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Maths · Geometry and Measures

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Pythagoras' Theorem

Using Pythagoras' theorem to find sides, test for right angles and solve problems.

  • 9 key terms
  • All boards

Learning Objectives

  1. 1State and use Pythagoras' theorem to find the hypotenuse or a shorter side of a right-angled triangle.
  2. 2Recognise Pythagorean triples and use them to avoid calculation.
  3. 3Use Pythagoras in context: ladders, diagonals of rectangles, and the height of an isosceles triangle.
  4. 4Leave answers as surds, and apply the theorem to coordinates and 3D shapes (Higher tier).

The longest side

Pythagoras' theorem links the three sides of a right-angled triangle, and it only works when the triangle has a right angle. It is one of the most reliable topics on the non-calculator paper because the numbers are almost always chosen to make square numbers, so you can square, add or subtract and then take a square root that comes out exactly. The secret to the marks is to decide, before you calculate, whether you are finding the longest side or a shorter side.

Finding the hypotenuse

Add the squares of the two shorter sides, then take the square root.

  • Step 1

    Label the sides: the hypotenuse is \(c\), the other two are \(a\) and \(b\).

  • Step 2

    Square each shorter side and add: \(a^2 + b^2\).

  • Step 3

    Take the square root of the total to get \(c\).

  • Check

    The hypotenuse must be the longest side, and shorter than \(a + b\).

Finding the hypotenuse

A right-angled triangle has shorter sides of 9 cm and 12 cm. Work out the length of the hypotenuse.

Show the solutionHide the solution
  1. 1 Write the theorem \(c^2 = a^2 + b^2\).
  2. 2 Substitute \(c^2 = 9^2 + 12^2 = 81 + 144\).
  3. 3 Add \(c^2 = 225\).
  4. 4 Square root \(c = \sqrt{225} = 15\) cm.

Answer15 cm

Finding a shorter side

This time you subtract. The hypotenuse is always the biggest, so it is the one you start with.

  • Rearrange

    \(a^2 = c^2 - b^2\).

  • Subtract the squares

    Square the hypotenuse, square the other short side, and subtract.

  • The mistake

    Adding the squares when you are looking for a shorter side gives an answer that is too big, bigger than the hypotenuse.

  • Check

    Your answer must be smaller than the hypotenuse.

Finding a shorter side

A right-angled triangle has hypotenuse 13 cm and one shorter side 5 cm. Work out the length of the other shorter side.

Show the solutionHide the solution
  1. 1 Rearrange \(a^2 = c^2 - b^2\).
  2. 2 Substitute \(a^2 = 13^2 - 5^2 = 169 - 25\).
  3. 3 Subtract \(a^2 = 144\).
  4. 4 Square root \(a = \sqrt{144} = 12\) cm.

Answer12 cm

Pythagorean triples

Some whole-number triangles come up again and again. If you recognise them, you save time and avoid mistakes.

  • Four to learn

    \(3, 4, 5\) and \(5, 12, 13\) and \(8, 15, 17\) and \(7, 24, 25\).

  • Multiples work too

    Doubling \(3, 4, 5\) gives \(6, 8, 10\), and tripling gives \(9, 12, 15\). Multiply every side by the same number.

  • Spotting them

    In a question with sides 6 and 8, the hypotenuse is 10 without any calculation, and a hypotenuse of 26 with a side of 10 gives 24.

  • Still show some working

    A bare answer might not get full marks unless the question allows it.

Is the triangle right-angled?

You can also use Pythagoras backwards to test a triangle, which is called the converse.

  • The test

    Check whether \(a^2 + b^2 = c^2\), where \(c\) is the longest side.

  • If it is equal

    The triangle is right-angled.

  • If it is not

    The triangle is not right-angled, and the answer is "no" with the two values as proof.

  • Example

    Sides 7, 24 and 25 give \(49 + 576 = 625 = 25^2\), so the triangle has a right angle.

Pythagoras in context

Many exam questions use a story, and the skill is to find the right-angled triangle hidden in it.

  • Ladders and walls

    The ladder is the hypotenuse, the wall is vertical, and the ground is horizontal.

