Do I need to memorise all these formulas for the exam?

No. Every volume and surface-area formula you need sits in the 'Prior learning' and 'Topic 3: Geometry and trigonometry' pages of the AI formula booklet. Your job is to recognise the solid, choose the matching formula, and know what each variable means in context. Keep the booklet open the whole time.

What is the difference between the slant height $l$ and the perpendicular height $h$?

The perpendicular height $h$ runs straight down from the apex to the centre of the base; the slant height $l$ runs along the sloping surface from the apex to the edge of the base. With the radius (for a cone) or half the base side (for a pyramid) they form a right-angled triangle, so $l^2 = h^2 + r^2$. Volume formulas use $h$; the curved-surface formula for a cone uses $l$. Mixing them up is one of the most common lost marks.

How do I handle units, and what about capacity?

Stay consistent: if lengths are in centimetres, areas come out in cm$^2$ and volumes in cm$^3$. Watch for a change of unit in the question — capacity is often wanted in litres, and $1000\text{ cm}^3 = 1$ litre, while $1\text{ m}^3 = 1000$ litres. Convert only at the end, from your full unrounded figure.

What if the shape is 'open', like a bucket or a pipe?

Only count the faces that actually exist. An open cylindrical bucket has a base and a curved wall but no top, so its surface area is $\pi r^2 + 2\pi r h$. A pipe open at both ends is just the curved wall, $2\pi r h$. Read the wording carefully — 'open', 'without a lid' and 'hollow' all change which faces you include.

How do I find an angle between a line and a plane inside a solid?

Project the line onto the plane: drop a perpendicular from the far end of the line down to the plane, then join that foot back to where the line meets the plane. The angle you want lives in the right-angled triangle made by the line, its projection and the perpendicular. Sketch that triangle, label the two sides you know, and use $\tan$, $\sin$ or $\cos$.

Distances, volumes and surface areas of 3D…

Revise Distances, volumes and surface areas of 3D solids for IB Mathematics: Applications and Interpretation (3.1)

Looking for free IB Mathematics: Applications and Interpretation notes on Distances, volumes and surface areas of 3D solids? This Higher Level lesson covers official syllabus point 3.1 on Maths AI HL. Use the visual steps, key terms, and quick checks below — then practise with IB criterion marking on Distances, volumes and surface areas of 3D solids. Built for students searching "IB Maths AI 3.1", "Distances, volumes and surface areas of 3D solids IB revision", or a free alternative to paid IB note sites.

Distances, volumes and surface areas of 3D solids

From packing boxes and pouring concrete to designing an ice-cream cone, three-dimensional measurement is everywhere. This lesson covers the full 3.1 syllabus in three moves. First, the mensuration formulas — volume and surface area of prisms, cylinders, pyramids, cones and spheres — read from the formula booklet and applied with the right units. Second, composite solids, where you combine simple shapes and are careful to paper only the faces that show. Third, distances and angles inside a solid, where the whole trick is to uncover a right-angled triangle and fall back on Pythagoras and basic trigonometry. Because 3.1 is a calculator topic, the GDC handles the arithmetic; what earns the marks is the correct formula, the method on the page, and honest units. As in every calculation topic, the worked examples below are laid out exactly the way our engine marks your own practice.

BY THE END, YOU CAN…

1Calculate the volume and surface area of prisms, cylinders, pyramids, cones and spheres, selecting the correct formula from the IB formula booklet and attaching the correct units.
2Solve problems involving composite solids by adding or subtracting volumes and by summing only the exposed surfaces.
3Find the distance between two points inside a 3D solid using Pythagoras' theorem, including the space diagonal of a cuboid.
4Find the size of an angle in a 3D solid, including the angle between a line and a plane, by identifying and solving a right-angled triangle.
5Apply 3D geometry to contextual problems with correct units, unit conversions and appropriate rounding.

In simple terms

A friendly intro before the formal notes — no formulas yet.

