The Pythagorean Theorem in 3D

The Pythagorean theorem also works in three-dimensional space. Most often we use it to compute the space diagonal of a cuboid or a cube – that is, the segment that connects two opposite vertices of the solid.

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Table of contents

• The space diagonal of a cuboid
• Formula for a cuboid
• Formula for a cube
• Solved examples

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The space diagonal of a cuboid

Imagine a cuboid with dimensions $a$, $b$, $c$. The space diagonal is the segment connecting two opposite vertices – the longest segment that fits inside the cuboid.

The computation is done in two steps:

1. First we compute the face diagonal $u_s$ on the base of the cuboid (a rectangle with sides $a$, $b$):
   
   $$u_s^2=a^2+b^2$$
2. Then we apply the Pythagorean theorem a second time in the right triangle formed by the face diagonal, the space diagonal, and the height of the cuboid $c$:
   
   $$u^2=u_s^2+c^2=a^2+b^2+c^2$$

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Formula for a cuboid

$$u=\sqrt{a^2+b^2+c^2}$$

> 💡 Just square all three dimensions, add them, and take the square root. It's essentially an "extended" Pythagorean theorem.

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Formula for a cube

A cube is a special case of a cuboid where $a=b=c$. The formula therefore simplifies to:

$$u=\sqrt{a^2+a^2+a^2}=\sqrt{3a^2}=a\sqrt{3}$$

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Solved examples

Example 1: A cuboid has dimensions $a=3$ cm, $b=4$ cm, $c=12$ cm. Compute its space diagonal.

$$u^2=3^2+4^2+12^2=9+16+144=169$$
$$u=\sqrt{169}=13\text{ cm}$$

Answer: The space diagonal of the cuboid is $13$ cm.

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Example 2: A cuboid has dimensions $a=6$ cm, $b=6$ cm, $c=7$ cm.

$$u^2=36+36+49=121$$
$$u=11\text{ cm}$$

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Example 3: A cube has edge $a=4$ cm. What is the length of its space diagonal?

$$u=a\sqrt{3}=4\sqrt{3}\approx 6.93\text{ cm}$$

Answer: The space diagonal of the cube is approximately $6.93$ cm.

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Example 4 (from real life): Will a wooden rod $1.4$ m long fit inside a box with dimensions $1.0\times 0.8\times 0.6$ m?

It is enough to compare the length of the rod with the space diagonal of the box:

$$u=\sqrt{1.0^2+0.8^2+0.6^2}=\sqrt{1+0.64+0.36}=\sqrt{2}\approx 1.41\text{ m}$$

A rod $1.4$ m long will fit in the box, because its longest diagonal is approximately $1.41$ m.

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Related articles

• The Pythagorean theorem – complete guide
• The Pythagorean theorem formula
• Distance between two points
• Word problems

Practice

• 📊 Word problems
• 🎯 Mixed problems