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Pythagoras links the three sides of a right-angled triangle. Find a missing side, check whether a triangle is right-angled, or calculate the distance between two points on a flat coordinate grid.
Key facts
\(a^2 + b^2 = c^2\), 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)
Notes
The theorem
diagram
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.
The hypotenuse c is always opposite the right angle.
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.
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own.
Work out the length of a side
3 marksGrade 4
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.
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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 marksGrade 5
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.
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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)
Don’t lose marks
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.
More tips
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.
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.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade 4
Find the hypotenuse of a right-angled triangleSquare the two shorter sides, add, then square root: \(c = \sqrt{a^2 + b^2}\)
Grade 4
Find a shorter side of a right-angled triangleSquare the hypotenuse, subtract the square of the other side, then square root.
Grade 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.
Grade 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.
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