3D Mensuration

SPEC G16 G17 — GEOMETRY & MEASURES

Volume and surface area of prisms, cylinders, pyramids, cones, spheres and composite solids — choosing the right formula and using exact form in π.

What You'll Learn

3D mensuration appears on every paper. The lesson covers six standard solids — prism, cylinder, pyramid, cone, sphere and composite combinations — each with its own formula and a typical trap to avoid. Spec G16 covers cuboid and prism volumes; spec G17 (additional content) extends to spheres, pyramids, cones and composite solids, including frustums.

By the end of this lesson, you will be able to:

Find the volume of any prism using Volume = cross-section area × length

Find the volume and surface area of a cylinder

Find the volume of a pyramid

Find the volume and total surface area of a cone, including the slant-height step

Use the sphere formulae (volume and surface area)

Combine shapes — cylinder + hemisphere, prism + pyramid — for composite solids

Tackle frustums by subtracting the removed tip from the whole cone

Distinguish perpendicular height from slant height in every formula that uses h

Keep π exact and use the right cubic vs square units

What's on the Higher Formula Sheet

Four of the formulae in this lesson are printed on the Higher formula sheet: cone curved SA = πrl, sphere SA = 4πr², sphere V = (4/3)πr³, cone V = (1/3)πr²h. Foundation papers may include the formula in the question stem if needed. Either way, you still have to memorise prism, cylinder and pyramid formulae, and you must know which formula to apply.

Prism Volume — Cross-Section × Length

A prism is any 3D shape with a constant cross-section running along its length — like a loaf of bread cut into identical slices.

Volume of a prism = cross-section area × length

Common prisms — find the right cross-section

Cuboid: rectangle cross-section (l × w) → V = l × w × h

Triangular prism: triangle cross-section (½ × b × h_tri) → V = ½ × b × h_tri × length

Cylinder: circle cross-section (πr²) → V = πr² × h

Hexagonal prism: hexagon cross-section → V = hex area × length

01Worked Example — Triangular Prism

A triangular prism has a right-angled triangular cross-section with legs 3 cm and 4 cm. The prism is 10 cm long. Find its volume. (3 marks)

MMethod. Volume of prism = cross-section area × length. The cross-section is a right triangle with legs 3 and 4, so its area is ½ × base × perpendicular height (using the two legs because the right angle is between them).

CCalculate.

Cross-section = ½ × 3 × 4 = 6 cm²

V = 6 × 10 = 60 cm³

EEvaluate. Cubic units (cm³) — three lengths multiplied. Sense-check: a cuboid 3 × 4 × 10 = 120 cm³; the triangular prism is exactly half of that, which matches the ½ factor.

Volume = 60 cm³

Cylinder — V and Surface Area

A cylinder is a prism with a circular cross-section. Its volume follows the prism rule; its surface area comes from "unrolling" the curved face.

V = πr²hSA = 2πr² + 2πrh

Why 2πr × h?

If you unroll the curved face of the cylinder, it becomes a rectangle. One side is the height h, the other is the circumference 2πr. So the curved area is 2πrh. Add the two circular ends (each πr²) for the total: 2πr² + 2πrh.

02Worked Example — Cylinder V and SA

A solid cylinder has radius 4 cm and height 10 cm. Find (a) its volume, (b) its total surface area. Give exact answers in terms of π. (4 marks)

MMethod. V = πr²h. Total SA = (two circular ends, each πr²) + (curved face, 2πrh) = 2πr² + 2πrh.

CCalculate.

(a) V = π × 4² × 10 = π × 16 × 10 = 160π cm³

(b) SA = 2π × 4² + 2π × 4 × 10

= 2π × 16 + 2π × 40

= 32π + 80π = 112π cm²

EEvaluate. Cubic units for volume; square units for surface area. Each part of the SA breakdown should make sense: 32π for the two end caps (two circles of area 16π each), 80π for the unrolled rectangle 8π × 10.

V = 160π cm³; SA = 112π cm²

Pyramid Volume

A pyramid has a polygon base (square, rectangle, triangle) and a single apex above. Its volume is exactly one-third of a corresponding prism.

V = ⅓ × base area × perpendicular height

Perpendicular Height — Not Slant

The h in (1/3) × base × h is the perpendicular height from base to apex (red dashed line) — straight up from the centre of the base. It is not the slant edge running from a base corner to the apex. Mixing them up is the most common error in pyramid questions.

03Worked Example — Pyramid

A square-based pyramid has a base of side 6 cm and a perpendicular height of 8 cm. Find its volume. (3 marks)

MMethod. V = (1/3) × base area × perpendicular height. The base is a 6 × 6 square, so base area = 36. Use h = 8 (perpendicular, not slant).

CCalculate.

