3D Shapes and Volume

Edexcel International GCSE Mathematics A· 4.10· 35 min read

1. Identifying 3D Solids and Key Terms★☆☆☆☆⏱ 5 min

📘 Definition

Face, Edge, Vertex

A face is a flat or curved surface of a 3D solid. An edge is the line where two faces meet. A vertex (plural: vertices) is the corner where three or more edges meet.

Example:

A cube has 6 square faces, 12 edges, and 8 vertices.

Cube: 6 identical square faces
Cuboid: 6 rectangular faces
Prism: Constant cross-section shape along its length
Cylinder: Circular cross-section, curved lateral surface
Sphere: Fully curved, all points equidistant from centre
Cone: Circular base, curved surface tapering to a single point

📐 Worked Example

Name the 3D solid that has a constant triangular cross-section and 5 faces.

1
Step 1: A solid with a constant cross-section is classified as a prism.
2
Step 2: The cross-section shape is a triangle, so it is a triangular prism.
3
Check: Triangular prisms have 2 triangular end faces + 3 rectangular side faces = 5 faces, which matches the description. The correct answer is triangular prism.

2. Surface Area Calculations★★☆☆☆⏱ 10 min

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📘 Definition

Total Surface Area (TSA)

The sum of the areas of all external faces of a 3D solid. For shapes with curved surfaces, include the area of curved faces plus any flat end faces.

Example:

The total surface area of a closed cylinder includes its curved lateral surface plus two circular end faces.

For simple prisms, calculate TSA by adding the area of the two cross-section ends to the area of all lateral side faces. For cylinders, use the given formula for curved surface area ( $2 π r h$ ) plus the area of the two circular ends ( $2 π r^{2}$ ) to get total surface area. Higher tier students will also use given formulae for sphere surface area ( $4 π r^{2}$ ) and cone curved surface area ( $π r l$ , where $l$ is the slant height).

📐 Worked Example

Calculate the total surface area of a closed cylinder with radius 3 cm and height 8 cm. Leave your answer in terms of $π$ .

1
Step 1: Calculate curved surface area using the given formula $2 π r h$
2
$2\pi \times 3 \times 8 = 48\pi \text{ cm}^2$
3
Step 2: Calculate area of the two circular ends: $2 \times π r^{2}$
4
$2 \times \pi \times 3^2 = 18\pi \text{ cm}^2$
5
Step 3: Add the two values for total surface area
6
$48\pi + 18\pi = 66\pi \text{ cm}^2$

📐 Worked Example

Higher only: Find the total surface area of a sphere with diameter 10 cm. Use $π = 3.14$ .

1
Step 1: Find radius = diameter / 2 = 10 / 2 = 5 cm
2
Step 2: Use given sphere surface area formula $4 π r^{2}$
3
$4 \times 3.14 \times 5^2 = 4 \times 3.14 \times 25 = 314 \text{ cm}^2$

Exam tip:

Always check if the question asks for total surface area or curved (lateral) surface area: missing flat end faces for cylinders and cones is one of the most common marking-point losses on this topic.

3. Volume Calculations★★★☆☆⏱ 12 min

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📘 Definition

Volume

The amount of 3D space occupied by a solid, measured in cubic units (e.g. cm³, m³).

Example:

A 10 cm × 10 cm × 10 cm cube has a volume of 1000 cm³, equal to 1 litre.

All volume formulae you need are given on your exam formula sheet, so you do not need to memorize them. The core rule for prisms is volume = area of cross-section × length, which applies to cuboids, triangular prisms, and cylinders (which are circular prisms). Higher tier students also use given formulae for sphere ( $\frac{4}{3} π r^{3}$ ) and cone ( $\frac{1}{3} π r^{2} h$ ) volume.

📐 Worked Example

Calculate the volume of a triangular prism with cross-sectional area 12 cm² and length 15 cm.

1
Step 1: Use given prism volume formula: volume = cross-sectional area × length
2
$12 \times 15 = 180 \text{ cm}^3$

📐 Worked Example

Higher only: A right circular cone has radius 6 cm and perpendicular height 10 cm. Calculate its volume, correct to 2 decimal places.

1
Step 1: Use given cone volume formula: $\frac{1}{3} π r^{2} h$
2
$\frac{1}{3} \times \pi \times 6^2 \times 10 = \frac{1}{3} \times \pi \times 36 \times 10 = 120\pi$
3
Step 2: Calculate the numerical value and round to 2 decimal places
4
$120 \times 3.14159... = 376.99 \text{ cm}^3 (2 d.p.)$

✓ Quick check

What is the volume of a cylinder with radius 2 m and height 5 m? Leave your answer in terms of π.
- 10π m³
- 20π m³
- 40π m³
Reveal answer
20π m³ —
Volume of cylinder = πr²h = π×2²×5 = 20π m³.

