The hypotenuse: add the squares
The sides 6 and 8 meet at the right angle
c² = 6² + 8² = 36 + 64 = 100
c = √100 = 10
Pythagoras' theorem Why do the squares on the two shorter sides fill the square on the longest?
Foundation and Higher · AQA · 45 minutes
Students count the squares on the sides of a 3, 4, 5 triangle on a grid and see why 9 + 16 = 25. Then they find the hypotenuse, a shorter side, test whether a triangle is right-angled and find the distance between two points. Slides to present, a vote that shows who's stuck, practice and homework.
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The curve is the points r away from (0, 0), above the x-axis. Move r from 1 to 13. For which r does it pass through whole-number points off the axes? Then check one: is 3² + 4² the same as 5²?
Whole-number points: (−4, 0), (0, 4), (4, 0)
The idea in one minute
In a right-angled triangle, the square drawn on the longest side, the hypotenuse, has the same area as the squares on the other two sides together: a² + b² = c². So the hypotenuse comes from adding the two squares and taking the square root, and a shorter side from taking one square from the other. It works only with a right angle, and the other way round too: if the squares add up, the triangle is right-angled. The distance between two points on a grid is the hypotenuse of the triangle that goes across and up between them.
The lesson goes picture first. Students count the squares on a 3, 4, 5 triangle drawn on the grid, the tilted one as a 7 by 7 box with four triangles taken away, before anyone says the rule. Then one habit for every question: find the hypotenuse first, then add (to find it) or subtract (to find a shorter side). The core is for Foundation and Higher; the stretch solves for a missing coordinate.
By the end, students can
- find the hypotenuse, opposite the right angle, and say why a² + b² = c²
- find the hypotenuse or a shorter side, rounding as asked
- test whether a triangle with three given sides is right-angled
- find the distance between two points from a triangle on a grid
- spot the right-angled triangle in a problem in words
Key words
- hypotenuse
- right angle
- square
- square root
- converse
- distance
Before this lesson
Students should square a number, use the square root key, round to 1 decimal place and plot points with coordinates. The theorem is taught here, from squares on a grid.
The lesson, step by step
45 minutes in total. The stretch is optional.
6 min 1 · Hook · students think
Count the squares
Students find the areas of the squares on a 3, 4, 5 triangle on a grid before anyone gives the rule.
Say“Here is a right-angled triangle on the grid, with a square on each side. Find the area of each square. Don't show anyone yet.”
Ask- The tilted square: how could you find its area without counting halves?
- What do you notice about the three areas?
Watch forCounting the tilted square square by square and giving up: suggest the 7 by 7 box around it, then the four triangles at its corners.
For the teacher
First slide: the squares on the sides of length 3 and 4 have areas 9 and 16, counted on the grid.
Second slide: the square on the longest side is tilted. It sits inside a 7 by 7 box, area 49. Each corner of the box holds a triangle with sides 3 and 4, area 6. So the tilted square is 49 − 4 × 6 = 25.
Don't name the rule yet. The question is the point: is 9 + 16 = 25 luck, or does it always happen?
Gather the three areas, 9, 16 and 25, with Ask the class. Keep the box method for the explanation: it is why the rule works.
Shorter: Show the three areas at once; skip the counting (−2 min).
12 min 2 · Explain · you explain
The rule, then four uses
a² + b² = c², with c the hypotenuse; then the hypotenuse, a shorter side, is it right-angled, and distance.
Say“Find the hypotenuse first: opposite the right angle, the longest side. Then the question is: add or subtract?”
Ask- Before we calculate: will the answer be longer or shorter than 13?
- Is a triangle with sides 6, 7 and 9 right-angled? How can you tell without drawing it?
Watch forAdding the squares when the hypotenuse is given: the answer then comes out longer than the hypotenuse, which can't be.
A shorter side: subtract
Hypotenuse 13, one side 5
a² = 13² − 5² = 169 − 25 = 144
a = √144 = 12
Is it right-angled?
Sides 7, 24 and 25: 7² + 24² = 49 + 576 = 625 = 25². Yes.
Sides 6, 7 and 9: 6² + 7² = 85, but 9² = 81. No.
Distance between two points
(1, 2) and (6, 4): across 5, up 2
d² = 5² + 2² = 29
d = √29 = 5.385… = 5.4 (1 decimal place)
For the teacher
First slide: the rule. The two shorter sides, a and b, meet at the right angle; c is the hypotenuse, opposite it and always the longest. a² + b² = c² says the two small squares fill the big one, as the grid showed: 9 + 16 = 25.