Base area = 6 × 6 = 36 cm²

V = (1/3) × 36 × 8 = 12 × 8 = 96 cm³

EEvaluate. Cross-check: a cuboid with the same base and height would have volume 36 × 8 = 288 cm³. The pyramid is exactly one-third of this — and 288 ÷ 3 = 96 ✓.

Volume = 96 cm³

Cone Formulae

A cone is a circular pyramid. Its volume is one-third of the matching cylinder; its curved surface area uses the slant length l.

V = ⅓πr²h (formula sheet)

Curved SA = πrl (formula sheet)

Total SA of solid cone = πrl + πr²

Slant Height by Pythagoras

The slant l, the perpendicular height h and the radius r form a right triangle. By Pythagoras:

l = √(r² + h²)

If a question gives r and h, find l first, then use πrl. If it gives r and l, find h via h = √(l² − r²) before using V = (1/3)πr²h.

04Worked Example — Cone

A solid cone has radius 5 cm and perpendicular height 12 cm. Find (a) its volume, (b) its total surface area. Give exact answers in terms of π. (5 marks)

MMethod. Volume uses h: V = (1/3)πr²h. Total SA needs the slant first via Pythagoras (l = √(r² + h²)), then SA = πrl + πr² (curved + base).

CCalculate.

(a) Volume.

V = (1/3) × π × 5² × 12 = (1/3) × π × 25 × 12 = (1/3) × 300π = 100π cm³

(b) Surface area.

Slant: l = √(5² + 12²) = √(25 + 144) = √169 = 13 cm (a 5-12-13 Pythagorean triple)

Curved SA = πrl = π × 5 × 13 = 65π

Base = πr² = π × 25 = 25π

Total SA = 65π + 25π = 90π cm²

EEvaluate. Common wrong answer: V = (1/3) × π × 25 × 13 = (325/3)π. That uses the slant l in place of h — wrong. Volume always uses the perpendicular height h, not the slant.

V = 100π cm³; SA = 90π cm²

Sphere Formulae

Spheres have just one variable: the radius. Both formulae are printed on the Higher formula sheet — useful for Higher tier candidates — but you still need to know which is volume and which is surface area, and Foundation students should learn to recognise both.

V = ⁴⁄₃πr³ (formula sheet)

Don't Mix Up the Two

The formulae look similar — both have π and r — so check which exponent each uses: r³ for volume (cubic units), r² for surface area (square units). Also: V has the (4/3) coefficient; SA has just 4 (no fraction).

Hemisphere — Half a Sphere

A hemisphere is half a sphere. Its volume is (1/2) × (4/3)πr³ = (2/3)πr³. Its curved surface area is half the full sphere SA: (1/2) × 4πr² = 2πr². For a solid hemisphere you also add the flat circular base: total SA = 2πr² + πr² = 3πr².

05Worked Example — Sphere

A solid sphere has radius 6 cm. Find (a) its volume, (b) its surface area. Give exact answers in terms of π. (3 marks)

MMethod. Look up V = (4/3)πr³ and SA = 4πr² on the formula sheet. Substitute r = 6.

CCalculate.

(a) V = (4/3) × π × 6³ = (4/3) × π × 216 = (4 × 216)/3 × π = 864/3 × π = 288π cm³

(b) SA = 4 × π × 6² = 4 × π × 36 = 144π cm²

EEvaluate. Cubic units (cm³) for V; square units (cm²) for SA. Sense-check the (4/3) factor: 4/3 ≈ 1.33, so V ≈ 1.33 × 216π — and indeed 288/216 = 1.33 ✓.

V = 288π cm³; SA = 144π cm²

Composite Solids

A composite 3D solid is built from two or more standard shapes — most commonly a cylinder topped with a hemisphere, or a cylinder with a cone on top.

Routine for volume

Identify each piece (e.g., cylinder + hemisphere)

Find each piece's volume separately

Add them — keeping each in terms of π — to get the total

Routine for surface area — the trap

Decide which surfaces are on the outside of the combined solid

Faces where the pieces meet are internal — do not count them

Sum only the external surfaces

Internal Faces Do Not Count

When a hemisphere sits on top of a cylinder, the flat face of the hemisphere and the top face of the cylinder are both internal — they are inside the solid where the two pieces meet. The total surface area is: cylinder bottom + cylinder curved + hemisphere curved. (Examiners' reports flag this as the most common error: students "omit the flat face of the hemisphere" or "double-count the join".)

06Worked Example — Cylinder + Hemisphere

A solid is made from a cylinder of radius 3 cm and height 10 cm with a hemisphere of radius 3 cm sitting on top. Find (a) its total volume, (b) its total surface area. Give exact answers in terms of π. (5 marks)

MMethod. Treat each piece separately. Volumes simply add. For surface area, count only external surfaces: cylinder bottom + cylinder curved + hemisphere curved. The cylinder top and the hemisphere flat face are internal and do not count.

CCalculate.

(a) Volume.

