IGCSE · Practice questions · Pythagoras' theorem

Square, add, root.

Six original Cambridge-style questions on Pythagoras' theorem, from a single missing side to a three-leg journey.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Find the hypotenuse first, then decide: add or subtract.

Sketch: draw the right-angled triangle and mark the right angle.
Label: the hypotenuse \(c\) is opposite the right angle.
Hypotenuse missing: \(c^2 = a^2 + b^2\), then square root.
Shorter side missing: \(a^2 = c^2 - b^2\), then square root.
Check: the hypotenuse must be the longest side; round to 3 s.f. at the end.

Calculation

[2 marks]

A right-angled triangle has shorter sides 8.5 cm and 6.2 cm. Calculate the length of the hypotenuse.

\(c^2 = 8.5^2 + 6.2^2 = 72.25 + 38.44 = 110.69\) ✓
\(c = \sqrt{110.69} = 10.5\) cm (3 s.f.) ✓

Calculation

[2 marks]

The hypotenuse of a right-angled triangle is 14.2 cm and one of the shorter sides is 11.3 cm. Calculate the length of the third side.

\(a^2 = 14.2^2 - 11.3^2 = 201.64 - 127.69 = 73.95\) ✓
\(a = \sqrt{73.95} = 8.60\) cm (3 s.f.) ✓

Calculation

[2 marks]

Without a calculator, find the length of the hypotenuse of a right-angled triangle whose shorter sides are 3 cm and 6 cm. Give your answer in the form \(k\sqrt{5}\).

\(c^2 = 3^2 + 6^2 = 9 + 36 = 45\) ✓
\(c = \sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}\) cm ✓

Proof

[2 marks]

A triangle has sides 20 cm, 21 cm and 29 cm. Show that the triangle is right-angled.

The longest side is 29 cm: \(29^2 = 841\). ✓
\(20^2 + 21^2 = 400 + 441 = 841\). ✓
Since \(20^2 + 21^2 = 29^2\), by the converse of Pythagoras' theorem the triangle is right-angled (the right angle is opposite the 29 cm side). ✓

Calculation

[2 marks]

Calculate the length of the line joining \(P(1, -3)\) and \(Q(6, 9)\).

Horizontal change \(6 - 1 = 5\); vertical change \(9 - (-3) = 12\). ✓
\(PQ = \sqrt{5^2 + 12^2} = \sqrt{169} = 13\) units ✓

Analysis

[3 marks]

A ship sails 12 km due north, then 7 km due east, then a further 5 km due north.

Calculate the straight-line distance of the ship from its starting point.

Total distance north: \(12 + 5 = 17\) km. Total distance east: 7 km. ✓
Distance\(^2 = 17^2 + 7^2 = 289 + 49 = 338\) ✓
Distance \(= \sqrt{338} = 18.4\) km (3 s.f.) ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.