Mathematics • Year 9 • Unit 4 • Lesson 6
Volume — Mixed Challenge
Pull together everything from Lesson 6: prism and cylinder volume, composite solids (early taste of L7), unit conversion, and rearranging formulas. You'll have to choose the right tool for each problem, spot a mistake in someone else's working, and tackle an open-ended challenge.
1. Mixed problems — choose the right rule
Each question uses a different combination of ideas from Lesson 6. Decide which formula to use before you start writing. Show your working. 3 marks each
1.1 A trapezoidal prism has a cross-section that is a trapezium with parallel sides $5$ cm and $9$ cm, and perpendicular height $4$ cm. The prism is $10$ cm long. Find its volume. (Area of trapezium $= \tfrac{1}{2}(a+b)h$.)
1.2 A solid is a cube of side $5$ cm with a cylindrical hole of radius $1$ cm drilled straight through it. Find the remaining volume, in terms of $\pi$.
1.3 A cylindrical hot-water cylinder holds exactly $250$ litres of water. The cylinder has radius $30$ cm. Find its height in cm, to the nearest cm. (1 L = 1000 cm³.)
1.4 A house-shaped solid has a rectangular prism base $6$ m × $4$ m × $3$ m with a triangular prism roof on top (the triangle has base $6$ m, perpendicular height $2$ m, and the same $4$ m length). Find the total volume.
1.5 If a cube of side $s$ has volume $1000$ cm³, find the side length $s$, and explain why doubling $s$ would make the volume $8000$ cm³, not $2000$ cm³.
1.6 A rectangular swimming pool $10$ m × $5$ m × $2$ m loses water through evaporation at $4$ mm per day during summer. How many litres does it lose per day? (Hint: model the lost water as a thin rectangular prism.)
2. Find the mistake
Another student has tried to find the capacity (in litres) of a cylindrical water tank with radius $2$ m and height $3$ m. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — capacity of cylinder $r = 2$ m, $h = 3$ m, in litres:
Line 1: $V = \pi r^2 h$
Line 2: $V = \pi \times 2^2 \times 3 = 12\pi \approx 37.7$ m³
Line 3: Convert to litres: $1$ m³ $= 100$ L
Line 4: Capacity $= 37.7 \times 100 \approx 3770$ L
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the corrected final answer.
Stuck? Revisit lesson § "Common Misconceptions" — the second listed misconception is exactly this trap. Check the lesson's stated conversion: $1$ m³ $= ?$ L.3. Open-ended challenge — design two tanks
This question has many valid answers. 4 marks
3.1 Design two different cylindrical water tanks, each holding very close to $1000$ litres ($1$ m³). The two tanks must have different radii (and therefore different heights). For each tank you design:
(i) State the radius and height (use whole-number cm or sensible decimals like $0.5$ m).
(ii) Calculate the volume in m³ and convert to litres.
(iii) Comment briefly on which tank would be easier to install in a narrow space.
Bonus: "Acceptable" means within $\pm 50$ litres of $1000$ L. Your two tanks must both pass this check.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Trapezoidal prism
Area of cross-section $= \tfrac{1}{2}(5 + 9) \times 4 = \tfrac{1}{2} \times 14 \times 4 = 28$ cm².
$V = 28 \times 10 = \mathbf{280}$ cm³.
1.2 — Cube with cylindrical hole
$V_{\text{cube}} = 5^3 = 125$ cm³.
Hole length = 5 cm (through the cube). $V_{\text{hole}} = \pi \times 1^2 \times 5 = 5\pi$ cm³.
Remaining $= \mathbf{125 - 5\pi}$ cm³ $\approx 109.3$ cm³.
1.3 — Hot-water cylinder
$V = 250$ L $= 250\,000$ cm³.
Rearrange $V = \pi r^2 h$: $h = \dfrac{V}{\pi r^2} = \dfrac{250\,000}{\pi \times 900} = \dfrac{250\,000}{2827.4} \approx 88.4$ cm.
Rounded: $\mathbf{88}$ cm tall.
1.4 — House-shaped solid
Rectangular part: $V_1 = 6 \times 4 \times 3 = 72$ m³.
Triangular roof: $A = \tfrac{1}{2} \times 6 \times 2 = 6$ m², so $V_2 = 6 \times 4 = 24$ m³.
Total $V = 72 + 24 = \mathbf{96}$ m³.
This is exactly the "decompose into parts" strategy from the lesson, applied early.
1.5 — Cube and the cube of the scale factor
$s^3 = 1000$, so $s = \sqrt[3]{1000} = \mathbf{10}$ cm.
If $s$ doubles to $20$ cm, then $V = 20^3 = 8000$ cm³ — that's $\mathbf{8}\times$ the original, not $2\times$. The reason: volume scales as the cube of the linear factor, so doubling each of the three dimensions multiplies the volume by $2 \times 2 \times 2 = 8$. (This idea is formalised in Lesson 8.)
1.6 — Pool evaporation
Depth lost $= 4$ mm $= 0.004$ m.
$V_{\text{lost}} = 10 \times 5 \times 0.004 = 0.2$ m³ $= \mathbf{200}$ L per day.
That's a lot — most outdoor pools in Sydney lose around 150-300 L per summer day, mostly to evaporation.
2 — Find the mistake
(a) The mistake is on Line 3.
(b) The conversion is wrong: $1$ m³ $= \mathbf{1000}$ L, not $100$ L. This is the second misconception listed in the lesson — mixing up the conversion factor between cubic metres and litres.
(c) Corrected working:
$V = \pi \times 2^2 \times 3 = 12\pi \approx 37.7$ m³
Convert: $1$ m³ $= 1000$ L
Capacity $= 37.7 \times 1000 \approx \mathbf{37\,700}$ L (or exactly $12\,000\pi$ L).
That's a factor-of-ten error — the kind that costs civil engineers (and Year 9 students) marks.
3 — Open-ended challenge (sample solutions)
We need $\pi r^2 h \approx 1$ m³, so $h \approx \dfrac{1}{\pi r^2}$. The target window is $0.95$ m³ to $1.05$ m³.
Tank A — tall and narrow: $r = 0.4$ m, $h = 2.0$ m.
$V = \pi \times 0.16 \times 2.0 \approx 1.005$ m³ $\approx \mathbf{1005}$ L ✓
Fits in a narrow corner — about 80 cm wide.
Tank B — short and wide: $r = 0.6$ m, $h = 0.9$ m.
$V = \pi \times 0.36 \times 0.9 \approx 1.018$ m³ $\approx \mathbf{1018}$ L ✓
Much shorter — fits under a low eave but takes up $1.2$ m of floor.
Comment: Tank A is better when space is narrow (e.g. tucked beside a house). Tank B is better when there is a height limit.
Marking: 2 marks per tank (1 for sensible $r$ and $h$ chosen, 1 for showing the calculation lands inside $\pm 50$ L of $1000$ L). Other valid answers: $r=0.5, h\approx1.27$; $r=0.7, h\approx0.65$; many more.