Calculating the volume of prisms
🎯 In this topic you will
- Derive the formula for the volume of prisms.
- Derive the formula for the volume of cylinders.
- Apply the formulae to calculate volumes of different prisms and cylinders.
You already know that a prism is a 3D shape that has the same cross-section along its length. Here are some examples of prisms. The cross-section of each shape is shaded.
You can work out the volume of a prism using the formula:
❓ EXERCISES
In this exercise when you need to use $\pi$, use the $\pi$ button on your calculator.
🧠 Reasoning Tip
Remember to use the formula: volume = area of cross-section $\times$ length.
1. Work out the volume of each prism.
a.
b.
c.
👀 Show answer
a.$V = A \times l = 15 \times 8 = 120\,\text{cm}^3$.
b.$V = 20 \times 6.5 = 130\,\text{cm}^3$.
c.$V = 32 \times 4.2 = 134.4\,\text{cm}^3$.
❓ EXERCISES
2. Copy and complete this table.
🧠 Reasoning Tip
In parts b and c, you will need to use inverse operations.
| Area of cross-section | Length of prism | Volume of prism |
|---|---|---|
| $12\,\text{cm}^2$ | $10\,\text{cm}$ | $\square\,\text{cm}^3$ |
| $24\,\text{cm}^2$ | $\square\,\text{cm}$ | $204\,\text{cm}^3$ |
| $\square\,\text{m}^2$ | $6.2\,\text{m}$ | $114.7\,\text{m}^3$ |
👀 Show answer
a. $12 \times 10 = 120\,\text{cm}^3$
b. $204 \div 24 = 8.5\,\text{cm}$
c. $114.7 \div 6.2 = 18.5\,\text{m}^2$
3. This is part of Yusaf’s homework.
a. Explain the mistakes Yusaf has made.
b. Write out the correct solution.
👀 Show answer
a. Yusaf calculated the area of the wrong face (a rectangle instead of the triangular cross-section). He also used the wrong dimension for the prism’s length.
b. Correct solution:
Area of triangle = $\tfrac{1}{2} \times 8 \times 5 = 20\,\text{cm}^2$
Volume = $20 \times 20 = 400\,\text{cm}^3$
🧠 Think like a Mathematician
4. Work on your own to answer these questions. The diagram shows a cylinder.
👀 Show Answer
- a: Strictly, a cylinder is not a prism in the usual school definition because prisms have polygonal cross-sections, whereas a cylinder has a circular cross-section. However, it behaves like a “circular prism” because the cross-section is constant along the height.
- b: Find the area of the circular base and multiply by the height: area of circle $=\pi r^{2}$, then volume $=$ base area $\times h$.
- c:$V=\pi r^{2}h$.
- d: The method uses the general prism rule $V=\text{area of cross-section}\times \text{length}$; for a cylinder the cross-section is a circle with area $\pi r^{2}$, so multiplying by $h$ gives $V=\pi r^{2}h$, which matches both geometric reasoning and dimensional units $(\text{length})^{3}$.
❓ EXERCISES
5. This is part of Sara’s homework.
Sara has got the answer wrong.
Explain the mistake Sara has made and work out the correct answer.
👀 Show answer
6. Work out the volume of each cylinder.
Give your answers correct to one decimal place (1 d.p.).
a.
b.
c.
👀 Show answer
a. $V = \pi r^2 h = \pi \times 5^2 \times 12 = \pi \times 25 \times 12 = 942.5\,\text{cm}^3$ (1 d.p.)
b. $V = \pi r^2 h = \pi \times 2.5^2 \times 18 = \pi \times 6.25 \times 18 = 353.4\,\text{cm}^3$ (1 d.p.)
c. Convert to $\text{cm}$: $r = 2.0\,\text{cm}$, $h = 1.4\,\text{cm}$ $V = \pi \times 2^2 \times 1.4 = \pi \times 5.6 = 17.6\,\text{cm}^3$ (1 d.p.)
