3. Right-angled triangles
A right-angled triangle contains a right angle.
The longest side opposite the right angle is called the hypotenuse.
Teacher notes
Review the definition of a right-angled triangle.
Tell pupils that in any triangle the angle opposite the longest side will always be the largest angle and vice-versa.
Ask pupils to explain why no other angle in a right-angled triangle can be larger than or equal to the right angle.
Ask pupils to tell you the sum of the two smaller angles in a right-angled triangle.
Recall that the sum of the angles in a triangle is always equal to 180°. Recall also, that two angles that add up to 90° are called complementary angles. Conclude that the two smaller angles in a right-angled triangle are complementary angles.

4. Identify the hypotenuse
Teacher notes
Use this exercise to identify the hypotenuse in right-angled triangles in various orientations.

5. The history of Pythagoras’ Theorem
Pythagoras’ Theorem concerns the relationship between the sides of a right- angled triangle.
The theorem is named after the Greek mathematician and philosopher, Pythagoras of Samos.
Teacher notes
Explain that the theorem is named after Pythagoras because he, or a member of his society, was the first person known to have formally proven the result.
Although the Theorem is named after Pythagoras, the result was known to many ancient civilizations including the Babylonians, Egyptians and Chinese, at least 1000 years before Pythagoras was born.

6. Pythagoras’ Theorem
Pythagoras’ Theorem states that the square formed on the hypotenuse of a right-angled triangle…
…has the same area as the sum of the areas of the squares formed on the other two sides.
Teacher notes
Pythagoras’ Theorem may be expressed as a relationship between areas, as shown here, or a relationship between side lengths.
In fact, the area of any similar shapes may by drawn on the sides of a right-angled triangle. The area of the shape drawn on the hypotenuse, will be equal to the sum of the areas of the shapes drawn on the two shorter sides.

7. Showing Pythagoras’ Theorem
Teacher notes
Drag the vertices of the triangle to change the lengths of the sides and rotate the right-angled triangle.
Ask a volunteer to come to the board and use the pen tool to demonstrate how to find the area of each square. For tilted squares this can be done by using the grid to divide the squares into triangles and squares. Alternatively, a larger square can be drawn around the tilted square and the areas of the four surrounding triangles subtracted.
Reveal the areas of the squares and verify that the area of the largest square is always equal to the sum of the areas of the squares on the shorter sides.

8. Pythagoras’ Theorem c2 a2 b2
Label the length of the sides of a right-angled triangle a, b and c.
The area of the largest square is c × c or c2. c2
The areas of the smaller squares are a2 and b2. c a2 a
Pythagoras’ Theorem can be written as:
Teacher notes
This slide shows how Pythagoras’ Theorem can be written as a relationship between the side lengths of a triangle with sides a, b and c, where c is the hypotenuse.
Ask pupils to tell you what a2 is equal to. (c2 – b2).
Ask pupils to tell you what b2 is equal to. (c2 – a2). b b2
c2 = a2 + b2

9. A proof of Pythagoras’ Theorem
Teacher notes
There are many proofs of Pythagoras’s Theorem. Pupils could be asked to research these on the internet and make posters to show them.
In this proof we can see that the area of the two large squares is the same: (a + b)2
Both of these squares contain four identical triangles with side lengths a, b and c. In each of the large squares the four triangles are arranged differently to show visually that the area of the square with side length c is equal to the sum of the areas of the squares with side lengths a and b.
This can be shown more formally considering the first arrangement.
The area the square with side length c is equal to the area of the large square with side length (a + b) minus the area of the four right-angled triangles.
The area of the four right-angled triangles is 4 × ½ab = 2ab.
c2 = (a + b)2 – 2ab
Expanding,
c2 = a2 + 2ab + b2 – 2ab
c2 = a2 + b2

10. Pythagoras’ Theorem
Pythagoras’ Theorem states that for a right-angled triangle with a hypotenuse of length c and the shorter sides of lengths a and b: c a
c2 = a2 + b2 b
Pythagoras’ Theorem is used to:
check whether a triangle is right-angled given the lengths of all the sides
Teacher notes
Stress that Pythagoras’ Theorem is only true for right-angled triangles.
If we are given the lengths of all three sides of a triangle, we can use Pythagoras’ Theorem to check whether or not the triangle is a right-angled triangle by squaring the length of the two shorter sides, adding the squares together and checking whether or not this is equal to the hypotenuse squared.
If we are given the lengths of two of the sides in a right-angled triangle, we can also use Pythagoras’ Theorem to find the lengths of the unknown side.
find the length of a missing side in a right-angled triangle given the lengths of the other two sides.

12. Identifying right-angled triangles
If we are given the lengths of the three sides of a triangle, we can use Pythagoras’ Theorem to find out whether or not the triangle contains a right angle.
A triangle has sides of length 4 cm, 7 cm and 9 cm.
Is this a right-angled triangle?
If the sum of the squares on the two shorter sides is equal to the square on the longest side, the triangle has a right angle. =
Photo credit: © Phoandrzej , Shutterstock.com
= 65
65  92
No, this is not a right-angled triangle.