3D Shapes: The Architect's Toolkit

This topic is about measuring three-dimensional objects: the space they enclose (volume) and the skin that wraps them (surface area), plus the lengths and angles hidden inside them. Two habits carry every question — pick the right formula from the booklet, and find the right-angled triangle buried in the solid.

THINK OF IT LIKE…

Think of wrapping a present and then filling the box. Surface area is the wrapping paper you need to cover the outside; volume is how much you can pack inside. For a complicated present — say a toy rocket that is a cone stacked on a cylinder — you deal with each part separately and only paper the faces that actually show. And when you need a length across the box, from one bottom corner to the opposite top corner, you slice out a flat right-angled triangle and use Pythagoras.

1
Identify the solid, or the simple solids that a composite is built from (cylinder, cone, sphere, prism, pyramid).
2
Choose the correct formula from the booklet for volume or for the surface you need — curved only, or total.
3
Substitute the given measurements; for slant heights and diagonals, first find the missing length from a right-angled triangle.
4
Compute on the GDC, attach cubic units to volumes and square units to areas, and round as instructed (usually 3 s.f.).

Explore the concept

Use the live diagram and synced steps — play it or tap a step card to walk through.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

$V = \pi r^2 h$

$a^2 + b^2 = c^2$

Full topic notes

Formal explanation with the rigour you need for the exam.

Using the formula booklet

For this topic the formula booklet is your most valuable tool. It lists every volume and surface-area formula you need, so nothing has to be memorised. The skill being examined is choosing the right formula and knowing what each letter stands for. Before substituting anything, decide three things: which solid you are dealing with, whether you want volume or surface area, and — for surface area — whether the question wants the curved surface only or the total surface.

Cylinder: $V = \pi r^2 h$ , \quad curved surface $= 2\pi r h$ , \quad total surface $= 2\pi r h + 2\pi r^2$ . \n Cone: $V = \frac{1}{3}\pi r^2 h$ , \quad curved surface $= \pi r l$ , \quad total surface $= \pi r l + \pi r^2$ . \n Sphere: $V = \frac{4}{3}\pi r^3$ , \quad surface $= 4\pi r^2$ . \quad Pyramid: $V = \frac{1}{3} A_b h$ .

Identify the solid first, then locate the matching formula in the Topic 3 pages of the booklet.
Know your letters: $r$ radius, $h$ perpendicular height, $l$ slant height, $A_b$ base area.
Curved or total? For cones and cylinders decide whether a base (or two) is included before you write anything.
Units follow the powers: lengths give areas in square units and volumes in cubic units.

Composite solids: combining shapes

Many real objects are not single solids but combinations — a silo (cylinder plus hemisphere), a pencil (cylinder plus cone), a capsule. For VOLUME you add the parts, or subtract when one solid is hollowed out of another. For SURFACE AREA the rule is subtler: count only the surfaces exposed on the outside. Wherever two solids are joined, both faces at the join vanish, so never include them. Decide face by face what a real observer would actually see.

Distances and angles in 3D

Often you must find a length or angle that is not given directly — a diagonal across a box, the slant edge of a pyramid, the angle a support wire makes with the ground. The universal method is to uncover a right-angled triangle inside the solid that contains what you want. Slice the solid, draw that 2D triangle on its own, label the sides you know, and the 3D problem collapses into ordinary Pythagoras or trigonometry.

Pythagoras: $a^2 + b^2 = c^2$ . \n Space diagonal of a cuboid $l \times w \times h$ : $d = \sqrt{l^2 + w^2 + h^2}$ . \n Right-angled trig: $\sin\theta = \dfrac{\text{opp}}{\text{hyp}}, \quad \cos\theta = \dfrac{\text{adj}}{\text{hyp}}, \quad \tan\theta = \dfrac{\text{opp}}{\text{adj}}.$

The angle between a line and a plane needs one extra idea: the line's projection, or 'shadow', on the plane. Drop a perpendicular from the top of the line straight down to the plane, then join that foot back to where the line touches the plane. The angle between the line and its shadow is the angle you want, and it lives in a right-angled triangle. The most common error is to measure to a convenient edge instead of to the true projection.

