Cambridge IGCSE Mathematics 0580: Pythagoras’ theorem – Study Guide

This study guide teaches subtopic 6.1, Pythagoras’ theorem, from topic 6 (Trigonometry) of the Cambridge IGCSE Mathematics 0580 syllabus for 2025, 2026 and 2027. The learning outcome is the same in both columns of the syllabus: C6.1 (Core) and E6.1 (Extended) both say “Know and use Pythagoras’ theorem.” So everything here is needed whichever tier you sit. Core candidates take Paper 1 (non-calculator) and Paper 3 (calculator); Extended candidates take Paper 2 (non-calculator) and Paper 4 (calculator). Pythagoras can appear on any of the four papers, so you need it both with and without a calculator. The only Extended-only material on this page is giving an answer in surd form, which belongs to E1.18 Surds, and it is labelled where it appears.

Use this page with the Cambridge IGCSE Mathematics hub and the printable 0580 checklist. For the whole of topic 6, see the Trigonometry study guide.

What this subtopic covers

0580 reference	What you must be able to do	Tier
C6.1 / E6.1	Know Pythagoras’ theorem: recall it without being given it, and recognise when it applies	Core and Extended
C6.1 / E6.1	Use Pythagoras’ theorem: find the hypotenuse, find a shorter side, and use it inside larger problems	Core and Extended
E1.18 (linked)	Leave a length as an exact surd, such as 4√5, when a question asks for an exact answer	Extended only

Two points from the syllabus shape how you prepare:

• The theorem is not on the List of formulas. The formulas sheet on page 2 of every paper lists areas, volumes and (for Extended) the sine rule, cosine rule and quadratic formula. Pythagoras’ theorem is not there, so “know” really means you must recall it.
• It feeds later subtopics. C6.2/E6.2 asks you to solve problems in two dimensions “using Pythagoras’ theorem and trigonometry”. E6.6 (Extended only) uses it in three dimensions, and E3.4 (Extended only) uses it to find the length of a line segment. This page sticks to 6.1 itself and points you to those pages at the end.

The theorem

In a right-angled triangle, the longest side is called the hypotenuse. It is always the side opposite the right angle. If the hypotenuse is c and the two shorter sides are a and b:

a² + b² = c²

In words: the square on the hypotenuse equals the sum of the squares on the other two sides.

Three facts to hold on to:

1. It only works for a right-angled triangle.
2. The hypotenuse is always the longest side, so a shorter side must come out less than the hypotenuse.
3. The letters do not matter. What matters is which side is opposite the right angle.

Find the hypotenuse first, every time

Before you write anything, find the right angle and mark the side opposite it. Then you know which of two jobs you are doing:

• Finding the hypotenuse: square both shorter sides, add, then square root.
• Finding a shorter side: square both known sides, subtract (hypotenuse² minus the other side²), then square root.

Finding the hypotenuse

Worked example 1 (calculator). Triangle PQR has a right angle at Q. PQ = 7.4 cm and QR = 5.9 cm. Calculate PR.

The right angle is at Q, so the hypotenuse is PR, the side opposite Q.

PR² = PQ² + QR²
PR² = 7.4² + 5.9²
PR² = 54.76 + 34.81
PR² = 89.57
PR  = √89.57
PR  = 9.464...
PR  = 9.46 cm (3 s.f.)

Check: 9.46 is longer than both 7.4 and 5.9, as a hypotenuse must be.

On Papers 3 and 4, non-exact answers go to 3 significant figures unless the question says otherwise. Write a few more figures (9.464…) before you round.

Finding a shorter side

Worked example 2 (calculator). In triangle ABC, angle B = 90°. The hypotenuse AC = 9.2 cm and BC = 4.7 cm. Calculate AB.

AB² + BC² = AC²
AB² = AC² − BC²
AB² = 9.2² − 4.7²
AB² = 84.64 − 22.09
AB² = 62.55
AB  = √62.55
AB  = 7.908...
AB  = 7.91 cm (3 s.f.)

Check: 7.91 is less than 9.2. If you had added instead of subtracted, you would get √106.73 = 10.3 cm, which is longer than the hypotenuse. That is impossible, and spotting it saves the marks.

Pythagoras without a calculator

On Paper 1 and Paper 2 you cannot use a calculator, so questions are set with numbers that work out cleanly, or they ask you to estimate or leave a root.

Pythagorean triples

A Pythagorean triple is a set of three whole numbers that fit a² + b² = c². Learning the common ones turns many non-calculator questions into quick recognition.

Triple	Check
3, 4, 5	9 + 16 = 25
5, 12, 13	25 + 144 = 169
8, 15, 17	64 + 225 = 289
7, 24, 25	49 + 576 = 625

Any multiple of a triple is also a triple: 6, 8, 10 comes from 3, 4, 5. Recognising a triple is a short cut, but still show the Pythagoras working.

Worked example 3 (non-calculator). A right-angled triangle has hypotenuse 34 cm and one shorter side 16 cm. Find the third side.

x² = 34² − 16²
x² = 1156 − 256
x² = 900
x  = 30 cm

You could also spot that 16, 30, 34 is 8, 15, 17 multiplied by 2. Squaring 34 by hand: 34² = 30² + 2 × 30 × 4 + 4² = 900 + 240 + 16 = 1156.

Square roots that are not whole numbers

If the answer is not a whole number on a non-calculator paper, the question will tell you what to do. Two common instructions:

• Estimate. Place the root between two consecutive square numbers. For example √52 lies between √49 = 7 and √64 = 8, and 52 is much nearer 49, so √52 is a little more than 7.
• Give an exact answer (Extended only). Simplify the surd, using E1.18. Look for the largest square factor.

