Skip to content
Cambridge IGCSE0607

Pythagoras’ theorem

International Mathematics 0607 Chapter Notes

What this chapter covers

Pythagoras’ theorem
ShareWhatsAppPost
Pythagoras’ theorem notes

Read online free. Download a watermarked copy with a free account.

Read the notes

The full Pythagoras’ theorem notes as text: skim, search, and jump between subtopics.

~6 min read

In this chapter

  1. 1The theorem and the hypotenuse
  2. 2Finding a shorter side
  3. 3Distance between two points on a grid
  4. 4A chord and its distance from the centre
  5. 5A tangent and a radius
  6. 6Accuracy and showing working

1. The theorem and the hypotenuse

Pythagoras' theorem links the three sides of a right-angled triangle. The hypotenuse is the side opposite the right angle.

Square the two shorter sides, add the squares, then square root the total. The answer is the hypotenuse.

c² = a² + b²

c = √(a² + b²)

Examiner insight

“Some candidates found the length of the hypotenuse only.” “Some candidates truncated their answer for the hypotenuse resulting in an inaccurate final answer.”

Worked example 13 marks

The diagram shows a right-angled triangle. Calculate the perimeter of the triangle.

  1. 1

    hypotenuse² = 8.5² + 5² [1]

  2. 2

    = 97.25, so hypotenuse = √(97.25) = 9.862… cm [1]

  3. 3

    perimeter = 8.5 + 5 + 9.862… = 23.4 cm [1]

2. Finding a shorter side

When the hypotenuse is given, you are finding a shorter side. Rearrange the theorem so that you subtract the squares.

Square the hypotenuse, then subtract the square of the side you know. Square root the result.

a² = c² - b²

a = √(c² - b²)

Examiner insight

“Many candidates failed to recognise the need to use Pythagoras’ theorem. Of those who did, it was still common to find them simply squaring and adding, whereas they needed to subtract their squares before square rooting because they were calculating the length of one of the shorter sides.”

Worked example 14 marks

Work out the perimeter of this triangle.

  1. 1

    the hypotenuse is 29 cm, so 29² = 20² + x² [1]

  2. 2

    x² = 29² - 20² = 841 - 400 = 441 [1]

  3. 3

    x = √(441) = 21 cm [1]

  4. 4

    perimeter = 29 + 20 + 21 = 70 cm [1]

3. Distance between two points on a grid

The straight line between two points is the hypotenuse of a right-angled triangle. Its shorter sides are the difference in the x-coordinates and the difference in the y-coordinates.

Find both differences first, then use Pythagoras' theorem. A negative difference becomes positive when you square it.

d = √((x₂ - x₁)² + (y₂ - y₁)²)

Examiner insight

“The question said ‘calculate’ indicating that measuring was not appropriate.” “Many candidates just measured the length of AB and gained no marks.”

Worked example 13 marks

Do not use a calculator. A is the point (-2, 3) and B is the point (4, -5). Calculate the length of AB.

  1. 1

    differences: 4 - (-2) = 6 and -5 - 3 = -8 [1]

  2. 2

    AB² = 6² + (-8)² = 36 + 64 = 100 [1]

  3. 3

    AB = √(100) = 10 [1]

4. A chord and its distance from the centre

A chord is a straight line joining two points on a circle. The distance of a chord from the centre is measured at right angles to the chord.

That line meets the chord at its middle. It makes a right-angled triangle with a radius as the hypotenuse. One shorter side is half the chord.

d² + (chord/2)² = r²

Examiner insight

“The diagram showed that there was a right-angled triangle but that they needed to use a half of 16 as the triangle’s base.” “Also, those who used Pythagoras’ Theorem with 16 and 10 did not get any marks.”

Worked example 13 marks

A circle has centre O and radius 13 cm. A chord of the circle is 24 cm long. Calculate the distance of the chord from O.