Second slide: to find the hypotenuse, add the squares and take the square root. To find a shorter side, take the smaller square from the hypotenuse's square, then the root. Ask first whether the answer should be longer or shorter than the sides given.
Third slide: the converse. If the two smaller squares add up to the largest, the triangle is right-angled; if not, it isn't. Test the longest side against the other two.
Fourth slide: the distance between two points is a hypotenuse. Draw across, then up: those are the shorter sides. Take the coordinates from each other; a minus sign doesn't matter once squared.
Vote before each answer (Ask the class): longer or shorter than the side you have? It is the habit that stops the adding-for-a-shorter-side mistake.
Shorter: The hypotenuse and a shorter side only; the converse and distance in the practice (−4 min).
8 min 3 · Your turn · students write
Worked, then yours
Each worked example beside a twin that students solve themselves.
Say“Watch me do one. Then you do its twin, which is almost the same.”
Ask- Where is the hypotenuse? Add or subtract?
Watch forStopping at c², such as 169: the length is its square root.
Find the hypotenuse
The sides 5 and 12 meet at the right angle
c² = 25 + 144 = 169
c = 13
Then theirs
The sides 9 and 12 meet at the right angle: find the hypotenuse
Find a shorter side
Hypotenuse 10, one side 6
a² = 100 − 36 = 64
a = 8
Then theirs
Hypotenuse 17, one side 8: find the other side
Round at the end
The sides 4 and 7 meet at the right angle
c² = 16 + 49 = 65
c = √65 = 8.062… = 8.1 (1 decimal place)
Then theirs
Hypotenuse 9, one side 5: find the other side to 1 decimal place
For the teacher
Work the first example aloud, writing each line. Then students solve its twin while you watch who finds the hypotenuse before writing anything.
Do the same for each pair. The second pair catches students who always add; the third, students who round c² before the root.
Open the answers slide once most have finished their twins.
Shorter: Do the first two pairs only (−3 min).
14 min 4 · Who's stuck? · students vote
Vote, then practise
One question shows who's stuck; then practice in levels, and a stretch.
Say“Choose an answer on your own. Nobody talks until everyone has chosen.”
Ask- Which side is the hypotenuse? So should the third side be longer or shorter than it?
Watch forAnswers longer than the hypotenuse, or answers with no square root taken.
Show your working on paper. Find the hypotenuse first in every question. Round only at the end, as each question says. The stretch is optional.
The question that shows who's stuck
- A right-angled triangle has hypotenuse 15 cm. One of its other sides is 9 cm. How long is the third side?
In your Classroom, as they answer (an example) - A8
- B5
- C11
- D4
Each wrong answer is a different mistake. While they answer, your Classroom shows who chose which, so you know in seconds whom to help with what.
- The two sides that meet at the right angle are 8 cm and 15 cm. Find the hypotenuse, in cm.
- A right-angled triangle has hypotenuse 25 cm and one other side 7 cm. Find the third side, in cm.
- The two sides that meet at the right angle are 3 cm and 8 cm. Find the hypotenuse in cm, to 1 decimal place.
- Without a calculator: the two sides that meet at the right angle are both 5 cm. Mark the length of the hypotenuse, in cm, on the number line.They drag a marker on a number line from 0 to 12.
- Each line gives the three sides of a triangle. Sort them: is the triangle right-angled? Test the longest side against the other two.They sort 8 items into Right-angled and Not right-angled.
- Triangle PQR has its right angle at Q. Which side is the hypotenuse?
- Find the distance between the points (2, 3) and (8, 11).
- A ladder 5 m long leans against a wall. Its foot is 1.4 m from the wall, on flat ground. How high up the wall does it reach, in m?
- An isosceles triangle has sides 10 cm, 10 cm and 12 cm. Find its area, in cm².
- Stretch: the points (1, k) and (4, 7) are 5 apart. So (k − 7)² + 3² = 5². Solve it line by line, and give both values of k.They solve (k − 7)² + 3² = 5² line by line, each line checked as they write it.
Support · For question 1: the two sides you are given meet at the right angle, so you want the hypotenuse. Square both, add, then take the square root. For a shorter side, square both, subtract the smaller from the larger, then take the root. Check: the hypotenuse is always the longest side.