EXAM TIP

Whenever a question asks for a length or angle inside a 3D shape, redraw the relevant right-angled triangle on its own, away from the clutter of the solid. Mark the right angle, label the known sides, and only then choose Pythagoras or a trig ratio. This single habit turns most 3D questions into routine 2D ones.

Applications and units

Contextual questions dress these formulas in real quantities: litres of water, kilograms of metal, square metres of paint. Two disciplines protect the marks. First, keep the units consistent throughout and convert only at the very end, from the full unrounded figure — remember $1000\text{ cm}^3 = 1$ litre and $1\text{ m}^3 = 1000$ litres. Second, read whether the object is solid or hollow, open or closed, before deciding which faces and which volume actually count. A watering can, a concrete pipe and a solid ball are three different surface-area problems even if they share a radius.

Common mistakes examiners penalise

Using the slant height where the perpendicular height belongs (or the reverse) — volume formulas need the vertical height $h$ ; the cone's curved-surface formula $\pi r l$ needs the slant height $l$ . Find $l$ from $l^2 = h^2 + r^2$ and keep the two straight.
Quoting $\pi r l$ as the total surface of a cone — that is the curved surface only; add $\pi r^2$ for a closed cone. State clearly whether the question wants curved or total.
Mixing up the sphere and cone formulas — a sphere is $V = \frac{4}{3}\pi r^3$ and $A = 4\pi r^2$ ; a cone is $V = \frac{1}{3}\pi r^2 h$ . Halve the sphere formulas (correctly) for a hemisphere.
Getting the units wrong — areas are in square units and volumes in cubic units; a length squared is not a length. When a cube of side $a$ is asked for, its surface is $6a^2$ and its volume is $a^3$ , not $a^2$ or $6a$ .
Measuring the wrong angle between a line and a plane — project the line onto the plane and measure to that shadow, not to a nearby edge; the correct angle sits in the triangle made by the line, its projection and the perpendicular.
Including hidden faces in a composite surface area — the two faces at a join disappear; count only what is exposed. Equally, do not drop a real face on an 'open' solid you should have kept.
Converting capacity incorrectly — $1000\text{ cm}^3 = 1$ litre and $1\text{ m}^3 = 1000$ litres. Convert once, at the end, from the unrounded value.
Over-rounding mid-calculation — carry the GDC's full figures (and keep $\pi$ exact where you can) and round only the final answer, to 3 significant figures unless told otherwise.

Where this leads

The right-angled-triangle habit you build here is the backbone of the rest of Topic 3: it reappears in the sine and cosine rules for non-right triangles, in bearings and navigation, and in the trigonometry of the unit circle. The mensuration formulas feed directly into optimisation and modelling questions later in the course, where you minimise the surface area of a can or maximise the volume of a box under a constraint. And the discipline this topic drills — name the solid, choose the formula, show the method, keep the units honest — is exactly what every Paper 2 geometry question rewards.

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

A cone has base radius 6 cm and vertical height 8 cm. Find its slant height, its total surface area and its volume. (Give answers to 3 s.f.) [6]

