Calculating the volume of triangular prisms
🎯 In this topic you will
- Derive and use the formula for the volume of a triangular prism.
🧠 Key Words
- cross-section
- prism
Show Definitions
- cross-section: The flat shape you see when a solid is cut straight through, usually perpendicular to its length.
- prism: A solid 3D shape with two identical, parallel ends (called bases) and flat sides that connect them.
📦 Volume of a Cuboid
You already know how to work out the volume of a cuboid by multiplying the length ($l$) by the base ($b$) by the height ($h$).
You can multiply the dimensions in any order.
🔺 Prisms and Cross-Sections
A cuboid is a rectangular prism.
A prism is a 3D shape which has the same 2D shape throughout its length.
This 2D shape is called the cross-section of the prism.
📐 Volume Using Cross-Sections
When you work out the volume of a cuboid, if you start with $b \times h$, you find the area of the rectangular cross-section of the cuboid.
When you multiply this area by the length $l$, you get the volume of the cuboid.
🔻 Triangular Prism
The diagram shows a triangular prism.
You can see that the cross-section of the prism is a triangle.
You can work out the volume of the prism using the formula:
volume = area of cross-section $\times$ length
❓ EXERCISES
1. Copy and complete the workings to find the volume of each triangular prism.
a.
👀 Show answer
Volume = area of cross-section $\times$ length = $24 \times 10 = 240 \,\text{cm}^3$
b.
👀 Show answer
Volume = area of cross-section $\times$ length = $6 \times 7 = 42 \,\text{m}^3$
2. Work out the volume of each triangular prism.
a.
👀 Show answer
Volume = area of cross-section $\times$ length = $54 \times 30 = 1620 \,\text{cm}^3$
b.
👀 Show answer
Volume = area of cross-section $\times$ length = $6 \times 7 = 42 \,\text{m}^3$
🧠 Think like a Mathematician
3. Yari and Mike use different methods to work out the volume of this triangular prism. This is what they write:
Follow-up Questions:
a. How does Mike’s method work? Why does it give the same answer as Yari’s method?
👀 Show Answer
- a: Mike’s method works by calculating the volume of the full cuboid that would enclose the triangular prism, then dividing it by 2. This works because the triangular prism takes up exactly half the volume of the cuboid.
- b: Answers may vary. Some may prefer Yari’s method because it directly uses the triangle’s area, while others may prefer Mike’s method as it is similar to the standard cuboid formula.
- c: Another method is to calculate the volume by considering stacking triangular layers of area $\tfrac{1}{2} \times b \times h$, repeated through the prism’s length. This leads back to the same formula.
❓ EXERCISES
4. This is part of Vin’s homework.
Work out the volume of this triangular prism.
Vin has got the answer wrong. Explain Vin’s mistake and work out the correct answer.
👀 Show answer
Convert $120\,\text{mm}$ to centimetres: $120\,\text{mm}=12\,\text{cm}$.
Cross-section area: $A=\tfrac{1}{2}bh=\tfrac{1}{2}\times 8 \times 7=28\,\text{cm}^2$.
Volume: $V=A\times \text{length}=28\times 12=336\,\text{cm}^3$.
🧠 Reasoning Tip
Make sure the length, width (base) and height are all in the same units before you work out the volume.
5. The table shows the base, perpendicular height and length of four triangular prisms. Copy and complete the table.
| Row | Base | Height | Length | Volume |
|---|---|---|---|---|
| a. | $4\,\text{cm}$ | $8\,\text{cm}$ | $5\,\text{mm}$ | $\square\,\text{cm}^3$ |
| b. | $2\,\text{cm}$ | $15\,\text{mm}$ | $8\,\text{mm}$ | $\square\,\text{mm}^3$ |
| c. | $7\,\text{m}$ | $9\,\text{m}$ | $10\,\text{cm}$ | $\square\,\text{m}^3$ |
| d. | $30\,\text{mm}$ | $6\,\text{cm}$ | $200\,\text{mm}$ | $\square\,\text{cm}^3$ |
👀 Show answer
Use $V=\text{area of cross-section} \times \text{length}$ with triangle area $A=\tfrac{1}{2}bh$, making units consistent first.
