Educerie · IB Diploma · Mathematics: analysis and approaches
Topic 3 Geometry and trigonometry · 3.1 Three-dimensional geometry
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Find the distance between two points in three dimensions | SL, HL | "Find AB" (2 marks), often exact, Paper 1 |
| Find the midpoint of a segment in three dimensions, or an endpoint from the midpoint | SL, HL | "Write down the coordinates of M" (1 to 2 marks) |
| Find the volume of a right pyramid, a right cone, a sphere, a hemisphere and combinations of these | SL, HL | "Find the volume of the solid" (3 to 4 marks), often in terms of π |
| Find the surface area of the same solids, counting only the faces on the outside | SL, HL | "Find the total surface area of the toy" (3 to 5 marks) |
| Work backwards from a volume or area to a missing length | SL, HL | "The sphere has volume 500 cm³. Find its radius" (2 to 3 marks), Paper 2 |
| Identify a right-angled triangle inside a solid and use it to find a length | SL, HL | "Find the height of the pyramid" (2 to 3 marks) |
| Find the angle between two intersecting lines, or between a line and a plane | SL, HL | "Find the angle between AG and the base ABCD" (2 to 3 marks) |
Before you start
You need Pythagoras' theorem, the plane distance and midpoint formulas, volumes of prisms and cylinders, and right-angled trigonometry (SOH CAH TOA, revisited in 3.2). The formula booklet gives every formula on this page except the hemisphere, which is half a sphere. It cannot spot the right-angled triangle for you, and that is where most of the marks are.
1The idea in one paragraph
A point in space needs three numbers, not two, so it is written (x, y, z). Nearly everything else on this page comes from one move: find a right-angled triangle and use Pythagoras or trigonometry on it. The distance between two points is Pythagoras used twice. The height of a pyramid, the slant height of a cone and the angle a line makes with a plane all come from a right-angled triangle standing inside the solid. Volumes and surface areas are booklet formulas; the skill is putting the right length into them, and remembering that when two solids are joined the face where they meet is no longer on the outside.
2Points in three dimensions
In space you need a third axis at right angles to the x- and y-axes, the z-axis, and a point is written (x, y, z). Figure 1 draws the three axes the way they are usually drawn on a page: x coming out towards you, y to the right, z up. To reach P(3, 4, 2) from the origin O, go 3 along the x-axis, then 4 in the y-direction, then 2 up. The dashed box shows that P is one corner of a cuboid whose opposite corner is O.
A flat drawing of a solid distorts lengths and angles: a right angle in a 3D sketch rarely looks like 90°. Reason about the diagram; never measure it.
3Distance and midpoint
The distance between (x₁, y₁, z₁) and (x₂, y₂, z₂) is in the formula booklet:
d = √((x₁ − x₂)² + (y₁ − y₂)² + (z₁ − z₂)²) · and the midpoint is ((x₁ + x₂)/2, (y₁ + y₂)/2, (z₁ + z₂)/2).
It is the 2D formula with one more squared term, and Figure 2 shows why. Take A(1, −2, 3) and B(5, 2, 1). Across the floor of the box, from the point directly below A to B, the change in x is 4 and the change in y is 4, so that floor diagonal has length √(4² + 4²) = √32. That diagonal is at right angles to the vertical edge of height 2, so Pythagoras again gives AB.
The order of subtraction in each bracket does not matter, since every difference is squared; the brackets around negatives do: 2 − (−2) is 4.
An unknown coordinate. Questions like to hide a coordinate and give you the distance. If A(2, k, −1) and B(4, 1, 5) are 7 units apart:
Two answers is right, not a mistake: two points of that form are exactly 7 from A. Give both unless the question rules one out, for instance by saying k > 0.
An endpoint from the midpoint. If M(1, 3, −2) is the midpoint of AB and A is (4, −1, 0), each coordinate of M is the average of the two ends, so each coordinate of B is 2 × (M's coordinate) − (A's coordinate). That gives B = (2 − 4, 6 + 1, −4 − 0) = (−2, 7, −4). Check by averaging back.
4Volume and surface area of solids
Prisms and cylinders are prior learning. This subtopic adds four solids, drawn in Figure 3, and combinations of them.
| Solid | Volume | Surface area | In the booklet? |
|---|---|---|---|
| Right pyramid (apex directly above the centre of the base) | V = ⅓Ah, A the area of the base | add the base and each triangular face | volume yes; faces you work out |
| Right cone | V = ⅓πr²h | curved surface πrl, plus the base πr² if it is closed | yes, both |
| Sphere | V = (4/3)πr³ | 4πr² | yes, both |
| Hemisphere | V = (2/3)πr³ | curved 2πr², plus the flat circle πr² if solid | no: halve the sphere |
Name the lengths before you use them. The height h is the perpendicular distance from the apex to the base. The slant height l of a cone is the distance from the apex to the edge of the base, along the surface. For a pyramid the slant height of a face runs from the apex to the midpoint of a base edge, down the middle of a triangular face, and the slant edge runs from the apex to a corner. Volume always uses the height. Surface area uses a slant height.