  1. 1

    half the chord = 24 ÷ 2 = 12 cm [1]

  2. 2

    d² + 12² = 13² [1]

  3. 3

    d² = 169 - 144 = 25, so d = 5 cm [1]

5. A tangent and a radius

A tangent to a circle meets the radius at the point of contact at a right angle. The reason to give is angle between tangent and radius = 90°.

That right angle makes a right-angled triangle. The line from the centre to the outside point is its hypotenuse.

OB² = OA² + AB²

Worked example 12 marks

AB and CB are tangents to the circle with centre O and radius 6.4 cm. AB is 11.3 cm. Work out the length of OB.

  1. 1

    angle OAB = 90^(°), so OB² = 6.4² + 11.3² = 40.96 + 127.69 = 168.65 [1]

  2. 2

    OB = √(168.65) = 12.986… = 13.0 cm [1]

6. Accuracy and showing working

Keep the full calculator value until the last line. Round only the final answer, to the accuracy the question asks for.

When a question says calculate, work it out. Measuring from the diagram gains no marks. The diagrams are not drawn to scale.

Examiner insight

“Almost all candidates gave a correct Pythagoras statement but often lost a mark by only showing 22.36 and not working to at least 3 decimal places.” “Some candidates showed their working out of Pythagoras’ Theorem and were awarded both marks if they gave their answer to 3 significant figures but only one mark if they wrote 4.5 as their answer.”

Worked example 13 marks

A right-angled triangle has shorter sides 10 cm and 20 cm. Show that the hypotenuse is 22.4 cm, correct to 3 significant figures.

  1. 1

    hypotenuse² = 10² + 20² = 500 [1]

  2. 2

    hypotenuse = √(500) = 22.3607… cm [1]

  3. 3

    = 22.4 cm to 3 significant figures [1]

Recap

  • Pythagoras' theorem is for right-angled triangles. The hypotenuse is opposite the right angle.
  • Hypotenuse: square, add, square root.
  • Shorter side: square, subtract, square root.
  • Distance between two points: subtract the coordinates first, then use Pythagoras.
  • A chord: use half the chord and the radius to find its distance from the centre.
  • A tangent meets the radius at 90°.
  • Keep the full value until the end, then round.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. State which side of a right-angled triangle is the hypotenuse.1 mark
  2. Do not use a calculator. A right-angled triangle has shorter sides 6 cm and 8 cm. Find the length of the hypotenuse.2 marks
  3. Do not use a calculator. A right-angled triangle has hypotenuse 13 cm and one shorter side 5 cm. Find the other shorter side.2 marks
  4. A right-angled triangle has shorter sides 7 cm and 9 cm. Calculate the length of the hypotenuse, correct to 3 significant figures.3 marks
  5. Do not use a calculator. A ladder 2.5 m long leans against a vertical wall. The foot of the ladder is 1.5 m from the wall on level ground. Calculate how high up the wall the ladder reaches.3 marks
  6. Do not use a calculator. P is the point (-3, 2) and Q is the point (5, -4). Calculate the distance PQ.3 marks
  7. A right-angled triangle has hypotenuse 17 cm and one shorter side 9 cm. Calculate the perimeter of the triangle, correct to 3 significant figures.4 marks
  8. Do not use a calculator. A circle has centre O and radius 10 cm. A chord of the circle is 16 cm long. Calculate the distance of the chord from O.3 marks
  9. Do not use a calculator. A pipe has a circular cross-section of radius 6.5 cm. Water lies in the bottom of the pipe. At its deepest point the water is 4 cm deep. Calculate the width of the surface of the water.4 marks
  10. ABCD is a garden in the shape of a trapezium. AB is parallel to DC. Angle DAB and angle ADC are both 90°. AB = 4.2 m, DC = 7.5 m and AD = 5.8 m. A fence goes all the way round the garden. Another straight fence joins A to C. Calculate the total length of fencing, correct to 3 significant figures.6 marks

Free notes are just the start.

Unlock every Workbook and Chapter at a Glance, and generate your own worksheets and predicted papers.

Explore plans

Related chapters

Types of numberSetsPowers and rootsFractions, decimals and percentagesOrderingThe four operations