3D Mensuration | FractionRush

Welcome to 3D Mensuration!

Mensuration is the branch of mathematics that deals with measuring lengths, areas and volumes of geometric shapes. In 3D Mensuration, we calculate the surface area (total outer area) and volume (space inside) of solid shapes. These skills are essential for Cambridge IGCSE 0580 and appear in real-world engineering, design and science.

Surface Area = total area of all outer faces | Volume = amount of space inside a solid

Always state your units! Area uses cm², m², etc. Volume uses cm³, m³, etc. Forgetting units costs marks on IGCSE papers.

Cube & Cuboid

6 rectangular faces, straightforward formulas

Prism

Any cross-section: V = Area × length

Cylinder

V = πr²h, SA = 2πr² + 2πrh

Cone

V = ⅓πr²h, slant height l = √(r²+h²)

Sphere & Hemisphere

V = ⁴⁄₃πr³, SA = 4πr²

Pyramid

V = ⅓ × base area × h

Composite Shapes

Combine or subtract simpler solids

Density Problems

mass = density × volume

Key Formulas — Quick Reference

Shape	Volume	Surface Area
Cube (side a)	a³	6a²
Cuboid (l × w × h)	lwh	2(lw + lh + wh)
Prism	A × l (cross-section area × length)	2A + perimeter × l
Cylinder (r, h)	πr²h	2πr² + 2πrh
Cone (r, h, l)	⅓πr²h	πr² + πrl
Sphere (r)	⁴⁄₃πr³	4πr²
Hemisphere (r)	²⁄₃πr³	3πr²
Pyramid	⅓ × base area × h	base + triangular faces

Cone slant height: l = √(r² + h²) — always use Pythagoras to find l before computing the curved surface area of a cone.

1. Cube and Cuboid

A cuboid has three pairs of rectangular faces. Its length is l, width is w, and height is h. A cube is a special cuboid where l = w = h = a.

Cuboid
Imagine a cardboard box. It has 6 faces:
Top & Bottom: l × w (×2) | Front & Back: l × h (×2) | Left & Right: w × h (×2)

Volume = l × w × h
Surface Area = 2(lw + lh + wh)

Cube: V = a³ SA = 6a²

Cuboid: V = lwh SA = 2(lw + lh + wh)

Example: Cuboid with l = 5 cm, w = 3 cm, h = 4 cm
V = 5 × 3 × 4 = 60 cm³
SA = 2(5×3 + 5×4 + 3×4) = 2(15 + 20 + 12) = 2(47) = 94 cm²

Example: Cube with side 6 cm
V = 6³ = 216 cm³
SA = 6 × 6² = 6 × 36 = 216 cm²

2. Prism

A prism is a solid with a uniform cross-section. The cross-section is the same shape throughout the length of the prism. The cross-section can be any 2D shape: triangle, trapezium, L-shape, hexagon — anything.

Triangular Prism
Cross-section = triangle (base b, height h_t)
Length of prism = l

Volume = Area of cross-section × length
Surface Area = 2 × (area of cross-section) + perimeter of cross-section × length

V = A × l where A = area of cross-section, l = length of prism

SA = 2A + P × l where P = perimeter of cross-section

Example: Triangular prism, triangle base = 6 cm, triangle height = 4 cm, prism length = 10 cm. Triangle sides: 5, 5, 6 cm.
Cross-section area A = ½ × 6 × 4 = 12 cm²
V = 12 × 10 = 120 cm³
Perimeter P = 5 + 5 + 6 = 16 cm
SA = 2(12) + 16 × 10 = 24 + 160 = 184 cm²

The cuboid is actually a special rectangular prism. The cylinder is also a type of prism — a circular prism! This is why the cylinder volume formula is just πr² × h (area × length).

3. Cylinder

A cylinder has a circular cross-section of radius r and a height (length) h. Its curved surface, when unrolled, forms a rectangle of width 2πr and height h.

