Mathematics Standard • Year 11 • Module 2 • Lesson 10
Volume of Pyramids, Cones, and Spheres
Practise HSC-style writing on pyramid, cone and sphere volumes — multi-mark short answers plus one structured extended response with explicit marking criteria.
1. Short-answer questions
1.1 A cone has slant height 10 m and base radius 6 m.
(a) Find the perpendicular height of the cone.
(b) Find the volume of the cone in exact form (in terms of π). 3 marks Band 3
1.2 A decorative paperweight consists of a hemisphere sitting on top of a cone. Both have radius 4 cm. The cone has perpendicular height 9 cm.
(a) Find the volume of the hemisphere in exact form (in terms of π).
(b) Find the volume of the cone in exact form.
(c) Find the total volume to the nearest cm³. 3 marks Band 3-4
1.3 A sphere has volume 256π/3 cm³.
(a) Find its radius.
(b) A student writes "V = ⅔πr²" and gets r² = 64, then r = 8 cm. Identify the student's error in one sentence, and give the correct radius. 4 marks Band 4
2. Extended response
2.1 A grain silo at a regional NSW farm is modelled as a cylinder with a cone on top.
Cylindrical section: diameter 4 m, height 10 m.
Conical roof: same diameter (4 m), perpendicular height 2.5 m.
Sorghum has density 720 kg/m³.
(a) Find the volume of the cylindrical section in exact form (in terms of π) and to 2 d.p.
(b) Find the volume of the conical section in exact form and to 2 d.p.
(c) Find the total silo volume to 2 d.p.
(d) Find the maximum mass of sorghum the silo can hold, to the nearest tonne. State a clear conclusion sentence including the total volume and the tonne figure. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — correct r = 2 m AND correct substitution into πr²h.
• 1 mark — correct exact form (40π m³) AND decimal to 2 d.p.
Part (b) — 1 mark
• 1 mark — correct cone volume in exact form (10π/3 m³ or equivalent) AND decimal.
Part (c) — 1 mark
• 1 mark — total volume correctly summed (130π/3 m³ or equivalent) to 2 d.p.
Part (d) — 3 marks
• 1 mark — correct mass calculation (volume × 720 kg/m³).
• 1 mark — correct conversion to tonnes (nearest tonne).
• 1 mark — explicit conclusion sentence stating total volume AND tonne figure.
Your response:
Stuck on (d)? Mass = volume (in m³) × 720 (kg/m³), then divide by 1000 to convert kg to tonnes.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Cone ℓ = 10, r = 6 (3 marks)
(a) Sample response. h = √(ℓ² − r²) = √(100 − 36) = √64 = 8 m.
(b) Sample response. V = ⅓π(36)(8) = 288π/3 = 96π m³.
Marking notes. (a) 1 mark — correct Pythagoras using √(ℓ² − r²), not √(ℓ² + r²). (b) 1 mark — correct substitution into ⅓πr²h with h (not ℓ). 1 mark — correct exact form 96π. Common error: using ℓ = 10 instead of h = 8 in the volume formula gives 120π, scoring 1/3.
1.2 — Paperweight (hemisphere + cone) (3 marks)
(a) Sample response. Vhemi = ⅙π(4)³ = ⅙π(64) = 128π/3 cm³.
(b) Sample response. Vcone = ⅓π(16)(9) = 144π/3 = 48π cm³.
(c) Sample response. Total = 128π/3 + 48π = 128π/3 + 144π/3 = 272π/3 ≈ 285 cm³.
Marking notes. (a) 1 mark — correct hemisphere using ⅙πr³. (b) 1 mark — correct cone using ⅓πr²h. (c) 1 mark — correct sum with a common denominator and final decimal. Common error: forgetting the ⅙ factor for the hemisphere (writing ⅔πr³) gives twice the value.
1.3 — Sphere V = 256π/3 (4 marks)
(a) Sample response.
256π/3 = ⅔πr³. Multiply both sides by 3: 256π = 4πr³. Divide by 4π: r³ = 64. Cube root: r = 4 cm.
(b) Sample response. The student wrote r² instead of r³ in the sphere volume formula. The correct formula is V = ⅔πr³ (not r²). With r³ = 64, the cube root is r = 4 cm, not 8 cm. (The student's 8 cm is √64, the square root.)
Marking notes. (a) 1 mark — correct setup. 1 mark — correct r = 4. (b) 1 mark — identifies the r² vs r³ error specifically. 1 mark — gives the correct r = 4 cm. A vague "they made a mistake" with no identification of r² vs r³ earns 1/2 on (b).
2.1 — Grain silo (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Cylinder (d = 4 m, h = 10 m).
r = 2 m. Vcyl = π(4)(10) = 40π ≈ 125.66 m³. [2 marks — r found and full formula used; correct exact + decimal.]
(b) Cone (same d = 4 m, h = 2.5 m).
Vcone = ⅓π(4)(2.5) = 10π/3 ≈ 10.47 m³. [1 mark.]
(c) Total volume.
Total = 40π + 10π/3 = 120π/3 + 10π/3 = 130π/3 ≈ 136.14 m³. [1 mark.]
(d) Mass of sorghum.
Mass = 136.14 × 720 = 98 020.78 kg. [1 mark — mass in kg.]
Convert to tonnes: 98 020.78 / 1000 = 98.02 t ≈ 98 t. [1 mark.]
Conclusion: the silo holds a total volume of approximately 136.14 m³, equivalent to a maximum sorghum load of about 98 tonnes. [1 mark — explicit conclusion stating both the volume and tonne figure.]
Total: 7/7.
Band descriptors for marker.
Band 3: Calculates cylinder correctly but uses V = πr²h for the cone (forgets the ⅓ factor), OR computes both volumes but does not multiply by density. ≈ 2-3 marks.
Band 4: Both volumes correct (with ⅓ factor for cone), mass calculation done, but tonne conversion missing or rounded incorrectly. ≈ 4-5 marks.
Band 5: Full numerical solution to tonnes, but conclusion sentence does not state both the total volume AND the tonne figure. ≈ 6 marks.
Band 6: Complete solution with correct cone ⅓ factor, total volume in m³, mass conversion to tonnes, AND a clear conclusion sentence with both figures. 7/7.