Worked Example: Cuboid Diagonal

Part of Pythagoras in 3D · Section 4 of 5

Study NotesUnit: Geometry & MeasuresGCSE

This study notes covers Worked Example: Cuboid Diagonal within Pythagoras in 3D for GCSE Mathematics. Revise Pythagoras in 3D in Geometry & Measures for GCSE Mathematics with 12 exam-style questions and 2 flashcards. Use this page as part of a wider topic revision path rather than treating it as an isolated fact. It is section 4 of 5 in this topic. Use this study notes to connect the idea to the wider topic before moving on to questions and flashcards.

Worked Example: Cuboid Diagonal

Find the space diagonal of a cuboid 3cm × 4cm × 12cm.

Method 1: Two steps

Base diagonal: √(3² + 4²) = √25 = 5

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

Method 2: Direct

d² = 3² + 4² + 12² = 9 + 16 + 144 = 169

d = √169 = 13 cm

Practice questions for Pythagoras in 3D

A cuboid has length l, width w, and height h. Which formula gives the length of the space diagonal d?

  • A. d = l + w + h
  • B. d² = l² + w²
  • C. d² = l² + w² + h²
  • D. d = √(l + w + h)
1 markfoundation

Describe the two-step method for finding the space diagonal of a cuboid with dimensions l, w, and h.

2 marksstandard

Quick recall flashcards

3D Pythagoras
Use Pythagoras twice: once for base diagonal, once for 3D diagonal
3D Pythagoras
d² = x² + y² + z² for space diagonal. Or use 2D Pythagoras twice: first on base, then with height.

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