Cylinder — Visualise the Net
2 circles (top and bottom): each has area πr²
Curved surface unrolled = rectangle of width 2πr and height h

Volume = πr²h
Total Surface Area = 2πr² + 2πrh = 2πr(r + h)

V = πr²h SA = 2πr² + 2πrh

Example: Cylinder with r = 5 cm, h = 12 cm
V = π × 5² × 12 = π × 25 × 12 = 300π ≈ 942 cm³
SA = 2π × 25 + 2π × 5 × 12 = 50π + 120π = 170π ≈ 534 cm²

Leaving the answer in terms of π (e.g. 300π) is exact. Converting using π ≈ 3.14159 gives a decimal approximation. IGCSE questions often ask for one or the other — read carefully!

4. Cone

A cone has a circular base of radius r, a vertical height h, and a slant height l. The slant height is the distance along the sloping side from base edge to apex. By Pythagoras:

Slant height: l = √(r² + h²)

Cone Anatomy
r = radius of circular base
h = vertical height (perpendicular from base to apex)
l = slant height (along the sloping surface from base rim to tip)
l² = r² + h² (Pythagoras — r, h, l form a right triangle inside the cone)

Volume = ⅓πr²h (one third of a cylinder with same base and height)
Curved Surface Area = πrl
Total Surface Area = πr² + πrl = πr(r + l)

V = ⅓πr²h CSA = πrl Total SA = πr² + πrl

Example: Cone with r = 3 cm, h = 4 cm
Slant height: l = √(3² + 4²) = √(9 + 16) = √25 = 5 cm
V = ⅓ × π × 9 × 4 = 12π ≈ 37.7 cm³
Total SA = π × 9 + π × 3 × 5 = 9π + 15π = 24π ≈ 75.4 cm²

The 3–4–5 right triangle is the classic Pythagorean triple. Many IGCSE cone questions use r = 3, h = 4 so that l = 5 comes out neatly.

5. Sphere and Hemisphere

A sphere is a perfectly round solid. Every point on the surface is the same distance r (the radius) from the centre. A hemisphere is exactly half a sphere.

Sphere
All surface points are at distance r from the centre
Volume = ⁴⁄₃πr³ | Surface Area = 4πr²

Hemisphere
Half a sphere + a flat circular base
Volume = ½ × ⁴⁄₃πr³ = ²⁄₃πr³
Total Surface Area = 2πr² (curved) + πr² (flat base) = 3πr²

Sphere: V = ⁴⁄₃πr³ SA = 4πr²

Hemisphere: V = ²⁄₃πr³ Total SA = 3πr²

Example: Sphere with r = 6 cm
V = ⁴⁄₃ × π × 6³ = ⁴⁄₃ × π × 216 = 288π ≈ 905 cm³
SA = 4 × π × 36 = 144π ≈ 452 cm²

Example: Hemisphere with r = 4 cm
V = ²⁄₃ × π × 64 = 128π/3 ≈ 134 cm³
Total SA = 3π × 16 = 48π ≈ 151 cm²

6. Pyramid

A pyramid has a polygon base and triangular faces that meet at a single apex (tip). The most common IGCSE pyramid has a square base. The volume of any pyramid is one-third of the volume of a prism with the same base and height.

Square-based Pyramid
Base: square with side a
Vertical height: h (from base centre to apex, perpendicular)
Slant height of triangular face: l_f

Volume = ⅓ × a² × h
Surface Area = a² + 4 × (½ × a × l_f) = a² + 2al_f

V = ⅓ × Base Area × h (works for ANY pyramid, any base shape)

Example: Square pyramid with base 8 cm, height 6 cm, slant height of face = 7 cm
V = ⅓ × 64 × 6 = ⅓ × 384 = 128 cm³
SA = 64 + 4 × (½ × 8 × 7) = 64 + 4 × 28 = 64 + 112 = 176 cm²

For a square pyramid, use Pythagoras to find the slant height of a triangular face: l_f = √(h² + (a/2)²) where a/2 is the distance from the base centre to the midpoint of a base edge.