Why a third? A pyramid or cone holds exactly one third of the prism or cylinder with the same base and height. You are not asked to prove it, but it is a useful check on an answer.
A pyramid, exactly (Paper 1). A right pyramid has a square base of side 6 cm and height 4 cm. Find its volume and its total surface area.
Figure 4 shows the right-angled triangle that unlocks the surface area. O is the centre of the base and M the midpoint of an edge, so OM is half the side, 3 cm, and the height VO stands at right angles to the base.
The other triangle in Figure 4, VOC, gives the slant edge. OC is half the diagonal of the base: the diagonal is 6√2, so OC = 3√2. Then VC = √((3√2)² + 4²) = √(18 + 16) = √34 cm. The triangular face needs VM, not VC. Draw the triangle you are using before you put numbers in.
A cone, exactly (Paper 1). A cone has radius 5 cm and height 12 cm. The slant height comes from the right-angled triangle formed by the height, the radius and the slant side: l = √(5² + 12²) = 13 cm.
On Paper 1 leave answers like 100π exactly, unless the question asks for a decimal.
Spheres and hemispheres. A sphere of radius 3 cm has V = (4/3)π(27) = 36π cm³ and surface area 4π(9) = 36π cm² (the same number only because r = 3). A solid hemisphere of radius 3 cm has half the volume, 18π cm³, but not half the surface: its curved surface is 18π and it also has a flat circular face of π(3²) = 9π, so its total surface area is 27π cm².
Working backwards (Paper 2). A sphere has volume 500 cm³. Find its radius.
Keep the unrounded r for later parts: rounding to 4.9 first gives a surface area of 302 cm² instead of 305 cm², and loses the accuracy mark.
5Combinations of solids
A composite solid is two or more of these solids joined. The rules are short.
Volumes add. Surfaces add only where they are on the outside. A face where two solids are glued together is inside the new solid, so it is not part of the surface.
Figure 5 is a grain silo: a cylinder of radius 2.5 m and height 8 m with a hemispherical roof. Find its volume, and the area of metal needed for its walls and roof (it stands on a concrete floor, so the base is not included).
The circle where roof meets wall is inside the silo, so it is not counted, and the floor is left out because the question said so. Always read which surface is wanted: a closed box, an open tank, "the area to be painted".
6Right-angled triangles inside solids
The guide asks you to identify right-angled triangles in three-dimensional objects and use them to find unknown lengths and angles. In an SL examination every angle in a solid will come from a right-angled triangle, so the whole job is finding the right one. Three places they hide:
- a vertical line meeting anything horizontal: the height of a pyramid or cone meets every line in its base at 90°;
- two edges of a cuboid that meet at a corner;
- an edge of a cuboid meeting a whole face it is perpendicular to: then it is perpendicular to every line in that face through its foot.
The angle between two intersecting lines. Two lines that meet at a point lie in one plane, so the angle between them is an ordinary angle in a triangle. Find the triangle that contains both lines, check whether it has a right angle, and use trigonometry.
The angle between a line and a plane. A line that hits a plane at a slant makes many different angles with the lines in the plane, so the angle needs a definition. Take a point P on the line, drop a perpendicular PN to the plane, and join the point A where the line meets the plane to N. The line AN is the projection of the line onto the plane: its shadow if the light shone straight down.
The angle between a line and a plane is the angle between the line and its projection on the plane. It is angle PAN in the right-angled triangle PAN, with the right angle at N.
Figure 6 shows the construction.
Worked example: a cuboid (Paper 2). Figure 7 is a cuboid ABCDEFGH with AB = 8 cm, BC = 6 cm and AE = 5 cm. The base is ABCD and E, F, G, H sit directly above A, B, C, D.
(a) Find the length of the space diagonal AG.
(b) Find the angle between AG and the base ABCD. G sits directly above C, so C is the foot of the perpendicular from G, and the projection of AG on the base is AC.
(c) Find the angle between AG and the edge AB. These two lines meet at A, so they sit in triangle ABG. Is there a right angle? AB is perpendicular to the whole face BCGF, because it meets both BC and BF at 90°, so AB is perpendicular to BG too. The right angle is at B, and AG is the hypotenuse.
The trap in (c) is to put the right angle wherever the drawing makes one look square. It is at B, for the reason given; write that reason down, since it earns an R mark in a "show that".
Worked example: angles in the pyramid. Return to the pyramid of Figure 4, base 6 cm, height 4 cm.
The angle between the slant edge VC and the base: VO is perpendicular to the base, so the projection of VC is OC = 3√2, and tan θ = 4 ÷ 3√2 = 0.9428…, so θ = 43.3°.
