Edexcel GCSE Maths Foundation (1MA1), Foundation tier · Geometry and measures
Practise Pythagoras' theorem (incl. 3D). unlimited generated questions on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Pythagoras' theorem links the three sides of a right-angled triangle. You use it to find a missing side, the distance between two points and, on Higher, lengths inside 3D shapes such as cuboids and pyramids. It appears on both tiers, often inside a longer problem.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
4
Find the hypotenuse of a right-angled triangleSquare the two shorter sides, add, then square root: \(c = \sqrt{a^2 + b^2}\).
4
Find a shorter side of a right-angled triangleSquare the hypotenuse, subtract the square of the other side, then square root.
5
Use Pythagoras in problems and on coordinatesSpot right-angled triangles in rectangles, isosceles triangles and on grids, e.g. the distance between two points.
5
Test whether a triangle is right-angledCheck whether the square of the longest side equals the sum of the squares of the other two sides.
Notes
The theorem
In a right-angled triangle, the hypotenuse is the longest side. It is opposite the right angle.
Pythagoras' theorem: \(a^2 + b^2 = c^2\), where c is the hypotenuse and a and b are the two shorter sides.
It works only in right-angled triangles.
Finding a missing side
Finding the hypotenuse: add the squares, then square root. Sides 6 cm and 8 cm give \(c = \sqrt{36 + 64} = \sqrt{100} = 10\) cm.
Finding a shorter side: subtract the squares (hypotenuse squared minus the other side squared), then square root. Hypotenuse 13 cm and side 5 cm give \(\sqrt{169 - 25} = \sqrt{144} = 12\) cm.
Sense check: the hypotenuse must be longer than each of the other two sides.
Using it in problems
Sketch the right-angled triangle on its own and label the sides you know.
The diagonal of a rectangle makes two right-angled triangles. An isosceles triangle splits into two right-angled triangles along its line of symmetry.
Distance between two points: find the horizontal and vertical differences and use them as a and b. From (1, 2) to (7, 10) the differences are 6 and 8, so the distance is 10 units.
Converse: if \(a^2 + b^2 = c^2\) where c is the longest side, the triangle is right-angled. If not, it is not right-angled.
Cheatsheet
\(a^2 + b^2 = c^2\), where c is the hypotenuse (opposite the right angle)
Hypotenuse: \(c = \sqrt{a^2 + b^2}\)
Shorter side: \(a = \sqrt{c^2 - b^2}\)
Distance between two points = \(\sqrt{(\text{difference in } x)^2 + (\text{difference in } y)^2}\)
Right-angled test: (longest side)² = sum of the squares of the other two sides
Common whole-number sides: 3, 4, 5; 5, 12, 13; 8, 15, 17 (and multiples, e.g. 6, 8, 10)
How to answer each type of question
Work out the length of a side
3 marks4
Find the hypotenuse: the side opposite the right angle.
Square the two known sides. Add them to find the hypotenuse; subtract them to find a shorter side.
Square root, then round as the question asks.
Example. ABC is a triangle with a right angle at B. AB = 7.2 cm and AC = 11.5 cm. Work out the length of BC. Give your answer correct to 1 decimal place.
Show the model answer
AC is the hypotenuse, so BC² = 11.5² − 7.2² = 132.25 − 51.84 = 80.41 (M1) BC = √80.41 (M1) = 8.967… = 9.0 cm (A1)
Problem in context: decide and justify
3 to 4 marks5
Draw the right-angled triangle and label the lengths from the question.
Use Pythagoras to find the missing length, keeping extra decimal places.
Compare with the value in the question and write a clear conclusion.
Example. A ladder 4.5 m long leans against a vertical wall. The foot of the ladder is on horizontal ground, 1.2 m from the wall. The ladder is safe to use if its top reaches at least 4.3 m up the wall. Is the ladder safe to use? You must show your working.
Show the model answer
Height² = 4.5² − 1.2² = 20.25 − 1.44 = 18.81 (M1) Height = √18.81 (M1) = 4.337… m (A1) 4.337 m is more than 4.3 m, so yes, the ladder is safe (C1)
Shortcuts and memory tricks
Hypotenuse? Add the squares. Shorter side? Subtract the squares.
Learn 3-4-5, 5-12-13 and 8-15-17 and their multiples; non-calculator questions often use them.
Sense check: the hypotenuse is always the longest side.
Type the whole calculation into your calculator in one go, e.g. √(11.5² − 7.2²), so you do not round too early.
Where marks are lost
Adding the squares when finding a shorter side (the answer comes out longer than the hypotenuse).
Forgetting to square root at the end.
Doubling instead of squaring: 7² is 49, not 14.
Using Pythagoras in a triangle that has no right angle.
Rounding the first answer in a two-step problem and then using the rounded value.
Exam technique
Write down the squares and the addition or subtraction before the answer; this earns method marks even if you slip later.
In 'show that' or 'is it safe' questions, give more decimal places than the value you compare with, then write a conclusion.
Round to the accuracy asked for and keep any final zero: 9.0 cm to 1 decimal place, not 9 cm.
Higher: in 3D problems, sketch the right-angled triangle you are using as a flat 2D triangle.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy3 marks
The diagram shows a right-angled triangle. Diagram NOT accurately drawn
[object Object]
Work out the value of \(x\).[3]
Show the answer and mark scheme
Answer: \(x = 15\)
M1 for \(17^{2} - 8^{2}\) (\(= 225\))
M1 for \(\sqrt{225}\)
A1 for 15 cao
Worked solution: \(x^{2} = 17^{2} - 8^{2} = 289 - 64 = 225\), so \(x = \sqrt{225} = 15\).
Question 2Medium3 marks
\(LMN\) is a right-angled triangle. Diagram NOT accurately drawn
[object Object]
Work out the length of \(MN\).[3]
Show the answer and mark scheme
Answer: \(45\) cm
M1 for \(51^2 - 24^2\) (= 2025)
M1 for \(\sqrt{2025}\)
A1 for \(45\)
Worked solution: \(MN^2 = 51^2 - 24^2 = 2025\) \(MN = \sqrt{2025} = 45\) cm