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.