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:
- find the length of the cube/cuboid.
- find the base area of the cube/cuboid.
- 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 \(=\) length^{3}
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. !
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