Study P6 Mathematics Volume of Cubes and Cuboid - Geniebook

# 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 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}

$$17 \;cm$$

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}

$$11 \;cm$$

## 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}

$$205 \;cm^2$$

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}

$$211 \;cm^2$$

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}

$$8\;cm$$

## 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}

$$25 \;cm$$

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}

$$24 \;cm$$

## 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}

$$11 \;m​$$

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}

$$6 \;cm$$

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}

$$9\;cm$$

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}

$$6912 \;cm^3$$

## 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

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