3D Pythagoras: Space Diagonals, Pyramids and Cones, with Examples (Year 10A)

3D Pythagoras: space diagonals, pyramids and cones

In 3D, Pythagoras’ theorem still works: you just find the right-angled triangle hiding inside the shape. Often that takes two steps. Here’s how, with worked examples and practice questions with answers.

By Ryan, founder of Bayside Academics, 15 years in private tutoring · Updated 6 November 2026

Quick answer

For a box l by w by h, the space diagonal is d = √(l² + w² + h²). Or work in two steps: find the diagonal of the base, then use it with the height in a second right-angled triangle. For a cone, slant height s = √(r² + h²).

The formula for a box

d = \sqrt{l^2 + w^2 + h^2}

Worked example

Find the space diagonal of a box 3 cm by 4 cm by 12 cm.

d = \sqrt{3^2 + 4^2 + 12^2}

= \sqrt{9 + 16 + 144}

= \sqrt{169} = 13 \text{ cm}

New to Pythagoras? Start with Pythagoras’ theorem in 2D.

The two-step method

Showing both triangles earns method marks, and works for shapes other than boxes.

Worked example

A box is 6 cm by 8 cm by 5 cm high. Find the space diagonal, to 2 decimal places.

Step 1: the base diagonal: $\sqrt{6^2 + 8^2} = \sqrt{100} = 10$ cm.

Step 2: the base diagonal and the height make a right-angled triangle:

d = \sqrt{10^2 + 5^2} = \sqrt{125}

= 5\sqrt{5} \approx 11.18 \text{ cm}

Pyramids and cones

Square pyramid

A square pyramid has a base of side 10 cm and a vertical height of 12 cm. Find the slant height of a face.

From the centre of the base to the midpoint of an edge is half the side, 5 cm:

s = \sqrt{12^2 + 5^2} = \sqrt{169} = 13 \text{ cm}

Cone

A cone has radius 5 cm and height 12 cm. The slant height is $\sqrt{5^2 + 12^2} = 13$ cm.

Practice questions with answers

Draw the right-angled triangle first. Try each one before you open the answer.

A box measures 2 cm by 3 cm by 6 cm. Find the length of its space diagonal.
Show the answer
$d = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{49} = 7$ cm
Find the space diagonal of a cube with sides of 5 cm.
Show the answer
$d = \sqrt{25 + 25 + 25} = \sqrt{75} = 5\sqrt{3} \approx 8.66$ cm
A box is 10 m by 6 m by 4 m. Find the space diagonal to 2 decimal places.
Show the answer
$d = \sqrt{100 + 36 + 16} = \sqrt{152} \approx 12.33$ m
A cone has radius 8 cm and vertical height 15 cm. Find its slant height.
Show the answer
$s = \sqrt{8^2 + 15^2} = \sqrt{289} = 17$ cm
A square pyramid has a base of side 6 cm and a vertical height of 4 cm. Find the slant height of a face (apex to the midpoint of a base edge).
Show the answer
Half the base is 3 cm, so $s = \sqrt{4^2 + 3^2} = 5$ cm
What's the longest pencil that fits in a pencil case 20 cm by 4 cm by 3 cm, to 2 decimal places?
Show the answer
$\sqrt{400 + 16 + 9} = \sqrt{425} \approx 20.62$ cm

By year level

Following the Victorian Curriculum 2.0. Schools sequence topics differently, so check your course.

Year 9

Pythagoras (applied)

Pythagoras in 2D problems, the skill 3D builds on.

Worksheets and topic test · Free PDF with answers

Year 10A

Pythagoras and trigonometry in 3D

Pythagoras and trigonometry in 3D shapes.

Worksheets and topic test · Free PDF with answers

3D Pythagoras: common questions

What is the 3D Pythagoras formula?

For a rectangular box with length l, width w and height h, the space diagonal is d = √(l² + w² + h²).

What's the two-step method?

Find a diagonal across the base with Pythagoras, then use it with the height in a second right-angled triangle to find the space diagonal.

How do I find the slant height of a cone?

The radius, the vertical height and the slant height make a right-angled triangle, with the slant height as the hypotenuse: s = √(r² + h²).

What year level learns 3D Pythagoras?

It's part of Year 10A in the Victorian Curriculum 2.0, alongside trigonometry in 3D.

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