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3D Pythagoras & TrigonometryCambridge IGCSE Maths: Revision notes

Section 1

How do I extend Pythagoras' theorem into three dimensions?

3D problems (cuboids, pyramids, prisms) are solved by finding a right-angled triangle hidden inside the solid and applying 2D Pythagoras and trigonometry to it, often twice in sequence.

A common approach: first find a diagonal across a face using 2D Pythagoras, then use that diagonal as one side of a second right-angled triangle running through the solid.

Key termsspace diagonal
Example

Cuboid with dimensions 3 cm x 4 cm x 12 cm: base diagonal =32+42=5= \sqrt{3^2+4^2} = 5 cm; space diagonal =52+122=13= \sqrt{5^2+12^2} = 13 cm.

Section 2

What is the method for solving 3D problems step by step?

  1. Sketch (mentally or on paper) the solid and identify the right-angled triangle needed for the question
  2. If the triangle isn't immediately visible, find an intermediate length first (e.g. a base diagonal or the midpoint of an edge) using a triangle that IS visible
  3. Apply Pythagoras or the trigonometric ratio to that intermediate triangle
  4. Use the result as one side of the next right-angled triangle, and solve for the final answer

Always work with full calculator accuracy between steps to avoid rounding errors compounding.

Section 3

How do I calculate the angle between a line and a plane? (Extended)

The angle between a line and a plane is measured between the line and its projection onto the plane — this is the smallest possible angle between them.

Method:

  1. Identify the point where the line meets the plane
  2. Drop a perpendicular from the other end of the line down to the plane
  3. The angle between the original line and the line joining the foot of the perpendicular to the meeting point is the angle required
  4. This forms a right-angled triangle you can solve with trigonometry
Key termsangle between a line and a planeprojection
Example

A vertical edge of length 10 cm meets a horizontal base at a point 6 cm from where a sloping line reaches the base. The angle between the sloping line and the base is tan⁡−1(106)≈59.0°\tan^{-1}\left(\frac{10}{6}\right) \approx 59.0°.

Section 4

Worked example: pyramid problem

A square-based pyramid has a base of side 8 cm and vertical height 15 cm from the centre of the base to the apex.

  1. Half the diagonal of the base: base diagonal =82+82=82= \sqrt{8^2+8^2} = 8\sqrt{2}, so half diagonal =42≈5.66= 4\sqrt{2} \approx 5.66 cm
  2. Slant edge from a base corner to the apex: 152+(42)2=225+32=257≈16.0\sqrt{15^2 + (4\sqrt2)^2} = \sqrt{225+32} = \sqrt{257} \approx 16.0 cm
Exam tip

For a pyramid with a square base, the vertical height meets the base at the CENTRE — use half the diagonal, not half the side length.

Must Know

  • 3D problems are solved using right-angled triangles hidden within the solid, often two in sequence
  • Find intermediate lengths (base diagonals, midpoints) before tackling the final triangle
  • The angle between a line and a plane is measured to the line's projection onto that plane
  • Keep full accuracy between calculation steps — do not round until the final answer
  • For a square-based pyramid, the apex sits above the centre of the base — use half the base diagonal
  • Space diagonal of a cuboid with sides p,q,rp, q, r: p2+q2+r2\sqrt{p^2+q^2+r^2}

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