Step-by-step worked examples and graded practice questions on Pythagoras' Theorem — finding the hypotenuse, finding a shorter side, coordinate geometry, word problems and 3D applications.
📚 Foundation & Higher✅ 15 Practice Questions🔍 Full Worked Examples⚠️ Common Mistakes
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Pythagoras' Theorem works for any right-angled triangle. The hypotenuse is always the longest side, and it's always the side directly opposite the right angle.
a² + b² = c², where c is the hypotenuse
To find the hypotenuse: square both shorter sides, add them, then square root the result
Worked Example 1
A right-angled triangle has shorter sides of 3cm and 4cm. Find the length of the hypotenuse.
1
a² + b² = c², so 3² + 4² = c²
2
9 + 16 = c², so c² = 25
3
c = √25 = 5cm
Answer5cm
The hypotenuse (marked ?) is always opposite the right angle, and is always the longest side.
Finding a shorter side
If the hypotenuse is already known, the formula is rearranged to find a shorter side instead: subtract the squares rather than adding them.
a² = c² − b² (rearranged to find a shorter side)
Worked Example 2
A right-angled triangle has a hypotenuse of 10cm and one shorter side of 6cm. Find the length of the other shorter side.
1
a² + b² = c², so a² + 6² = 10²
2
a² + 36 = 100, so a² = 100 − 36 = 64
3
a = √64 = 8cm
Answer8cm
Here the hypotenuse (10cm) is already known — subtract the known shorter side's square from it, rather than adding.
Applications: coordinates, word problems and 3D
Pythagoras' Theorem also applies wherever a right angle is hiding — between two points on a grid, in a real-world diagram, or inside a 3D shape.
Worked Example 3 — Coordinate geometry
Find the distance between the points (1, 1) and (4, 5).
1
Find the horizontal and vertical distances: 4 − 1 = 3, and 5 − 1 = 4
2
These form the two shorter sides of a right-angled triangle: 3² + 4² = 9 + 16 = 25
3
Distance = √25 = 5
Answer5 units
The dashed horizontal and vertical lines form a right-angled triangle — the distance between the points is the hypotenuse.
Worked Example 4 — Word problem
A ladder 13m long leans against a vertical wall. The foot of the ladder is 5m from the wall. How high up the wall does the ladder reach?
1
Sketch the situation: the ladder is the hypotenuse, the wall and ground are the two shorter sides.
2
height² + 5² = 13², so height² = 169 − 25 = 144
3
height = √144 = 12m
Answer12m
The ladder forms the hypotenuse — sketching this triangle first prevents mixing up which measurement is which.
Practice questions
Work through each question before checking the answers. Where a question describes a real-world, coordinate, or 3D scenario, the diagram shown is required to answer it.
Foundation (Grade 3–5)
Q1A right-angled triangle has shorter sides of 6cm and 8cm. Find the hypotenuse.Foundation
Diagram for Question 1 (not to scale).
Q2A right-angled triangle has shorter sides of 5cm and 12cm. Find the hypotenuse.Foundation
Q3A right-angled triangle has shorter sides of 9cm and 12cm. Find the hypotenuse.Foundation
Q4A triangle has sides of 5cm, 12cm and 13cm. Show that it is right-angled.Foundation
Q5A right-angled triangle has a hypotenuse of 13cm and one shorter side of 5cm. Find the other shorter side.Foundation
Diagram for Question 5 (not to scale).
Higher (Grade 5–7)
Q6A right-angled triangle has a hypotenuse of 17cm and one shorter side of 8cm. Find the other shorter side.Higher
Q7Find the distance between the points (0, 0) and (5, 12).Higher
Diagram for Question 7 (not to scale).
Q8A right-angled triangle has shorter sides of 7cm and 24cm. Find the hypotenuse.Higher
Q9A ladder 10m long leans against a vertical wall. The foot of the ladder is 6m from the wall. How high up the wall does the ladder reach?Higher
Diagram for Question 9 (not to scale).
Q10A right-angled triangle has shorter sides of x cm and (x + 7)cm, and a hypotenuse of 13cm. Find x.Higher
Higher — Hard (Grade 8–9)
Q11The diagram shows a cuboid. Find the length of the space diagonal shown.Grade 8–9
Diagram for Question 11 (not to scale).
Q12The diagram shows an isosceles triangle. Find its area.Grade 8–9
Diagram for Question 12 (not to scale). The dashed line splits the triangle into two right-angled triangles.
Q13Find the distance between the points (−2, 3) and (4, −5).Grade 8–9
Q14A right-angled triangle has shorter sides of (3x)cm and (4x)cm, and a hypotenuse of 20cm. Find x.Grade 8–9
Q15A triangle has sides of 7cm, 24cm and 25cm. Show that it is right-angled, and state which two sides meet at the right angle.Grade 8–9
Answers
Foundation (Q1–Q5)
Q110cm(√(36+64))
Q213cm(√(25+144))
Q315cm(√(81+144))
Q4Yes — 5² + 12² = 25 + 144 = 169 = 13², so the triangle is right-angled.
Q512cm(√(169−25))
Higher (Q6–Q10)
Q615cm(√(289−64))
Q713 units(√(25+144))
Q825cm(√(49+576))
Q98m(√(100−36))
Q10x = 5(x² + (x+7)² = 169 → x² + 7x − 60 = 0)
Higher — Hard (Q11–Q15)
Q1113cm(√(3²+4²+12²) = √169)
Q1260cm²(height = √(13²−5²) = 12, area = ½ × 10 × 12)
Q1310 units(√(6²+8²))
Q14x = 4(9x²+16x² = 400 → 25x² = 400)
Q15Yes — 7² + 24² = 49 + 576 = 625 = 25², so the right angle is between the sides of 7cm and 24cm.
Common mistakes
Common Mistake 1
Adding instead of subtracting (or vice versa)
Add the squares to find the hypotenuse. Subtract to find a shorter side. Mixing these up is the single most common Pythagoras error at GCSE.
Common Mistake 2
Forgetting to square root at the end
Calculating a² + b² (or c² − b²) gives the squared length, not the side length itself. Always take the square root as the final step.
Common Mistake 3
Misidentifying the hypotenuse in word problems
In real-world questions (like ladders leaning against walls), the hypotenuse is the sloped or diagonal length — not necessarily the first number mentioned. Sketch the triangle before starting.
Common Mistake 4
Rounding too early in multi-step problems
For 3D or multi-stage problems, keep exact values (like √169) through the working and only round the final answer — early rounding compounds errors.
Exam tips
💡 Exam Tip 1
Always sketch the triangle first
Even for word problems with no diagram given, draw a quick right-angled triangle and label the two shorter sides and the hypotenuse before writing any formula.
💡 Exam Tip 2
Write the formula before substituting
Writing "a² + b² = c²" before putting numbers in secures method marks even if the final calculation goes wrong.
💡 Exam Tip 3
Check whether you're finding the hypotenuse or a shorter side
If the missing side is the longest one, add the squares. If it's one of the two shorter sides, subtract.
💡 Exam Tip 4
Use exact values in multi-step problems
For 3D diagonals or problems with two stages, carry surds or exact square roots through your working rather than rounding until the final line.
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