Cylinder: V = πr²h = π × 9 × 10 = 90π

Hemisphere: V = ½ × (4/3)πr³ = (2/3) × π × 27 = 18π

Total = 90π + 18π = 108π cm³

(b) Surface area.

Cylinder bottom (circle): πr² = 9π

Cylinder curved: 2πrh = 2π × 3 × 10 = 60π

Hemisphere curved: ½ × 4πr² = 2π × 9 = 18π

Total = 9π + 60π + 18π = 87π cm²

EEvaluate. The cylinder top (πr² = 9π) and hemisphere flat face (πr² = 9π) are not in the surface area — they are inside the solid where the two pieces meet. If you mistakenly added 9π for the cylinder top you'd get 96π — wrong by 9π.

V = 108π cm³; SA = 87π cm²

Frustum — A Cone with the Tip Cut Off

A frustum is a cone (or pyramid) with the top sliced off parallel to the base. The AQA spec explicitly mentions frustums under G17.

Method — by subtraction

The cleanest way to handle a frustum is to subtract:

Frustum volume = (whole cone volume) − (removed tip volume)

Both the whole cone and the removed tip are similar — they share the same apex angle. If the question gives you the heights or you can find them via similar triangles (Lesson 4.07), find each cone volume with V = (1/3)πr²h, then subtract.

Surface Area of a Frustum

Total SA = (large base) + (small top) + (curved frustum side). The curved side is "large cone curved SA − small cone curved SA" using slant heights, with each curved face being πrl. Frustum questions are rare but appear at the top end of Higher papers.

Pick the Right Formula First

3D mensuration questions usually fail at the formula-choice stage, not the arithmetic. Spot the shape, write down the formula before substituting any numbers.

ShapeVolumeSurface AreaFormula Source

Cuboidl × w × h2(lw + lh + wh)Memorise

Prism (general)cross-section × lengthsum of face areasMemorise

Cylinderπr²h2πr² + 2πrhMemorise

Pyramid⅓ × base × hsum of base + triangular facesMemorise

Cone⅓πr²hπrl + πr²Higher formula sheet (V, πrl)

Sphere⁴⁄₃πr³4πr²Higher formula sheet

Hemisphere⅔πr³3πr² (solid)Halve sphere formulae

Reverse Problems (cross-reference 2.25)

If a question gives you the volume and asks for a missing length, set the formula equal to the given volume and solve as an equation. Example: a cylinder has volume 100π cm³ and radius 5; find h. Solve πr²h = 100π → 25h = 100 → h = 4 cm.

Examiner Alerts

The same handful of errors lose marks on 3D questions every series:

Slant Height Used in Volume

What students write: V = (1/3)πr²l, with l (slant) substituted for h. What to write instead: volumes always use the perpendicular height. The slant l is for curved surface area only (πrl).

Forgetting (1/3) for Cone or Pyramid

What students write: V = πr²h for a cone (which is the cylinder volume). What to write instead: V = (1/3)πr²h. The cone is one-third of the matching cylinder; the pyramid is one-third of the matching prism.

Wrong Sphere Formula — Volume vs SA

What students write: Sphere volume = 4πr², or sphere SA = (4/3)πr³. What to write instead: Volume has the cubed (r³) and the (4/3); surface area has the squared (r²) and the plain 4. The cubic vs square dimension is what tells you which is which — V is in cm³, SA in cm².

Composite SA — Internal Faces Counted

What students write: SA of cylinder + hemisphere = (full cylinder SA) + (full sphere SA / 2) — including the cylinder top and the hemisphere flat face. What to write instead: the join is internal. SA = cylinder bottom + cylinder curved + hemisphere curved = πr² + 2πrh + 2πr².

Cubic vs Square Units

What students write: Volume = 100π cm² or surface area = 90π cm³. What to write instead: volume always cm³ (or m³, mm³); surface area always cm² (or m², mm²). One small letter, one easy mark.

Lesson Checkpoint

You should now be able to:

Must Memorise (not on the formula sheet)

Prism: V = cross-section × length

Cuboid: V = l × w × h; SA = 2(lw + lh + wh)

Cylinder: V = πr²h; SA = 2πr² + 2πrh

Pyramid: V = ⅓ × base × h

On the Higher Formula Sheet

Cone: V = ⅓πr²h; curved SA = πrl. For solid cone, total SA = πrl + πr²

Sphere: V = ⁴⁄₃πr³; SA = 4πr²

Hemisphere: V = ⅔πr³; solid SA = 3πr² (curved face + flat circle)

Frustum: handle by subtraction (whole cone − removed tip)

Discipline

Always use perpendicular height , not slant, in any volume formula

Composite SA: count external surfaces only — internal faces don't count

Volume is in cm³ ; surface area is in cm²

Keep π exact unless a decimal answer is explicitly required

Up Next

Lesson 4.15 covers Pythagoras' Theorem in 2D — the tool you used here to find the slant height of a cone, now applied to general right triangles in coordinate and shape problems.