| Row | Working (units aligned) | Volume |
|---|---|---|
| a. | Convert length: $5\,\text{mm}=0.5\,\text{cm}$. $A=\tfrac{1}{2}\times 4 \times 8=16\,\text{cm}^2$, $V=16\times 0.5=8\,\text{cm}^3$. |
$8\,\text{cm}^3$ |
| b. | Convert base: $2\,\text{cm}=20\,\text{mm}$. $A=\tfrac{1}{2}\times 20 \times 15=150\,\text{mm}^2$, $V=150\times 8=1200\,\text{mm}^3$. |
$1200\,\text{mm}^3$ |
| c. | Convert length: $10\,\text{cm}=0.10\,\text{m}$. $A=\tfrac{1}{2}\times 7 \times 9=31.5\,\text{m}^2$, $V=31.5\times 0.10=3.15\,\text{m}^3$. |
$3.15\,\text{m}^3$ |
| d. | Convert: $30\,\text{mm}=3\,\text{cm}$, $200\,\text{mm}=20\,\text{cm}$. $A=\tfrac{1}{2}\times 3 \times 6=9\,\text{cm}^2$, $V=9\times 20=180\,\text{cm}^3$. |
$180\,\text{cm}^3$ |
🧠 Think like a Mathematician
6. The diagram shows a compound prism. The compound prism is made of a triangular prism and a cuboid.
👀 Show Answer
- a: Split the solid into a cuboid and a triangular prism with the same length $20\,\text{cm}$ and base width $8\,\text{cm}$.
Cuboid volume: $V_{\text{cuboid}}=8 \times 7 \times 20 = 1120\,\text{cm}^3$.
Triangular prism volume: cross-section area $=\tfrac{1}{2}\times 8 \times 10 = 40\,\text{cm}^2$, so $V_{\triangle}=40 \times 20 = 800\,\text{cm}^3$.
Total: $1120+800=1920\,\text{cm}^3$. - b: I decomposed the compound prism into two simpler prisms, computed each volume using $V=\text{area of cross-section} \times \text{length}$, and added the results.
- c: The easiest method is usually to decompose the shape into familiar prisms (cuboids/triangular prisms), because formulas are straightforward and you avoid working with irregular cross-sections. Always check units and that the chosen parts exactly cover the solid without overlap.
❓ EXERCISES
7. Work out the volume of each compound prism.
a.
b.
👀 Show answer
a. Split the cross-section into a rectangle and a right triangle (all in $\text{m}$):
Rectangle area: $A_r = 2.5 \times 3 = 7.5\,\text{m}^2$
Triangle area: $A_t = \tfrac{1}{2}\times 4 \times 3 = 6\,\text{m}^2$
Cross-section area: $A = 7.5 + 6 = 13.5\,\text{m}^2$
Volume: $V = A \times \text{length} = 13.5 \times 6 = 81\,\text{m}^3$.
b. Use consistent units ($\text{mm}$): rectangle plus roof triangle.
Rectangle area: $A_r = 15 \times 7 = 105\,\text{mm}^2$
Triangle area: $A_t = \tfrac{1}{2}\times 15 \times 8 = 60\,\text{mm}^2$
Cross-section area: $A = 105 + 60 = 165\,\text{mm}^2$
Volume: $V = 165 \times 12 = 1980\,\text{mm}^3$.
❓ EXERCISES
🧠 Reasoning Tip
Choose your own values for the base and height of the triangle that will give the area you found in part a.
8. The diagram shows a triangular prism. The volume of the prism is $96\,\text{cm}^3$.
a. Work out the area of the shaded triangle.
b. Copy and complete these possible dimensions for the shaded triangle:
Option 1: base = $\square$ cm and height = $\square$ cm
Option 2: base = $\square$ cm and height = $\square$ cm
c. Compare your answers to part b with those of a partner. Did you have the same base and height measurements, or were they different? Discuss the number of different possible combinations.
