3D Pythagoras

Statement

In three dimensions, the Pythagoras theorem extends to find the length of the space diagonal of a cuboid or right-angled box:

\[ d = \sqrt{a^2 + b^2 + c^2} \]

where \(a, b, c\) are the side lengths and \(d\) is the diagonal running corner-to-corner through the solid.

Why it’s true

• In 2D: right-angled triangle gives \(d^2=a^2+b^2\).
• In 3D: apply Pythagoras twice.
• First, diagonal on base: \(\sqrt{a^2+b^2}\).
• Then include height: \(d^2=(a^2+b^2)+c^2\).

Recipe (how to use it)

1. Square each edge length \(a, b, c\).
2. Add them together.
3. Take the square root.
4. Answer in the same units as the edges.

Spotting it

Look for problems asking for the longest diagonal of a cuboid, box, or 3D shape made of right angles.

Common pairings

• Often paired with standard Pythagoras and trigonometry.
• Can be linked to vector magnitude in 3D: \(\sqrt{x^2+y^2+z^2}\).

Mini examples

1. Cuboid 3×4×12: \(d=\sqrt{3^2+4^2+12^2}=\sqrt{9+16+144}=\sqrt{169}=13\).
2. Cube side 5: \(d=\sqrt{5^2+5^2+5^2}=\sqrt{75}=5\sqrt{3}\).

Pitfalls

• Forgetting to include all three dimensions.
• Confusing face diagonal with space diagonal.
• Leaving answer as squared, not square-rooted.

Exam strategy

• Always sketch the cuboid to see the 3D right-angled triangle.
• Double-check whether question asks for face diagonal or space diagonal.
• Keep answers exact unless decimals requested.

Summary

The 3D Pythagoras formula finds the space diagonal of a cuboid: square each side length, add, and take the square root.