The science idea
Pythagoras’ theorem is one of the most useful facts in mathematics. It says that in a right-angled triangle, the square on the hypotenuse (the longest side, opposite the right angle) has the same area as the two squares on the other two sides added together. If the sides are a and b and the hypotenuse is c, then a² + b² = c². It is used to find distances and lengths, in building, navigation, surveying, computer graphics and in the distance formula of coordinate geometry.
The rule is named after the Greek mathematician Pythagoras, but it was known long before him. It appears in the Sulba Sutras of ancient India (in the work of Baudhayana, composed many centuries before the Common Era), on Babylonian clay tablets, and in Chinese texts. The Indian mathematician Bhaskaracharya, in the 12th century, gave a famous proof in which one drawing is accompanied by only one word, “Behold!”. The model in this project is of that kind: a proof by rearrangement of pieces.
Here is how the puzzle works. Take a big square with side a + b. Place four identical right-angled triangles with legs a and b inside it, in one way, so that their hypotenuses form a tilted square of side c in the middle. The area that is not covered by triangles is c². Now rearrange the same four triangles into two rectangles inside the same big square, so that what is left is a square of side a and a square of side b. The area not covered by triangles is now a² + b². Since the big square and the four triangles have not changed, the leftover areas must be equal: c² = a² + b².
This can also be written in algebra. The big square has area (a + b)² = a² + 2ab + b². Each triangle has area ½ab, so four triangles have area 2ab. In the first arrangement, (a + b)² = 2ab + c². Equating them, a² + 2ab + b² = 2ab + c², so a² + b² = c². The puzzle and the algebra are the same proof in two languages, and writing both in your report is a good way to show you understand it.
The theorem also works backwards: if the sides of a triangle satisfy a² + b² = c², the triangle has a right angle. Sets of whole numbers that work, such as 3, 4, 5 and 5, 12, 13, are called Pythagorean triples. Builders use the 3, 4, 5 triple to make right angles: a rope with twelve equal spaces, held out into a triangle of 3, 4 and 5 spaces, always gives an exact right angle.
Materials and tools
Cost: Free
Cardboard, used coloured paper, glue, a ruler and a pencil. A 1.2 m string for the rope is from the house.
| What you need | How many | Free substitute or where to find it |
|---|---|---|
| Thick cardboard for the tray base | 1 piece, about 20 cm × 20 cm | The side of a carton. Glue two layers together if it is soft. |
| Card strips for the tray walls (about 1 cm high) | 4 strips, 14 cm and 16 cm long as needed | From the same carton. |
| Coloured paper or card in three colours for the pieces | 3 sheets (A4 or notebook size) | Used gift wrap, magazine pages or coloured chart paper. Stick it to thin card for strength. |
| Glue stick and fevicol | 1 each | From the stationery box. |
| Ruler (cm and mm), pencil and set-square | 1 each | From the geometry box. |
| String or soft rope, 1.2 m long | 1 | From the house. Mark it with a pen every 10 cm. |
| A sheet of squared paper (graph paper) | 1 | From an old maths notebook. |
Tools
- Scissors
- Craft knife (adult) for the thick card
- Pencil and a sharp eraser
- Compass and protractor (optional)
- Marker pen
Safety first
Read this before you start
- An adult should use the craft knife for the thick card. Always cut on a board and away from your hand.
- Keep the small pieces away from small children and do not put them in the mouth.
- Cut slowly and accurately. A cut that is 2 mm out will stop the pieces from fitting.
- Wash your hands after using glue, and keep the glue cap closed.
- Do not hang the rope from a high place; use it flat on the floor or a table.
Parts and layout
These drawings show the parts and how they fit together. Keep them beside you while you build.
Scroll sideways to see the whole diagram.
Scroll sideways to see the whole diagram.
Step-by-step build
12 steps. Do them in order, and test each part before you go on to the next.
Choose the triangle
We use legs a = 6 cm and b = 8 cm. Check that 6² + 8² = 36 + 64 = 100 = 10², so the hypotenuse c should be 10 cm. Write this on a card.
Tip: This triangle is 2 times the 3, 4, 5 triangle, so the numbers are easy to check.
Make the tray
Draw a 14 cm × 14 cm square on the thick card (14 = 6 + 8). Stick strips of card about 1 cm high along its four edges, outside the square, to make a shallow tray, so the pieces cannot slip.
Common mistake: Making the tray larger than 14 cm × 14 cm inside. The pieces then will not fill it, and you cannot see whether they fit.