  • Diagonals

    The diagonal of a rectangle is the hypotenuse of a triangle whose shorter sides are the length and the width.

  • Distances on a grid

    Use the horizontal and vertical difference as the two shorter sides.

  • Always draw it

    Sketch the triangle and label the sides you know, because it makes the method clear.

A ladder against a wall

A ladder 6.5 m long leans against a vertical wall. The foot of the ladder is 2.5 m from the wall. How high up the wall does the ladder reach?

Show the solutionHide the solution
  1. 1 Draw the triangle The ladder is the hypotenuse (6.5 m) and the distance from the wall is a shorter side (2.5 m).
  2. 2 Subtract the squares \(h^2 = 6.5^2 - 2.5^2 = 42.25 - 6.25\).
  3. 3 Calculate \(h^2 = 36\), so \(h = 6\) m.
  4. 4 Spot the triple Doubling the sides gives 13, 5 and 12, which is a \(5, 12, 13\) triangle.

Answer6 m

Surds, coordinates and 3D (Higher tier)

The numbers do not always give a perfect square, and Higher tier questions may ask for an exact answer.

  • Leave it as a surd

    If \(c^2 = 20\), the exact answer is \(\sqrt{20} = 2\sqrt{5}\), because \(20 = 4 \times 5\).

  • Distance between two points

    The distance from \((1, 2)\) to \((7, 10)\) uses the differences 6 and 8: \(\sqrt{6^2 + 8^2} = \sqrt{100} = 10\).

  • Diagonal of a cuboid

    Use Pythagoras twice: first for the diagonal of the base, then for the diagonal of the box.

  • Show the exact value

    Write \(\sqrt{20}\) or \(2\sqrt{5}\) and only give a decimal if asked.

A surd answer

(Higher tier) A right-angled triangle has shorter sides of 2 cm and 4 cm. Work out the length of the hypotenuse. Give your answer in the form \(a\sqrt{b}\).

Show the solutionHide the solution
  1. 1 Square and add \(c^2 = 2^2 + 4^2 = 4 + 16 = 20\).
  2. 2 Square root \(c = \sqrt{20}\).
  3. 3 Simplify \(\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}\) cm.

Answer\(2\sqrt{5}\) cm

What the OCR exam gives you

OCR prints a formulae sheet with the paper, so some formulae are given to you.

  • Given

    Pythagoras' theorem, \(a^2 + b^2 = c^2\), where \(c\) is the hypotenuse, is on the formulae page.

  • Not given

    Which side is the hypotenuse, when to add and when to subtract, and the Pythagorean triples are not given, and they are what the marks depend on.

Test yourself

  1. 1

    What is Pythagoras' theorem?

    Show answerHide answer

    \(a^2 + b^2 = c^2\), where \(c\) is the hypotenuse.

  2. 2

    Which side is the hypotenuse?

    Show answerHide answer

    The longest side, opposite the right angle.

  3. 3

    How do you find a shorter side?

    Show answerHide answer

    Subtract: \(a^2 = c^2 - b^2\).

  4. 4

    What is the hypotenuse of a triangle with shorter sides 6 and 8?

    Show answerHide answer

    10.

  5. 5

    How can you tell if a triangle is right-angled from its sides?

    Show answerHide answer

    Check whether the squares of the two shorter sides add up to the square of the longest.

Exam technique: Pythagoras

Most mistakes come from adding when you should subtract.

  • Hypotenuse first

    Decide whether you are finding the longest side before you begin.

  • Square root at the end

    Finish with \(\sqrt{\ }\) and do not forget to take it.

  • Reasonable answer

    The hypotenuse is always the longest side.

  • Show every step

    \(c^2 = 9^2 + 12^2\) earns a mark before you reach the answer.

Summary and exam focus

  • Pythagoras' theorem only works in right-angled triangles: \(a^2 + b^2 = c^2\).
  • To find the hypotenuse, add the squares and square root. To find a shorter side, subtract.
  • Learn the triples \(3, 4, 5\), \(5, 12, 13\), \(8, 15, 17\) and \(7, 24, 25\), and their multiples.
  • Split an isosceles triangle down the middle to find its height.
  • Exact answers may be surds at Higher tier.

