Mathematics Standard • Year 11 • Module 2 • Lesson 10
Volume of Pyramids, Cones, and Spheres
Apply the one-third rule and sphere formula to real Australian contexts: grain silos, ice-cream, sports balls and composite structures.
Problem 1 — Grain silo capacity
A grain silo on a rural NSW farm is modelled as a cylinder (diameter 6 m, height 8 m) with a cone on top (same diameter, perpendicular height 3 m). Wheat has a typical density of 750 kg/m³.
Set up: What are we solving for?
(i) Find the volume of the cylindrical section in exact form (in terms of π). 1 mark
(ii) Find the volume of the conical section in exact form. 1 mark
(iii) Find the total volume in m³, correct to 1 d.p. 1 mark
(iv) Find the maximum mass of wheat the silo can hold, to the nearest tonne (1 tonne = 1000 kg). 2 marks
Stuck? Revisit lesson § Pyramids and Cones — cone volume is ⅓ that of the cylinder of same r and h.Problem 2 — Ice-cream scoop on a cone
A child orders an ice cream: a hemisphere of ice-cream (radius 3.5 cm) sitting on top of a waffle cone (same radius, perpendicular height 12 cm).
Set up: What are we solving for?
(i) Find the volume of the ice-cream hemisphere in exact form (in terms of π) and to 2 d.p. 2 marks
(ii) Find the volume of the cone in exact form and to 2 d.p. 2 marks
(iii) If the ice cream completely melts down into the cone (assume it fits without overflowing — check whether it does), determine in one sentence whether the melted volume would overflow. Justify with a numerical comparison. 2 marks
Stuck? Compare the hemisphere volume with the cone volume directly — if hemisphere > cone, it would overflow.Problem 3 — Comparing sports ball volumes
A regulation soccer ball has diameter 22 cm. A tennis ball has diameter 6.7 cm.
Set up: What are we solving for?
(i) Find the volume of the soccer ball in cm³, to 2 d.p. 2 marks
(ii) Find the volume of the tennis ball in cm³, to 2 d.p. 2 marks
(iii) Find approximately how many tennis balls would fit (by volume only, ignoring packing inefficiency) into the same space as a soccer ball, to the nearest whole number. State your conclusion. 2 marks
Stuck? Soccer ball volume ÷ tennis ball volume — then round to the nearest whole.Problem 4 — Cone given slant height (find h first)
A traffic cone (often used at Sydney roadworks) is solid plastic with base radius 14 cm and slant height 50 cm. Plastic has density 1.35 g/cm³.
Set up: What are we solving for?
(i) Use Pythagoras to find the vertical height h (whole-number cm). 2 marks
(ii) Find the volume of plastic in the cone in cm³, to 2 d.p. 2 marks
(iii) Find the mass of plastic per cone in kg, to 2 d.p. 2 marks
Stuck? Revisit lesson § Cones (Card 3) — h = √(ℓ² − r²) = √(50² − 14²). Hint: 50, 14 fits a 25-7-... scaled triple.Problem 5 — Composite paperweight (cone + hemisphere)
A glass paperweight is a cone of radius 4 cm and perpendicular height 9 cm, capped with a glass hemisphere (radius 4 cm) on top.
Set up: What are we solving for?
(i) Find the volume of the cone in exact form (in terms of π). 1 mark
(ii) Find the volume of the hemisphere in exact form. 1 mark
(iii) Find the total volume to the nearest cm³. 2 marks
(iv) Glass has density 2.5 g/cm³. Find the mass of the paperweight in grams to the nearest gram. 2 marks
Stuck? Volumes add directly for composite solids. Multiply by density at the end for mass.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Grain silo
Set up. Cylinder + cone (both r = 3 m). Volume gives mass via density.
(i) r = 3 m. Vcyl = π(9)(8) = 72π m³.
(ii) Vcone = ⅓π(9)(3) = 27π/3 = 9π m³.
(iii) Total = 72π + 9π = 81π ≈ 254.5 m³.
(iv) Mass = 254.5 × 750 = 190 851.75 kg ≈ 190.9 t ≈ 191 t.
Problem 2 — Ice-cream scoop on cone (overflow check)
Set up. Hemisphere volume vs cone volume.
(i) Vhemi = ⅙π(3.5)³ = ⅙π(42.875) = 85.75π/3 ≈ 89.80 cm³.
(ii) Vcone = ⅓π(3.5²)(12) = ⅓π(12.25)(12) = 49π ≈ 153.94 cm³.
(iii) The hemisphere (89.80 cm³) is less than the cone capacity (153.94 cm³), so the melted ice cream would NOT overflow — about 64.14 cm³ of cone capacity remains.
Problem 3 — Sports balls
Set up. Use V = ⅔πr³ for each ball, then divide.
(i) Soccer ball: r = 11 cm. V = ⅔π(1331) = 5324π/3 ≈ 5575.28 cm³.
(ii) Tennis ball: r = 3.35 cm. V = ⅔π(37.5953...) = (4 × 37.5953)/3 × π = 50.127π ≈ 157.48 cm³.
(iii) Tennis balls per soccer ball (by volume) = 5575.28 / 157.48 ≈ 35.40 ≈ 35 tennis balls. About 35 tennis balls have the same total volume as one soccer ball.
Problem 4 — Traffic cone
Set up. ℓ given, h via Pythagoras, then V, then mass.
(i) h = √(50² − 14²) = √(2500 − 196) = √2304 = 48 cm.
(ii) V = ⅓π(196)(48) = (196 × 48)/3 × π = 9408/3 × π = 3136π ≈ 9852.04 cm³.
(iii) Mass = 9852.04 × 1.35 = 13 300.25 g = 13.300 kg ≈ 13.30 kg.
Problem 5 — Paperweight (cone + hemisphere)
Set up. Add the two volumes, then multiply by density.
(i) Vcone = ⅓π(16)(9) = 144π/3 = 48π cm³.
(ii) Vhemi = ⅙π(64) = 128π/3 cm³.
(iii) Total = 48π + 128π/3 = 144π/3 + 128π/3 = 272π/3 ≈ 285 cm³ (nearest cm³).
(iv) Mass = 285 × 2.5 = 713 g (nearest g, using 284.85... cm³ gives 712 g; either acceptable to the nearest gram).