Mathematics • Year 10 • Unit 1 • Lesson 17
Curved Surface Area — Mixed Challenge
Pull together every idea from Lesson 17: cylinder TSA, the slant-height step for cones (l = √(r² + h²)), sphere/hemisphere SA, composite curved solids and unit-rate problems (paint/fabric/insulation). Choose the right tool for each problem, spot another student's mistake, then design your own minimum-material curved container.
1. Mixed problems — choose the right tool
Each question uses a different idea from Lesson 17. Decide which formula applies before you start writing. Give exact answers in π unless told otherwise. 3 marks each
1.1 Find the total surface area of a closed cylinder with radius 8 cm and height 14 cm. Leave your answer in terms of π.
1.2 Find the total surface area of a cone with radius 9 cm and perpendicular height 12 cm.
1.3 A sphere has diameter 14 cm. Find its surface area in exact form and as a decimal (1 d.p.).
1.4 A cylindrical concrete pipe is open at both ends. It has radius 30 cm and length 2 m. Find the area of the curved surface (inside + outside count as the same area for this question — give the outside curved area only). Answer in m².
1.5 A composite solid is a cylinder of radius 5 cm and height 8 cm with a hemisphere of the same radius (5 cm) sealed on top. Find the total external surface area (curved cylinder side + curved hemisphere + one circular base at the bottom).
1.6 A spherical hot-air balloon has radius 4 m. The fabric for the balloon costs $42 per m². Calculate the total cost of the fabric.
2. Find the mistake
Another Year 10 student has tried to find the total surface area of a cone with radius 6 cm and perpendicular height 8 cm. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — find TSA of a cone with r = 6 cm and h = 8 cm:
Line 1: CSA = πrh = π × 6 × 8 = 48π cm²
Line 2: Base = πr² = π × 36 = 36π cm²
Line 3: TSA = 48π + 36π = 84π ≈ 263.9 cm²
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the corrected final answer.
Stuck? CSA of a cone uses the SLANT height l, not the perpendicular height h. Find l with Pythagoras first.3. Open-ended challenge — design a low-surface-area can
This question has many valid answers. Be creative but show every number. 4 marks
3.1 A drinks company wants to design a closed cylindrical can with internal volume about 375 mL (= 375 cm³, the standard Australian can size). Aluminium costs roughly $0.20 per 100 cm². The company wants to minimise material.
Design two different cans that both have volume close to 375 cm³ (within ±5 cm³ is fine):
(i) Can A: a tall thin can (e.g., radius about 3 cm).
(ii) Can B: a short wide can (e.g., radius about 5 cm).
For each can:
(a) Choose the radius. Calculate the required height using V = πr²h ⇒ h = V ÷ (πr²). Round h to 1 d.p.
(b) Verify the actual volume is within ±5 cm³ of 375 cm³.
(c) Calculate the total surface area (closed cylinder: 2πrh + 2πr²).
(d) Calculate the aluminium cost.
Bonus: One sentence at the end identifying which can uses less aluminium and why.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Closed cylinder (r = 8, h = 14)
TSA = 2π(8)(14) + 2π(64) = 224π + 128π = 352π cm² (≈ 1105.8 cm²).
1.2 — Cone (r = 9, h = 12)
l = √(81 + 144) = √225 = 15 cm.
TSA = πrl + πr² = π(9)(15) + π(81) = 135π + 81π = 216π cm² (≈ 678.6 cm²).
1.3 — Sphere (d = 14)
r = 7. SA = 4π(7)² = 4π(49) = 196π cm² ≈ 615.8 cm² (1 d.p.).
1.4 — Pipe open at both ends (r = 30 cm = 0.3 m, L = 2 m)
Curved outside = 2πrL = 2π(0.3)(2) = 1.2π ≈ 3.77 m².
No ends because it's open at both ends.
1.5 — Cylinder + hemisphere top (r = 5, cylinder h = 8)
Cylinder side = 2π(5)(8) = 80π.
Hemisphere curved = 2π(25) = 50π.
One base (bottom) = π(25) = 25π.
Total = 80π + 50π + 25π = 155π ≈ 486.9 cm².
1.6 — Hot-air balloon (r = 4 m)
SA = 4π(16) = 64π ≈ 201.06 m².
Cost = 201.06 × $42 = $8444.60 (to nearest cent; or exact 2688π ≈ $8444.60).
2 — Find the mistake
(a) The mistake is on Line 1.
(b) The curved surface area of a cone is πrl (slant height), not πrh (perpendicular height). The student multiplied by h = 8 instead of finding l first.
(c) Corrected working:
l = √(6² + 8²) = √(36 + 64) = √100 = 10 cm.
CSA = πrl = π(6)(10) = 60π cm².
Base = π(6)² = 36π cm².
TSA = 60π + 36π = 96π ≈ 301.6 cm².
Common trap — always run Pythagoras to find l before πrl.
3 — Open-ended challenge (sample solutions)
We need two closed cylindrical cans with volume ≈ 375 cm³, then compare surface areas.
Can A — tall and thin (r = 3 cm)
h = 375 ÷ (π × 9) ≈ 13.3 cm.
Check: π × 9 × 13.3 ≈ 376.0 cm³ ✓
SA = 2π(3)(13.3) + 2π(9) = 79.8π + 18π = 97.8π ≈ 307.2 cm².
Cost = (307.2 ÷ 100) × $0.20 ≈ $0.61.
Can B — short and wide (r = 5 cm)
h = 375 ÷ (π × 25) ≈ 4.8 cm.
Check: π × 25 × 4.8 ≈ 377.0 cm³ ✓
SA = 2π(5)(4.8) + 2π(25) = 48π + 50π = 98π ≈ 307.9 cm².
Cost = (307.9 ÷ 100) × $0.20 ≈ $0.62.
Bonus: The two are very close — both about $0.61–$0.62 — because the optimum surface area for a closed cylinder of given volume occurs at h = 2r (which gives r ≈ 3.9 cm and h ≈ 7.8 cm and SA ≈ 287 cm², the true minimum). Tall thin and short wide both overshoot the optimum by a similar amount.
Marking: 1 for two valid radius/height pairs giving ≈ 375 cm³; 1 for correct surface area on each can; 1 for correct cost on each can; 1 for the comparison sentence. Other valid radius choices accepted.