac {3}{8} \;\text{ of total volume } &= 2250 \;cm^3 \times 8\\ &= 18\,000 \;cm^3 \\ \\ \text{Height Of The Container} &= \text{Volume} \div \text{Base Area} \\ &= 18\,000 \;cm^3 ÷ 750 \;cm^2 \\ &= 24 \;cm \end{align} \)

Answer:

\(24 \;cm\)

Practice Questions

Question 1:

The shaded face of the cuboid is a square. The length of the cuboid is \(12 \;m\) and its volume is \(1452 \;m^3\). Find the length of one side of the square face.

Solution:

\(\begin{align}​​​​ \text{Volume of cuboid} &= \text{Length} \times \text{Breadth} \times \text{Height} \\ \text{Volume of cuboid} &= \text{Area of the shaded face} \times \text{Height} \\ \\ ​\text{Area of the square face} &= \text{Volume} \div \text{Height}\\ &= 1452 \;m^3 ÷ 12 \;m\\ &= 121 \;m^2 \\ \\ \text{Since, the shaded face of } &\text{the cuboid is a square, then }\\ \text{Length} &= \text{Breadth.}\\\\ ​ \text{Area of the square} &= \text{Length} \times \text{length} \\ \text{Length} &= \sqrt{121} \;m^2 \\ &= 11 \;m​ ​ \end{align}\)

Answer:

\(11 \;m​ \)

Question 2:

The shaded face of a cuboid is a square. The length of a cuboid is \(28 \;cm\) and its volume is \(1008 \;cm^3\). Find the length of one side of the square face.

Solution:

\(\begin{align}​​​​​ \text{Volume of a cuboid} &=  \text{Length} \times \text{Breadth} \times \text{Height}\\ &=\text{Length} \times \text{Area of the shaded square} \\ \\ \text{Area of the square base} &= \text{Volume} \div \text{Length} \\ &= 1008 \;cm^3 \div 28 \;cm \\ &= 36 cm² \\ \\ \text{Side of the square} &= \sqrt{36} \;cm^2 \\ &= 6 \;cm​ \end{align}\)

Answer:

\(6 \;cm\)

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 \(972 \;cm^3\). Find the length of the rectangular tank.

Solution:

\(\begin{align}​​​ \text{Volume of water} &= 972 \;cm^3 \\ \\ \textbf{Volume of the water} &= \textbf{Base area} \times \textbf{Height}\\ \\ \text{Base area of the tank} &= \text{Volume} \div \text{height} \\ &= 972 \;cm^3 ÷ 12 \;cm \\ &= 81 \;cm^2 \\ \\ \text{Side of the square base} &= \sqrt{81} \;cm^2 \\ &= 9 \;cm \\ \\ \text{Length of the rectangular tank} &= 9 \;cm ​​ \end{align}\)

Answer:

\(9\;cm\)

Question 4:

The area of one of the faces of a cube is \(144 \;cm^2\). What is the volume of four such cubes ?

Solution:

\(\begin{align}​​​ \text{​Side of the cube} &= \sqrt{144} \;cm \\ &= 12 \;cm \\\\ \text{​Volume of each cube} &= \text{​Length} \times \text{​Breadth} \times \text{​Height}\\ &= 12 \;cm \times 12 \;cm \times 12 \;cm\\ &= 1728 \;cm^3 \\\\ \text{​Volume of four cubes} &= 4 \times 1728 \;cm^3 \\   &= 6912 \;cm^3 ​​\end{align}\)

Answer:

\(6912 \;cm^3\)

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. !