1
Model answer — full working.\n\nSlant height. The radius, the vertical height and the slant height form a right-angled triangle, so by Pythagoras\n $l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ cm}.$ \n\nTotal surface area. For a closed cone, $TSA = \pi r^2 + \pi r l$ :\n $TSA = \pi(6)^2 + \pi(6)(10) = 36\pi + 60\pi = 96\pi \approx 301.59\ldots$ \n $TSA \approx 302 \text{ cm}^2 \ (3\text{ s.f.}).$ \n\nVolume. $V = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\pi(6)^2(8) = \tfrac{1}{3}\pi(288) = 96\pi \approx 301.59\ldots$ \n $V \approx 302 \text{ cm}^3 \ (3\text{ s.f.}).$ \n\n(The surface area and volume share the value $96\pi$ here only by coincidence — note the different units, cm $^2$ against cm $^3$ .)\n\n---\nHow our marking engine awards the 6 marks:\n\n- M1 — Pythagoras for the slant height. Awarded for using $l = \sqrt{r^2 + h^2}$ (or equivalent) with the radius and vertical height. It is the method that scores, so it survives an arithmetic slip.\n- A1 — $l = 10$ cm. The accuracy mark for the correct slant height. It depends on the M1 above.\n- M1 — total-surface formula. Awarded for $\pi r^2 + \pi r l$ with the slant height (not the vertical height) substituted. Using $\pi r l$ alone — the curved surface only — would miss this method.\n- A1 — $TSA \approx 302$ cm $^2$ . The dependent accuracy mark. Follow-through (FT) applies: a candidate whose slant height was slightly wrong but who substitutes it correctly into $\pi r^2 + \pi r l$ still earns this on their own figure. Correct square units are part of the mark.\n- M1 — volume formula. Awarded for $\tfrac{1}{3}\pi r^2 h$ with the VERTICAL height 8 (not the slant height 10).\n- A1 — $V \approx 302$ cm $^3$ . The dependent accuracy mark, with cubic units.\n\n**'Accept equivalent forms and correct rounding.'** The engine accepts $96\pi$ , $301.6$ or $302$ for either answer, and accepts $l$ written as $\sqrt{100}$ or $10$ . Once a correct value appears, subsequent working is ignored (ISW). The two method marks are what protect your score if a number slips — but only if the working is on the page.

✓ FULLY WORKED

EXAMPLE 2

A grain silo is a cylinder of radius 4 m and height 15 m, topped by a hemisphere of the same radius, and standing on a flat circular base. Find (a) the total volume in m $^3$ , and (b) the total exterior surface area, including the base. (3 s.f.) [6]

1
(a) Volume.\nCylinder: $V_{\text{cyl}} = \pi r^2 h = \pi(4)^2(15) = 240\pi$ m $^3$ .\nHemisphere: $V_{\text{hemi}} = \tfrac{1}{2}\cdot\tfrac{4}{3}\pi r^3 = \tfrac{2}{3}\pi(4)^3 = \tfrac{128}{3}\pi$ m $^3$ .\nTotal $= 240\pi + \tfrac{128}{3}\pi = \tfrac{848}{3}\pi \approx 888.02\ldots$ [M1: add the two correct volumes]\nTotal volume $\approx 888$ m $^3$ (3 s.f.). [A1]\n\n**(b) Surface area.** The exposed faces are the circular base, the curved cylinder wall, and the curved hemisphere; the flat top of the cylinder and the flat face of the hemisphere are hidden at the join.\nBase $= \pi r^2 = 16\pi$ m $^2$ .\nCylinder wall $= 2\pi r h = 2\pi(4)(15) = 120\pi$ m $^2$ .\nHemisphere curved $= \tfrac{1}{2}\cdot 4\pi r^2 = 2\pi r^2 = 32\pi$ m $^2$ .\nTotal $= 16\pi + 120\pi + 32\pi = 168\pi \approx 527.79\ldots$ [M1: add only the exposed areas]\nTotal surface area $\approx 528$ m $^2$ (3 s.f.). [A1]\n\nMarks: M1 A1 for the volume, M1 A1 for the surface area. The M-marks reward the correct combination — adding both volumes, and adding only the three exposed areas. Including the hidden join faces would forfeit the surface-area M1.

✓ FULLY WORKED

EXAMPLE 3

A cuboid-shaped room is 8 m long, 6 m wide and 3 m high. A wire runs from one bottom corner to the diagonally opposite top corner. Find (a) the length of the wire, and (b) the angle the wire makes with the floor. (3 s.f.) [5]

