N.B.Nota Bene

Home  /  Extended Maths  /  MYP 4

Pythagorean Theorem

In every right-angled triangle the three sides obey one tidy rule. Master it and you can find any side from the other two, in flat shapes and in three dimensions alike.

MYP 4Extended MathsGeometryCriteria A · C · D~10 min read

Written by Fayek Jawad Haider · MYP student

The rule

Pythagoras' theorem links the three sides of a right-angled triangle. If the two shorter sides are \(a\) and \(b\), and the hypotenuse (the longest side, opposite the right angle) is \(c\), then:

\[ a^2 + b^2 = c^2 \]

In words: the square on the hypotenuse equals the sum of the squares on the other two sides. It only holds for right-angled triangles, so a right angle must be present before you use it.

Hypotenuse
The longest side of a right-angled triangle, always opposite the right angle. It is the \(c\) in \(a^2 + b^2 = c^2\).

Finding the hypotenuse

When the two shorter sides are known, substitute them and square-root the total.

Worked example

A right-angled triangle has shorter sides of 5 cm and 12 cm. Find the hypotenuse.

1
Write the rule with the shorter sides as \(a\) and \(b\): \(c^2 = 5^2 + 12^2\).
2
Square and add: \(c^2 = 25 + 144 = 169\).
3
Square-root: \(c = \sqrt{169}\).
\(c = 13\) cm

The hypotenuse is always the biggest

Your answer for \(c\) should come out longer than either shorter side. If it does not, you have probably subtracted when you should have added.

Finding a shorter side

If the hypotenuse is one of the known sides, rearrange the rule to subtract instead of add:

\[ a^2 = c^2 - b^2 \]
Worked example

A right-angled triangle has a hypotenuse of 13 cm and one shorter side of 5 cm. Find the other shorter side.

1
The hypotenuse is \(c = 13\), so rearrange: \(a^2 = c^2 - b^2 = 13^2 - 5^2\).
2
Square and subtract: \(a^2 = 169 - 25 = 144\).
3
Square-root: \(a = \sqrt{144}\).
\(a = 12\) cm

Add or subtract? Check the hypotenuse

Look for the longest side. If you are looking for the hypotenuse, add the squares. If the hypotenuse is already given, subtract. Getting this the wrong way round is the most common Pythagoras slip.

A 3D application

The theorem stretches into three dimensions. The longest diagonal of a cuboid (corner to opposite corner) follows the same idea, applied twice, which gives \(d^2 = l^2 + w^2 + h^2\).

Worked example

Find the length of the longest straight rod that fits inside a box measuring 3 cm by 4 cm by 12 cm.

1
First find the diagonal across the base: \(3^2 + 4^2 = 9 + 16 = 25\), so the base diagonal is \(\sqrt{25} = 5\) cm.
2
Now use that base diagonal (5 cm) with the height (12 cm) as the two shorter sides of a new right-angled triangle: \(d^2 = 5^2 + 12^2 = 25 + 144 = 169\).
3
Square-root: \(d = \sqrt{169}\). (This matches \(d^2 = 3^2 + 4^2 + 12^2 = 169\).)
\(d = 13\) cm

Where this is assessed

Substituting and computing cleanly is Criterion A (Knowing and understanding). A labelled diagram with each step shown earns Criterion C (Communicating). Ladders against walls, cable lengths and box diagonals are Criterion D (Applying maths in context), where the 3D version above is a favourite.

Check yourself

1. Find the hypotenuse of a right-angled triangle with shorter sides 8 cm and 15 cm. +

\(c^2 = 8^2 + 15^2 = 64 + 225 = 289\), so \(c = \sqrt{289} = \mathbf{17}\) cm.

2. A right-angled triangle has hypotenuse 10 cm and one shorter side 6 cm. Find the other shorter side. +

The hypotenuse is given, so subtract: \(a^2 = 10^2 - 6^2 = 100 - 36 = 64\). Then \(a = \sqrt{64} = \mathbf{8}\) cm.

3. Is a triangle with sides 5 cm, 6 cm and 8 cm right-angled? +

Test the longest side as the hypotenuse: \(5^2 + 6^2 = 25 + 36 = 61\), but \(8^2 = 64\). Since \(61 \neq 64\), the theorem fails, so no, it is not right-angled.

Part of the Extended Maths library. Spotted an error or want a topic added? That feedback makes the notes better.