Mathematics Standard • Year 11 • Module 2 • Lesson 8
Surface Area of Pyramids, Cones, and Spheres
Practise HSC-style writing on pyramid, cone and sphere SA — multi-mark short answers plus one structured extended response with explicit marking criteria.
1. Short-answer questions
1.1 A cone has base radius 6 cm and vertical height 8 cm. Find the total surface area in exact form (in terms of π). 3 marks Band 3
1.2 A square pyramid has base side 12 cm and vertical height 8 cm.
(a) Find the slant height ℓ.
(b) Find the total surface area in cm². 3 marks Band 3-4
1.3 A hemisphere has radius 7 cm.
(a) Find its total surface area in exact form (in terms of π) and to 2 d.p.
(b) A student writes "SA = half of sphere = ½ × 4π(49) = 98π cm²". Explain in one sentence what error they have made. 4 marks Band 4
2. Extended response
2.1 A community centre is constructing a decorative concrete landmark in a regional NSW town. The landmark is a composite solid:
Base: a square pyramid with base side 2 m and vertical height 1.5 m, sitting upside-down on the ground (apex pointing into the ground). The square base of the pyramid is at the top (1.5 m off the ground) and acts as the platform.
Top: a hemisphere with radius 0.6 m sits centrally on top of the square platform (it is small enough not to overhang).
The exposed concrete surfaces (to be sealed with a waterproof coating) are: the four triangular faces of the pyramid, the visible part of the square platform around the hemisphere's flat base, and the hemisphere's dome. The buried apex is NOT sealed.
(a) Find the slant height ℓ of the pyramid using Pythagoras.
(b) Find the total area of the four triangular faces.
(c) Find the area of the square platform around the hemisphere (square area minus the hemisphere's circular footprint).
(d) Find the area of the hemisphere's curved dome.
(e) Find the total area to be sealed, to 2 d.p., and state a conclusion sentence. Sealer covers 6 m² per litre; state how many full 1-litre tins are required. 8 marks Band 5-6
Explicit marking criteria
Part (a) — 1 mark
• 1 mark — correct ℓ = √(h² + (b/2)²) = √(1.5² + 1²) = √3.25 m.
Part (b) — 1 mark
• 1 mark — correct lateral SA = 2bℓ = 4√3.25 m² (or decimal).
Part (c) — 2 marks
• 1 mark — identifies platform area = 2² = 4 m².
• 1 mark — subtracts hemisphere footprint πr² = 0.36π m².
Part (d) — 1 mark
• 1 mark — correct curved dome = 2πr² = 0.72π m².
Part (e) — 3 marks
• 1 mark — correct sum of the three contributions to 2 d.p.
• 1 mark — correct number of tins (round UP).
• 1 mark — explicit conclusion sentence stating sealed area and tin count.
Your response:
Stuck on (c)? The hemisphere covers a circle of radius 0.6 m on the square — subtract that circle, then add the dome.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Cone, r = 6, h = 8 (3 marks)
Sample response.
Step 1: ℓ = √(6² + 8²) = √(36 + 64) = √100 = 10 cm. (6-8-10 triple.)
Step 2: SA = π(6)(10) + π(6²) = 60π + 36π = 96π cm².
Marking notes. 1 mark — correct ℓ with Pythagoras shown. 1 mark — correct substitution into πrℓ + πr². 1 mark — correct exact form 96π cm². Common error: using h = 8 instead of ℓ = 10 gives 84π, scoring at most 1/3.
1.2 — Square pyramid, b = 12, h = 8 (3 marks)
(a) Sample response. ℓ² = 8² + 6² = 64 + 36 = 100. ℓ = 10 cm. (b/2 = 6, not 12 — this is the standard trap.)
(b) Sample response. SA = b² + 2bℓ = 144 + 2(12)(10) = 144 + 240 = 384 cm².
Marking notes. (a) 1 mark — correct ℓ with b/2 = 6 used (not 12). (b) 1 mark — correct substitution into b² + 2bℓ. 1 mark — correct final answer with units.
1.3 — Hemisphere, r = 7 (4 marks)
(a) Sample response. SA = 3πr² = 3π(49) = 147π cm² ≈ 461.81 cm².
(b) Sample response. The student halved the sphere's curved surface (giving 98π) but forgot to add the flat circular base πr² = 49π that appears when the sphere is cut. The correct total includes both: 98π + 49π = 147π.
Marking notes. (a) 1 mark — uses 3πr² (or 2πr² + πr²). 1 mark — correct exact AND decimal. (b) 1 mark — identifies the missing flat face. 1 mark — sentence shows the correct breakdown 98π + 49π = 147π. A response of just "they forgot a face" with no maths = 1 mark only.
2.1 — Composite landmark (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Pyramid slant height.
b = 2 m, h = 1.5 m. ℓ = √(h² + (b/2)²) = √(2.25 + 1) = √3.25 ≈ 1.8028 m. [1 mark.]
(b) Four triangular faces.
Lateral SA = 2bℓ = 2(2)(1.8028) ≈ 7.21 m². [1 mark.]
(c) Square platform minus hemisphere footprint.
Platform area = 2² = 4 m². [1 mark.]
Hemisphere footprint = π(0.6)² = 0.36π ≈ 1.131 m².
Platform around hemisphere = 4 − 0.36π ≈ 2.87 m². [1 mark.]
(d) Hemisphere curved dome.
Curved dome = 2πr² = 2π(0.36) = 0.72π ≈ 2.26 m². [1 mark.]
(e) Total sealed area and tin count.
Total = 7.21 + 2.87 + 2.26 = 12.34 m². [1 mark — sum to 2 d.p.]
Tins = 12.34 / 6 = 2.06... ⇒ 3 tins (round up). [1 mark.]
Conclusion: the total area to seal is approximately 12.34 m², requiring 3 full 1-litre tins of sealer. [1 mark — explicit conclusion sentence stating both the area and the tin count.]
Total: 8/8.
Band descriptors for marker.
Band 3: Calculates pyramid lateral SA correctly but treats the platform as the full 4 m² (forgets to subtract the hemisphere footprint), OR forgets the dome. ≈ 3-4 marks.
Band 4: All three component areas calculated correctly but the tin count is wrong (rounds down to 2 instead of up to 3) OR the conclusion sentence is missing. ≈ 5-6 marks.
Band 5: Full numerical solution with correct rounding, but the conclusion sentence does not state both the area AND the tin count. ≈ 7 marks.
Band 6: Complete solution with platform-minus-footprint correctly handled, three tins (round up), and a clear conclusion sentence stating both the sealed area and tin count. 8/8.