7. Composite 3D Shapes

Many IGCSE questions combine two or more basic shapes. To find the volume of a composite shape, either add the volumes of its parts, or subtract one from another (for a hollow shape).

Example: A cylinder with a hemisphere on top, r = 4 cm, cylinder height = 10 cm
V(cylinder) = π × 16 × 10 = 160π
V(hemisphere) = ²⁄₃π × 64 = 128π/3
Total V = 160π + 128π/3 = 480π/3 + 128π/3 = 608π/3 ≈ 636 cm³

Example: A cone drilled out of a cylinder. Cylinder r = 5, h = 12. Cone r = 5, h = 12.
V(cylinder) = π × 25 × 12 = 300π
V(cone) = ⅓ × π × 25 × 12 = 100π
Remaining volume = 300π − 100π = 200π ≈ 628 cm³

For composite surface areas, be careful: internal faces where two shapes join are NOT included in the total surface area. Only count visible outer surfaces.

8. Units Conversion

Volume units scale by the cube of the length conversion factor.

1 cm = 10 mm ⟹ 1 cm³ = 1000 mm³ | 1 m = 100 cm ⟹ 1 m³ = 1,000,000 cm³

Litres and cm³: 1 litre = 1000 cm³ 1 ml = 1 cm³
So 500 cm³ = 0.5 litres and 2.4 litres = 2400 cm³

Area units:
1 cm = 10 mm ⟹ 1 cm² = 100 mm²
1 m = 100 cm ⟹ 1 m² = 10,000 cm²
1 km = 1000 m ⟹ 1 km² = 1,000,000 m²

Quick memory aid: for volume, cube the scale factor (×10 for mm→cm means ÷1000 for mm³→cm³). For area, square it. For length, just use it directly.

9. Density Problems

Density is mass per unit volume. The relationship between mass, density and volume is the famous triangle formula:

mass = density × volume density = mass ÷ volume volume = mass ÷ density

Common units:
Density in g/cm³ ⟹ mass in grams, volume in cm³
Density in kg/m³ ⟹ mass in kilograms, volume in m³
Water: density = 1 g/cm³ | Iron: ~7.9 g/cm³ | Aluminium: ~2.7 g/cm³

Example: A solid cylinder has r = 4 cm, h = 10 cm. Density = 8.9 g/cm³. Find the mass.
V = π × 16 × 10 = 160π ≈ 502.65 cm³
mass = 8.9 × 502.65 ≈ 4474 g ≈ 4.47 kg

Always check your units match. If density is in g/cm³, make sure volume is in cm³ before multiplying. Converting beforehand avoids errors.

Example 1 — Volume of a Cylinder (IGCSE style)

A cylindrical tank has radius 3.5 m and height 8 m. Calculate the volume. Give your answer correct to 3 significant figures.

Formula: V = πr²h

Substitute: V = π × (3.5)² × 8 = π × 12.25 × 8 = 98π

Calculate: V = 98 × 3.14159… = 307.876…

Answer: V = 308 m³ (3 s.f.)

Example 2 — Total Surface Area of a Cone

A cone has base radius 6 cm and vertical height 8 cm. Find the total surface area in terms of π.

Find slant height first: l = √(r² + h²) = √(36 + 64) = √100 = 10 cm

Curved SA: πrl = π × 6 × 10 = 60π cm²

Base area: πr² = π × 36 = 36π cm²

Total SA: 60π + 36π = 96π cm²

Example 3 — Sphere Volume and Units

A spherical ball has diameter 9 cm. Find its volume in cm³ and in litres, correct to 3 s.f.

Radius: r = 9 ÷ 2 = 4.5 cm

Volume: V = ⁴⁄₃ × π × (4.5)³ = ⁴⁄₃ × π × 91.125 = 121.5π

Calculate: V = 381.703… ≈ 382 cm³ (3 s.f.)

