3D Pythagoras
Practise finding lengths in three-dimensional shapes by spotting the right-angled triangles hidden in their diagrams. This GCSE Higher worksheet moves from face and space diagonals to prisms, cones and pyramids, with careful attention to choosing the correct lengths and rounding only at the end. Jump to the questions
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Worksheet preview and key skills
Worksheet preview
Eight diagram-based questions build from a diagonal across one face to two-stage length problems in prisms, cones and pyramids.
Each question has one numerical answer. Give non-integer answers to 1 decimal place and keep full calculator values during intermediate steps.
What you’ll practise
- Distinguishing face diagonals from true space diagonals.
- Breaking a 3D length problem into one or two right-angled triangles.
- Using forward and reverse Pythagoras without rounding too early.
Use the interactive worksheet below, or read the Topic guide for the method and worked example.
3D Pythagoras
Find each requested length. Give non-integer answers to 1 decimal place, keeping full calculator values until the final step.
Topic guide
Start with a right-angled triangle
Pythagoras’ theorem says that a2 + b2 = c2 in a right-angled triangle, where c is the hypotenuse: the side opposite the right angle. A perspective drawing can make other angles look like right angles, so use only angles guaranteed by the shape or marked on the diagram.
A 3D problem is usually solved by finding one right-angled triangle, then using its result in a second right-angled triangle. Keep the full calculator value from the first step.
Face and space diagonals
For a cuboid with perpendicular dimensions x, y and z, first find a face diagonal: d2 = x2 + y2. The space diagonal s then satisfies s2 = d2 + z2. Substituting for d2 gives s2 = x2 + y2 + z2. This shortcut is simply two applications of Pythagoras.
Example: a cuboid measuring 3 cm by 4 cm by 12 cm has face diagonal √(32 + 42) = 5 cm, so its space diagonal is √(52 + 122) = 13 cm.
Other 3D shapes
- Triangular prism: use the right-angled end face to find its missing diagonal, then combine that length with the prism length.
- Cone: in an axial cross-section, the radius and perpendicular height meet at 90°, and the slant height is the hypotenuse.
- Square or rectangular pyramid: find the base diagonal. The distance from the centre to a corner is half that diagonal. It forms a right-angled triangle with the perpendicular height and the sloping corner edge.
When the hypotenuse is known, use reverse Pythagoras: subtract the square of the known shorter side before taking the square root.
Common mistakes
- Using a face diagonal when the question asks for a line through the solid.
- Adding lengths instead of adding their squares, or forgetting the final square root.
- Assuming an unmarked angle in a perspective drawing is 90°.
- Rounding an intermediate diagonal before the second step.
- Using the full base diagonal instead of half of it in a pyramid.
- Adding squares in a reverse problem when the hypotenuse is already known.
- Confusing a cone’s radius with its diameter.
Keep full calculator values until the final answer, then round once to the requested accuracy. Trigonometry and measurements such as area and volume are separate topics; they are not needed here.