Key Idea: 3D geometry turns a solid into something you can measure: the distance between corners, the volume and surface area of a shape, and the angle a line makes with a face. It runs across both papers — by hand on Paper 1, with the GDC doing the arithmetic on Paper 2.
📏 Distance & midpoint in 3D
- distance between two 3D points (x, y, z)
Distance = subtract each coordinate, square, add, root (point order doesn't matter — the squaring kills the sign). Midpoint = add each coordinate pair and halve — $M = \left( \tfrac{x₁+ₓ₂}{2}, \tfrac{y₁+y₂}{2}, \tfrac{z₁+z₂}{2} \right)$. Given the midpoint and one end, the missing endpoint is B = 2M − A.
🧊 Volume, surface area & angles
| Solid | Volume (booklet) | Surface area |
|---|---|---|
| Cuboid / prism | base area × length | sum of all faces |
| Cylinder | $\pi r² h$ | $2\pi r² + 2\pi r h$ (closed) |
| Cone | $\tfrac{1}{3}\pi r² h$ | $\pi r² + \pi r l$, with $l=\sqrt{r²⁺ʰ²}$ |
| Sphere | $\tfrac{4}{3}\pi r³$ | $4\pi r²$ |
Composite (e.g. a cylinder topped by a hemisphere) → add the part volumes, but for surface area count only exposed faces — never the shared join. Backwards (given a volume, find r or h) → set the formula equal and solve; cancel the π on both sides first to keep the numbers friendly.
Tip: Almost every 3D angle becomes a 2D right-angled triangle — find it, redraw it flat, use SOH-CAH-TOA. The angle between a line and a plane is measured to the line's projection (shadow) on the plane, and you usually need a face or base diagonal (Pythagoras) first — so it's a two-step problem: length, then angle.
✏️ IB-style worked examples
IB-style question — distance & midpoint in 3D
Points are P(1, 3, 2) and Q(7, 11, 26). Find (a) the distance PQ and (b) the midpoint of PQ.
Step by step:
(a) Coordinate gaps are 6, 8 and 24.
Square, add, root.
(b) Midpoint = average each coordinate.
PQ = 26 and the midpoint is (4, 7, 14).
Switch between 2D distance, the midpoint, and 3D distance — the 3D distance just adds the z-gap under the root.
🔒 Interactive diagram
Explore the labelled diagram, charts and maps for this topic in study mode.
IB-style question — volume, surface area and a composite solid
A solid is a cone of base radius 5 and height 12, sitting on top of a cylinder of the same radius and height 8. Find (a) the cone's curved (slant) surface area and (b) the total volume of the solid. Leave answers in terms of π.
Step by step:
(a) Slant height first, then curved area = πrl.
(b) Add the two volumes — cone is ⅓πr²h.
Cylinder is πr²h; total = sum.
Curved area = 65π; total volume = 300π ≈ 942.
Each standard solid with its volume and surface-area formulas — here the cone (V = ⅓πr²h, curved area πrl) and cylinder (V = πr²h); for a composite, add the volumes.
🔒 Interactive diagram
Explore the labelled diagram, charts and maps for this topic in study mode.
IB-style question — angle between a line and a plane (Paper 2)
A cuboid has a square base of side 6 and height 8. Find the angle the space diagonal (corner to opposite corner) makes with the base.
Step by step:
Length first — the base diagonal by Pythagoras.
Right triangle: opposite = height 8, adjacent = base diagonal.
Inverse tan on the GDC.
The diagonal makes about 43.3° with the base.
The space diagonal and the angle it makes with the base: the right triangle is base-diagonal, height, space-diagonal — so tan θ = height ÷ base diagonal.
🔒 Interactive diagram
Explore the labelled diagram, charts and maps for this topic in study mode.
🔒 GDC walkthrough
Step through the exact calculator keystrokes, screen by screen, in study mode.
Important: Two traps dominate this topic. Geometry: in $V = \tfrac{1}{3}\pi r² h$ only the radius is squared — never square h. And for an angle, find the right diagonal first (the adjacent side is often a base diagonal, not an edge). Calculator: on Paper 2 an inverse-trig answer is only right if the GDC is in DEGREE mode.
Tap each card to reveal the answer.
Distance from A(0, 0, 0) to B(2, 3, 6) 7 — √(4 + 9 + 36) = √49.
Midpoint of (1, 5, −2) and (7, 1, 8) (4, 3, 3) — average each coordinate.
M(5, 0, 4) is the midpoint of A(1, −2, 1) and B. Find B B = (9, 2, 7) — use B = 2M − A.
Volume of a cone, r = 3, h = 7 21π ≈ 66.0 — ⅓π(3²)(7); only r is squared.
A sphere has volume 36π. Find r r = 3 — cancel π: r³ = 27.
Slant height of a cone, r = 9, h = 12 l = 15 — √(9² + 12²) = √225.
Exam Tips
- Distance: subtract, square, add, root. Midpoint: add and halve. Missing end: B = 2M − A.
- Volume/area formulas are in the booklet — pick the right one and read the correct r and h.
- Composite solids: add the parts (subtract holes); count only exposed faces for surface area.
- Angle problems are two-step — find the diagonal with Pythagoras first, then SOH-CAH-TOA.
- On Paper 2, check the GDC is in DEGREE mode before any inverse-trig angle.