1
(a) Length of the wire (the space diagonal).\nFirst the diagonal of the floor: $f = \sqrt{8^2 + 6^2} = \sqrt{100} = 10$ m. [M1: Pythagoras on the base]\nNow the space diagonal, using the floor diagonal and the height:\n $d = \sqrt{f^2 + 3^2} = \sqrt{10^2 + 3^2} = \sqrt{109} \approx 10.44$ m.\n(Equivalently $d = \sqrt{8^2 + 6^2 + 3^2} = \sqrt{109}$ .)\nWire length $\approx 10.4$ m (3 s.f.). [A1]\n\n**(b) Angle with the floor.** The wire, its shadow on the floor (the floor diagonal, 10 m) and the vertical height (3 m) form a right-angled triangle. The angle $\theta$ at the floor satisfies\n $\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{3}{10}.$ [M1: correct triangle and ratio]\n $\theta = \arctan\!\left(\dfrac{3}{10}\right) = 16.699\ldots^\circ.$ [A1]\nAngle with the floor $\approx 16.7^\circ$ (3 s.f.). [A1]\n\nMarks: M1 A1 for the diagonal, then M1 A1 A1 for the angle. Follow-through applies — a candidate who uses their own (slightly different) floor diagonal in $\tan\theta = 3/f$ still earns the angle marks. Note the angle uses the vertical height opposite and the floor diagonal adjacent, NOT an edge of the room.

✓ FULLY WORKED

EXAMPLE 4

A right pyramid has a square base of side 16 cm and perpendicular height 15 cm. Find (a) its volume, (b) its slant height, and (c) its total surface area. [6]

1
(a) Volume. Base area $A_b = 16 \times 16 = 256$ cm $^2$ .\n $V = \tfrac{1}{3} A_b h = \tfrac{1}{3}(256)(15) = 1280$ cm $^3$ . [M1 A1]\n\n**(b) Slant height.** The slant height reaches the MIDPOINT of a base edge, so the right-angled triangle uses the perpendicular height and HALF the base side, 8 cm:\n $l = \sqrt{h^2 + 8^2} = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17$ cm. [M1 A1]\n\n**(c) Total surface area.** Base plus four identical triangular faces, each of area $\tfrac{1}{2} \times \text{base} \times \text{slant height}$ :\none triangle $= \tfrac{1}{2}(16)(17) = 136$ cm $^2$ ,\n $TSA = 256 + 4(136) = 256 + 544 = 800$ cm $^2$ . [M1 A1]\n\nMarks: M1 A1 three times. The slant-height M1 rewards using half the base side (8), not the full side (16) and not the radius idea from a cone; the surface M1 rewards adding the base to the four sloping faces.

✓ FULLY WORKED

How it all connects

The big idea sits in the middle — tap a linked idea to explore the link.

MAIN IDEADistances, volumes and surface areas of 3D solids

Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.

Glossary

Try to recall each definition before you reveal it.

Quick check

Answer in your head first — then tap to check. No pressure.

Revision flashcards

Flip the card. Test yourself before the exam.

QUESTION · 1 / 13

Volume of a pyramid

TAP OR SPACE TO FLIP

ANSWER

$V = \frac{1}{3} A_b h$ , where $A_b$ is the area of the base and $h$ is the perpendicular (vertical) height — not the slant height.

TAP OR SPACE TO FLIP BACK

1 / 13 · ← → space

Key takeaways

Review these before you close the topic — retrieval beats re-reading.

✓
Identify the solid first, then locate the matching formula in the Topic 3 pages of the booklet.
✓
Know your letters: $r$ radius, $h$ perpendicular height, $l$ slant height, $A_b$ base area.
✓
Curved or total? For cones and cylinders decide whether a base (or two) is included before you write anything.
✓
Units follow the powers: lengths give areas in square units and volumes in cubic units.

Practice — then mark it

The whole point: a real Cambridge question, marked mark-by-mark.

Real past paper[0]

Get a Paper 2 solid-geometry problem marked: show the formula, the right-angled triangle and the units with full working

Do it on paper → mark itMARKED MARK-BY-MARK · B1 / M1 / A1 · OFFICIAL SCHEME

Extra simulations & links

PhET, GeoGebra and other curated tools — open in a new tab.

Frequently asked

✓

Checkpoint

One marked question is worth ten re-reads — close the loop before you move on.

Checkpoint

Reading it isn’t knowing it — prove it.

Before you move on: do Get a Paper 2 solid-geometry problem marked: show the formula, the right-angled triangle and the units with full working on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.