Exam focus

A rectangle is 15 cm long and 8 cm wide. Work out the length of its diagonal. (3 marks) (3 marks)

Draw the rectangle and the diagonal to make a right-angled triangle, with the diagonal as the hypotenuse. Then \(15^2 + 8^2 = 225 + 64 = 289\), and \(\sqrt{289} = 17\) cm. This is an \(8, 15, 17\) triple.

Key terms

The words this lesson expects you to use. Each one is linked from the first place it appears above.

Right-angled triangle
A triangle with one angle of 90 degrees.
Hypotenuse
The longest side of a right-angled triangle, opposite the right angle.
Pythagoras' theorem
In a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.
Pythagorean triple
Three whole numbers that fit Pythagoras' theorem, such as 3, 4, 5.
Square root
The number that gives a stated number when multiplied by itself.
Converse
The reverse of a statement, used here to test whether a triangle has a right angle.
Diagonal
A line joining two opposite corners of a shape.
Surd
A root that cannot be written as a whole number or a fraction, such as the square root of 5.
Isosceles triangle
A triangle with two equal sides and two equal angles.

Questions and answers

16 questions set on this lesson, with the mark schemes and model answers open.

1. Exam question Calculate 4 marks Core

The diagram shows a right-angled triangle \(ABC\). (a) Calculate the length of \(AC\). [3 marks] (b) Calculate the perimeter of the triangle. [1 mark]

A right-angled triangle ABC with the right angle at B, BC = 12 cm and AB = 9 cm.

Mark scheme — 4 marks available

  • (a) \(9^2 + 12^2\) — M1
  • (a) 225 — M1
  • (a) 15 cm — A1
  • (b) 36 cm — B1

Model answer

(a) \(AC^2 = 9^2 + 12^2 = 81 + 144 = 225\), so \(AC = 15\) cm. (b) The perimeter is \(9 + 12 + 15 = 36\) cm.

2. Exam question Calculate 3 marks Core

A right-angled triangle has shorter sides of length 8 cm and 15 cm. Calculate the length of the hypotenuse.

Mark scheme — 3 marks available

  • \(8^2 + 15^2\) — M1
  • 289 — M1
  • 17 cm — A1

Model answer

\(8^2 + 15^2 = 64 + 225 = 289\), and \(\sqrt{289} = 17\) cm.

3. Exam question Calculate 3 marks Core

A right-angled triangle has a hypotenuse of 41 cm and one shorter side of 9 cm. Calculate the length of the other shorter side.

Mark scheme — 3 marks available

  • \(41^2 - 9^2\) — M1
  • 1600 — M1
  • 40 cm — A1

Model answer

\(41^2 - 9^2 = 1681 - 81 = 1600\), and \(\sqrt{1600} = 40\) cm.

4. Exam question Calculate 3 marks Core

A vertical flagpole stands on level ground. A wire 13 m long joins the top of the flagpole to a point on the ground 5 m from the base of the pole. Calculate the height of the flagpole.

Mark scheme — 3 marks available

  • \(13^2 - 5^2\) — M1
  • 144 — M1
  • 12 m — A1

Model answer

\(h^2 = 13^2 - 5^2 = 169 - 25 = 144\), so \(h = 12\) m.

5. Exam question Calculate 3 marks Stretch

A rhombus has diagonals of length 16 cm and 12 cm. Calculate the length of one side of the rhombus.

Mark scheme — 3 marks available

  • Half-diagonals 8 and 6 used — M1
  • \(8^2 + 6^2 = 100\) — M1
  • 10 cm — A1

Model answer

The diagonals of a rhombus cross at right angles and bisect each other, so a right-angled triangle has shorter sides of 8 cm and 6 cm. The side is \(\sqrt{8^2 + 6^2} = \sqrt{100} = 10\) cm.

6. Exam question Calculate 4 marks Stretch

An isosceles triangle has a base of 12 cm and two equal sides of 10 cm. Calculate the area of the triangle.