The angle between VM, the slant height of a face, and the base: the projection of VM is OM = 3, so tan θ = 4 ÷ 3 and θ = 53.1°. It is steeper, because M is nearer the centre than C.
A cone's angle. For the cone with r = 5 and h = 12, the slant side meets the base at θ with tan θ = 12 ÷ 5, so θ = 67.4°. The angle with the axis, the semi-vertical angle, is 90° − 67.4° = 22.6°.
7Where marks are lost
Using the slant height in a volume formula. V = ⅓πr²h needs the perpendicular height. If the question gives the slant height, find h with Pythagoras first.
Forgetting the third. A pyramid or cone is a third of its prism or cylinder. Leaving out ⅓ triples the answer.
Using the diameter as the radius. Read "diameter 10 cm" and write r = 5 before you touch a formula. A factor-of-8 error in a sphere volume comes from this.
Mis-counting the surface of a hemisphere. A solid hemisphere has a curved surface 2πr² and a flat face πr², total 3πr². A bowl (an open hemisphere) has only the 2πr². Say which you are using.
Counting the join in a composite solid. The circle where a cone sits on a hemisphere is inside the solid; it is not surface area.
Measuring the angle to the wrong line. The angle between a line and a plane is the angle with its projection, which runs from where the line meets the plane to the foot of the perpendicular. It is not the angle with whichever edge of the base happens to be nearby.
Assuming where the right angle is. A 3D drawing distorts angles, so state why the right angle is where you say.
Rounding too early, and wrong units. Carry full calculator values through each part, and give areas in cm² and volumes in cm³.
8Work it right
- Redraw the solid, or the part of it you need, and mark every right angle you are sure of with the reason.
- Before a formula, write down r, h and l separately. Never let a diameter or a slant height slip into the wrong place.
- For a distance in 3D, write all three differences in brackets, then square.
- For an angle in a solid, draw the right-angled triangle flat on its own, label its three sides, and only then choose sin, cos or tan.
- For the angle between a line and a plane, name the foot of the perpendicular and the projection in words: "the projection of AG on ABCD is AC".
- For a composite solid, list the parts, add the volumes, and tick off each outside face for the surface area.
- Paper 1: leave π and surds in the answer unless told otherwise. Paper 2: give 3 significant figures, with units, from unrounded working.
9Try it
Marks in brackets. Q1 and Q2 are Paper 1 style, no calculator. Q3 to Q6 are Paper 2 style, with a GDC.
Q1. The points A(−1, 4, 2) and B(3, 0, −2) are given.
(a) Find the coordinates of the midpoint M of AB. 2 marks
(b) Find the exact distance AB, giving your answer in the form p√3, where p ∈ ℤ. 3 marks
(c) The point C(t, 0, 3) is such that AC = √26. Find the possible values of t. 3 marks
Q2. A right cone has base radius 9 cm and slant height 15 cm.
(a) Find the height of the cone. 2 marks
(b) Find the volume of the cone, giving your answer in terms of π. 2 marks
(c) Find the total surface area of the cone, including its base, in terms of π. 2 marks
Q3. A storage box is a cuboid ABCDEFGH, with base ABCD, AB = 12 cm, BC = 5 cm and height AE = 9 cm. E, F, G, H lie directly above A, B, C, D.
(a) Find AC. 2 marks
(b) Find the length of the longest thin rod that fits inside the box, AG. 2 marks
(c) Find the angle between AG and the base ABCD. 2 marks
(d) Find the angle GÂB, stating which angle of triangle ABG is a right angle and why. 3 marks
Q4. A spinning top is made of a solid hemisphere of radius 3 cm with a right cone of height 7 cm on its flat face, the cone's base exactly covering that face.
(a) Find the volume of the top. 3 marks
(b) Find the total surface area of the top. 4 marks
Q5. A metal sphere of radius 4.5 cm is melted down and all the metal is recast as a right cone of base radius 6 cm. Find the height of the cone. 4 marks
Q6. A glass roof is a right pyramid VABCD with a square base ABCD of side 10 m. Each slant edge, such as VA, is 13 m long.
(a) Find the height of the pyramid. 3 marks
(b) Find the angle between the slant edge VA and the base. 2 marks
(c) Find the volume of air enclosed by the roof. 2 marks
10In one breath
A point in space is (x, y, z), and the distance between two points is the plane formula with a third squared term, which is just Pythagoras used twice; the midpoint averages each coordinate, and two answers for an unknown coordinate are normal. A pyramid is ⅓Ah and a cone ⅓πr²h, both with the perpendicular height; a cone's curved surface is πrl with the slant height; a sphere is (4/3)πr³ with surface 4πr², and a hemisphere is half of that volume with a curved surface 2πr² plus a flat πr² if it is solid. In a composite solid the volumes add but a glued face is not surface. Every length and angle inside a solid comes from a right-angled triangle you have to find and justify, and the angle between a line and a plane is the angle between the line and its projection, the shadow it casts straight down.
