Maths explained · Years 8 to 10

Pythagoras’ theorem: the formula, explained with examples

Pythagoras’ theorem finds a missing side of a right-angled triangle when you know the other two. Here’s the formula, how to use it step by step, the mistakes that cost marks, and practice questions with answers.

Quick answer

In a right-angled triangle, a² + b² = c², where c is the hypotenuse (the longest side, opposite the right angle). To find the hypotenuse, square the two shorter sides, add them, then take the square root. To find a shorter side, square the hypotenuse, subtract the square of the other side, then take the square root.

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The formula

a2+b2=c2a^2 + b^2 = c^2
  • cc is the hypotenuse: the longest side, opposite the right angle.
  • aa and bb are the two shorter sides, which meet at the right angle. It doesn't matter which is which.
  • It only works in right-angled triangles, and you need two sides to find the third.

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

Pythagoras calculator (with working)

Fill in any two sides. You’ll get the answer and every step, set out the way your teacher wants to see it.

abc

The hypotenuse is 10 cm

  1. You're finding the hypotenuse cc, the longest side, opposite the right angle.
  2. Write Pythagoras' theorem: a2+b2=c2a^2 + b^2 = c^2
  3. Substitute the two shorter sides: 62+82=c26^2 + 8^2 = c^2
  4. Square each side length: 36+64=c236 + 64 = c^2
  5. Add: c2=100c^2 = 100
  6. Take the square root of both sides:
  7. c=100=10 cmc = \sqrt{100} = 10 \text{ cm}

Fill in any two sides and leave the one you want blank. The hypotenuse is the side opposite the right angle.

Finding the hypotenuse

Square the two shorter sides, add them, then take the square root.

Worked example

A right-angled triangle has shorter sides of 6 cm and 8 cm. Find the hypotenuse.

c2=62+82c^2 = 6^2 + 8^2 c2=36+64=100c^2 = 36 + 64 = 100 c=100=10 cmc = \sqrt{100} = 10 \text{ cm}

Finding a shorter side

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

Worked example

The hypotenuse is 17 m and one shorter side is 8 m. Find the other side.

a2+82=172a^2 + 8^2 = 17^2 a2=289−64=225a^2 = 289 - 64 = 225 a=225=15 ma = \sqrt{225} = 15 \text{ m}

Quick check: the hypotenuse must be the longest side. If your "shorter side" comes out longer than the hypotenuse, you added when you should have subtracted.

Word problems

Draw the triangle first, mark the right angle, and label the hypotenuse. Ladders, ramps, diagonals of rectangles and screens, and distances on a map are all Pythagoras problems.

Worked example

A 4 m ladder leans against a wall. Its foot is 1.2 m from the wall. How high up the wall does it reach, to 2 decimal places?

The ladder is the hypotenuse (it's opposite the right angle between the wall and the ground).

h2+1.22=42h^2 + 1.2^2 = 4^2 h2=16−1.44=14.56h^2 = 16 - 1.44 = 14.56 h=14.56≈3.82 mh = \sqrt{14.56} \approx 3.82 \text{ m}

Is a triangle right-angled?

Square the longest side and compare it with the sum of the squares of the other two (this is the converse of Pythagoras' theorem).

  • Sides 8, 15, 17: 82+152=64+225=289=1728^2 + 15^2 = 64 + 225 = 289 = 17^2, so yes, it's right-angled.
  • Sides 6, 7, 9: 62+72=36+49=856^2 + 7^2 = 36 + 49 = 85, but 92=819^2 = 81, so no.

Pythagorean triples are whole-number sets that work: 3, 4, 5; 5, 12, 13; 8, 15, 17; 7, 24, 25. Any multiple works too, like 6, 8, 10. Spotting one saves time in a test.

In 3D (Year 10A): use Pythagoras twice. For a box 3 by 4 by 12, the space diagonal is 32+42+122=169=13\sqrt{3^2 + 4^2 + 12^2} = \sqrt{169} = 13.