👀 Show answer
a. Let the prism length be $8\,\text{cm}$. Volume $V=\text{area}\times \text{length}$, so cross-section area $A=V/\text{length}=96/8=12\,\text{cm}^2$.
b. For a triangle, $A=\tfrac{1}{2}bh$ so $\tfrac{1}{2}bh=12 \Rightarrow bh=24$. Examples:
Option 1: $b=4\,\text{cm}$, $h=6\,\text{cm}$ (since $\tfrac{1}{2}\times 4 \times 6=12$)
Option 2: $b=3\,\text{cm}$, $h=8\,\text{cm}$ (since $\tfrac{1}{2}\times 3 \times 8=12$)
c. Many different pairs satisfy $bh=24$, so answers can differ. Any positive pair that makes $\tfrac{1}{2}bh=12\,\text{cm}^2$ is valid.
9. The diagram shows a triangular prism. The volume of the prism is $168\,\text{m}^3$.
a. Work out the height of the triangle.
b. Compare the method you used to answer part a with other learners in the class. Which method do you think is best to use to answer this type of question? Explain why.
👀 Show answer
a. Use $V=\tfrac{1}{2}bh \times \ell$ with base $b=4\,\text{m}$ and length $\ell=12\,\text{m}$: $168=\tfrac{1}{2}\times 4 \times h \times 12 = 24h$, so $h=168/24=7\,\text{m}$.
b. A clear method is to rearrange $V=\tfrac{1}{2}bh\ell$ to $h=\dfrac{2V}{b\ell}$. It directly substitutes known values and avoids extra steps or rounding.
10. A triangular prism has a base of $10\,\text{cm}$, a height of $6\,\text{cm}$ and a length of $15\,\text{cm}$.
a. Work out the volume of the triangular prism.
b. Work out the dimensions of three other triangular prisms with the same volume.
👀 Show answer
a.$V=\tfrac{1}{2}\times 10 \times 6 \times 15=450\,\text{cm}^3$.
b. Any sets with $\tfrac{1}{2}bhl=450$ (so $bhl=900$) work. Examples (all in $\text{cm}$):
• $b=12$, $h=5$, $l=15$ → $\tfrac{1}{2}\times 12 \times 5 \times 15=450$
• $b=9$, $h=10$, $l=10$ → $\tfrac{1}{2}\times 9 \times 10 \times 10=450$
• $b=5$, $h=6$, $l=30$ → $\tfrac{1}{2}\times 5 \times 6 \times 30=450$
❓ EXERCISES
11. The diagram shows a triangular prism made from silver. Jan is going to melt the prism and make the silver into cubes. The side length of each cube is $8\,\text{mm}$. Jan thinks he can make nine cubes from this prism. Is Jan correct? Explain your answer. Show all your working.
👀 Show answer
Volume of the prism
Cross-section (triangle) area: $A=\tfrac{1}{2}bh=\tfrac{1}{2}\times 25 \times 12=150\,\text{mm}^2$.
Prism length: $30\,\text{mm}$ ⇒ $V=A\ell=150\times 30=4500\,\text{mm}^3$.
Volume of one cube
Side $=8\,\text{mm}$ ⇒ $V_{\text{cube}}=8^3=512\,\text{mm}^3$.
How many cubes?
$\dfrac{4500}{512}\approx 8.79$ so at most $8$ whole cubes.
Check: $8\times 512=4096\,\text{mm}^3$; leftover $=4500-4096=404\,\text{mm}^3$ (not enough for another cube).
To make $9$ cubes would need $9\times 512=4608\,\text{mm}^3$, which is $108\,\text{mm}^3$ more than available.
Conclusion: Jan is not correct — he can make $8$ cubes, with some silver left over.