Use Pythagoras’ theorem in two dimensions
Find a right triangle before using a² + b² = c² for missing sides, chords and distances on a grid.
Use this pathway for Cambridge IGCSE International Mathematics 0607, 2028–2030. Core and Extended objectives are labelled in each lesson; Extended adds requirements beyond Core. Paper 1 and Paper 2 are non-calculator. Papers 3–6 require a graphic display calculator (GDC). These original lessons use self-checked practice; they are not a Cambridge grade, full timed paper or teacher review.
Before you begin
- Check whether your school has entered you for Core or Extended. Core and Extended share most Number ideas, but some notation and methods are Extended only.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which side is c in a² + b² = c²?
What you will learn
- Identify the hypotenuse opposite the right angle
- Find a missing side in a right triangle
- Apply the theorem to circle chords and coordinate grids
Square, add and take the square root
For legs a and b and hypotenuse c, a² + b² = c². The hypotenuse is always the longest side and lies opposite the right angle.
To find a leg, subtract the other leg's square from c² before taking the square root. Preserve units and reject a negative result as evidence that the measurements do not form a right triangle.
Look for a perpendicular construction
A perpendicular from a circle's centre to a chord bisects that chord. This makes a right triangle with the radius as hypotenuse. On a grid, horizontal and vertical changes are the perpendicular legs between two points.
The same distance formula is Pythagoras on the coordinate differences: distance = √((x₂−x₁)² + (y₂−y₁)²).
Pause and explain
The perpendicular from a circle's centre meets a chord. What does it do to the chord?
Worked example
A right triangle has legs 9 cm and 12 cm. Find its hypotenuse.
Show the worked solution
- The side opposite the right angle is the hypotenuse.
- c² = 9² + 12² = 81 + 144 = 225.
- c = √225 = 15 cm.
Answer 15 cm.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A right triangle has legs 6 cm and 8 cm. Find its hypotenuse.
Give me a hint
Square both perpendicular legs, add and take the positive square root.
Compare my reasoning
- c² = 6² + 8².
- c² = 36 + 64 = 100.
- c = 10 cm.
10 cm.
Look for these in your work
- Right angle and hypotenuse identified
- Squares added
- Length unit stated
A circle has radius 13 cm. A chord is 5 cm from the centre. Find the chord length.
Give me a hint
Draw the perpendicular from the centre; it bisects the chord.
Compare my reasoning
- The half-chord, centre distance 5 and radius 13 form a right triangle.
- Half-chord² = 13² − 5² = 169 − 25 = 144.
- Half-chord = 12 cm, so the full chord is 24 cm.
24 cm.
Look for these in your work
- Perpendicular bisector used
- Missing leg found
- Half-length doubled
Find the straight-line distance between grid points A(−2, 1) and B(4, 9).
Give me a hint
Use the horizontal and vertical coordinate changes as perpendicular legs.
Compare my reasoning
- Horizontal change = 4 − (−2) = 6.
- Vertical change = 9 − 1 = 8.
- Distance = √(6² + 8²) = √100 = 10 units.
10 units.
Look for these in your work
- Coordinate differences found
- Distance relation applied
- Units included
Common mistakes
- Choosing a side adjacent to the right angle as the hypotenuse
- Subtracting when finding the hypotenuse
- Forgetting to halve a chord after drawing the perpendicular
- Giving a squared unit for a length
What can you explain now?
A right triangle has hypotenuse 13 cm and one leg 5 cm. Find the other leg.
Compare with the explanation
12 cm
The missing leg squared is 13² − 5² = 144.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Why is a perpendicular construction useful when applying Pythagoras to a chord or two grid points?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.