IB Maths AA HL Topic 3 — Geometry & Trigonometry Paper 1 & 2 ~8 min read

Volume & Surface Area

Volume measures how much space a 3D solid occupies; surface area measures how much “skin” it has on the outside. The IB formula booklet gives you almost all the formulas you need — for prisms, cylinders, pyramids, cones, and spheres. Your job is to recognise the shape, pick the right formula, substitute carefully, and (for composite shapes) split the solid into recognisable parts.

📘 What you need to know

Prism: a 3D shape with two identical parallel bases joined by rectangles. Cross-section is the same all the way through. Cuboids and triangular prisms are the standard examples.
Pyramid: a base shape connected to a single apex by triangular faces.
Cylinder & cone: not strictly prisms or pyramids, but obey the same volume rules with a circular base.
Sphere: every point on the surface is the same distance (radius) from the centre.
Volume formulas: V = Ah for prisms, V = ⅓Ah for pyramids — the cone and sphere formulas follow the same pattern with circles.
Surface area = sum of all the outer face areas. For prisms and pyramids, build it up face by face.
Most formulas are in the formula booklet (prior learning + Geometry & Trigonometry sections). The total surface areas of cylinders and cones are not given — but the curved-surface parts are.
Composite shapes: split into recognisable solids, find each piece, then add (or subtract for hollow regions).

Volume formulas

The key idea: for a “constant cross-section” solid (prism, cylinder), volume is cross-section area times length. For a tapered solid (pyramid, cone), divide that by 3. For the sphere, there’s a separate formula.

Shape	Volume formula	Notes
Prism (general)	V = Ah	A = cross-section area; h = length
Cuboid	V = lwh	special case: cross-section is a rectangle
Cylinder	V = πr²h	cross-section is a circle
Pyramid (general)	V = ⅓Ah	A = base area; h = vertical height to apex
Cone	V = ⅓πr²h	circular base with apex
Sphere	V = (4/3)πr³	only the radius needed

The pattern: prism & cylinder use full base area; pyramid & cone use ⅓ of base area. The factor of ⅓ is the same for any pointed solid — it follows from a calculus argument you’ll meet in Topic 5.

Surface area formulas

Surface area is the total area of every face. For prisms and pyramids, work it out face by face — there’s no single shortcut formula. For cylinders, cones, and spheres, dedicated formulas exist.

Sphere — total surface area A = 4πr²

Cylinder

curved: 2πrh
total: 2πr² + 2πrh

two circles + rectangle wrapped around

Cone

curved: πrl
total: πr² + πrl

l = slant height (not vertical)

Slant height vs vertical height: in a cone, h is the perpendicular height from base to apex; l is the slant — the distance from the apex to a point on the base edge. They’re related by l = √(r² + h²) (Pythagoras). Use h for volume; use l for curved surface.

Prisms and pyramids — net-based approach

For a prism, sum: 2 × (cross-section area) + (perimeter of cross-section) × (length). For a pyramid, sum: base area + each triangular face. The slant heights of the triangular faces almost always need a Pythagorean calculation first.

🧭 Recipe — solving volume & surface area problems

Identify the shape: prism, cylinder, pyramid, cone, sphere — or composite of these.
Sketch a labelled diagram. Mark every given length and decide which is radius, which is height, which is slant height.
Pick the right formula from the volume or surface-area list. For composites, split into pieces first.
For surface area: list every external face. Don’t double-count internal faces in composites; don’t miss any external face that appears at a join.
For pyramids/cones: check whether the height given is vertical (use for volume) or slant (use for SA). Convert via Pythagoras if needed.
Substitute and simplify. Leave answers in terms of π unless decimals are requested.
Sanity-check units: volume = (length)³; surface area = (length)². If the units don’t match, something is wrong.

Worked examples

WE 1

Volume of a triangular prism

A triangular prism has a right-angled triangular cross-section with legs 6 cm and 8 cm. The prism is 15 cm long. Find its volume.

Step 1: Find the cross-section area A = ½ × base × height = ½ × 6 × 8 A = 24 cm² Step 2: Apply V = Ah V = 24 × 15 V = 360 cm³ two-step process for any prism: find the cross-section first, then multiply by length

WE 2

Cylinder reverse problem — find the height

A cylinder has volume 250π cm³ and radius 5 cm. Find its height.

Step 1: Set up V = πr²h 250π = π × 5² × h 250π = 25πh Step 2: Solve for h h = 250π / (25π) = 10 h = 10 cm leaving V in exact form (250π not 785.4) keeps the algebra clean — the π cancels

WE 3

Volume of a cone

A right cone has radius 6 cm and vertical height 10 cm. Find its volume in (a) exact form, and (b) correct to 3 s.f.

