IB Maths AA SLTopic 3 — Geometry & TrigPaper 1 & 2~12 min read
Volume & Surface Area
Cuboids, cylinders, pyramids, cones, spheres — every 3D shape you’ll meet at SL has a clean formula for its volume and surface area. Most are in the formula booklet. Recognise the shape, pick the right formula, plug in the numbers.
📘 What you need to know
Prisms & cylinders use V = Ah, where A is the area of the cross-section and h is the length.
Pyramids & cones use V = 13Ah — exactly 13 the volume of the equivalent prism.
Sphere: V = 43πr3 and A = 4πr2.
For prisms and pyramids, surface area = add up the areas of all the faces. No single formula.
Most volume formulas are in the formula booklet — but full surface area formulas usually aren’t (only the curved-surface ones are).
Recognising the shapes
Five shapes cover everything you’ll meet. Two work as “stacked cross-sections” (prism, cylinder), two work as “base + apex” (pyramid, cone), and one stands alone (sphere).
Cuboid
a “box”
Cylinder
circular prism
Pyramid
base + apex
Cone
circular pyramid
Sphere
all points = r from centre
Triangular prism
any prism shape
A prism has the same cross-section all the way through. A pyramid tapers from a base to a single apex. A cylinder is a “circular prism” and a cone is a “circular pyramid” — they follow the same patterns.
Volume formulas
Prisms and cylinders — volume = base × length
For any shape that has the same cross-section all the way through, you just multiply the cross-section area by the length.
Volume of a prism
V = Ah |
Cuboid: V = lwh |
Cylinder: V = πr2h✓ all in formula booklet
Pyramids and cones — volume = ⅓ × base × height
Pyramids and cones have exactly one third the volume of the prism or cylinder they would fit inside. Same base, same height, just one-third as much volume.
Volume of a pyramid
V = 13Ah |
Cone: V = 13πr2h✓ in formula booklet
🧠 Memory trick — the “⅓” rule
Cones and pyramids always carry a 13. Imagine a cone or pyramid sitting inside a cylinder or prism with the same base and height. The cone/pyramid takes up exactly one third of the box — the rest is empty space.
Sphere — the special one
Volume of a sphere
V = 43πr3✓ in formula booklet
Cuboid
V = lwh
length × width × height
✓ booklet
Cylinder
V = πr2h
r = radius, h = height
✓ booklet
Pyramid
V = 13Ah
A = base area
✓ booklet
Cone
V = 13πr2h
h = vertical height
✓ booklet
Sphere
V = 43πr3
just one number: r
✓ booklet
Prism (general)
V = Ah
A = cross-section area
✓ booklet
Surface area formulas
Prisms and pyramids — add up the faces
For prisms and pyramids, there’s no shortcut formula — you have to identify each face, work out its area, and add them all together. Drawing a quick net (the shape unfolded flat) makes this much easier.
Cylinder, cone, sphere — use the formulas
For curved shapes, you’ve got proper formulas. The cylinder and cone formulas split into a “flat part” + a “curved part”:
Cylinder
A = 2πr2 + 2πrh
two circles + curved side
✓ curved part only
Cone
A = πr2 + πrl
base circle + curved side l = slant height
✓ curved part only
Sphere
A = 4πr2
no flat part!
✓ booklet
🤔 Why is the cylinder’s curved side 2πrh?
Imagine peeling the label off a soup can and laying it flat. You get a rectangle. Its height is just the cylinder’s height (h), and its width is the circumference of the circle (2πr). So the area of the curved surface is height × circumference = 2πrh.
Slant height vs vertical height (cone trap)
The cone formulas use two different heights. Don’t mix them up — this is one of the most common errors in this topic.
Vertical height (h)
used in volume formula
Slant height (l)
used in surface area formula
Pythagoras link: for a cone, l2 = r2 + h2. So if the question gives you only two of the three, you can always find the third.
Full reference table
Shape
Volume
Surface area
Cuboid
lwh
add 6 rectangles
Cylinder
πr2h
2πr2 + 2πrh
Cone
13πr2h
πr2 + πrl
Pyramid
13Ah
base + triangles
Sphere
43πr3
4πr2
All volume formulas are in the formula booklet. For surface area, only the curved-surface formulas (cylinder side, cone slant, sphere) are given — full surface area formulas usually aren’t.
Worked examples
WE 1
Volume of a cylinder
A water tank is a cylinder of radius 4 m and height 10 m. Find its volume in m3, leaving your answer as a multiple of π.
