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).