Pythagoras: revision notes and practice

What you will learn

• Identify a right angle and the hypotenuse reliably.
• Find either the hypotenuse or a shorter side.
• Use Pythagoras in diagrams, coordinates and practical contexts.
• Higher extensions: work with exact surds and a cuboid's space diagonal.

Key rules

Find the hypotenuse

The hypotenuse c is opposite the right angle and is the longest side. If the other sides are a and b, c² = a² + b².

Find a shorter side

Rearrange to a² = c² − b². Subtract the squares, then take the positive square root.

Use consistent lengths

Convert all lengths to the same unit first. Keep calculator accuracy until the end and round only as requested.

Worked examples

Example 1: A right-angled triangle has perpendicular sides 6 cm and 8 cm. Find its hypotenuse.

1. c² = 6² + 8² = 36 + 64 = 100.
2. c = √100 = 10. This is an exact value.

10 cm

Example 2: A right-angled triangle has hypotenuse 13 cm and one shorter side 5 cm. Find the other side.

1. b² = 13² − 5² = 169 − 25 = 144.
2. b = √144 = 12. This is shorter than 13 cm.

12 cm

A practice question

A right-angled triangle has perpendicular sides 25 cm and 60 cm. Find its hypotenuse.

1. Find the right triangle first

Pythagoras links the three side lengths of a right-angled triangle. The hypotenuse is the side opposite the 90° angle, and it is always the longest side. The other two sides meet at the right angle. Label the hypotenuse c before choosing an operation.

A sloping side is not automatically the hypotenuse, because diagrams can be rotated. Look for the right-angle marker or information that establishes perpendicular lines. Do not assume a triangle is right-angled just because it looks that way; an unlabelled sketch may not be drawn to scale.

2. Understand the squares relationship

If a and b are the perpendicular side lengths, then a² + b² = c². A geometric interpretation is that the areas of squares built on the shorter sides add to the area of the square built on the hypotenuse. The rule concerns squares of lengths, not a direct sum of the lengths.

For sides 3, 4 and 5, the relationship is 9 + 16 = 25. This explains why adding 3 + 4 does not give the hypotenuse. The equation can also check whether three positive lengths form a right triangle: assign the longest to c and test the equality.

a² + b² = c²

3. Find the hypotenuse by adding squares

When both shorter sides are known, square them, add the results, and take the positive square root. For perpendicular lengths 5 cm and 12 cm, c² = 25 + 144 = 169, so c = 13 cm.

The square root is essential: 169 is the value of c², not the side length. A physical length is positive, so use the positive root. Check that the answer exceeds both shorter sides. If your result is smaller than 12 cm in this example, you have chosen or entered the calculation incorrectly.

c = √(a² + b²)

4. Find a shorter side by subtracting squares

If the hypotenuse and one shorter side are known, rearrange the equation. With c = 10 m and a = 6 m, b² = 10² − 6² = 64, so b = 8 m. Subtract the known shorter-side square from the hypotenuse square.

Use brackets around the whole difference on a calculator. The answer must be shorter than the hypotenuse. A negative value under the square root in this setting usually indicates swapped labels, inconsistent units, or impossible measurements; do not remove the negative sign to force an answer.

b = √(c² − a²)

5. Build triangles in larger diagrams

A rectangle's diagonal forms a right triangle with its length and width. A ladder leaning against a vertical wall forms one with the ground distance and wall height, assuming the wall and ground are perpendicular. Identify the three connected lengths before substituting.

In an isosceles triangle, a perpendicular line from the apex to the base also bisects the base. If the equal sides are 13 cm and the base is 10 cm, use half the base, 5 cm, with hypotenuse 13 cm. The height is √(169 − 25) = 12 cm, not the value found using the full base.

6. Use coordinates to find a distance

For two points, the horizontal difference and vertical difference form the shorter sides of a right triangle. From (−2, 1) to (4, 9), those differences are 6 and 8, so the straight-line distance is √(6² + 8²) = 10 coordinate units.

Subtract coordinates consistently. Either point can be first, because squaring removes the sign of each difference. Do not add the horizontal and vertical distances: that finds a route along two perpendicular segments rather than the direct distance between the points.

Distance = √((x₂ − x₁)² + (y₂ − y₁)²)

7. Keep units and accuracy consistent

All side lengths in one calculation need the same unit. If a triangle has perpendicular sides 0.3 m and 40 cm, convert 0.3 m to 30 cm first. Then c = √(30² + 40²) = 50 cm. Mixing 0.3 and 40 as though they share a unit changes the problem.

Keep an unrounded square root for any later calculation, especially an area or another triangle. If a result is √65 cm and a decimal answer is requested, it is about 8.06 cm to two decimal places. Higher extension: exact surd answers can be simplified, such as √72 = 6√2, without replacing them by decimals.

8. Higher extension: work in three dimensions

A cuboid's space diagonal can be found using Pythagoras twice. First find a diagonal of the base rectangle. Then that base diagonal and the vertical height form another right triangle whose hypotenuse is the space diagonal.

For a cuboid 3 cm by 4 cm by 12 cm, the base diagonal is √(3² + 4²) = 5 cm. The space diagonal is √(5² + 12²) = 13 cm. Equivalently, combine the steps as √(3² + 4² + 12²). These dimensions must be mutually perpendicular; the shortcut is not a general formula for any solid.

Cuboid space diagonal = √(length² + width² + height²)

Check your understanding

How to earn the marks

• Label the hypotenuse before substituting numbers.
• Show the squared equation, then the positive square root.
• Explain any half-base or coordinate differences used to form the triangle.
• Check the size, units and requested accuracy of the final length.

A common mistake

The equation only applies directly to right-angled triangles. Do not add the two side lengths instead of their squares.

Course context and official specifications

This guide covers selected skills. Some extensions are Higher content; choose your official tier separately from practice difficulty in the app. These are original questions and explanations.

• AQA GCSE Mathematics — G20

  Right-angled triangle calculations are available at both tiers. Higher courses also include wider geometric applications.

• Edexcel GCSE Mathematics — G20

  Right-angled triangle calculations are available at both tiers. Higher courses also include wider geometric applications.

• OCR GCSE Mathematics — 10.05a

  Right-angled triangle calculations are available at both tiers. Higher courses also include wider geometric applications.

Keep your next step small

Try five questions, read the feedback, and revisit one thing you found difficult. The app tracks a 20-answer practice goal separately from accuracy.

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