For the teacher
Everyone chooses before anyone explains. Then ask: which side is the hypotenuse, and is your answer shorter than it? Did you square the sides before adding or subtracting? Did you take the square root at the end?
Questions 1–3 build fluency; 4–5 change the look (an estimate on a number line, sides sorted into right-angled or not); 6 asks for a reason; 7 is distance; 8–9 are problems in words. The stretch solves for a missing coordinate, line by line.
Students show their working on paper or the board; the quiz keeps their final answers.
Ask two students who chose different answers to say how they got them, before you say which is right. Then go through the sense check: the hypotenuse is the longest side.
Shorter: After the vote, questions 1–5 only (−6 min).
Longer: The stretch, then: why does it have two answers? (+10 min).
5 min 5 · Exit · students show
One length, one test
A shorter side found, and one triangle tested, show who subtracts and who checks for a right angle.
Say“Two questions. Show your working for both.”
Watch forAdding for the shorter side; testing the converse with a shorter side as the hypotenuse.
Show your working for both.
- A right-angled triangle has hypotenuse 20 cm and one other side 12 cm. Find the third side, in cm.
- A triangle has sides 6 cm, 9 cm and 11 cm. Is it right-angled? Show how you know.
For the teacher
Find a shorter side from the hypotenuse and one other side.
Then test whether a triangle with three given sides is right-angled, and say how you know.
The tests show who uses the longest side as c: start the next lesson with two triangles to test if few do. Set the homework.
Longer: Make up a right-angled triangle with whole-number sides that isn't in the lesson (+5 min).
One to one: before any calculation, ask the student to point to the hypotenuse and say whether the answer will be longer or shorter than the sides they have; check each answer against that guess.
Homework: lengths, tests and distances
For each triangle, say first which side is the hypotenuse, and whether you add or subtract. Round only at the end, as the question says. For question 5, sketch the two points on a grid with the triangle between them.
- The two sides that meet at the right angle are 20 cm and 21 cm. Find the hypotenuse.
- A right-angled triangle has hypotenuse 26 cm and one other side 10 cm. Find the third side.
- The two sides that meet at the right angle are 7 cm and 9 cm. Find the hypotenuse to 1 decimal place.
- Is a triangle with sides 11 cm, 60 cm and 61 cm right-angled? Show how you know.
- Find the distance between the points (−3, 2) and (5, 8).
- A boat sails 9 km north, then 5 km east. How far is it from where it started? Give your answer to 1 decimal place.
Print the worksheet with working space and homework. Teacher answers come with your Library copy.
Teaching pythagoras' theorem: common questions
How do you find the hypotenuse?
It is the side opposite the right angle: the one that doesn't touch the little square in the corner. It is always the longest side. Find it before anything else; then you know whether to add the squares (to find it) or subtract (to find a shorter side).
How do you show why Pythagoras' theorem works?
Draw a 3, 4, 5 triangle on squared paper with a square on each side. The two small squares are 9 and 16. The tilted one sits in a 7 by 7 box with four copies of the triangle at its corners: 49 − 4 × 6 = 25. The same box, drawn with letters a and b, proves it for every right-angled triangle.
Why do students add when they should subtract?
They learn a² + b² = c² as 'square, add, root'. Ask them to say before any calculation whether the answer will be longer or shorter than the sides they have. A shorter side found by adding comes out longer than the hypotenuse, which can't be.
How is the distance between two points Pythagoras?
Join the two points with a triangle that goes across and then up. Its two shorter sides are the differences of the x-coordinates and of the y-coordinates; the distance is its hypotenuse. A difference that comes out negative doesn't matter: it is squared.
Is this Pythagoras lesson suitable for Foundation GCSE?
Yes. The hypotenuse, a shorter side, the converse and distance between points are for Foundation and Higher alike. The stretch, solving for a missing coordinate with two answers, suits Higher. Pythagoras in three dimensions is a later, Higher, lesson.
Can I use the lesson without creating an account?
Yes. Read the lesson, print the worksheet and open its slides without signing in; nothing you draw there is kept. A free teacher account keeps the whole lesson in your Library to change and teach, with the quizzes, the answers and the answer slides.
Curriculum reference: AQA GCSE Mathematics 8300: G20 (Pythagoras' theorem, a² + b² = c², to find lengths in right-angled triangles). The stretch, a missing coordinate with two answers, suits Higher.
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