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Cambridge IGCSE Mathematics — 0580 Extended

Topic 6.1: Trigonometry — Pythagoras’ Theorem

The theorem

In a right-angled triangle the square on the hypotenuse equals the sum of the squares on the other two sides.

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

The hypotenuse \(c\) is the side opposite the right angle — the longest side, and the only side that does not touch the right angle.

Right-angled 3-4-5 triangle with squares built on each side, areas 9, 16 and 25, showing a squared plus b squared equals c squared
On a \(3\)-\(4\)-\(5\) triangle the squares have areas \(9\), \(16\) and \(25\). Pythagoras is the statement \(9 + 16 = 25\).

Method

  1. Mark the right angle. The hypotenuse is the side that does not touch it.
  2. Call that side \(c\). The other two sides are \(a\) and \(b\) (either way round).
  3. Write \(a^2 + b^2 = c^2\), substitute, then rearrange for the unknown.

Finding the hypotenuse

When the unknown is the hypotenuse you add the squares of the two legs, then square-root.

A right-angled triangle has legs \(6\,\text{cm}\) and \(8\,\text{cm}\). Find the hypotenuse.

Right triangle with legs 6 cm and 8 cm and unknown hypotenuse c, working showing c equals 10 cm
\(c^2 = 6^2 + 8^2 = 100\), so \(c = 10\,\text{cm}\). This is a \(3\)-\(4\)-\(5\) triangle scaled by \(2\).

Finding a shorter side

When the unknown is a leg, rearrange first: \(a^2 = c^2 - b^2\). You subtract the known leg from the hypotenuse — you never add a leg to the hypotenuse.

A right-angled triangle has hypotenuse \(13\,\text{cm}\) and one leg \(5\,\text{cm}\). Find the other leg.

Right triangle with hypotenuse 13 cm, one leg 5 cm and unknown leg a, working showing a equals 12 cm
\(a^2 = 13^2 - 5^2 = 144\), so \(a = 12\,\text{cm}\). This is the \(5\)-\(12\)-\(13\) triple.

Integer triples and isosceles

Learn the two triples that appear on Paper 2: \(3\)-\(4\)-\(5\) and \(5\)-\(12\)-\(13\), and their scalings (\(6\)-\(8\)-\(10\), \(9\)-\(12\)-\(15\), \(10\)-\(24\)-\(26\)).

If the two legs are equal, the hypotenuse is the leg times \(\sqrt{2}\). Leave the surd — do not decimalise it on Paper 2.

Two right triangles side by side: a 9-12-15 scaled 3-4-5, and an isosceles right triangle with legs 7 and hypotenuse 7 root 2
Left: \(3\)-\(4\)-\(5\) scaled by \(3\). Right: equal legs \(7\) give hypotenuse \(7\sqrt{2}\).

Paper 2 (non-calculator)

\(\sqrt{7^2 + 7^2} = \sqrt{98} = \sqrt{49 \times 2} = 7\sqrt{2}\). Writing \(9.9\) (or similar) from memory of \(\sqrt{2}\) is not exact and will lose the mark if exact form is required.

Try this

A rectangle is \(9\,\text{cm}\) by \(12\,\text{cm}\). Find the length of a diagonal.

Show answer
Answer

  1. The diagonal is the hypotenuse of a \(9\)-\(12\) right triangle

    \[ d^2 = 9^2 + 12^2 = 81 + 144 = 225 \]

  2. A scaled \(3\)-\(4\)-\(5\)

    \[ d = 15\,\text{cm} \]

Exam Traps

  • The hypotenuse is opposite the right angle. Squaring and adding all three given sides, or treating a labelled leg as \(c\), is the standard Pythagoras error.
  • To find a leg, subtract: \(a^2 = c^2 - b^2\). Computing \(\sqrt{c^2 + b^2}\) makes the unknown longer than the hypotenuse, which is impossible.

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