Mathematics • Year 9 • Unit 4 • Lesson 6
Volume in the Real World
Use $V = A_{\text{cross}} \times h$ and $V = \pi r^2 h$ in everyday contexts: a backyard pool, a concrete pour, a soft drink can, a school water tank and a packed shipping container. Then explain your method in your own words.
1. Word problems
Each problem uses a volume formula from Lesson 6 and almost always needs a unit conversion. Show your working — a single final answer with no working only earns half marks.
1.1 — Backyard pool. A rectangular backyard pool is 8 m long, 4 m wide and 1.5 m deep.
(a) Find its volume in m³.
(b) Convert to litres (1 m³ = 1000 L).
(c) If the local water tariff is $\$0.0028$ per litre, what would it cost to fill the empty pool, to the nearest dollar? 4 marks
1.2 — Concrete slab pour. A builder needs to pour a concrete slab 6 m long, 4 m wide and 12 cm thick. Concrete is sold by the cubic metre.
(a) Convert the thickness to metres before substituting.
(b) Calculate the volume of concrete needed in m³.
(c) If concrete costs $\$240$ per m³, find the total cost. 4 marks
1.3 — Soft drink can. A 375 mL soft drink can is a cylinder with diameter 6.5 cm. (1 mL = 1 cm³.)
(a) Find the radius of the can.
(b) Use $V = \pi r^2 h$ to find the height of the can (to 1 decimal place). 3 marks
1.4 — School rainwater tank. A school installs a cylindrical rainwater tank of radius 1.2 m and height 2.5 m. The tank manufacturer says it holds "about 11 000 litres".
(a) Calculate the volume in m³ to 2 decimal places.
(b) Convert to litres and decide whether the manufacturer's claim is reasonable. Justify your answer in one sentence. 3 marks
1.5 — Shipping container packing. A standard 20-foot shipping container has internal dimensions $5.9$ m × $2.35$ m × $2.39$ m. A logistics company wants to fill it with boxes that are $30$ cm × $30$ cm × $30$ cm.
(a) Find the volume of the container in m³.
(b) Find the volume of one box in m³.
(c) Assuming perfect packing (no gaps), what is the maximum number of whole boxes that fit? 3 marks
2. Explain your thinking
This question is about communication, not just answers. Use full sentences. 4 marks
2.1 A classmate is given a cylinder with diameter 8 cm and height 5 cm and writes $V = \pi \times 8^2 \times 5 = 320\pi$ cm³. In your own words, explain (i) what mistake they have made, (ii) which key word in the question they ignored, and (iii) what the correct volume is. Refer to the formula $V = \pi r^2 h$ somewhere in your explanation.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Backyard pool
(a) $V = 8 \times 4 \times 1.5 = \mathbf{48}$ m³.
(b) $48 \times 1000 = \mathbf{48\,000}$ L.
(c) Cost $= 48\,000 \times 0.0028 = \$134.40 \approx \mathbf{\$134}$.
Real-world warning: that is just the water cost. Heating the pool would cost much more.
1.2 — Concrete slab pour
(a) Thickness $= 12$ cm $= \mathbf{0.12}$ m.
(b) $V = 6 \times 4 \times 0.12 = \mathbf{2.88}$ m³.
(c) Cost $= 2.88 \times 240 = \mathbf{\$691.20}$.
If you forgot to convert and used 12 instead of 0.12, you would get 288 m³ — a hundred times too much, costing $\$69\,120$. Always convert units before substituting.
1.3 — Soft drink can
(a) $r = 6.5 \div 2 = \mathbf{3.25}$ cm.
(b) Rearrange: $h = \dfrac{V}{\pi r^2} = \dfrac{375}{\pi \times (3.25)^2} = \dfrac{375}{\pi \times 10.5625} \approx \dfrac{375}{33.18} \approx \mathbf{11.3}$ cm.
A real 375 mL can is about 11.5 cm tall — our answer matches.
1.4 — School rainwater tank
(a) $V = \pi \times (1.2)^2 \times 2.5 = \pi \times 1.44 \times 2.5 = 3.6\pi \approx \mathbf{11.31}$ m³.
(b) $11.31 \times 1000 = \mathbf{11\,310}$ L. The manufacturer's claim of "about 11 000 L" is reasonable — they have rounded down slightly, which is honest practice for tank capacity (you cannot fill right to the rim).
1.5 — Shipping container packing
(a) $V_{\text{container}} = 5.9 \times 2.35 \times 2.39 \approx \mathbf{33.13}$ m³.
(b) Box side $= 0.3$ m, so $V_{\text{box}} = 0.3^3 = \mathbf{0.027}$ m³.
(c) Max boxes $= 33.13 \div 0.027 \approx 1227$, so round DOWN to $\mathbf{1227}$ whole boxes.
In reality, walls of boxes don't divide evenly into the dimensions, so the real number would be lower (about $19 \times 7 \times 7 = 931$ boxes).
2.1 — Explain your thinking (sample response)
My classmate has used the diameter ($8$ cm) directly in the formula $V = \pi r^2 h$, when the formula requires the radius. The key word they ignored is diameter. To use $V = \pi r^2 h$, you must first halve the diameter to get the radius: $r = 8 \div 2 = 4$ cm. Then $V = \pi \times 4^2 \times 5 = \pi \times 16 \times 5 = \mathbf{80\pi}$ cm³ — exactly one quarter of their answer, because $4^2 = 16$ is one quarter of $8^2 = 64$.
Marking: 1 mark for naming the mistake (diameter vs radius); 1 for naming the key word; 1 for the correct value $80\pi$; 1 for clear sentences referring to the formula.