Worked example 4 (non-calculator, Extended only). A right-angled triangle has shorter sides 4 cm and 8 cm. Find the exact length of the hypotenuse, in its simplest form.

c² = 4² + 8²
c² = 16 + 64
c² = 80
c  = √80
c  = √(16 × 5)
c  = 4√5 cm

Writing 8.94 here would lose the final mark, because 8.94 is a rounded decimal, not the exact value. The syllabus says that when a question asks for an exact value, the answer may need to be in surd form.

Using Pythagoras inside other shapes

Many questions do not show a right-angled triangle directly. You have to find one inside the shape. Look for:

• a rectangle split by a diagonal (two right-angled triangles);
• an isosceles triangle split by its line of symmetry (the height meets the base at 90° and cuts it in half);
• a square split by a diagonal;
• a horizontal and a vertical distance, such as a ladder against a wall or a walk “east then north”.

Worked example 5 (non-calculator). An isosceles triangle has two equal sides of 17 cm and a base of 16 cm. Find its perpendicular height and its area.

The line of symmetry is perpendicular to the base and cuts it into two halves of 8 cm. Each half is a right-angled triangle with hypotenuse 17 cm (the equal side) and one shorter side 8 cm.

h² = 17² − 8²
h² = 289 − 64
h² = 225
h  = 15 cm

Area = ½ × base × height
     = ½ × 16 × 15
     = 120 cm²

The usual slip is to use the full base, 16, in the Pythagoras step. Only half the base is in the right-angled triangle.

Worked example 6 (calculator, problem in context). A rectangular field is 85 m long and 62 m wide. Amira walks from one corner to the opposite corner along two edges of the field. Ben walks straight across the diagonal. How much further does Amira walk? Give your answer to 3 significant figures.

Diagonal² = 85² + 62²
          = 7225 + 3844
          = 11069
Diagonal  = √11069 = 105.209... m

Amira's route = 85 + 62 = 147 m
Difference    = 147 − 105.209...
              = 41.790...
              = 41.8 m (3 s.f.)

Notice what happens if you round too early. Rounding the diagonal to 105 m first gives 147 − 105 = 42 m, which is wrong to 3 significant figures. The syllabus says you should avoid rounding until you have your final answer, and should use the unrounded value when a result is carried forward. Keep 105.209… in your calculator memory.

“Show that” questions

Sometimes the answer is given and you have to prove it. The syllabus says that when a question asks you to show a given answer to a stated accuracy, you must show full working, intermediate answers and a final answer to at least one degree of accuracy more than the one asked for.

Worked example 7 (calculator). A ladder 7.2 m long leans against a vertical wall. Its foot is on level ground, 2.1 m from the wall. Show that the top of the ladder is 6.9 m above the ground, correct to 1 decimal place.

The wall and the ground meet at a right angle, so the ladder is the hypotenuse.

h² = 7.2² − 2.1²
h² = 51.84 − 4.41
h² = 47.43
h  = √47.43
h  = 6.886... m
h  = 6.89 m (2 d.p.)
   = 6.9 m (1 d.p.)

Writing only “h = 6.9” earns little, because 6.9 was given. The 6.886… line is what proves it.

Is the triangle right-angled?

Knowing the theorem also means knowing it only holds when there is a right angle. So you can test three side lengths: if the square of the longest side equals the sum of the squares of the other two, the triangle is right-angled; if not, it is not.

• Sides 20, 21 and 29: 20² + 21² = 400 + 441 = 841 and 29² = 841. Equal, so the triangle is right-angled, with the right angle opposite the 29 side.
• Sides 9, 11 and 14: 9² + 11² = 81 + 121 = 202 and 14² = 196. Not equal, so the triangle is not right-angled.

Always compare with the longest side squared.

Common errors

• Adding when you should subtract. When the hypotenuse is given, subtract. A “shorter side” longer than the hypotenuse tells you this has happened.
• Choosing the wrong side as the hypotenuse. In a rotated triangle, the hypotenuse may not be the side drawn at the top or on the slant. It is the side opposite the right angle.
• Forgetting the square root. Stopping at c² = 89.57 and writing 89.57 cm as the length.
• Square-rooting each term. √(7.4² + 5.9²) is not 7.4 + 5.9. Add the squares first, then take one square root.
• Using a full base instead of a half. In an isosceles triangle, only half the base is in the right-angled triangle.
• Rounding before the end. Rounding the diagonal to 105 changed the final answer in example 6.
• Giving a decimal when “exact” is asked (Extended). Leave √80 as 4√5, not 8.94.

Where this leads

Once Pythagoras is secure, the next steps in the 0580 syllabus are:

• C6.2/E6.2 Right-angled triangles: sine, cosine and tangent, and two-dimensional problems that combine them with Pythagoras (bearings may be needed). The Trigonometry study guide covers this.
• E3.4 Length and midpoint (Extended only): the length of a line segment between two coordinates is Pythagoras on the horizontal and vertical differences. See the Coordinate geometry study guide.
• E6.6 Pythagoras’ theorem and trigonometry in 3D (Extended only): find a flat right-angled triangle inside a solid, then use the same method as on this page.

Test yourself with the Trigonometry practice questions and the Extended trigonometry practice, recap with the Trigonometry revision notes, and check where you stand with the Core diagnostic or Extended diagnostic.

Official syllabus

Cambridge International, Cambridge IGCSE Mathematics 0580 syllabus for 2025, 2026 and 2027 (Version 3), Subject content, topic 6 Trigonometry: C6.1 and E6.1 Pythagoras’ theorem. Linked references: E1.18 Surds, E3.4 Length and midpoint, C6.2/E6.2 Right-angled triangles and E6.6 Pythagoras’ theorem and trigonometry in 3D.