Draw the triangle template
On stiff paper, draw a line 8 cm long. At one end draw a line at exactly 90 degrees with a set-square, 6 cm long. Join the two ends. Measure the third side: it must be 10 cm.
Tip: If the third side is not 10 cm, your angle is not 90 degrees. Fix it before cutting.
Cut four triangles
Use the template to cut four identical triangles from card of one colour (clay or orange). Mark the sides 6, 8 and 10 cm on each, and check that they are all the same by stacking them.
Common mistake: Cutting inside or outside the line inconsistently. Cut exactly on the line, every time.
Cut the squares
Cut three squares from other colours: one 10 × 10 cm (for c²), one 6 × 6 cm (for a²), and one 8 × 8 cm (for b²). Write the area on each: 100, 36 and 64 cm².
Tip: Use the set-square to check that every corner is a right angle. A square is only a square if all the sides are equal and all angles are 90 degrees.
Arrangement 1: leave the tilted square
Place a triangle in each corner of the tray, with its right angle in the corner. Turn them so that their hypotenuses make a tilted square in the middle. Fit the 10 × 10 square in the gap. It should fit exactly.
Common mistake: Forcing the pieces. If the square does not fit, check your cutting and the triangle sizes rather than pressing it in.
Arrangement 2: leave two squares
Take the pieces out. Place the 8 × 8 square in the top-left corner and the 6 × 6 square in the bottom-right corner. Pair the triangles into two rectangles in the other two corners. Check that all pieces fill the tray with no gaps.
Tip: Each of the two rectangles (6 × 8) is made of two triangles. Put the long sides together.
Compare the leftover areas
In arrangement 1 the triangles cover 96 cm² and the leftover is 100 cm². In arrangement 2 the same triangles are used, and the leftover is 64 + 36 = 100 cm². Write this in your notebook.
Tip: Take a photo of both arrangements side by side for your report.
Check with other triangles
Draw a 5-12-13 triangle on squared paper and check 25 + 144 = 169. Then draw a 6-7-10 triangle: 36 + 49 = 85, which is not 100, and measure the angle opposite the 10 cm side with a protractor. It is not 90 degrees.
Common mistake: Believing the rule works for every triangle. It works only for right-angled triangles.
Make the 12-knot rope
Mark a string with 12 equal spaces (for example 10 cm each) and tie it into a loop. Hold it at the knots to make a triangle with sides of 3, 4 and 5 spaces, and check that the corner fits exactly into the corner of a book.
Tip: Ask two friends to hold the corners while you pull the rope tight.
Use it to check a right angle
Use a 60 cm, 80 cm, 100 cm triangle on a table corner or a door frame to see whether the corner is exactly 90 degrees. Measure along the edges and the diagonal.
Common mistake: Measuring from the wrong places. Start from the corner itself and measure along each edge.
Write the proof and present
Write the algebra: (a + b)² = 4 × ½ab + c² = 2ab + c², and also = a² + 2ab + b², so a² + b² = c². Display the model, the photographs and the rope with a short explanation.
Tip: Practise explaining it to a friend in one minute with the pieces.
How it should look when it is finished
A flat cardboard tray with a coloured frame, and nine neat pieces: four orange right-angled triangles and three squares in different colours. In one arrangement a tilted yellow square sits in the middle of four triangles, and in the other a blue and a green square sit in two corners. Next to it is a knotted string loop and a notebook page with the algebra written out.
This is a description, not a photograph. Compare your own model with it and with the figures above.
How it works
The tray holds a big square of side a + b, which is 14 cm. Four identical triangles take up 4 × ½ × 6 × 8 = 96 cm² in either arrangement. In the first arrangement the triangles are placed with their hypotenuses facing inwards, leaving a tilted square whose side is the hypotenuse c, so the empty area is c².
In the second arrangement the triangles are paired into two rectangles, and the remaining area is a square of side a and a square of side b, so the empty area is a² + b². As the empty area is the big square minus the same four triangles in both cases, the two empty areas are equal, so c² = a² + b².
The knotted rope uses the numbers 3, 4 and 5, which satisfy 9 + 16 = 25. By the converse of the theorem, a triangle with these sides must have a right angle opposite the longest side.
Scroll sideways to see the whole diagram.