In litres: 382 ÷ 1000 = 0.382 litres (3 s.f.)

Example 4 — Composite Shape (Cone + Cylinder)

A pencil is modelled as a cylinder of radius 0.4 cm and height 17 cm, with a cone on top of radius 0.4 cm and height 2 cm. Find the total volume to 3 s.f.

Cylinder volume: π × (0.4)² × 17 = π × 0.16 × 17 = 2.72π

Cone volume: ⅓ × π × (0.4)² × 2 = ⅓ × π × 0.32 = 0.32π/3 ≈ 0.10667π

Total: (2.72 + 0.10667)π = 2.82667π ≈ 8.879… ≈ 8.88 cm³

Example 5 — Density and Mass Problem

A solid hemisphere is made of lead. Radius = 5 cm. Density of lead = 11.3 g/cm³. Find the mass to 3 s.f.

Volume of hemisphere: V = ²⁄₃πr³ = ²⁄₃ × π × 125 = 250π/3 ≈ 261.8 cm³

Mass = density × volume: m = 11.3 × 261.8 ≈ 2958.3 g

Answer: mass ≈ 2960 g (3 s.f.) or 2.96 kg

Example 6 — Water Filling Problem

A rectangular tank is 120 cm long, 80 cm wide, 90 cm tall. Water is poured in from a cylindrical pipe of radius 2 cm at a flow speed of 150 cm/s. How long (in minutes) to fill the tank to a depth of 60 cm? Give answer to 1 d.p.

Volume needed: V = 120 × 80 × 60 = 576,000 cm³

Flow rate: volume per second = πr²v = π × 4 × 150 = 600π cm³/s

Time in seconds: t = 576,000 ÷ (600π) = 960/π ≈ 305.6 s

Time in minutes: 305.6 ÷ 60 ≈ 5.1 minutes

Interactive 3D Shape Calculator

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Units Converter

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Exercise 1 — Volume of Cuboids & Cubes

Calculate the volume of each shape. Give integer answers in cm³.

1. Cube with side 3 cm. Volume in cm³.

2. Cube with side 5 cm. Volume in cm³.

3. Cuboid: l = 4, w = 3, h = 2 cm. Volume in cm³.

4. Cuboid: l = 6, w = 5, h = 4 cm. Volume in cm³.

5. Cuboid: l = 10, w = 8, h = 3 cm. Volume in cm³.

6. Cube with side 4 cm. Volume in cm³.

Exercise 2 — Surface Area of Cuboids

Calculate the total surface area. Give answers in cm².

1. Cube with side 3 cm. Surface area in cm².

2. Cube with side 5 cm. Surface area in cm².

3. Cuboid: l = 4, w = 3, h = 2 cm. Surface area in cm².

4. Cuboid: l = 5, w = 5, h = 5 cm (i.e. a cube). Surface area in cm².

5. Cuboid: l = 6, w = 4, h = 3 cm. Surface area in cm².

6. Cuboid: l = 10, w = 2, h = 2 cm. Surface area in cm².

Exercise 3 — Cylinder Volume & Surface Area

Calculate volume and surface area of cylinders. Round answers to 1 decimal place.

1. Cylinder r = 3, h = 10 cm. Volume in cm³ (1 d.p.).

2. Cylinder r = 5, h = 8 cm. Volume in cm³ (1 d.p.).

3. Cylinder r = 2, h = 7 cm. Total Surface Area in cm² (1 d.p.).

4. Cylinder r = 4, h = 6 cm. Total Surface Area in cm² (1 d.p.).

5. Cylinder r = 1, h = 20 cm. Volume in cm³ (1 d.p.).

6. Cylinder r = 6, h = 6 cm. Volume in cm³ (1 d.p.).

Exercise 4 — Cone & Sphere

Calculate volumes and surface areas. Round to 1 decimal place.

