Mathematics Standard • Year 11 • Module 2 • Lesson 9
Volume of Prisms and Cylinders
Apply V = Ah and V = πr²h to real Australian contexts: pools, rainwater tanks, concrete works and garden beds.
Problem 1 — Backyard rainwater tank capacity
A cylindrical rainwater tank has internal diameter 2.4 m and internal height 3.0 m. A typical Sydney household uses about 450 L of water per day.
Set up: What are we solving for?
(i) Find the volume of the tank in m³, correct to 2 d.p. 2 marks
(ii) Express the capacity in litres. 1 mark
(iii) Estimate the number of full days a full tank can supply the household, rounded down to the nearest whole day. Justify the rounding choice in one sentence. 2 marks
Stuck? Revisit lesson § Worked Example 3 — cylinder, then convert m³ to litres. Round DOWN for "full days you can supply".Problem 2 — Concrete for a retaining wall (trapezoidal prism)
A concrete retaining wall has a trapezoidal cross-section: parallel sides of 0.5 m (top) and 0.8 m (bottom), with a perpendicular height of 1.2 m. The wall is 15 m long. Ready-mix concrete costs $220 per m³.
Set up: What are we solving for?
(i) Find the area of the trapezoidal cross-section in m². 1 mark
(ii) Find the volume of concrete required in m³. 1 mark
(iii) Find the cost of the concrete to the nearest dollar. 1 mark
(iv) The supplier only sells in whole-cubic-metre orders. Find the smallest whole number of m³ that must be ordered, and the new cost. 2 marks
Stuck? Revisit lesson § Worked Example 4 — Trapezoidal Prism. Round UP for ordering material so you don't run short.Problem 3 — Backyard swimming pool capacity
A backyard pool is 20 m long, 8 m wide, with a depth that slopes uniformly from 1.0 m at the shallow end to 2.5 m at the deep end (the side-on cross-section is a trapezium).
Set up: What are we solving for?
(i) Identify the cross-section (viewed from the long side), state its shape, and find its area in m². 2 marks
(ii) Find the volume of the pool in m³. 1 mark
(iii) Express the capacity in kilolitres and in litres. 2 marks
Stuck? Revisit lesson § Worked Example 4 — the trapezium is viewed from the long side (parallel sides = depths, "width" of trapezium = pool length). Multiply by the OTHER pool dimension (the 8 m width).Problem 4 — Composite garden bed (rectangle + trapezium)
A garden bed has a composite cross-section: a rectangle 1.2 m wide and 0.4 m deep, sitting on top of a trapezium with parallel sides 1.2 m and 2.0 m and perpendicular height 0.3 m. The garden bed is 8 m long. Soil costs $58 per bag of 0.1 m³.
Set up: What are we solving for?
(i) Find the area of the rectangular portion of the cross-section. 1 mark
(ii) Find the area of the trapezoidal portion of the cross-section. Then add to give total cross-section area. 2 marks
(iii) Find the total volume of soil required in m³. 1 mark
(iv) Find the number of bags needed (rounded up) and the total cost. 2 marks
Stuck? Revisit lesson § Composite Volumes — calculate each component separately then add. Bags must round UP.Problem 5 — Copper pipe metal volume
A length of copper plumbing pipe has outer diameter 22 mm and inner diameter 18 mm. The pipe is 3.0 m long. Copper density is 8 940 kg/m³ (you'll use this in part (iii)).
Set up: What are we solving for?
(i) Find the outer and inner radii in mm. Then state both in metres. 2 marks
(ii) Find the volume of copper metal in the pipe, in m³ to 5 significant figures (use the annular cross-section formula). 3 marks
(iii) Multiply by the density 8 940 kg/m³ to find the mass of copper in the pipe, to the nearest gram. 2 marks
Stuck on (ii)? Convert mm to m FIRST (divide by 1000). Then annular cross-section = π(R² − r²), multiply by length.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Rainwater tank
Set up. Cylinder volume, then convert m³ → L, then divide by daily use.
(i) r = 1.2 m. V = π(1.44)(3.0) = 4.32π ≈ 13.57 m³.
(ii) Capacity = 13.57 × 1000 = 13 570 L.
(iii) Days = 13 570 ÷ 450 = 30.155... ⇒ 30 full days. Round DOWN: on day 31 the household has only 13 570 − 30 × 450 = 70 L left, not a full day's supply.
Problem 2 — Concrete retaining wall
Set up. Trapezoidal cross-section × length, then cost, then round-up for ordering.
(i) A = ½(0.5 + 0.8)(1.2) = ½(1.3)(1.2) = 0.78 m².
(ii) V = 0.78 × 15 = 11.7 m³.
(iii) Cost = 11.7 × $220 = $2574.
(iv) Order 12 m³ (smallest whole-m³ ≥ 11.7). New cost = 12 × $220 = $2640.
Problem 3 — Swimming pool
Set up. Side-on view: trapezium with parallel sides 1.0 m and 2.5 m, width 20 m. Then multiply by the pool's 8 m width.
(i) Cross-section is a trapezium. A = ½(1.0 + 2.5)(20) = ½(3.5)(20) = 35 m².
(ii) V = 35 × 8 = 280 m³.
(iii) Capacity = 280 kL = 280 000 L.
Problem 4 — Composite garden bed
Set up. Total cross-section = rectangle + trapezium, then × length, then round up for bags.
(i) Arect = 1.2 × 0.4 = 0.48 m².
(ii) Atrap = ½(1.2 + 2.0)(0.3) = ½(3.2)(0.3) = 0.48 m². Total cross-section = 0.48 + 0.48 = 0.96 m².
(iii) V = 0.96 × 8 = 7.68 m³.
(iv) Bags = 7.68 ÷ 0.1 = 76.8 ⇒ 77 bags. Cost = 77 × $58 = $4466.
Problem 5 — Copper pipe
Set up. Convert mm → m. Annular cross-section × length gives volume, then multiply by density for mass.
(i) Outer radius = 22 / 2 = 11 mm = 0.011 m. Inner radius = 18 / 2 = 9 mm = 0.009 m.
(ii) Annular cross-section = π(0.011² − 0.009²) = π(0.000121 − 0.000081) = π(0.00004) = 0.0000400π m².
V = 0.0000400π × 3.0 = 0.00012π ≈ 3.7699 × 10−4 m³ (5 s.f.).
(iii) Mass = 3.7699 × 10−4 × 8940 = 3.3703 kg = 3370.3 g ≈ 3370 g.