Cuboids, prisms and exposed faces
Multiply a uniform cross-section by length for volume, then count exposed faces for surface area.
Section Paper 4 · Extended · Calculator
Core and Extended. This is the separate 9–1 qualification 0980. Cambridge intends Core for grades 4–1 and Extended for grades 9–5; the eligible award bands are grades 5–1 for the Core papers and grades 9–3 for the Extended papers.
Before you begin
- Find areas of rectangles and triangles.
- Interpret a solid's net.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the area of a right triangle with perpendicular sides 3 and 4 cm?
What you will learn
- Find volume and surface area of cuboids and right prisms.
- Interpret a uniform cross-section and distinguish open from closed solids.
Stack the same cross-section
A prism has a uniform cross-section along its length. Its volume is cross-sectional area times perpendicular length. A right triangular prism with a 3–4–5 cm cross-section and length 10 cm has cross-sectional area 3 × 4/2 = 6 cm² and volume 60 cm³. The length between its matching ends is not a triangle side.
A cuboid is a rectangular prism: volume abc for three perpendicular edge lengths. If volume and cross-sectional area are known, divide to find the length. The uniform cross-section may be more complex than a polygon, including a sector-shaped end; find that end's area first. For a 90° sector of radius 6 cm, end area is 9π cm²; a 5 cm long prism then has volume 45π cm³. Its end perimeter is 3π + 12 cm, so closed surface area is 2(9π) + 5(3π + 12) = 33π + 60 cm². Use matching length units before multiplying.
Surface area needs each boundary face
A closed right prism has two matching ends plus rectangles along every cross-section edge. Its area is 2A + pL, where A is end area, p its perimeter and L the prism length. For the 3–4–5 prism above, this is 12 + (3 + 4 + 5) × 10 = 132 cm². A cuboid gives 2(ab + ac + bc).
For an open-top box, remove only the missing top face from the closed area. Do not remove the base or all side faces. A painted or wrapped solid may expose fewer surfaces than a closed net; draw a face list before calculating and distinguish a face area from its boundary length.
Pause and explain
A prism has cross-sectional area 12 cm² and length 7 cm. Find its volume.
Worked example
Find volume and closed surface area of a right triangular prism with 3–4–5 cm cross-section and length 10 cm.
Show the worked solution
- Cross-section area is 3 × 4/2 = 6 cm²; volume is 6 × 10 = 60 cm³.
- The three side rectangles total (3 + 4 + 5) × 10 = 120 cm².
- Add two triangular ends, 12 cm², giving surface area 132 cm².
Answer Volume 60 cm³; surface area 132 cm².
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A cuboid has edges 2, 3 and 5 cm. Find volume and closed surface area.
Give me a hint
List its three pairs of equal faces.
Compare my reasoning
- Volume is 2 × 3 × 5 = 30 cm³.
- One of each face pair has area 6, 10 and 15 cm².
- Double their sum: surface area 62 cm².
30 cm³; 62 cm².
Look for these in your work
- Used three dimensions for volume.
- Included all six faces.
A prism has cross-sectional area 18 cm² and volume 216 cm³. Its end perimeter is 20 cm. Find its length and closed surface area.
Give me a hint
Recover length before the side areas.
Compare my reasoning
- Length is 216/18 = 12 cm.
- Side area is 20 × 12 = 240 cm².
- The two ends add 36 cm², giving 276 cm².
Length 12 cm; surface area 276 cm².
Look for these in your work
- Divided volume by cross-sectional area.
- Included both ends.
An open-top rectangular box has base 8 by 5 cm and height 3 cm. Find capacity and material area, ignoring thickness.
Give me a hint
The top is missing but the base remains.
Compare my reasoning
- Capacity is 8 × 5 × 3 = 120 cm³.
- The base is 40 cm²; side pairs are 2 × 24 and 2 × 15 cm².
- Material area is 40 + 48 + 30 = 118 cm².
Capacity 120 cm³; material area 118 cm².
Look for these in your work
- Included exactly the exposed five faces.
- Distinguished capacity from material area.
Common mistakes
- Using a triangle's perimeter as its area.
- Omitting one prism end.
- Including an absent lid in an open box's material area.
What can you explain now?
What must be uniform along a prism?
Compare with the explanation
Its cross-section.
Equal cross-sectional areas and shape at every position let volume equal area times length.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Tomorrow, sketch a right prism net and derive its surface area from two ends and the side strip.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.