Answers
Q1. (a) M = ((−1 + 3)/2, (4 + 0)/2, (2 + (−2))/2) = (1, 2, 0). M1 for averaging the coordinates, A1 for (1, 2, 0). Award A0 if any one coordinate is wrong.
(b) AB = √(4² + (−4)² + (−4)²) = √48 = √(16 × 3) = 4√3. M1 for substituting into the 3D distance formula, A1 for √48, A1 for 4√3. Leaving √48 scores A1 A0, since the form was asked for.
(c) (t + 1)² + (0 − 4)² + (3 − 2)² = 26, so (t + 1)² + 17 = 26, (t + 1)² = 9 and t + 1 = ±3. So t = 2 or t = −4. M1 for a correct distance equation with the square root removed, A1 for (t + 1)² = 9, A1 for both values. Giving only t = 2 scores A0 for the last mark.
Q2. (a) h = √(15² − 9²) = √(225 − 81) = √144 = 12 cm. M1 for Pythagoras with the slant height as hypotenuse, A1 for 12.
(b) V = ⅓π(9²)(12) = ⅓π(81)(12) = 324π cm³. M1 for substituting their h into ⅓πr²h, A1 for 324π. Using l = 15 in place of h gives 405π and scores M0.
(c) Curved surface = π(9)(15) = 135π, base = π(9²) = 81π, total = 216π cm². M1 for πrl with l = 15 added to πr², A1 for 216π.
Q3. (a) AC = √(12² + 5²) = √169 = 13 cm. M1 for Pythagoras in triangle ABC, A1 for 13.
(b) AG = √(13² + 9²) = √250 = 15.8 cm (3 s.f.). M1 for Pythagoras in triangle ACG or the 3D formula √(12² + 5² + 9²), A1 for 15.8.
(c) The projection of AG on the base is AC, so tan θ = 9 ÷ 13 and θ = 34.7° (3 s.f.). M1 for a correct ratio in triangle ACG using their AC, A1 for 34.7°. An angle found with AB or BC as the adjacent side scores M0.
(d) The right angle is at B, because AB is perpendicular to the face BCGF and so to the line BG in that face. cos(GÂB) = 12 ÷ √250, so GÂB = 40.6° (3 s.f.). R1 for the right angle at B with a reason, M1 for cos = 12 ÷ their AG (or an equivalent ratio using BG = √106), A1 for 40.6°.
Q4. (a) Hemisphere: (2/3)π(3³) = 18π. Cone: ⅓π(3²)(7) = 21π. Total = 39π = 123 cm³ (3 s.f.). M1 for half the sphere formula, M1 for the cone volume, A1 for 123. Using the full sphere, 36π, is M0 for that part.
(b) Slant height l = √(3² + 7²) = √58. Curved surface of cone = π(3)(√58) = 71.8 cm². Curved surface of hemisphere = 2π(3²) = 18π = 56.5 cm². Total = 128 cm² (3 s.f.). M1 for l = √58, M1 for πrl with their l, M1 for 2πr² for the hemisphere, A1 for 128. Adding a flat circle πr² anywhere is wrong, because the join is inside the solid: that loses the final A1.
Q5. Volume of metal = (4/3)π(4.5³) = 121.5π = 381.70… cm³. Cone: ⅓π(6²)h = 12πh. So 12πh = 121.5π and h = 10.1 cm (3 s.f.; exactly 10.125 cm). M1 for the sphere volume, M1 for the cone volume in terms of h, M1 for equating the two, A1 for 10.1.
Q6. (a) The diagonal of the base is √(10² + 10²) = 10√2, so the distance from the centre O to A is 5√2 = 7.07… m. Triangle VOA has its right angle at O, so h = √(13² − (5√2)²) = √(169 − 50) = √119 = 10.9 m (3 s.f.). M1 for half the diagonal of the base, M1 for Pythagoras with 13 as the hypotenuse, A1 for 10.9. Using half the side, 5, gives √144 = 12, which is the slant height of a face, not the height: M0 for that step.
(b) The projection of VA on the base is OA, so cos θ = 5√2 ÷ 13, giving θ = 57.0° (3 s.f.). Equivalently tan θ = √119 ÷ 5√2. M1 for a correct ratio in triangle VOA, A1 for 57.0°.
(c) V = ⅓ × 100 × √119 = 364 m³ (3 s.f.). M1 for ⅓ × base area × their height, A1 for 364. Follow-through from their (a).
Educerie · written from the published IB Diploma Programme Mathematics: analysis and approaches guide, first assessment 2021, section 3.1 Three-dimensional geometry. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.