Mathematics Standard • Year 11 • Module 2 • Lesson 7
Surface Area of Prisms and Cylinders
Build fluency with the three core surface area formulas: rectangular prism SA = 2(ℓw + ℓh + wh), triangular prism SA = 2A△ + (a+b+c)L, and cylinder SA = 2πr² + 2πrh.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Complete each surface area formula.
Rectangular prism (6 faces): SA = ______________________
Closed cylinder: SA = ______________________
Curved surface area of a cylinder only: SA = ______________________
Q1.2 Tick the units that surface area uses. (One only.)
Q1.3 A cylinder has diameter 12 cm. What is the very first line you should write before using SA = 2πr² + 2πrh? ____________________________
2. Worked example — rectangular prism (six-face checklist)
Follow each line of working. Every step has a reason on the right.
Problem. Find the surface area of a rectangular prism with length 8 cm, width 5 cm, and height 3 cm.
Step 1 — List the three pairs of faces.
ℓ = 8, w = 5, h = 3. Top/bottom: ℓ × w. Front/back: ℓ × h. Left/right: w × h.
Reason: write all three pairs before calculating — prevents missing a face.
Step 2 — Compute each pair.
Top/bottom: 2 × (8 × 5) = 80. Front/back: 2 × (8 × 3) = 48. Left/right: 2 × (5 × 3) = 30.
Reason: each pair gets its own line — method marks are awarded for visible structure.
Step 3 — Add and label units.
SA = 80 + 48 + 30 = 158 cm²
Reason: always cm² (square), not cm (linear) or cm³ (cubic).
Conclusion. Surface area = 158 cm².
3. Faded example — closed cylinder
Find the total surface area of a closed cylinder with radius 7 cm and height 12 cm. Give your answer in exact form (in terms of π) and to 2 decimal places. Fill in each blank line. 4 marks
Step 1 — State the formula.
SA = 2πr² + 2πrh. r = ____ cm, h = ____ cm
Step 2 — Substitute: SA = 2π( ____ )² + 2π( ____ )( ____ )
Step 3 — Compute each piece, keep π exact: SA = ____ π + ____ π
Step 4 — Add and evaluate: SA = ____ π cm² ≈ ____________ cm²
Conclusion. Closed cylinder SA = ______ π cm² ≈ ______ cm².
4. Graduated practice — prisms and cylinders
Show your working in the space below each part. Decimals to 2 d.p. unless told otherwise.
Foundation — single-step SA (4 questions)
| Q | Problem | Answer |
|---|---|---|
| 4.1 1 | Find SA of a cube with side 5 m. (Hint: 6 identical faces.) | |
| 4.2 1 | Find SA of a rectangular prism with ℓ = 10 cm, w = 4 cm, h = 6 cm. | |
| 4.3 1 | Find the curved surface area only of a cylinder with r = 3 m and h = 7 m. Give in exact form (terms of π). | |
| 4.4 1 | Find SA of a closed cylinder with r = 4 cm and h = 9 cm. Exact form. |
Standard — typical HSC difficulty (6 questions)
List every face before substituting. Label the final answer with cm² or m².
4.5 A rectangular box with no lid has base 12 cm × 8 cm and height 5 cm. Find the area of material needed. 2 marks
4.6 A cylindrical tin can has diameter 10 cm and height 14 cm with top and bottom lids. Find total SA to 2 d.p. 2 marks
4.7 A cylindrical pipe (no end caps) has radius 5 cm and length 30 cm. Find the curved SA to 2 d.p. 2 marks
4.8 A triangular prism has a right-angled triangular cross-section with legs 6 cm and 8 cm (hypotenuse 10 cm). The prism is 15 cm long. Find the total SA. 3 marks
4.9 A water tank, open at the top, is cylindrical with diameter 3 m and height 4 m. Find the area of sheet metal needed (curved surface + base), to 2 d.p. 2 marks
4.10 A rectangular prism has ℓ = 6 m, w = 4 m, h = 3 m. One face measuring 6 × 4 m is open (no lid). Find the SA. 2 marks
Extension — Pythagoras + composite (2 questions)
4.11 A triangular prism has a right-angled cross-section with legs 5 m and 12 m. The prism is 8 m long. (a) Use Pythagoras to find the hypotenuse first. (b) Then calculate total SA. 3 marks
4.12 A section of pipe has outer radius 5 cm, inner radius 4 cm, and length 20 cm. Find the total SA including the inner curved surface, the outer curved surface, and the two annular end faces. Give 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 — Three formulas
Rectangular prism: SA = 2ℓw + 2ℓh + 2wh (or 2(ℓw + ℓh + wh)). Closed cylinder: SA = 2πr² + 2πrh. Curved SA only: 2πrh.
