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Pythagoras

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How to use Pythagoras’ theorem to find a missing side of a right-angled triangle, whether it is the longest side or a shorter one, how to round the answer, and how to find right-angled triangles inside ladders, rectangles, isosceles triangles and trapeziums.

What Pythagoras’ theorem is for

Pythagoras’ theorem links the three sides of a right-angled triangle. If you know two of the sides, you can work out the third. It finds lengths, not angles.

In the exam you are usually shown a right-angled triangle with two lengths marked and asked to work out the third, for 2 or 3 marks. You often have to round, for example correct to 1 decimal place or to 3 significant figures.

Harder questions, for 4 or 5 marks, hide the triangle inside another shape, such as a rectangle, a trapezium or an isosceles triangle. You then use the length you find to work out a perimeter or an area.

The hypotenuse and the formula

The hypotenuse is the side opposite the right angle. It is always the longest side. Find the right angle first: the hypotenuse is the one side that does not touch it.

Call the two shorter sides a and b, and call the hypotenuse c. Then:

a^2+b^2=c^2

The squares of the two shorter sides add up to the square of the hypotenuse. Since 2022 a formulae sheet has been given in the exam, and this formula is on it, but learn it anyway.

To square a number, multiply it by itself: 6^2=6\times6=36. A square root undoes squaring: \sqrt{36}=6

Worked example 1: Finding the hypotenuse

A right-angled triangle has shorter sides of length 12 cm and 16 cm.

Work out the length of the hypotenuse.

  1. The hypotenuse is the missing side, so add the squares: c^2=12^2+16^2

  2. Your turn. Work out 12^2+16^2

    c^2=144+256=400

  3. 400 is c^2, not c. Take the square root: c=\sqrt{400}=20
  4. Check: 20 cm is longer than 12 cm and 16 cm, as the hypotenuse must be.
  5. Answer: 20 cm

Finding a shorter side

If you know the hypotenuse, the missing side is one of the shorter sides. Start with the square of the hypotenuse and take away the square of the other side:

a^2=c^2-b^2

For example, if the hypotenuse is 2.5 m and one side is 1.5 m, work out 2.5^2-1.5^2=6.25-2.25=4. Then \sqrt{4}=2, so the missing side is 2 m.

Quick check: if you add when you should subtract, your answer comes out longer than the hypotenuse. That cannot be right, because the hypotenuse is the longest side.

Worked example 2: Finding a shorter side

Triangle PQR has a right angle at Q.

PR=29 cm and PQ=20 cm.

Work out the length of QR.

  1. The right angle is at Q, so the hypotenuse is the side opposite it, PR. The missing side QR is a shorter side.
  2. Subtract the squares: QR^2=29^2-20^2

  3. Your turn. Work out 29^2-20^2

    QR^2=841-400=441

  4. QR=\sqrt{441}=21, which is shorter than the hypotenuse, as it should be.
  5. Answer: 21 cm

Rounding the answer

Most square roots are not whole numbers. Write down several digits from your calculator, for example \sqrt{40}=6.324555\ldots, and round only at the very end.

Do not round c^2, or any length you will use again. Rounding too early can move your final answer away from the right one.

Give the accuracy the question asks for. Here, 6.3 is correct to 1 decimal place and 6.32 is correct to 3 significant figures.

Worked example 3: Rounding the hypotenuse

A right-angled triangle has shorter sides of length 4 cm and 7 cm.

Work out the length of the hypotenuse. Give your answer correct to 1 decimal place.

  1. c^2=4^2+7^2=16+49=65
  2. c=\sqrt{65}=8.062257\ldots

  3. Your turn. Round 8.062257\ldots to 1 decimal place.

    The second decimal place is 6, so round up to 8.1

  4. Answer: 8.1 cm

Finding the triangle inside a shape

A diagonal cuts a rectangle into two right-angled triangles. The width and the height are the shorter sides, and the diagonal is the hypotenuse.

A ladder leaning against a vertical wall on horizontal ground makes a right angle where the wall meets the ground. The ladder is the hypotenuse.

In an isosceles triangle, a line from the top, where the two equal sides meet, straight down to the base cuts the base in half. Each half of the triangle is a right-angled triangle.

Mark the triangle you are using and write its lengths on it. Then read the question again: the length you find is often only one step towards a perimeter, an area or a difference.

Worked example 4: A ladder against a wall

A ladder of length 5.5 m leans against a vertical wall. The ground is horizontal. The foot of the ladder is 1.8 m from the wall.

Work out how far up the wall the ladder reaches. Give your answer correct to 1 decimal place.

  1. The wall meets the ground at a right angle. The ladder is the hypotenuse, so the height h up the wall is a shorter side: h^2=5.5^2-1.8^2

  2. Your turn. Work out 5.5^2-1.8^2

    h^2=30.25-3.24=27.01

  3. h=\sqrt{27.01}=5.197114\ldots, which is 5.2 correct to 1 decimal place.
  4. Check: 5.2 m is less than the length of the ladder, 5.5 m, as it must be.
  5. Answer: 5.2 m

Worked example 5: The area of an isosceles triangle

An isosceles triangle has two sides of length 17 cm and a base of length 16 cm.

Work out the area of the triangle.

  1. To find the area you need the height, h. Draw it from the top to the middle of the base. It cuts the base into two halves of 16\div2=8 cm.
  2. In each right-angled triangle, the 17 cm side is the hypotenuse. Use half the base, 8 cm, not the whole base.

  3. Your turn. Work out 17^2-8^2

    h^2=289-64=225

  4. h=\sqrt{225}=15, so the height is 15 cm.
  5. Area =\frac{1}{2}\times\text{base}\times\text{height}=\frac{1}{2}\times16\times15=120
  6. Answer: 120 cm²

Worked example 6: The perimeter of a trapezium

ABCD is a trapezium. Angle DAB and angle ADC are right angles.

AB=9 cm, DC=15 cm and AD=8 cm.

Work out the perimeter of the trapezium.

  1. You need the sloping side BC. Draw a line from B straight down to DC, and call the point where it meets DC point E. This makes a right-angled triangle BEC, with hypotenuse BC.
  2. BE is the same length as AD, so BE=8 cm. Also EC=15-9=6 cm.

  3. Your turn. Work out 6^2+8^2

    BC^2=36+64=100

  4. So BC=\sqrt{100}=10 cm
  5. The perimeter is the total of all four sides: 9+15+8+10=42
  6. Answer: 42 cm

Common mistakes

  • Adding the squares when a shorter side is missing. If you know the hypotenuse, start with its square and subtract the other square.
  • Using the lengths without squaring them. Square each length first, then add or subtract.
  • Subtracting the squares the wrong way round. The square of the hypotenuse always comes first.
  • Forgetting the square root. Adding or subtracting the squares gives the missing side squared, so finish with a square root.
  • Rounding too early. Keep the full calculator value until the last step, then round as the question asks.
  • Stopping once you have found the side. Read the question again: you may still need a perimeter, an area or a difference.

Practice questions

More on this topic: Pythagoras worksheet with full solutions