Surface Area of Prisms and Cylinders
Unfold the solid into a net. Every face appears exactly once. Add them all — then subtract any faces that are missing.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
You are wrapping a rectangular gift box in paper. You need to estimate how much paper you need.
Without any formula — what information would you use? What would you calculate? How is this different from finding the volume of the box?
Three solids — three formulas. One key insight links them all: surface area = sum of all face areas. A net diagram makes every face visible so you cannot accidentally omit one.
Rectangular prism: Three pairs of rectangles — top/bottom, front/back, left/right. Triangular prism: Two triangular ends plus three rectangular side faces (one per triangle edge). The lateral SA = perimeter of cross-section × length for any right prism.
Key facts
- What surface area means and how net diagrams represent it
- The SA formula for a cylinder: $\text{SA} = 2\pi r^2 + 2\pi rh$
- How to handle open or partial surface area problems
Concepts
- Why surface area = sum of all face areas — and why a net makes every face visible
- Why the cylinder's curved surface "unrolls" into a rectangle of width $2\pi r$
- Why removing a face means subtracting its area from the total
Skills
- Draw or describe the net of any prism or cylinder and use it to find total SA
- Calculate SA of any right prism or cylinder
- Adjust calculations for open-top containers, pipes, and partial solids
Surface area answers: if you peeled off every face of this solid and laid them flat, what would the total area be?
This is directly useful: calculating how much material to build a container, how much paint to cover a surface, how much foil to wrap a package.
The Method — Always Three Steps
- Identify every face of the solid
- Find the area of each face
- Add them all together
The net diagram makes step 1 reliable — when you unfold the solid, every face is visible and you cannot accidentally omit one.
Net Diagrams
A net is what you get when you cut along some edges of a solid and unfold it completely flat. Every face appears exactly once.
Net components
Rectangular prism: Three pairs of rectangles (top/bottom, front/back, left/right) — 6 faces.
Triangular prism: Two identical triangular ends + three rectangles (one per side) — 5 faces.
Cylinder: Two circles (top and bottom) + one rectangle (curved surface unrolled) — 3 components.
The Cylinder Key Insight
When you unroll the curved surface of a cylinder, you get a rectangle. The width wraps once around the circle — so its width equals the circumference:
$\text{Curved SA} = 2\pi r \times h$
This is simply: rectangle area = width × height, where width = circumference.
What to write in your book
- Surface area = total area of all outer faces — measured in cm², m² (square units).
- Method: (1) List every face. (2) Find each area. (3) Add all together. Use a net to avoid missing faces.
- Rectangular prism: $\text{SA} = 2\ell w + 2\ell h + 2wh$ — six faces in three pairs.
- Triangular prism: $\text{SA} = 2A_\triangle + (a+b+c) \times L$ — if right-angled triangle, find hypotenuse first.
- Cylinder: closed $= 2\pi r^2 + 2\pi rh$; open top $= \pi r^2 + 2\pi rh$; pipe $= 2\pi rh$.
- If diameter is given: write $r = d \div 2$ as the first line of working — every time.
Did you get this? True or false: surface area is measured in cubic units (cm³) because solids are three-dimensional.
Worked examples · 3 in a row, reveal as you go
Find the surface area of a rectangular prism with length 8 cm, width 5 cm, and height 3 cm.
3 pairs: top/bottom, front/back, left/right
Front/back: $2 \times (8 \times 3) = 48$
Left/right: $2 \times (5 \times 3) = 30$
A triangular prism has a right-angled triangular cross-section with legs 6 cm and 8 cm. The prism is 15 cm long. Find the total surface area.
$c^2 = 6^2 + 8^2 = 36 + 64 = 100$
$c = 10\text{ cm}$
$A_\triangle = \tfrac{1}{2} \times 6 \times 8 = 24\text{ cm}^2$
Two ends: $2 \times 24 = 48\text{ cm}^2$
Face 1 (leg 6): $6 \times 15 = 90$
Face 2 (leg 8): $8 \times 15 = 120$
Face 3 (hyp 10): $10 \times 15 = 150$
Total: $\text{SA} = 48 + 90 + 120 + 150 = 408\text{ cm}^2$
(a) Find the total SA of a closed cylinder with radius 7 cm and height 12 cm, correct to 2 decimal places.
(b) A cylindrical water tank is open at the top with diameter 3 m and height 4 m. Find the material needed, correct to 2 decimal places.
$\text{SA} = 2\pi r^2 + 2\pi rh$
$= 2\pi(49) + 2\pi(7)(12) = 98\pi + 168\pi = 266\pi$
$\text{SA} = 835.66\text{ cm}^2$
$\text{SA} = \pi r^2 + 2\pi rh = 2.25\pi + 12\pi = 14.25\pi$
$= 44.77\text{ m}^2$
What to write in your book
- Rectangular prism: List all 6 faces with their dimensions and tick each one. The most common multi-mark error is missing one or two faces.
- Triangular prism: If right-angled cross-section, find hypotenuse first with Pythagoras — this step is mandatory and earns a mark.
- Cylinder diameter trap: Write $r = d \div 2$ as the very first line of working — every time without exception. Substituting diameter into $\pi r^2$ gives an area four times too large.
- Keep $\pi$ exact: Compute $2\pi r^2 + 2\pi rh$, factor out $\pi$, evaluate at the very last step.
Quick check: A closed cylinder has radius 3 cm and height 8 cm. Its total SA is:
Common errors · the 3 traps that cost marks
What to write in your book
- Misconception to fix: metric conversions for area use powers of 100 (not 10), and volume uses powers of 1000.