Common mistakes that cost marks

  • Adding when finding a shorter side. To find a shorter side you subtract: c² − b².
  • Forgetting the square root. c² = 100 means c = 10, not 100.
  • Picking the wrong hypotenuse. It's opposite the right angle, not just the side drawn at a slant.
  • Squaring the sum. 6² + 8² is 36 + 64, not (6 + 8)².
  • Rounding too early. Keep the exact value until the last line, then round.
  • Using it without a right angle. No right angle, no Pythagoras.

Practice questions with answers

Try each one before you open the answer.

  1. The two shorter sides of a right-angled triangle are 9 cm and 12 cm. Find the hypotenuse.

    Show the answer

    c2=92+122=81+144=225c^2 = 9^2 + 12^2 = 81 + 144 = 225, so c=225=15c = \sqrt{225} = 15 cm.

  2. The hypotenuse is 13 m and one shorter side is 5 m. Find the other side.

    Show the answer

    a2=132−52=169−25=144a^2 = 13^2 - 5^2 = 169 - 25 = 144, so a=12a = 12 m.

  3. The shorter sides are 7 cm and 10 cm. Find the hypotenuse, to 2 decimal places.

    Show the answer

    c2=49+100=149c^2 = 49 + 100 = 149, so c=149≈12.21c = \sqrt{149} \approx 12.21 cm.

  4. The hypotenuse is 20 mm and one side is 11 mm. Find the third side, to 2 decimal places.

    Show the answer

    a2=400−121=279a^2 = 400 - 121 = 279, so a=279≈16.70a = \sqrt{279} \approx 16.70 mm.

  5. A 5 m ladder leans against a wall with its foot 1.5 m from the wall. How high up the wall does it reach, to 2 decimal places?

    Show the answer

    h2=52−1.52=25−2.25=22.75h^2 = 5^2 - 1.5^2 = 25 - 2.25 = 22.75, so h≈4.77h \approx 4.77 m.

  6. Is a triangle with sides 7, 24 and 25 right-angled?

    Show the answer

    Yes. 72+242=49+576=625=2527^2 + 24^2 = 49 + 576 = 625 = 25^2.

  7. A rectangle is 16 cm long and 12 cm wide. How long is its diagonal?

    Show the answer

    d2=162+122=256+144=400d^2 = 16^2 + 12^2 = 256 + 144 = 400, so d=20d = 20 cm.

  8. Find the diagonal of a square with sides of 6 cm. Give an exact answer and a decimal to 2 places.

    Show the answer

    d2=62+62=72d^2 = 6^2 + 6^2 = 72, so d=72=62≈8.49d = \sqrt{72} = 6\sqrt{2} \approx 8.49 cm.

Want more? Each year level below has a free printable PDF of 10 questions with answers, and a full 20-question test. Or check your working with the calculator.

Pythagoras by year level

Following the Victorian Curriculum 2.0. Schools sequence topics differently, so check your course.

Year 8

Pythagoras' theorem

Introducing the theorem: finding the hypotenuse and shorter sides, and checking for right angles.

Year 9

Pythagoras (applied)

Applying it to problems, alongside trigonometry: ladders, ramps, diagonals and distances.

Year 10A

Pythagoras and trigonometry in 3D

Pythagoras and trigonometry in 3D: space diagonals in boxes, pyramids and cones.

Pythagoras' theorem: common questions

What is Pythagoras' theorem?

In a right-angled triangle, the square of the hypotenuse (the longest side, opposite the right angle) equals the sum of the squares of the other two sides: a² + b² = c².

When can you use Pythagoras' theorem?

Only in right-angled triangles, and only when you know two sides and want the third. If you know an angle, or want to find one, use trigonometry (SOH CAH TOA) instead.

How do you know which side is the hypotenuse?

It's the side opposite the right angle, and always the longest side. It doesn't touch the right angle.

What year level learns Pythagoras' theorem?

In the Victorian Curriculum 2.0 it's introduced in Year 8 and applied in Year 9 alongside trigonometry. Year 10A extends it to 3D shapes. Schools sequence topics differently, so check your child's course.

Should I round my answer?

Leave it exact (for example √149) unless the question asks you to round. If it says 'to 2 decimal places', round only at the very end.

Tuterly · Years 7 to 10 maths

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