4. Volume Unit Conversions★★☆☆☆⏱ 5 min

Volume units are cubic, so you convert them using the cube of the linear conversion factor. For example, 1 m = 100 cm, so 1 m³ = 100³ = 1 000 000 cm³. You also need to recall that 1 litre = 1000 cm³, a conversion that is not given on the formula sheet.

📐 Worked Example

Convert 2.5 m³ to cm³, and then to litres.

1
Step 1: Linear conversion: 1 m = 100 cm, so cubic conversion factor is 100³ = 1 000 000
2
$2.5 \times 1 000 000 = 2 500 000 \text{ cm}^3$
3
Step 2: Convert cm³ to litres: divide by 1000
4
$2 500 000 \div 1000 = 2500 \text{ litres}$

5. Common Pitfalls

Wrong move:

Forgetting to add flat end faces when calculating total surface area of a cylinder or cone

Why:

Questions often specify 'curved surface area' or 'total surface area', and mixing them up loses easy marks

Correct move:

Check question wording carefully: add area of circular ends for total surface area of cylinders, add base area for total surface area of cones

Wrong move:

Using diameter instead of radius in sphere, cylinder or cone formulae

Why:

All 3D shape formulae use radius, not diameter, so substituting diameter directly gives incorrect results

Correct move:

Always divide diameter by 2 to get radius before substituting into any 3D shape formula

Wrong move:

Using linear conversion factors for volume (e.g. 1 m³ = 100 cm³)

Why:

Volume is a 3-dimensional measure, so conversions use the cube of the linear factor, not the linear factor itself

Correct move:

Cube the linear conversion factor before multiplying/dividing for volume unit changes, e.g. 1 m³ = 100³ cm³ = 1 000 000 cm³

Wrong move:

Using slant height instead of perpendicular height in cone volume calculations

Why:

Cone volume formula uses perpendicular height, while curved surface area uses slant height l

Correct move:

Confirm which height you are given: use h (perpendicular height) for volume, l (slant height) for curved surface area

Wrong move:

Rounding values partway through a calculation instead of at the end

Why:

Intermediate rounding introduces errors that make your final answer inaccurate

Correct move:

Keep all values exact (using π in calculations) until the final step, then round as required by the question

6. Quick Reference Cheatsheet

Shape	Foundation Surface Area	Foundation Volume	Higher Only Surface Area	Higher Only Volume
Cube/Cuboid	Sum of area of all 6 faces	Width × Height × Length	Same as Foundation	Same as Foundation
Prism	2×cross-section area + sum of side face areas	Cross-section area × Length	Same as Foundation	Same as Foundation
Cylinder	$2 π r h + 2 π r^{2}$	$π r^{2} h$	Same as Foundation	Same as Foundation
Sphere	Not required	Not required	$4 π r^{2}$	$\frac{4}{3} π r^{3}$
Cone	Not required	Not required	$π r l + π r^{2}$	$\frac{1}{3} π r^{2} h$
Unit Conversions	1 m³ = 1 000 000 cm³; 1 litre = 1000 cm³		Same as Foundation

7. Frequently Asked

Do I need to memorize the volume formula for a prism?

No, the volume of a prism, cylinder, and curved surface area of a cylinder are given on both Foundation and Higher formula sheets. You only need to recall how to apply them and unit conversion rules.

Is the volume of a pyramid covered in this topic?

No, volume of pyramids is not part of this sub-topic for Edexcel IGCSE Maths A 4MA1; you only need to work with prisms, cylinders, and (Higher only) spheres and cones.

Going deeper

formula sheetEdexcel IGCSE Maths A 4MA1 Formula SheetOfficial formula sheet for all 4MA1 exams

What's Next

Now that you have mastered 3D shape surface area and volume calculations, you are ready to move on to more advanced geometry topics in Edexcel IGCSE Maths A. The next related sub-topic is similar shapes, where you will learn to apply scale factors to area and volume of similar 3D solids. You will also use these volume skills when solving real-world application problems, such as calculating the capacity of containers or the mass of solid objects given their density. Practicing past paper questions on this topic will help you avoid common errors and get comfortable applying the given formulae quickly under exam conditions. Make sure you familiarize yourself with the official formula sheet to avoid wasting time memorizing formulae that are provided for you.