Mark scheme — 4 marks available

  • Half of the base \(= 6\) used — M1
  • \(10^2 - 6^2 = 64\) — M1
  • \(h = 8\) — A1
  • 48 cm\(^2\) — A1

Model answer

The height splits the base into two lots of 6 cm, so \(h = \sqrt{10^2 - 6^2} = \sqrt{64} = 8\) cm. The area is \(\dfrac{1}{2} \times 12 \times 8 = 48\) cm\(^2\).

7. Exam question Calculate 3 marks Stretch

\(P\) is the point \((2, -1)\) and \(Q\) is the point \((9, 23)\). Calculate the length of \(PQ\).

Mark scheme — 3 marks available

  • Differences 7 and 24 found — M1
  • \(7^2 + 24^2 = 625\) — M1
  • 25 — A1

Model answer

The horizontal difference is 7 and the vertical difference is \(23 - (-1) = 24\). So \(PQ = \sqrt{7^2 + 24^2} = \sqrt{625} = 25\).

8. Multiple choice 1 mark Easier

A right-angled triangle has shorter sides of 9 cm and 12 cm. What is the hypotenuse?

  1. A 15 cm Correct
  2. B 21 cm
  3. C 225 cm
  4. D 10 cm

Why: \(9^2 + 12^2 = 81 + 144 = 225\), and \(\sqrt{225} = 15\).

9. Multiple choice 1 mark Core

A right-angled triangle has hypotenuse 13 cm and one shorter side 5 cm. What is the other shorter side?

  1. A 8 cm
  2. B 18 cm
  3. C 144 cm
  4. D 12 cm Correct

Why: \(13^2 - 5^2 = 169 - 25 = 144\), and \(\sqrt{144} = 12\).

10. Multiple choice 1 mark Core

Which set of lengths makes a right-angled triangle?

  1. A 5 cm, 7 cm, 9 cm
  2. B 6 cm, 7 cm, 10 cm
  3. C 6 cm, 8 cm, 10 cm Correct
  4. D 4 cm, 5 cm, 7 cm

Why: \(6^2 + 8^2 = 36 + 64 = 100 = 10^2\). The other sets do not satisfy \(a^2 + b^2 = c^2\).

11. Multiple choice 1 mark Easier

Which side of a right-angled triangle is the hypotenuse?

  1. A The shortest side
  2. B The longest side, opposite the right angle Correct
  3. C The side along the bottom
  4. D The side next to the right angle

Why: The hypotenuse is always the longest side, and it is opposite the right angle.

12. Multiple choice 1 mark Core

A rectangle is 15 cm long and 8 cm wide. How long is its diagonal?

  1. A 17 cm Correct
  2. B 23 cm
  3. C 7 cm
  4. D 289 cm

Why: \(15^2 + 8^2 = 225 + 64 = 289\), and \(\sqrt{289} = 17\).

13. Multiple choice 1 mark Core

A ladder 6.5 m long leans against a wall. Its foot is 2.5 m from the wall. How high up the wall does it reach?

  1. A 9 m
  2. B 4 m
  3. C 36 m
  4. D 6 m Correct

Why: \(6.5^2 - 2.5^2 = 42.25 - 6.25 = 36\), and \(\sqrt{36} = 6\).

14. Multiple choice 1 mark Stretch

An isosceles triangle has base 10 cm and equal sides of 13 cm. What is its height?

  1. A 8 cm
  2. B 9 cm
  3. C 12 cm Correct
  4. D 7 cm

Why: The height splits the base into two lots of 5 cm. \(13^2 - 5^2 = 144\), so the height is 12 cm.

15. Multiple choice 1 mark Stretch

A right-angled triangle has shorter sides of 2 cm and 4 cm. What is the hypotenuse?

  1. A \(2\sqrt{3}\) cm
  2. B \(2\sqrt{5}\) cm Correct
  3. C 6 cm
  4. D \(\sqrt{6}\) cm

Why: \(2^2 + 4^2 = 20\), and \(\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}\).

16. Multiple choice 1 mark Stretch

What is the distance between the points \((1, 2)\) and \((7, 10)\)?

  1. A 10 Correct
  2. B 14
  3. C 100
  4. D 8

Why: The horizontal difference is 6 and the vertical difference is 8, so the distance is \(\sqrt{6^2 + 8^2} = \sqrt{100} = 10\).