Attempt & get marked →Takes about a minute · you’ll land back on this lesson

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Distances, volumes and surface areas of 3D solids

BY THE END, YOU CAN…

1Calculate the volume and surface area of prisms, cylinders, pyramids, cones and spheres, selecting the correct formula from the IB formula booklet and attaching the correct units.
2Solve problems involving composite solids by adding or subtracting volumes and by summing only the exposed surfaces.
3Find the distance between two points inside a 3D solid using Pythagoras' theorem, including the space diagonal of a cuboid.
4Find the size of an angle in a 3D solid, including the angle between a line and a plane, by identifying and solving a right-angled triangle.
5Apply 3D geometry to contextual problems with correct units, unit conversions and appropriate rounding.

In simple terms

A friendly intro before the formal notes — no formulas yet.

3D Shapes: The Architect's Toolkit

THINK OF IT LIKE…

1
Identify the solid, or the simple solids that a composite is built from (cylinder, cone, sphere, prism, pyramid).
2
Choose the correct formula from the booklet for volume or for the surface you need — curved only, or total.
3
Substitute the given measurements; for slant heights and diagonals, first find the missing length from a right-angled triangle.
4
Compute on the GDC, attach cubic units to volumes and square units to areas, and round as instructed (usually 3 s.f.).

Explore the concept

Use the live diagram and synced steps — play it or tap a step card to walk through.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

$V = \pi r^2 h$

$a^2 + b^2 = c^2$

Full topic notes

Formal explanation with the rigour you need for the exam.

Using the formula booklet

Identify the solid first, then locate the matching formula in the Topic 3 pages of the booklet.
Know your letters: $r$ radius, $h$ perpendicular height, $l$ slant height, $A_b$ base area.
Curved or total? For cones and cylinders decide whether a base (or two) is included before you write anything.
Units follow the powers: lengths give areas in square units and volumes in cubic units.

Composite solids: combining shapes

Distances and angles in 3D

EXAM TIP

Applications and units

Common mistakes examiners penalise

Using the slant height where the perpendicular height belongs (or the reverse) — volume formulas need the vertical height $h$ ; the cone's curved-surface formula $\pi r l$ needs the slant height $l$ . Find $l$ from $l^2 = h^2 + r^2$ and keep the two straight.
Quoting $\pi r l$ as the total surface of a cone — that is the curved surface only; add $\pi r^2$ for a closed cone. State clearly whether the question wants curved or total.
Mixing up the sphere and cone formulas — a sphere is $V = \frac{4}{3}\pi r^3$ and $A = 4\pi r^2$ ; a cone is $V = \frac{1}{3}\pi r^2 h$ . Halve the sphere formulas (correctly) for a hemisphere.
Getting the units wrong — areas are in square units and volumes in cubic units; a length squared is not a length. When a cube of side $a$ is asked for, its surface is $6a^2$ and its volume is $a^3$ , not $a^2$ or $6a$ .
Measuring the wrong angle between a line and a plane — project the line onto the plane and measure to that shadow, not to a nearby edge; the correct angle sits in the triangle made by the line, its projection and the perpendicular.
Including hidden faces in a composite surface area — the two faces at a join disappear; count only what is exposed. Equally, do not drop a real face on an 'open' solid you should have kept.
Converting capacity incorrectly — $1000\text{ cm}^3 = 1$ litre and $1\text{ m}^3 = 1000$ litres. Convert once, at the end, from the unrounded value.
Over-rounding mid-calculation — carry the GDC's full figures (and keep $\pi$ exact where you can) and round only the final answer, to 3 significant figures unless told otherwise.