- Before calculating: write SA = or Vol = to commit to the correct formula type.
- Check your answer's units — cm² confirms surface area, cm³ confirms volume.
Fill the gap: A closed cylinder has radius 4 cm and height 9 cm. $\text{SA} = 2\pi($$) + 2\pi(4)(9) = 32\pi +$ $\pi = 104\pi \approx 326.73\text{ cm}^2$.
Quick-fire practice · 9 calculations
Find the SA of a rectangular prism with $\ell = 10$ cm, $w = 4$ cm, $h = 6$ cm.
Find the SA of a cube with side length 5 m.
A rectangular box has no lid. Its base is 12 cm × 8 cm and height is 5 cm. Find the SA of material needed.
A triangular prism has an equilateral triangular cross-section with side 6 cm and prism length 10 cm. Find the total SA. (Use $A = \frac{1}{2} \times 6 \times 6 \times \sin 60°$ for triangle area.)
A triangular prism has a right-angled triangular cross-section with legs 5 m and 12 m, and prism length 8 m. Find the total SA.
Find the total SA of a closed cylinder with $r = 4$ cm and $h = 9$ cm. Answer to 2 decimal places.
Find the curved surface area only of a cylinder with $r = 3$ m and $h = 7$ m. Answer to 2 decimal places.
A cylindrical tin can has diameter 10 cm and height 14 cm, with top and bottom lids. Find total SA to 2 decimal places.
A section of pipe has inner radius 4 cm, outer radius 5 cm, and length 20 cm. Find total SA (inner curved, outer curved, two annular ends) to 2 decimal places.
Match each solid to its surface area formula:
Earlier you described what you would need to wrap a rectangular gift box. Look back at your initial response.
You need the three dimensions ($\ell$, $w$, $h$) — and you calculate the total area of all six faces: $\text{SA} = 2\ell w + 2\ell h + 2wh$. This is different from volume ($\ell \times w \times h$) which measures the space inside in cubic units — surface area measures the outside skin in square units.
Odd one out: Three of these statements about surface area are correct. Which one is wrong?
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
SA 4. Find the total SA of a closed cylinder with diameter 10 cm and height 6 cm. Give your answer in terms of $\pi$. (2 marks)
1 mark correct $r$ + formula; 1 mark correct exact answer
SA 5. 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. Find the total SA. (3 marks)
1 mark hypotenuse; 1 mark all faces listed; 1 mark correct total
SA 6. A water trough is a triangular prism lying on its side. The cross-section is an isosceles triangle with two equal sides of 50 cm and a base of 60 cm. The trough is 120 cm long and open at the top.
(a) Find the perpendicular height of the triangular cross-section. (1 mark)
(b) Find the area of one triangular end face. (1 mark)
(c) Find the total SA of the material used. (The top is open.) (2 marks)
Comprehensive answers (click to reveal)
Drill 1: $2(40) + 2(60) + 2(24) = 80 + 120 + 48 = \mathbf{248\text{ cm}^2}$
Drill 2: $\text{SA} = 6 \times 25 = \mathbf{150\text{ m}^2}$
Drill 3: Full: $2(96) + 2(60) + 2(40) = 392$ cm²; subtract lid $96$: $\mathbf{296\text{ cm}^2}$
Drill 4: $A_\triangle = \frac{1}{2}(36)\sin 60° = 15.59$ cm²; $2 \times 15.59 + 3 \times (6 \times 10) = 31.18 + 180 = \mathbf{211.18\text{ cm}^2}$
Drill 5: Hyp $= 13$ m; $2 \times (\frac{1}{2}(5)(12)) = 60$ m²; Rects: $5(8)+12(8)+13(8) = 240$; Total $= \mathbf{300\text{ m}^2}$
Drill 6: $2\pi(16) + 2\pi(36) = 32\pi + 72\pi = 104\pi = \mathbf{326.73\text{ cm}^2}$
Drill 7: $2\pi(3)(7) = 42\pi = \mathbf{131.95\text{ m}^2}$
Drill 8: $r=5$; $2\pi(25) + 2\pi(5)(14) = 50\pi + 140\pi = 190\pi = \mathbf{596.90\text{ cm}^2}$
Drill 9: Outer: $200\pi$; Inner: $160\pi$; Annuli: $18\pi$; Total $= 378\pi = \mathbf{1187.52\text{ cm}^2}$
SA 4: $r = 5$; $\text{SA} = 2\pi(25) + 2\pi(5)(6) = 50\pi + 60\pi = \mathbf{110\pi\text{ cm}^2}$
SA 5: $c = 15$ cm; Triangles: $2 \times 54 = 108$ cm²; Rects: $9(20)+12(20)+15(20) = 720$; Total $= \mathbf{828\text{ cm}^2}$
SA 6(a): $h^2 = 50^2 - 30^2 = 1600$; $h = \mathbf{40\text{ cm}}$
SA 6(b): $A = \frac{1}{2}(60)(40) = \mathbf{1200\text{ cm}^2}$
SA 6(c): 2 triangles: 2400 cm²; 2 sloped faces: $2 \times 50 \times 120 = 12000$ cm²; base: $60 \times 120 = 7200$ cm²; Total $= \mathbf{21600\text{ cm}^2}$
Five timed questions on surface area of prisms and cylinders. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Pool: lessons 1–7. Replays welcome.
⚔ Enter the arenaClimb platforms by answering surface area questions. Pool: lesson 7.
Mark lesson as complete
Tick when you've finished the practice and review.