Mathematics • Year 10 • Unit 1 • Lesson 18

Volume and Capacity in the Real World

Apply prism and cylinder volume formulas to real Australian situations — sizing rainwater tanks, filling aquariums, calculating pool capacity for chlorine dosing, and pouring a concrete path. Then explain your method in your own words.

Apply · Real-World Maths

1. Word problems

Each problem uses a volume formula from Lesson 18: V = lwh (rectangular prism), V = ½bh × L (triangular prism), or V = πr²h (cylinder). Convert units carefully (1 cm³ = 1 mL, 1000 mL = 1 L, 1 m³ = 1000 L = 1 kL). Show every step.

1.1 — Rainwater tank for a rural property. A standard Australian poly rainwater tank is a cylinder with diameter 2.4 m and height 2.5 m.

(a) Calculate the volume of the tank in m³ to 2 d.p.
(b) Convert this to kilolitres (1 m³ = 1 kL).
(c) How many litres is that? 3 marks

Stuck? r = 1.2 m. V = πr²h = π(1.2)²(2.5).

1.2 — Filling a home aquarium. A rectangular aquarium measures 90 cm by 45 cm by 50 cm. The owner fills it to 4 cm from the top.

(a) Find the total capacity of the aquarium in litres.
(b) Find the volume of water it actually contains (filled to 4 cm from the top).
(c) The owner empties one bucket of 9 L; how many litres remain? 3 marks

Stuck? Water depth = 50 − 4 = 46 cm. Capacity = lwh ÷ 1000 (to get litres).

1.3 — Pool capacity for chlorine dosing. A backyard swimming pool is a rectangular prism 8 m by 4 m with depth 1.5 m (uniform depth). The label on the chlorine bottle says "add 80 mL per 10 kL of pool water" for a weekly dose.

(a) Calculate the pool's capacity in kilolitres.
(b) How many mL of chlorine should be added for the weekly dose? 3 marks

Stuck? Capacity in kL = volume in m³. Then chlorine = (kL ÷ 10) × 80 mL.

1.4 — Concrete for a garden path. A concrete path 1.2 m wide is laid around a rectangular garden bed measuring 8 m by 5 m. The concrete is 10 cm thick. Concrete costs $180 per m³.

(a) Find the outer dimensions of the concrete path (path goes all the way around).
(b) Find the area of the concrete path (outer area minus garden bed).
(c) Find the volume of concrete needed in m³.
(d) Find the total cost. 4 marks

Stuck? Path adds 1.2 m on each side, so outer dimensions are 8 + 2.4 by 5 + 2.4. Convert 10 cm = 0.1 m before multiplying.

1.5 — Triangular prism storage box. A timber storage box has a triangular cross-section: an isosceles triangle with base 60 cm and perpendicular height 40 cm. The box is 1.2 m long. How many litres can it hold?

3 marks

Stuck? Convert 1.2 m = 120 cm so all units match. V = ½ × 60 × 40 × 120 cm³ = ___ cm³ = ___ L.

2. Explain your thinking

This question is about communication, not just numbers. Use full sentences. 4 marks

2.1 Sydney's Warragamba Dam has a capacity of about 2 027 000 megalitres (1 ML = 1000 kL = 1000 m³). A friend says: "If we doubled the dam's depth, it would hold twice as much water." Using the lesson's universal volume formula, explain (i) when the friend's statement would be correct (what shape would the dam need to be?), (ii) why it is not generally true for a real dam shape, and (iii) what other change to the dam's dimensions would also double the capacity. Refer to the formula V = A_base × h in your answer.

Stuck? The formula V = A_base × h only doubles when you double h if A_base stays the same. Real dams are wider at the top.

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Answers — Do not peek before attempting

1.1 — Rainwater tank (d = 2.4 m, h = 2.5 m)

(a) r = 1.2 m. V = π(1.2)²(2.5) = 3.6π ≈ 11.31 m³ (2 d.p.).
(b) 11.31 m³ = 11.31 kL.
(c) = 11 310 L (1 kL = 1000 L).

1.2 — Aquarium 90 × 45 × 50 cm

(a) Capacity = 90 × 45 × 50 = 202 500 cm³ ÷ 1000 = 202.5 L.
(b) Water depth = 50 − 4 = 46 cm. V = 90 × 45 × 46 = 186 300 cm³ = 186.3 L.
(c) After removing 9 L: 186.3 − 9 = 177.3 L remain.

1.3 — Pool 8 m × 4 m × 1.5 m

(a) V = 8 × 4 × 1.5 = 48 m³ = 48 kL.
(b) Chlorine = (48 ÷ 10) × 80 = 4.8 × 80 = 384 mL.

1.4 — Concrete path around 8 × 5 m bed (1.2 m wide, 10 cm thick)

(a) Outer dimensions = (8 + 2 × 1.2) × (5 + 2 × 1.2) = 10.4 m × 7.4 m.
(b) Path area = 10.4 × 7.4 − 8 × 5 = 76.96 − 40 = 36.96 m².
(c) Concrete volume = 36.96 × 0.1 = 3.696 m³.
(d) Cost = 3.696 × $180 = $665.28.

1.5 — Triangular prism box (base 60 cm, ht 40 cm, length 120 cm)

A_base = ½ × 60 × 40 = 1200 cm².
V = 1200 × 120 = 144 000 cm³ = 144 000 mL = 144 L.

2.1 — Explain your thinking (sample response)

(i) The friend's statement would be correct if the dam were a right prism with vertical walls (a rectangular prism or a cylinder), because then V = A_base × h, and doubling h while A_base stays the same doubles V.
(ii) For a real dam, the cross-section is not constant — the walls slope outwards as you go up, so the surface area at the top is bigger than at the bottom. Doubling the depth more than doubles the volume because the upper layers are wider, so the friend's "× 2" rule under-estimates the actual capacity.
(iii) For a prism, doubling the base area would also double the capacity (V = A_base × h is symmetric: double either factor and V doubles). For a rectangular base, that could mean doubling the length, or doubling the width, but not doubling both (that would quadruple the capacity).

Marking: 1 for naming a prism/cylinder as the shape where the rule works, 1 for explaining why a real dam shape breaks it (sloping walls / wider at top), 1 for the alternative of doubling base area, 1 for explicit reference to V = A_base × h.