❓ EXERCISES
7. Copy and complete this table. Give your answers correct to two decimal places (2 d.p.).
| Radius of circle | Area of circle | Height of cylinder | Volume of cylinder |
|---|---|---|---|
| a. $2.50\,\text{m}$ | $\square\,\text{m}^2$ | $4.20\,\text{m}$ | $\square\,\text{m}^3$ |
| b. $6.00\,\text{cm}$ | $\square\,\text{cm}^2$ | $\square\,\text{cm}$ | $507.00\,\text{cm}^3$ |
| c. $\square\,\text{m}$ | $20.00\,\text{m}^2$ | $2.50\,\text{m}$ | $\square\,\text{m}^3$ |
| d. $\square\,\text{mm}$ | $\square\,\text{mm}^2$ | $16.00\,\text{mm}$ | $1044.00\,\text{mm}^3$ |
👀 Show answer
Use $A=\pi r^{2}$ and $V=A\times h$ (keep units consistent).
- a. $A=\pi(2.50)^2=19.63\,\text{m}^2$, $V=19.63\times 4.20=82.47\,\text{m}^3$.
- b. $A=\pi(6.00)^2=113.10\,\text{cm}^2$, $h=\dfrac{507.00}{113.10}=4.48\,\text{cm}$.
- c. $r=\sqrt{\dfrac{20.00}{\pi}}=2.52\,\text{m}$, $V=20.00\times 2.50=50.00\,\text{m}^3$.
- d. $A=\dfrac{1044.00}{16.00}=65.25\,\text{mm}^2$, $r=\sqrt{\dfrac{65.25}{\pi}}=4.56\,\text{mm}$.
(All values rounded to $2$ d.p.)
8. Each of these prisms has a volume of $256\,\text{cm}^3$. Work out the length marked $x$ in each diagram. Give your answers correct to one decimal place (1 d.p.).
a.
b.
c.
👀 Show answer
a. Rectangular cross-section: $A=3.8\times x$. $256 = (3.8x)\times 12.3 \Rightarrow x=\dfrac{256}{3.8\times 12.3}=5.5\,\text{cm}$ (1 d.p.).
b. Triangular cross-section: $A=\tfrac{1}{2}\times 9.8 \times x$. $256 = \big(\tfrac{1}{2}\cdot 9.8\cdot x\big)\times 12.4 \Rightarrow x=\dfrac{256}{0.5\cdot 9.8\cdot 12.4}=4.2\,\text{cm}$ (1 d.p.).
c. Cylinder with diameter $x$ ($r=\tfrac{x}{2}$): $256=\pi\left(\tfrac{x}{2}\right)^2\cdot 18.2 \Rightarrow x=\sqrt{\dfrac{256\cdot 4}{\pi\cdot 18.2}}=4.2\,\text{cm}$ (1 d.p.).
🧠 Think like a Mathematician
9. Work on your own to answer this question.
The diagram shows an empty cylindrical container.
Ana puts a solid cube of side length 8 cm into the container. She then pours 1.5 litres of water into the container.
Question: Will the water come over the top of the container?
Explain your answer and show all your working.
Tip: $1 \,\text{cm}^3 = 1 \,\text{mL}$
👀 Show Answer
- Step 1: Volume of container
Radius $r = 6 \,\text{cm}$, height $h = 18 \,\text{cm}$.
$V = \pi r^2 h = \pi \times 6^2 \times 18 = \pi \times 648 \approx 2036.0 \,\text{cm}^3$. - Step 2: Volume of cube
Side = $8 \,\text{cm}$.
$V = 8^3 = 512 \,\text{cm}^3$. - Step 3: Remaining capacity
$2036.0 - 512 = 1524.0 \,\text{cm}^3$. - Step 4: Water poured
$1.5 \,\text{litres} = 1500 \,\text{cm}^3$. - Step 5: Compare
Space available = $1524.0 \,\text{cm}^3$.
Water poured = $1500 \,\text{cm}^3$.
Since $1500 < 1524$, the water fits inside without overflowing. - Final Answer: The water will not come over the top. The container can just hold it with $24 \,\text{cm}^3$ of spare space.