Mathematics Standard • Year 11 • Module 2 • Lesson 8
Surface Area of Pyramids, Cones, and Spheres
Build fluency with three new SA formulas plus the critical slant-vs-vertical-height distinction: square pyramid SA = b² + 2bℓ, cone SA = πrℓ + πr², sphere SA = 4πr².
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Complete each surface area formula.
Square pyramid (base b, slant height ℓ): SA = __________________
Closed cone (base radius r, slant height ℓ): SA = __________________
Sphere (radius r): SA = __________________ Hemisphere (radius r): SA = __________________
Q1.2 Tick the line you ALWAYS write first when given a cone with r and vertical height h, before applying SA = πrℓ + πr²:
Q1.3 Why is a hemisphere's total SA equal to 3πr² (and not just 2πr²)? Write one sentence. ____________________________________________
2. Worked example — cone given vertical height (find slant first)
Follow each line of working. Every step has a reason on the right.
Problem. A cone has base radius r = 5 cm and vertical height h = 12 cm. Find the total surface area in exact form and to 2 d.p.
Step 1 — Find slant height ℓ first (Pythagoras).
ℓ² = r² + h² = 25 + 144 = 169. ℓ = 13 cm.
Reason: SA formula uses slant height, not vertical height. (5-12-13 is a Pythagorean triple.)
Step 2 — Apply SA formula.
SA = πrℓ + πr² = π(5)(13) + π(25) = 65π + 25π.
Reason: curved lateral surface + circular base. Keep π exact.
Step 3 — Combine and evaluate.
SA = 90π cm² ≈ 282.74 cm²
Reason: 65 + 25 = 90. Evaluate at the final step only.
Conclusion. Total SA = 90π cm² ≈ 282.74 cm².
3. Faded example — square pyramid given vertical height
A square pyramid has base side b = 8 cm and vertical height h = 3 cm. Find total SA. Fill in each blank line. 4 marks
Step 1 — Find slant height ℓ (right triangle: h, b/2, ℓ).
ℓ² = h² + (b/2)² = ____² + ____² = ____ + ____ = ____.
ℓ = ________ cm.
Step 2 — Apply SA formula: SA = b² + 2bℓ = ____² + 2( ____ )( ____ )
Step 3 — Compute each term: SA = ____ + ____ = ____
Conclusion. Total SA = ________ cm².
4. Graduated practice — pyramids, cones, spheres
Show your working. Decimals to 2 d.p. unless told otherwise.
Foundation — single-step substitution, ℓ given (4 questions)
| Q | Problem | Answer |
|---|---|---|
| 4.1 1 | Square pyramid: b = 10 m, slant height ℓ = 13 m. Find total SA. | |
| 4.2 1 | Cone: r = 6 m, slant height ℓ = 10 m. Find total SA in exact form. | |
| 4.3 1 | Sphere: r = 9 cm. Find SA in exact form. | |
| 4.4 1 | Hemisphere (closed flat face): r = 7 cm. Find total SA in exact form. |
Standard — typical HSC difficulty (6 questions)
For every pyramid or cone given vertical height: write Step 1 = find ℓ before any SA formula.
4.5 Square pyramid with base side 6 cm and vertical height 4 cm. Find total SA. 2 marks
4.6 Cone with r = 3 cm and vertical height 4 cm. Find total SA in exact form and to 2 d.p. 2 marks
4.7 Cone with r = 5 cm and vertical height 12 cm. Find the curved surface area only in exact form. 2 marks
4.8 Square pyramid with base side 8 cm and vertical height 3 cm. Find the lateral SA only (four triangular faces, no base). 2 marks
4.9 A sphere has diameter 14 m. Find its SA to 2 d.p. 2 marks
4.10 Hemisphere with r = 5 cm. State the formula you use and find the total SA in exact form. 2 marks
Extension — composite hidden faces (2 questions)
4.11 A cone (r = 4 cm, vertical height 3 cm) sits on top of a cylinder (r = 4 cm, height 6 cm). List the exposed faces, then find the total SA in exact form and to 2 d.p. 3 marks
4.12 A hemisphere (r = 4 cm) sits on top of a cylinder (r = 4 cm, height 10 cm). List the exposed faces, then find the total SA in exact form and to 2 d.p. 3 marks
5. Self-check the easy 3
Tick the first three once you've checked your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Four formulas
Square pyramid: SA = b² + 2bℓ. Closed cone: SA = πrℓ + πr² (or πr(ℓ + r)). Sphere: SA = 4πr². Hemisphere: SA = 3πr².