Where this leads

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

A cone has base radius 6 cm and vertical height 8 cm. Find its slant height, its total surface area and its volume. (Give answers to 3 s.f.) [6]

1
Model answer — full working.\n\nSlant height. The radius, the vertical height and the slant height form a right-angled triangle, so by Pythagoras\n $l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ cm}.$ \n\nTotal surface area. For a closed cone, $TSA = \pi r^2 + \pi r l$ :\n $TSA = \pi(6)^2 + \pi(6)(10) = 36\pi + 60\pi = 96\pi \approx 301.59\ldots$ \n $TSA \approx 302 \text{ cm}^2 \ (3\text{ s.f.}).$ \n\nVolume. $V = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\pi(6)^2(8) = \tfrac{1}{3}\pi(288) = 96\pi \approx 301.59\ldots$ \n $V \approx 302 \text{ cm}^3 \ (3\text{ s.f.}).$ \n\n(The surface area and volume share the value $96\pi$ here only by coincidence — note the different units, cm $^2$ against cm $^3$ .)\n\n---\nHow our marking engine awards the 6 marks:\n\n- M1 — Pythagoras for the slant height. Awarded for using $l = \sqrt{r^2 + h^2}$ (or equivalent) with the radius and vertical height. It is the method that scores, so it survives an arithmetic slip.\n- A1 — $l = 10$ cm. The accuracy mark for the correct slant height. It depends on the M1 above.\n- M1 — total-surface formula. Awarded for $\pi r^2 + \pi r l$ with the slant height (not the vertical height) substituted. Using $\pi r l$ alone — the curved surface only — would miss this method.\n- A1 — $TSA \approx 302$ cm $^2$ . The dependent accuracy mark. Follow-through (FT) applies: a candidate whose slant height was slightly wrong but who substitutes it correctly into $\pi r^2 + \pi r l$ still earns this on their own figure. Correct square units are part of the mark.\n- M1 — volume formula. Awarded for $\tfrac{1}{3}\pi r^2 h$ with the VERTICAL height 8 (not the slant height 10).\n- A1 — $V \approx 302$ cm $^3$ . The dependent accuracy mark, with cubic units.\n\n**'Accept equivalent forms and correct rounding.'** The engine accepts $96\pi$ , $301.6$ or $302$ for either answer, and accepts $l$ written as $\sqrt{100}$ or $10$ . Once a correct value appears, subsequent working is ignored (ISW). The two method marks are what protect your score if a number slips — but only if the working is on the page.

✓ FULLY WORKED

EXAMPLE 2

1
(a) Volume.\nCylinder: $V_{\text{cyl}} = \pi r^2 h = \pi(4)^2(15) = 240\pi$ m $^3$ .\nHemisphere: $V_{\text{hemi}} = \tfrac{1}{2}\cdot\tfrac{4}{3}\pi r^3 = \tfrac{2}{3}\pi(4)^3 = \tfrac{128}{3}\pi$ m $^3$ .\nTotal $= 240\pi + \tfrac{128}{3}\pi = \tfrac{848}{3}\pi \approx 888.02\ldots$ [M1: add the two correct volumes]\nTotal volume $\approx 888$ m $^3$ (3 s.f.). [A1]\n\n**(b) Surface area.** The exposed faces are the circular base, the curved cylinder wall, and the curved hemisphere; the flat top of the cylinder and the flat face of the hemisphere are hidden at the join.\nBase $= \pi r^2 = 16\pi$ m $^2$ .\nCylinder wall $= 2\pi r h = 2\pi(4)(15) = 120\pi$ m $^2$ .\nHemisphere curved $= \tfrac{1}{2}\cdot 4\pi r^2 = 2\pi r^2 = 32\pi$ m $^2$ .\nTotal $= 16\pi + 120\pi + 32\pi = 168\pi \approx 527.79\ldots$ [M1: add only the exposed areas]\nTotal surface area $\approx 528$ m $^2$ (3 s.f.). [A1]\n\nMarks: M1 A1 for the volume, M1 A1 for the surface area. The M-marks reward the correct combination — adding both volumes, and adding only the three exposed areas. Including the hidden join faces would forfeit the surface-area M1.

