Mathematics Standard • Year 11 • Module 2 • Lesson 7
Surface Area of Prisms and Cylinders
Practise HSC-style writing on prism and cylinder surface area — multi-mark short answers plus one structured extended response with full marking criteria.
1. Short-answer questions
1.1 Find the total surface area of a closed cylinder with diameter 10 cm and height 6 cm. Give your answer in exact form (in terms of π). 2 marks Band 3
1.2 A chocolate box is a triangular prism. The triangular ends are right-angled triangles with legs 9 cm and 12 cm. The box is 20 cm long.
(a) Find the hypotenuse of the triangular cross-section.
(b) Find the total surface area of the box. 3 marks Band 3-4
1.3 A rectangular plastic crate has external dimensions 60 cm × 40 cm × 30 cm and is open at the top (no lid).
(a) Find the area of plastic used in cm².
(b) Plastic is supplied in sheets of 1 m × 1 m (10 000 cm²). State whether one sheet is enough to construct the crate, and justify with a comparison sentence. 4 marks Band 4
2. Extended response
2.1 A school is ordering a new cylindrical rainwater tank for the kitchen garden. Two suppliers offer the same height but different shapes.
Tank A — closed cylinder: diameter 1.6 m, height 1.8 m.
Tank B — open-top cylinder: diameter 1.4 m, height 1.8 m.
(a) Find the total surface area of Tank A (both ends present), in exact form (in terms of π) and to 2 d.p.
(b) Find the total surface area of Tank B (open-top: one end only + curved surface), in exact form and to 2 d.p.
(c) Sheet metal costs $42 per m². Find the materials cost of each tank, to the nearest dollar, then state which tank is cheaper to build and by how much. Conclusion sentence required. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — correct r = 0.8 m AND correct substitution 2π(0.8)² + 2π(0.8)(1.8).
• 1 mark — correct simplification (4.16π m² or equivalent) AND decimal to 2 d.p.
Part (b) — 2 marks
• 1 mark — uses open-top formula πr² + 2πrh (NOT 2πr² + 2πrh) with r = 0.7 m.
• 1 mark — correct exact form (3.01π m² or equivalent) AND decimal to 2 d.p.
Part (c) — 3 marks
• 1 mark — correct cost calculation for Tank A.
• 1 mark — correct cost calculation for Tank B.
• 1 mark — explicit conclusion sentence naming the cheaper tank and stating the dollar difference.
Your response:
Stuck on (c)? After both SA values, multiply each by $42 and subtract. Write: "Tank ___ is cheaper by $___".How did this worksheet feel?
What I'll revisit before next class:
1.1 — Closed cylinder, d = 10, h = 6 (2 marks)
Sample response.
r = 10 ÷ 2 = 5 cm.
SA = 2π(5)² + 2π(5)(6) = 50π + 60π = 110π cm².
Marking notes. 1 mark — halves diameter correctly AND writes the full formula with substitution. 1 mark — simplifies to the exact form 110π cm². Common loss: using d = 10 directly gives 200π + 120π = 320π, an answer almost 3× too large.
1.2 — Triangular prism, legs 9 and 12, length 20 (3 marks)
(a) Sample response. c = √(9² + 12²) = √(81 + 144) = √225 = 15 cm.
(b) Sample response.
Two triangle ends: 2 × (½ × 9 × 12) = 108 cm².
Three rectangles: 9(20) + 12(20) + 15(20) = 180 + 240 + 300 = 720 cm².
Total SA = 108 + 720 = 828 cm².
Marking notes. (a) 1 mark — correct hypotenuse with Pythagoras shown. (b) 1 mark — both triangular ends calculated; 1 mark — all three rectangles calculated and added (including the hypotenuse face 15 × 20). Missing the hypotenuse face is the most common error; total then comes out to 528 cm², losing 1 mark.
1.3 — Open-top crate (4 marks)
(a) Sample response. Full SA = 2(60×40) + 2(60×30) + 2(40×30) = 2(2400) + 2(1800) + 2(1200) = 4800 + 3600 + 2400 = 10 800 cm².
Subtract the missing lid (60 × 40 = 2400 cm²): plastic used = 10 800 − 2400 = 8400 cm².
(b) Sample response. One sheet = 10 000 cm². Since 8400 cm² < 10 000 cm², one sheet is enough for the plastic needed (with 1600 cm² of off-cuts).
Marking notes. (a) 1 mark — correct full SA. 1 mark — correct subtraction of one 60×40 face. (b) 1 mark — numerical comparison 8400 vs 10 000. 1 mark — clear conclusion sentence stating one sheet IS enough. A response saying only "yes" without comparing the two numbers earns 0 on (b).
2.1 — Tank A vs Tank B (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Tank A — closed cylinder (d = 1.6 m, h = 1.8 m).
rA = 1.6 ÷ 2 = 0.8 m.
SAA = 2π(0.8)² + 2π(0.8)(1.8) = 1.28π + 2.88π = 4.16π m² ≈ 13.07 m². [2 marks — r found and full formula used; correct exact + decimal.]
(b) Tank B — open-top cylinder (d = 1.4 m, h = 1.8 m).
rB = 1.4 ÷ 2 = 0.7 m.
SAB = π(0.7)² + 2π(0.7)(1.8) = 0.49π + 2.52π = 3.01π m² ≈ 9.46 m². [2 marks — open-top formula (one end only); correct exact + decimal.]
(c) Comparison — cost at $42 per m².
Tank A cost = 13.07 × $42 ≈ $548.94 ≈ $549. [1 mark.]
Tank B cost = 9.46 × $42 ≈ $397.32 ≈ $397. [1 mark.]
Difference = $549 − $397 = $152.
Conclusion: Tank B is cheaper to build by $152 (Tank B costs $397 vs Tank A's $549). [1 mark — explicit conclusion naming cheaper tank with dollar difference.]
Total: 7/7.
Band descriptors for marker.
Band 3: Calculates Tank A SA correctly but uses the closed-cylinder formula for Tank B as well, OR finds both SA values but does not multiply by $42. ≈ 2-3 marks.
Band 4: Both SA values found correctly (including open-top formula for Tank B), but cost comparison is incomplete or the difference is stated without a conclusion sentence. ≈ 4-5 marks.
Band 5: Full numerical solution with both costs and the dollar difference, but the conclusion sentence does not name which tank is cheaper (e.g. just "$152" with no naming). ≈ 6 marks.
Band 6: Complete solution with correct open-top treatment for Tank B, both costs, dollar difference, AND an explicit conclusion sentence naming the cheaper tank. 7/7.