1. Cone: r = 3, h = 4 cm. Slant height l in cm (integer).

2. Cone: r = 3, h = 4 cm. Volume in cm³ (1 d.p.).

3. Cone: r = 6, h = 8 cm. Slant height l in cm (integer).

4. Sphere: r = 3 cm. Volume in cm³ (1 d.p.).

5. Sphere: r = 3 cm. Surface Area in cm² (1 d.p.).

6. Hemisphere: r = 6 cm. Volume in cm³ (1 d.p.).

Exercise 5 — Units Conversion & Density

Apply conversions and the density formula. Round to 1 d.p. where needed.

1. Convert 3500 cm³ to litres. Answer in litres.

2. Convert 0.45 litres to cm³. Answer in cm³.

3. Convert 2 m³ to cm³. Answer in cm³.

4. A cube has side 5 cm and density 2.7 g/cm³. Mass in grams.

5. Density = 8 g/cm³, volume = 50 cm³. Mass in grams.

6. Mass = 270 g, density = 2.7 g/cm³. Volume in cm³.

Practice — 20 Questions

Mixed practice covering all 3D mensuration skills. Round decimal answers to 1 d.p. unless told otherwise.

1. Cube side = 2 cm. Volume in cm³.

2. Cuboid l = 5, w = 4, h = 3. Volume in cm³.

3. Cube side = 4 cm. Surface area in cm².

4. Cuboid l = 3, w = 3, h = 3. Surface area in cm².

5. Cylinder r = 2, h = 5. Volume in cm³ (1 d.p.).

6. Cylinder r = 3, h = 4. Total Surface Area in cm² (1 d.p.).

7. Cone r = 3, h = 4. Slant height l in cm (integer).

8. Cone r = 3, h = 4. Volume in cm³ (1 d.p.).

9. Cone r = 3, h = 4. Total Surface Area in cm² (1 d.p.).

10. Sphere r = 3. Volume in cm³ (1 d.p.).

11. Sphere r = 6. Surface area in cm² (1 d.p.).

12. Hemisphere r = 3. Volume in cm³ (1 d.p.).

13. Square pyramid: base 6, height 4. Volume in cm³.

14. Convert 4200 cm³ to litres. Answer in litres.

15. Convert 2.5 litres to cm³. Answer in cm³.

16. Density = 5 g/cm³, volume = 40 cm³. Mass in grams.

17. Mass = 180 g, density = 9 g/cm³. Volume in cm³.

18. Cylinder r = 4, h = 5. Volume in cm³ (1 d.p.).

19. Sphere r = 1. Surface Area in cm² (1 d.p.).

20. Cone r = 5, h = 12. Volume in cm³ (1 d.p.).

Challenge — 8 Questions

Multi-step problems involving composite shapes, water filling, and density. Round to 1 d.p. unless stated.

1. A cylindrical tank has r = 50 cm and height = 120 cm. It is filled with water to 80% of its capacity. How many litres of water does it contain? (1 litre = 1000 cm³). Give answer in litres to 1 d.p.

2. A solid consists of a cylinder (r = 4 cm, h = 10 cm) with a hemisphere on top (same r). Find the total volume in cm³ to 1 d.p.

3. A cone has r = 5 cm and h = 12 cm. Find the total surface area in cm² to 1 d.p.

4. A spherical metal ball has radius 7 cm and is made of iron (density = 7.9 g/cm³). Find its mass to the nearest gram.

5. Water flows into a cuboid tank (length 60 cm, width 40 cm) through a pipe at 200 cm³/s. How many seconds to fill to a depth of 30 cm? Give integer answer.

6. A square pyramid has base side 10 cm and height 12 cm. The slant height of each triangular face is 13 cm. Find the total surface area in cm².

7. A cylinder of radius 3 cm and height 8 cm has a cone of the same radius and height 6 cm removed from its centre (sharing the base). Find the remaining volume in cm³ to 1 d.p.

8. A large sphere of radius 9 cm is melted down and recast into small spheres of radius 3 cm. How many small spheres can be made? (Give exact integer answer.)