Step 1: Pick the formulaV = πr2hStep 2: Substitute r = 4, h = 10V = π(4)2(10) = π × 16 × 10V = 160π m3≈ 503 m³ if a decimal is asked for
WE 2
Volume of a cone
A cone has radius 5 cm and vertical height 9 cm. Find its volume to 3 s.f.
Step 1: Pick the formulaV = 13πr2hStep 2: Substitute r = 5, h = 9V = 13π(5)2(9) = 13 × 225πStep 3: Compute= 75πV ≈ 236 cm3 (3 s.f.)don’t forget the 1/3 — it’s the #1 mistake on cones
WE 3
Composite — ice cream cone (SME-style)
A dessert is modelled as a right cone of radius 3 cm and height 12 cm with a scoop of ice cream as a sphere of radius 3 cm sitting on top. Find the total volume to 3 s.f.
Step 1: Volume of the sphereVsphere = 43π(3)3 = 43π × 27 = 36πStep 2: Volume of the coneVcone = 13π(3)2(12) = 13 × 9 × 12 × π = 36πStep 3: Add themVtotal = 36π + 36π = 72πV ≈ 226 cm3 (3 s.f.)nice symmetry — sphere and cone happen to have equal volumes here!
WE 4
Surface area — basic plug-in
A football is approximately a sphere of radius 11 cm. Find its surface area to 3 s.f.
Step 1: Pick the formulaA = 4πr2Step 2: Substitute r = 11A = 4π(11)2 = 484πStep 3: ComputeA ≈ 1520 cm2 (3 s.f.)remember units — area is cm²!
WE 5
Pyramid surface area (SME-style)
ABCD is the square base of a right pyramid with vertex V. The centre of the base is M. The sides of the square base are 3.6 cm and the vertical height VM is 8.2 cm. N is the midpoint of side BC. Find:
(i) the slant height VN (ii) the area of triangle ABV (iii) the surface area of the pyramid
(i) Slant height VN — use Pythagoras in triangle MNVM is the centre, N is midpoint of BC so MN = 3.6 ÷ 2 = 1.8 cmVN2 = 8.22 + 1.82 = 67.24 + 3.24 = 70.48VN = √70.48 ≈ 8.40 cm (3 s.f.)(ii) Area of triangle ABV
Base AB = 3.6 cm, height = VN = √70.48 cm
Area = 12(3.6)(√70.48) = 15.111…Area ≈ 15.1 cm2(iii) Total surface area
Net = 1 square base + 4 identical triangles.
SA = 3.62 + 4 × 15.111… = 12.96 + 60.44…SA ≈ 73.4 cm2 (3 s.f.)always sketch the net — it stops you missing faces
💡 Top tips
Identify the shape first. Is it a prism, pyramid, or sphere? That tells you which family of formulas to use.
For composite shapes (e.g. ice cream on cone), break into separate pieces, find each volume, then add or subtract.
Use Pythagoras to find slant height (cones, pyramids) when only vertical height and radius/half-base are given.
Sketch the net for prisms and pyramids — it forces you to count every face.
Leave answers in terms of π in your working, then convert to a decimal only at the end if asked.
Check units carefully: volume is cm3, surface area is cm2. Wrong units lose easy marks.
Volume formulas are in the booklet, but full surface area formulas usually aren’t — memorise A = 2πr2 + 2πrh and A = πr2 + πrl.
⚠ Common mistakes
Forgetting the ⅓ for cones and pyramids. Easy to write πr2h instead of 13πr2h — and you’ll get triple the right answer.
Confusing slant height (l) with vertical height (h). Volume uses h; cone surface area uses l. Don’t mix them up.
Cubing the wrong thing in the sphere volume. It’s r3, not r2 — small mistake, huge consequences.
Missing faces in surface area. Always draw the net and count all faces — for a pyramid that’s 1 base + 4 triangles, not just 4 triangles.
Forgetting one of the cylinder’s circular ends. Full surface area = 2 circles + curved side, not 1 circle + curved side.
Wrong units in the final answer. cm3 for volume, cm2 for surface area — never both the same.
Using diameter instead of radius. If the question gives you the diameter, halve it first. r = d ÷ 2.
That’s the full Geometry of 3D Shapes done — you can now find any volume or surface area in the syllabus. These come up in nearly every IB paper as part of context-rich problems, so practice with composite shapes (ice creams, lampshades, towers) where you combine two formulas.
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