Q1.2 — SA units
cm² (square units). cm is linear (length), cm³ is cubic (volume).
Q1.3 — Diameter trap
r = d ÷ 2 = 12 ÷ 2 = 6 cm. Substituting the diameter into the formula inflates each term by a factor of 4 (for πr²) or 2 (for 2πrh).
Q3 — Faded example (closed cylinder r = 7, h = 12)
Step 1: r = 7, h = 12.
Step 2: SA = 2π(7)² + 2π(7)(12).
Step 3: SA = 98π + 168π.
Step 4: SA = 266π cm² ≈ 835.66 cm².
Q4.1 — Cube of side 5 m
SA = 6 × 5² = 150 m².
Q4.2 — Rectangular prism 10 × 4 × 6
SA = 2(40) + 2(60) + 2(24) = 80 + 120 + 48 = 248 cm².
Q4.3 — Curved SA only, r = 3, h = 7
Curved SA = 2π(3)(7) = 42π m² ≈ 131.95 m².
Q4.4 — Closed cylinder r = 4, h = 9
SA = 2π(16) + 2π(4)(9) = 32π + 72π = 104π cm² ≈ 326.73 cm².
Q4.5 — Open-top box 12 × 8 × 5
Full SA = 2(96) + 2(60) + 2(40) = 192 + 120 + 80 = 392 cm². Subtract the lid (12 × 8 = 96 cm²): SA = 392 − 96 = 296 cm².
Q4.6 — Closed cylinder d = 10, h = 14
r = 5 cm. SA = 2π(25) + 2π(5)(14) = 50π + 140π = 190π ≈ 596.90 cm².
Q4.7 — Pipe (curved only) r = 5, L = 30
SA = 2π(5)(30) = 300π ≈ 942.48 cm².
Q4.8 — Triangular prism, right-angled (6, 8, 10), L = 15
Two triangle ends: 2 × (½ × 6 × 8) = 48 cm².
Three rectangles: 6×15 + 8×15 + 10×15 = 90 + 120 + 150 = 360 cm².
Total SA = 48 + 360 = 408 cm².
Q4.9 — Open-top tank d = 3 m, h = 4 m
r = 1.5 m. SA = π(1.5)² + 2π(1.5)(4) = 2.25π + 12π = 14.25π ≈ 44.77 m². (One end only because the top is open.)
Q4.10 — Open prism (6 × 4 face missing)
Full SA = 2(24) + 2(18) + 2(12) = 48 + 36 + 24 = 108 m². Subtract open face 6 × 4 = 24 m²: SA = 108 − 24 = 84 m².
Q4.11 — Triangular prism, legs 5 and 12
(a) Hypotenuse: c = √(5² + 12²) = √169 = 13 m.
(b) Triangles: 2 × (½ × 5 × 12) = 60 m². Rectangles: 5(8) + 12(8) + 13(8) = 40 + 96 + 104 = 240 m². Total = 60 + 240 = 300 m².
Q4.12 — Pipe with two annular ends, R = 5, r = 4, L = 20
Outer curved: 2π(5)(20) = 200π.
Inner curved: 2π(4)(20) = 160π.
Two annular ends: 2 × π(25 − 16) = 18π.
Total = 200π + 160π + 18π = 378π ≈ 1187.52 cm².
Common error: forgetting the two annular ring ends.