Troubleshooting
| What you see | Likely cause | What to do |
|---|---|---|
| The tilted square does not fit | The triangles or the square are cut inaccurately, or the right angle is not 90 degrees. | Recheck 6, 8 and 10 cm and the right angle with a set-square, and recut the wrong pieces. |
| There are gaps in arrangement 2 | The squares or triangles are slightly too small or large. | Measure each piece to the nearest millimetre and compare with the 14 cm tray. |
| The hypotenuse of my triangle measures 9.8 cm | The right angle is slightly less than 90 degrees. | Use a set-square or the corner of a book, and redraw. |
| The pieces slip about | The tray walls are too low or the card is too thin. | Make the walls higher and stick the pieces on thicker card. |
| I get 85 not 100 for the 6-7-10 triangle | That triangle is not right-angled. | This is correct. It shows the rule applies only to right-angled triangles. |
| The rope triangle is not a right angle | The rope is stretched unevenly or the spaces are not equal. | Mark the spaces again with a ruler and hold the rope tight. |
Test and record
Measure the sides in cm to the nearest millimetre (draw each triangle on squared paper). A triangle is right-angled if a² + b² equals c².
| Triangle | a (cm) | b (cm) | c, measured (cm) | a² + b² | c² | Right-angled? (yes or no) |
|---|---|---|---|---|---|---|
| 6, 8, 10 | 6 | 8 | ||||
| 3, 4, 5 | 3 | 4 | ||||
| 5, 12, 13 | 5 | 12 | ||||
| 6, 7, 10 (test) | 6 | 7 |
Copy the table into your notebook and write each reading as you take it, in pencil, with the date. Repeat every test at least three times and use the average. Then draw a graph or chart of your results.
Viva questions and answers
A teacher or a judge will ask you how your project works. Practise these eight questions and answer them in your own words.
- State Pythagoras’ theorem.
In a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides: c² = a² + b².
- What is the hypotenuse?
The longest side of a right-angled triangle, opposite the right angle.
- How does your model prove the theorem?
The same four triangles inside the same big square leave an area c² in one arrangement and a² + b² in the other. The leftover areas must be equal.
- What is the converse of the theorem?
If the sides of a triangle satisfy a² + b² = c², then the triangle is right-angled, with the right angle opposite c.
- What is a Pythagorean triple? Give two examples.
Three whole numbers that satisfy a² + b² = c², such as 3, 4, 5 and 5, 12, 13 (also 8, 15, 17 and 7, 24, 25).
- Who knew this theorem before Pythagoras?
It was known in the Sulba Sutras of ancient India (Baudhayana), in Babylon and in China. The name honours the Greek mathematician who is credited with a proof.
- Does the rule work for a triangle with sides 6, 7 and 10?
No. 6² + 7² = 85, not 100, so it is not right-angled.
- Name two uses of the theorem in real life.
Checking right angles in building with a 3-4-5 rope or a carpenter’s square, finding the length of a ladder or a diagonal, and calculating the distance between two points on a map.
Written report outline
Use these headings in your project file, and write each part in your own words. Add a title page, your diagrams, the precautions you took and the books or sites you used.
- Aim
- To verify Pythagoras’ theorem by making a rearrangement puzzle model.
- Apparatus
- Cardboard, coloured paper, glue, ruler, set-square, scissors, squared paper and string.
- Procedure
- Describe how you made the tray, the four triangles with legs 6 and 8 cm, and the three squares, then arranged the pieces in two ways and compared the leftover areas. Add the diagrams and the algebra.
- Observation
- In arrangement 1 the four triangles left a tilted square of side 10 cm (area 100 cm²). In arrangement 2 they left squares of 6 cm and 8 cm (areas 36 and 64 cm²). Include the table for the other triangles.
- Result
- c² = 100 = 36 + 64 = a² + b² for the 6, 8, 10 triangle. The same holds for 3, 4, 5 and for 5, 12, 13, but not for 6, 7, 10.
- Conclusion
- In a right-angled triangle the square on the hypotenuse equals the sum of the squares on the other two sides.
Take it further
- Make a square-root spiral from cardboard: start with a right-angled triangle with legs 1 and 1, then add a leg of 1 each time to make lengths √2, √3, √4 and so on.
- Draw semicircles on the three sides and show that the areas also satisfy the theorem.
- Find a Pythagorean triple from any two whole numbers m > n using (m² − n², 2mn, m² + n²). Check it with m = 2, n = 1 and m = 3, n = 2.
- Calculate the diagonal of your classroom floor and check it with a measuring tape.
- Look up another proof, such as President Garfield’s trapezoid proof, and make a model of it.
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