Step 1: Apply V = ⅓πr²h V = ⅓ × π × 6² × 10 V = ⅓ × π × 36 × 10 = ⅓ × 360π (a) V = 120π cm³ Step 2: Decimal form 120π ≈ 376.99… (b) V ≈ 377 cm³ (3 s.f.) make sure h is the vertical height, not the slant — the cone formula uses perpendicular distance

WE 4

Sphere — surface area to volume

A solid sphere has surface area 100π cm². Find its volume in exact form.

Step 1: Use the surface area formula to find r 4πr² = 100π r² = 25 → r = 5 cm (positive root) Step 2: Apply V = (4/3)πr³ V = (4/3) × π × 5³ V = (4/3) × π × 125 = 500π/3 V = 500π/3 cm³ ≈ 524 cm³ (3 s.f.) two-step problem: SA gives r, then r gives V — always discard the negative root

WE 5

Surface area of a closed cylinder

A closed cylinder has radius 4 cm and height 10 cm. Find the total surface area in (a) exact form, and (b) correct to 3 s.f.

Step 1: Use SA = 2πr² + 2πrh SA = 2π × 4² + 2π × 4 × 10 Step 2: Compute each part separately two circles: 2π × 16 = 32π curved surface: 2π × 4 × 10 = 80π Step 3: Add SA = 32π + 80π = 112π (a) SA = 112π cm² 112π ≈ 351.86… (b) SA ≈ 352 cm² (3 s.f.) “closed” means both end-circles are included — for an open cylinder, drop one πr² term

WE 6

Surface area of a right square pyramid

A right pyramid has a square base of side 6 cm and a vertical height of 4 cm. Find its total surface area.

Step 1: Sketch — let M be the centre of the base, N the midpoint of one base edge MN = ½ × 6 = 3 cm (half the base side) vertical height VM = 4 cm Step 2: Find the slant height VN by Pythagoras VN² = VM² + MN² = 4² + 3² VN² = 16 + 9 = 25 → VN = 5 cm Step 3: Area of one triangular face A_tri = ½ × 6 × 5 = 15 cm² Step 4: Total surface area = base + 4 triangles SA = 6² + 4 × 15 SA = 36 + 60 SA = 96 cm² the slant height of the triangular face uses MN (half the base side), not the diagonal — and a 3-4-5 triangle pops out for free

💡 Top tips

Sketch and label every solid. Mark the radius, vertical height, and slant height clearly — that single step kills most errors.
Use the formula booklet. Volumes of all standard solids and curved-surface areas of cylinders/cones are listed there. Total surface areas of cylinders and cones are not given — build them yourself.
Distinguish vertical height (h) from slant height (l). h is for volume; l is for curved surface area on a cone. Use l = √(r² + h²) to convert.
For pyramids, the slant height of a triangular face uses Pythagoras with half the base side, not the full base or the base diagonal.
Leave answers in terms of π for exact-form requests. 120π is more precise than 376.99.
Composite shapes — split, then sum: identify each recognisable component, find each piece, then add (or subtract for hollow regions). Don’t double-count internal faces.
Check units carefully: volume in cm³, area in cm² — and convert if the question mixes mm, cm, and m.

⚠ Common mistakes

Using slant height in the volume formula. Cone volume needs vertical height h, not the slant l.
Forgetting the ⅓ factor for pyramids and cones. Without it, you’ve calculated a prism or cylinder’s volume — three times too big.
Using the base diagonal instead of half the side when finding the triangular-face slant height of a square pyramid.
Forgetting that a square pyramid has 4 identical triangles, not 1. Add four triangle areas to the base.
Including internal faces in a composite shape’s surface area. The surface where two solids join doesn’t count — it’s not external.
Mixing up “closed” and “open” cylinders: a closed cylinder has two circular caps; an open one (like a tube or pipe) has zero or one.
Cubing instead of squaring (or vice versa). Sphere volume is (4/3)πr³; sphere surface area is 4πr². Confusing these is a classic exam slip.
Premature rounding. Carry exact values through; round only at the final step.

And that closes Section 3.6 — Geometry of 3D Shapes. You’ve now got two full sections of Topic 3 in the bag: the Geometry Toolkit (coordinate, arc, sector, radian) and the Geometry of 3D Shapes (3D coordinates, volume, surface area). The next major sections of Topic 3 build on this with right-angled and non-right-angled trigonometry — SOH CAH TOA, the sine rule, and the cosine rule — followed by the unit circle, identities, equations, and modelling with trigonometric functions. Pythagoras and basic 2D/3D geometry will keep showing up as the foundation underneath all of it.

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