Pythagoras in 3D –

Pythagoras in 3D extends the familiar a² + b² = c² into three dimensions. It is a Higher-only GCSE Maths topic tested by AQA, Edexcel, and OCR. Questions typically ask you to find the space diagonal of a cuboid, the longest rod that fits inside a box, or a slant length in a pyramid. The key skill is identifying the right-angled triangle hidden inside the 3D shape. This guide breaks down the method with three worked examples.

What Is Pythagoras in 3D?

In two dimensions, Pythagoras' theorem finds the hypotenuse of a right-angled triangle. In three dimensions, you apply the theorem twice — first on the base of the shape to find an intermediate length, then again using that length and the height to find the final distance.

Key Formulas

The space diagonal of a cuboid goes from one corner to the opposite corner, passing through the interior of the shape.

Step-by-Step Method

1. Draw or label the 3D shape with all given dimensions.
2. Identify the right-angled triangle on the base. Use Pythagoras to find the base diagonal.
3. Form a second right-angled triangle using the base diagonal and the height of the shape.
4. Apply Pythagoras again to find the space diagonal or the required length.
5. Round to the required degree of accuracy.

Worked Example 1 — Foundation Level

This topic is Higher only, but this example uses simpler numbers.

Question: A cuboid has dimensions 3 cm by 4 cm by 12 cm. Find the length of the space diagonal.

Working:

Step 1 — Find the diagonal of the base.

Base diagonal = sqrt(3² + 4²) = sqrt(9 + 16) = sqrt(25) = 5 cm

Step 2 — Use the base diagonal and the height (12 cm) to find the space diagonal.

Space diagonal = sqrt(5² + 12²) = sqrt(25 + 144) = sqrt(169) = 13 cm

Answer: The space diagonal is 13 cm.

Worked Example 2 — Higher Level

Question: A cuboid measures 6 cm by 8 cm by 10 cm. Find the space diagonal. Give your answer to 1 decimal place.

Working:

Using the direct formula:

d = sqrt(6² + 8² + 10²) = sqrt(36 + 64 + 100) = sqrt(200) = 14.142...

Answer: The space diagonal is 14.1 cm (1 d.p.)

Worked Example 3 — Exam Style

Question: A square-based pyramid has a base edge of 8 cm and a slant height of 10 cm. Find the vertical height of the pyramid.

Working:

Step 1 — The apex is directly above the centre of the square base. The distance from the centre of the base to the midpoint of a base edge is half the base edge = 8 ÷ 2 = 4 cm.

Step 2 — A right-angled triangle is formed by the vertical height (h), the distance from centre to edge midpoint (4 cm), and the slant height (10 cm).

h² + 4² = 10²

h² + 16 = 100

h² = 84

h = sqrt(84) = 9.165...

Answer: The vertical height is 9.2 cm (1 d.p.)

Common Mistakes

• Applying Pythagoras only once. 3D problems almost always require two applications of Pythagoras. Students who stop after the first triangle miss the final answer.
• Using the wrong lengths in the triangle. Carefully identify which lengths form the right-angled triangle. Sketch the triangle separately from the 3D shape to avoid confusion.
• Confusing the slant height with the vertical height. The slant height runs along the face of a pyramid; the vertical height runs straight up from the base to the apex. They are different measurements.

Exam Tips

• Always sketch the right-angled triangle you are working with separately — this prevents errors from misreading the 3D diagram.
• The formula d = sqrt(a² + b² + c²) is a shortcut for cuboids, but make sure you understand why it works (two applications of Pythagoras).
• "The longest rod that fits inside a box" means the space diagonal — this is a common exam phrasing.
• Leave your intermediate answers unrounded. Only round at the final step to avoid cumulative rounding errors.

Practice Questions

Q1 (Foundation): This topic is Higher only.

Q2 (Higher): A cuboid has dimensions 5 cm, 12 cm, and 8 cm. Find the space diagonal to 1 d.p.

Q3 (Higher): A square-based pyramid has a base of 10 cm and a vertical height of 12 cm. Find the slant height to 1 d.p.

Practise 3D Pythagoras with instant feedback free on GCSEMathsAI.

Related Topics

• Pythagoras' Theorem
• Trigonometry in 3D
• Volume of 3D Shapes

Summary

• Pythagoras in 3D involves applying a² + b² = c² twice: once on the base to find an intermediate diagonal, and once using that diagonal and the height to find the space diagonal or slant length. For a cuboid with dimensions a, b, and c, the space diagonal is sqrt(a² + b² + c²). Always sketch the right-angled triangle separately, use unrounded intermediate values, and identify the correct lengths from the 3D shape.