Mathematics Standard • Year 11 • Module 2 • Lesson 8
Surface Area of Pyramids, Cones, and Spheres
Apply pyramid, cone and sphere SA to realistic Australian contexts: party hats, marquees, beach balls and composite landmarks.
Problem 1 — Party hat (open cone)
A conical party hat has base radius 9 cm and vertical height 12 cm. The hat has no base (it sits on a person's head). Cardboard is sold by the m².
Set up: What are we solving for?
(i) Find the slant height ℓ using Pythagoras. 1 mark
(ii) Find the area of cardboard used for one hat (curved surface only, no base). Give in exact form and to 2 d.p. 2 marks
(iii) A party order needs 80 hats. Find the total cardboard area in m² (1 m² = 10 000 cm²), to 2 d.p. 2 marks
Stuck? Revisit lesson § Card 3 — cones use slant height, and "no base" means curved SA = πrℓ only.Problem 2 — Marquee with pyramid roof (lateral SA only)
A small marquee has a square base of side 5 m and a pyramid-shaped roof with apex 3 m above the centre of the base. The "roof" is the four triangular panels only (no base, since that is the marquee floor inside the building).
Set up: What are we solving for?
(i) Find the slant height ℓ of one triangular panel. (Right triangle: h, b/2, ℓ.) 2 marks
(ii) Find the lateral SA (4 triangular panels only) in m², to 2 d.p. 2 marks
(iii) Waterproof canvas costs $24 per m². Find the cost of the roof canvas to the nearest dollar. 1 mark
Stuck? Revisit lesson § Card 2 — lateral SA = 2bℓ (no base term).Problem 3 — Beach ball material
A beach ball is spherical with diameter 40 cm. A manufacturer cuts the surface from a PVC sheet, allowing 8% extra material for seams and waste.
Set up: What are we solving for?
(i) Find the radius. Then find the SA of one ball in cm², to 2 d.p. 2 marks
(ii) Add 8% for seams and waste. Find the PVC required per ball, to 2 d.p. 1 mark
(iii) A production run makes 500 beach balls. Find the total PVC required in m², to 2 d.p. 2 marks
Stuck? Revisit lesson § Card 4 — sphere SA = 4πr². Then add 8% and multiply by 500 before converting to m².Problem 4 — Storage tank: hemisphere on cylinder
A grain storage tank consists of a cylinder of radius 1.5 m and height 4 m, with a hemisphere of the same radius sealing the top. The tank sits on the ground (the cylinder base is also painted on the underside as it is visible during cleaning).
Set up: What are we solving for?
(i) List which faces are exposed and which are hidden at the join between the cylinder and hemisphere. 1 mark
(ii) Calculate the total external SA (cylinder curved + cylinder base + hemisphere dome), in exact form (in terms of π) and to 2 d.p. 3 marks
(iii) Anti-rust paint covers 8 m² per litre. Find the number of full 1-litre tins required. 1 mark
Stuck? Revisit lesson § Card 5 — the cylinder top and hemisphere flat face are hidden. Round UP for tins.Problem 5 — Ice-cream model (hemisphere on cone)
A display model in an ice-cream shop is a hemisphere of "ice cream" sitting on top of a cone (the waffle cone is plastic). Both have radius 4 cm. The cone has vertical height 9 cm.
Set up: What are we solving for?
(i) Find the slant height of the cone using Pythagoras. 1 mark
(ii) List the exposed surfaces: the cone's curved surface (no base — it is hidden at the join with the hemisphere) plus the hemisphere's curved dome (no flat face, also hidden at join). State both. 1 mark
(iii) Find the total external SA of the model, in exact form (in terms of π) and to 2 d.p. 3 marks
Stuck? Revisit lesson § Card 5 — for cone + hemisphere joined at the same-radius circle, BOTH flat faces at the join are hidden.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Party hat (open cone)
Set up. Open cone (no base) ⇒ curved SA = πrℓ only.
(i) ℓ = √(9² + 12²) = √(81 + 144) = √225 = 15 cm.
(ii) Curved SA = π(9)(15) = 135π cm² ≈ 424.12 cm².
(iii) 80 hats: 80 × 424.12 = 33 929.20 cm². Convert: 33 929.20 / 10 000 ≈ 3.39 m².
Problem 2 — Marquee pyramid roof
Set up. Lateral SA only of a square pyramid (4 triangles, no base). b/2 = 2.5 m, h = 3 m.
(i) ℓ² = 3² + 2.5² = 9 + 6.25 = 15.25. ℓ = √15.25 ≈ 3.9051 m ≈ 3.91 m.
(ii) Lateral SA = 2bℓ = 2(5)(3.9051) ≈ 39.05 m².
(iii) Cost = 39.05 × $24 = $937.20 ≈ $937.
Problem 3 — Beach ball
Set up. Sphere SA = 4πr², then add 8%, then scale to 500 balls, then convert cm² → m².
(i) r = 40 ÷ 2 = 20 cm. SA = 4π(400) = 1600π ≈ 5026.55 cm².
(ii) +8%: 5026.55 × 1.08 ≈ 5428.67 cm² per ball.
(iii) 500 balls: 500 × 5428.67 = 2 714 335 cm². Convert: 2 714 335 / 10 000 ≈ 271.43 m².
Problem 4 — Storage tank: hemisphere on cylinder
Set up. Hidden = cylinder top + hemisphere flat face (at the join). Exposed = cylinder curved + cylinder base + hemisphere dome.
(i) Hidden: cylinder top, hemisphere flat face. Exposed: cylinder curved (2πrh), cylinder base (πr²), hemisphere curved dome (2πr²).
(ii) r = 1.5, h = 4. Cylinder curved = 2π(1.5)(4) = 12π. Cylinder base = π(2.25) = 2.25π. Hemisphere dome = 2π(2.25) = 4.5π.
Total = 12π + 2.25π + 4.5π = 18.75π m² ≈ 58.90 m².
(iii) Tins = 58.90 ÷ 8 = 7.36... ⇒ 8 tins (always round up).
Problem 5 — Ice-cream model: hemisphere on cone
Set up. Both same radius (r = 4) ⇒ both flat faces at the join are hidden. Exposed: cone curved + hemisphere dome.
(i) ℓ = √(4² + 9²) = √(16 + 81) = √97 ≈ 9.85 cm.
(ii) Exposed: cone's curved surface (πrℓ) + hemisphere's curved dome (2πr²). Hidden: cone base + hemisphere flat face (both at the join).
(iii) Cone curved = π(4)(√97) = 4√97 · π. Hemisphere dome = 2π(16) = 32π.
Total = 4√97 · π + 32π = (4√97 + 32)π cm² ≈ (39.395 + 32)π ≈ 71.395π ≈ 224.30 cm².