Q1.2 — First line for a cone given h
ℓ = √(r² + h²) (Pythagoras). Substituting h directly into πrℓ gives a wrong answer that looks plausible.
Q1.3 — Hemisphere SA
The hemisphere has two surfaces: the curved dome (2πr², half the sphere) AND the flat circular base (πr²) that appears when the sphere is cut. Both must be added: 2πr² + πr² = 3πr².
Q3 — Faded example (square pyramid, b = 8, h = 3)
Step 1: ℓ² = 3² + 4² = 9 + 16 = 25. ℓ = 5 cm.
Step 2: SA = 8² + 2(8)(5) = 64 + 80.
Step 3: SA = 64 + 80 = 144.
Conclusion: Total SA = 144 cm².
Q4.1 — Pyramid b = 10, ℓ = 13
SA = 100 + 2(10)(13) = 100 + 260 = 360 m².
Q4.2 — Cone r = 6, ℓ = 10
SA = π(6)(10) + π(36) = 60π + 36π = 96π m² ≈ 301.59 m².
Q4.3 — Sphere r = 9
SA = 4π(81) = 324π cm² ≈ 1017.88 cm².
Q4.4 — Hemisphere r = 7
SA = 3π(49) = 147π cm² ≈ 461.81 cm².
Q4.5 — Pyramid b = 6, h = 4
ℓ² = 4² + 3² = 16 + 9 = 25 → ℓ = 5 cm. SA = 36 + 2(6)(5) = 36 + 60 = 96 cm².
Q4.6 — Cone r = 3, h = 4
ℓ² = 9 + 16 = 25 → ℓ = 5 cm (3-4-5 triple). SA = π(3)(5) + π(9) = 15π + 9π = 24π cm² ≈ 75.40 cm².
Q4.7 — Cone curved SA only, r = 5, h = 12
ℓ = 13 cm (5-12-13 triple). Curved SA = πrℓ = π(5)(13) = 65π cm² (≈ 204.20 cm²).
Q4.8 — Pyramid lateral SA only, b = 8, h = 3
ℓ = 5 cm (3-4-5). Lateral SA = 2bℓ = 2(8)(5) = 80 cm².
Q4.9 — Sphere d = 14 m
r = 7 m. SA = 4π(49) = 196π ≈ 615.75 m².
Q4.10 — Hemisphere r = 5
Formula: SA = 3πr². SA = 3π(25) = 75π cm² (≈ 235.62 cm²). (NOT 50π — that forgets the flat base.)
Q4.11 — Cone on cylinder (r = 4 throughout)
Exposed: cylinder curved, cylinder base, cone curved. Hidden: cylinder top, cone base.
ℓcone = √(16 + 9) = 5 cm.
Cylinder curved: 2π(4)(6) = 48π. Cylinder base: π(16) = 16π. Cone curved: π(4)(5) = 20π.
Total = 48π + 16π + 20π = 84π cm² ≈ 263.89 cm².
Q4.12 — Hemisphere on cylinder (r = 4)
Exposed: cylinder curved, cylinder base, hemisphere curved dome. Hidden: cylinder top, hemisphere flat face.
Cylinder curved: 2π(4)(10) = 80π. Cylinder base: π(16) = 16π. Hemisphere dome: 2π(16) = 32π.
Total = 80π + 16π + 32π = 128π cm² ≈ 402.12 cm².