✓ FULLY WORKED

EXAMPLE 3

1
(a) Length of the wire (the space diagonal).\nFirst the diagonal of the floor: $f = \sqrt{8^2 + 6^2} = \sqrt{100} = 10$ m. [M1: Pythagoras on the base]\nNow the space diagonal, using the floor diagonal and the height:\n $d = \sqrt{f^2 + 3^2} = \sqrt{10^2 + 3^2} = \sqrt{109} \approx 10.44$ m.\n(Equivalently $d = \sqrt{8^2 + 6^2 + 3^2} = \sqrt{109}$ .)\nWire length $\approx 10.4$ m (3 s.f.). [A1]\n\n**(b) Angle with the floor.** The wire, its shadow on the floor (the floor diagonal, 10 m) and the vertical height (3 m) form a right-angled triangle. The angle $\theta$ at the floor satisfies\n $\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{3}{10}.$ [M1: correct triangle and ratio]\n $\theta = \arctan\!\left(\dfrac{3}{10}\right) = 16.699\ldots^\circ.$ [A1]\nAngle with the floor $\approx 16.7^\circ$ (3 s.f.). [A1]\n\nMarks: M1 A1 for the diagonal, then M1 A1 A1 for the angle. Follow-through applies — a candidate who uses their own (slightly different) floor diagonal in $\tan\theta = 3/f$ still earns the angle marks. Note the angle uses the vertical height opposite and the floor diagonal adjacent, NOT an edge of the room.

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EXAMPLE 4

A right pyramid has a square base of side 16 cm and perpendicular height 15 cm. Find (a) its volume, (b) its slant height, and (c) its total surface area. [6]

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(a) Volume. Base area $A_b = 16 \times 16 = 256$ cm $^2$ .\n $V = \tfrac{1}{3} A_b h = \tfrac{1}{3}(256)(15) = 1280$ cm $^3$ . [M1 A1]\n\n**(b) Slant height.** The slant height reaches the MIDPOINT of a base edge, so the right-angled triangle uses the perpendicular height and HALF the base side, 8 cm:\n $l = \sqrt{h^2 + 8^2} = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17$ cm. [M1 A1]\n\n**(c) Total surface area.** Base plus four identical triangular faces, each of area $\tfrac{1}{2} \times \text{base} \times \text{slant height}$ :\none triangle $= \tfrac{1}{2}(16)(17) = 136$ cm $^2$ ,\n $TSA = 256 + 4(136) = 256 + 544 = 800$ cm $^2$ . [M1 A1]\n\nMarks: M1 A1 three times. The slant-height M1 rewards using half the base side (8), not the full side (16) and not the radius idea from a cone; the surface M1 rewards adding the base to the four sloping faces.

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How it all connects

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MAIN IDEADistances, volumes and surface areas of 3D solids

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QUESTION · 1 / 13

Volume of a pyramid

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ANSWER

$V = \frac{1}{3} A_b h$ , where $A_b$ is the area of the base and $h$ is the perpendicular (vertical) height — not the slant height.

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Key takeaways

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Identify the solid first, then locate the matching formula in the Topic 3 pages of the booklet.
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Know your letters: $r$ radius, $h$ perpendicular height, $l$ slant height, $A_b$ base area.
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Curved or total? For cones and cylinders decide whether a base (or two) is included before you write anything.
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Units follow the powers: lengths give areas in square units and volumes in cubic units.

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Get a Paper 2 solid-geometry problem marked: show the formula, the right-angled triangle and the units with full working

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Revise Distances, volumes and surface areas of 3D solids for IB Mathematics: Applications and Interpretation (3.1)

In simple terms

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Key formulas

Full topic notes

Using the formula booklet

Composite solids: combining shapes

Distances and angles in 3D

Applications and units

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Where this leads

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How it all connects

Glossary

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Practice — then mark it

Get a Paper 2 solid-geometry problem marked: show the formula, the right-angled triangle and the units with full working

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

In simple terms

Explore the concept

Key formulas

Full topic notes

Using the formula booklet

Composite solids: combining shapes

Distances and angles in 3D

Applications and units

Common mistakes examiners penalise

Where this leads

Worked examples

How it all connects

Glossary

Quick check

Revision flashcards

Key takeaways

Practice — then mark it

Get a Paper 2 solid-geometry problem marked: show the formula, the right-angled triangle and the units with full working

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

Discuss Distances, volumes and surface areas of 3D solids