Find a missing side in a right-angled triangle by squaring, adding or subtracting, then square-rooting. It is on Foundation and Higher, aimed at grade 4. Free questions with answers are on this page. Worked examples show each line before you try the questions.
In any right-angled triangle the two shorter sides (the ones that touch the right angle) and the hypotenuse (the side opposite the right angle) are locked together by a2+b2=c2. Square both shorter sides and add to get the hypotenuse squared. If you already know the hypotenuse, square it and subtract the other known square to get the missing shorter side.
The method is the same whether the triangle is drawn as a ladder against a wall, a diagonal of a rectangle, or two points on a coordinate grid. Sketch, mark the right angle, decide whether you are adding squares or subtracting them, then root at the end. Forgetting the square root is the classic way to throw the accuracy mark after doing the hard part.
Exact answers such as 65 are often wanted on the non-calculator paper. On a calculator paper, round only at the last line to 1 decimal place or 3 significant figures as asked, and write the units.
Objectives
a² + b² = c², where c is the hypotenuse.
Square both legs, add, then square-root.
Subtract the known square from the hypotenuse squared, then square-root.
Key points
Identify the hypotenuse — longest side, opposite the right-angle square.
Find the hypotenuse with c=a2+b2 (square, add, root).
Find a shorter side with a=c2−b2 (square, subtract, root).
Use Pythagoras on a coordinate grid, a rectangle diagonal, or a ladder against a wall.
Worked examples
Example 1
A right-angled triangle has shorter sides 3 and 4. How long is the hypotenuse?
3 marks
3² = 9 and 4² = 16.
9 + 16 = 25.
√25 = 5.
Answer: 5
Common slips
Slip: 5² = 9 and 4² = 16. That changes a number from the question, so it answers a different question.
Slip: 9 − 16 = −7. That uses subtraction, but this step needs addition.
Slip: √29 = 5. That does not match the working above it.
Example 2
A right-angled triangle has hypotenuse 25 and one shorter side 7. How long is the other shorter side?
3 marks
25² = 625 and 7² = 49.
625 − 49 = 576.
√576 = 24.
Answer: 24
Common slips
Slip: 27² = 625 and 7² = 49. That changes a number from the question, so it answers a different question.
Slip: 625 + 49 = 674. That uses addition, but this step needs subtraction.
Slip: √580 = 24. That does not match the working above it.
Example 3
A ladder 29 m long leans against a vertical wall. The foot of the ladder is 20 m from the wall. How high up the wall does the ladder reach?
3 marks
29² = 841 and 20² = 400.
841 − 400 = 441.
√441 = 21.
Answer: 21
Common slips
Slip: 31² = 841 and 20² = 400. That changes a number from the question, so it answers a different question.
Slip: 841 + 400 = 1241. That uses addition, but this step needs subtraction.
Slip: √445 = 21. That does not match the working above it.
Questions and answers
GCSE Maths pythagoras questions and answers. Try each one, then open the answer. Easy questions come first.
Question 1 Easy2 marks
A right-angled triangle has shorter sides 6 cm and 8 cm. How long is the hypotenuse?
Show answer
Answer: 10
6² + 8² = 100.
The square root is 10.
M1 — 6² + 8² = 100.
A1 — Answer 10
Question 2 Easy2 marks
The hypotenuse is 13 cm and one shorter side is 5 cm. How long is the other shorter side?
Show answer
Answer: 12
13² − 5² = 144.
The square root is 12.
M1 — 13² − 5² = 144.
A1 — Answer 12
Question 3 Medium2 marks
A right-angled triangle has shorter sides 3 cm and 3 cm. Give the hypotenuse as a surd.
Show answer
Answer: 3√2
3² + 3² = 18.
That is 3√2.
M1 — 3² + 3² = 18.
A1 — Answer 3√2
Question 4 Medium2 marks
An isosceles triangle has equal sides 5 cm and a base of 6 cm. What is the perpendicular height?
Show answer
Answer: 4
Half of the base is 3 cm.
5² − 3² = 16.
M1 — Half of the base is 3 cm.
A1 — Answer 4
Question 5 Hard2 marks
The hypotenuse is 17 cm and one shorter side is 8 cm. Give the other side and the area, side first.
Show answer
Answer: 15, 60
The other side is 15 cm.
The area is 1/2 × 8 × 15 = 60.
M1 — The other side is 15 cm.
A1 — Answer 15, 60
Question 6 Hard2 marks
The distance from (0, 0) to (5, y) is 13. Find the missing positive coordinate y.
Show answer
Answer: 12
13² − 5² = 144.
The positive coordinate is 12.
M1 — 13² − 5² = 144.
A1 — Answer 12
Question 7 Exam-style3 marks
The hypotenuse is 25 cm and one shorter side is 7 cm. Give the other side and the area, side first.
Show answer
Answer: 24, 84
The other side is 24 cm.
The area is 1/2 × 7 × 24 = 84.
M1 — The other side is 24 cm.
M1 — The area is 1/2 × 7 × 24 = 84.
A1 — Answer 24, 84
Common mistakes
Most lost marks on this topic are one slip.
Adding the squares when you were finding a shorter side.
Subtract from the hypotenuse squared.
Square-rooting before you add.
Add the squares first, then take one square root.
Using Pythagoras on a triangle with no right angle.
The theorem needs a right angle.
Spot the slip
The hypotenuse is
the longest side, opposite the right angle
either of the shorter sides
the right angle itself
Show the slip
the longest side, opposite the right angle It is the side across from the right angle.
Exam tips
This topic can appear on a calculator or a non-calculator paper. Show enough working for a method mark.
If the question says “show that”, the result is given. The marks are for the steps that reach it.
The method is the same on AQA-style and Edexcel-style papers.
The first method mark is almost always the squared equation written out: c2=6.42+8.72. Write it even on a calculator paper.
Keep the full square root on the calculator until the final rounding. Rounding 10.800 to 10.8 too early can look like 1 d.p. when 2 d.p. was asked.
Printable worksheet
The same questions are in a PDF you can print. Answers start on a later page.