13. Identifying right-angled triangles
Teacher notes
Use this activity to demonstrate how the relationship between the sum of the squares on the two shorter sides and the square on the hypotenuse changes as the right angle is modified.
If the right angle becomes obtuse, c2 gets bigger. If the right-angle becomes acute, c2 gets smaller.
We can use this to decide whether a non-right-angled triangle has three acute angles or whether it contains an obtuse angle.
Ask pupils to explain why it is impossible for a triangle to contain more than one obtuse angle.

14. Identifying right-angled triangles

15. Identifying right-angled triangles
A triangle has sides of length 5 cm, 8 cm and 10 cm.
Is the angle opposite the longest side acute, obtuse or right-angled? =
10 cm
8 cm
5 cm ?
= 89
102 = 100
100 > 89 so the angle opposite the longest side is an obtuse angle.

16. Obtuse, acute or right-angled triangle
Teacher notes
In this activity pupils must find the sum of the squares of the two shorter sides. If the square of the longest side is greater than the sum of the squares of the two shorter sides then the angle must be obtuse. If this is less than the sum of the squares of the two shortest sides, the angle opposite the longest side must be acute.
If this is the same as the square of the longest side, the triangle obeys Pythagoras’ Theorem and the angle opposite the longest side must therefore be a right angle.
Pupils may use their calculators for lengths involving decimals.

18. Pythagorean triples
A triangle has sides of length 3 cm, 4 cm and 5 cm.
Does this triangle have a right angle?
Using Pythagoras’ Theorem, if the sum of the squares on the two shorter sides is equal to the square on the longest side, the triangle has a right angle. =
= 25
Teacher notes
Explain that a Pythagorean triple is three whole numbers that obey Pythagoras’ Theorem. Many Pythagorean triples were known to people in ancient times.
Photo credit: © Anykaj , Shutterstock.com
= 52
Yes, the triangle has a right-angle.
The numbers 3, 4 and 5 form a Pythagorean triple.

19. Pythagorean triples
Ancient Egyptians used the fact that a triangle with sides of length 3, 4 and 5 contained a right-angle to mark out field boundaries and for building.

20. Pythagorean triples
Three whole numbers a, b and c, where c is the largest, form a Pythagorean triple if,
a2 + b2 = c2
3, 4, 5 is the simplest Pythagorean triple.
Write down every square number from 12 = 1 to 202 = 400.
Use these numbers to find as many Pythagorean triples as you can.
Teacher notes
Pupils should be able to find six Pythagorean triples using these numbers. These are shown on the next slide.
They may notice that some of these are multiples of the 3, 4, 5 triangle.
Write down any patterns that you notice.

21. Pythagorean triples How many of these did you find? 9 + 16 = 25
= 52
3, 4, 5
= 100
= 102
6, 8, 10
= 169
= 132
5, 12, 13
= 225
= 152
9,12, 15
= 289
= 172
8, 15, 17
Teacher notes
Pupils should notice that the Pythagorean triples that have not been circled are multiples of the the 3, 4, 5 Pythagorean triple.
Tell pupils that a primitive Pythagorean triple is a Pythagorean triple that is not a multiple of another Pythagorean triple.
Ask pupils to explain why any multiple of a Pythagorean triple must be another Pythagorean triple (using similar triangles).
= 400
= 202
12, 16, 20
The Pythagorean triples 3, 4, 5;
5, 12, 13
and 8, 15 17
are called primitive Pythagorean triples because they are not multiples of another Pythagorean triple.

23. Similar right-angled triangles
The following right-angled triangles are similar. 10 ? 15 6 9 8 12 ?
Teacher notes
Remind pupils that the corresponding lengths in two mathematically similar shapes are always in the same ratio. That is, one shape is an enlargement of the other.
Ask pupils if they can work out which Pythagorean triple both these triangles are based on.
For the first triangle, 6 and 8 share a common factor of 2. Dividing these lengths by 2, gives us 3 and 4.
For the second triangle, 9 and 15 share a common factor of 3. Dividing these lengths by 3, gives us 3 and 5.
These triangles are both enlargements of a right-angled triangle with sides of length 3, 4 and 5.
Use this to find the lengths of the missing sides.
Alternatively, we could use a scale factor of 3/2 to scale from the smaller triangle to the larger triangle, or a scale factor of 2/3 to scale from the larger triangle to the smaller triangle.
Find the lengths of the missing sides.
Check that Pythagoras’ Theorem holds for both triangles.

24. Similar right-angled triangles
The following right-angled triangles are similar. 10 ? 15 6 9 8 12 ?
Teacher notes
Check the lengths of the missing sides by showing that Pythagoras’ Theorem holds for both triangles. = =
= 100
= 225
= 102
= 152

25. Similar right-angled triangles
The following right-angled triangles are similar. 60 45 ? 32 ? 24 68 51
Teacher notes
Ask pupils if they can work out which Pythagorean triple both these triangles are based on.
For the first triangle, 60 and 68 share a common factor of 4. Dividing these lengths by 4, gives us 15 and 17.
For the second triangle, 24 and 51 share a common factor of 3. Dividing these lengths by 3, gives us 8 and 17.
These triangles are both enlargements of a right-angled triangle with sides of length 8, 15 and 17.
Use this to find the lengths of the missing sides.
Alternatively, we could use a scale factor of 3/4 to scale from the larger triangle to the smaller triangle, or a scale factor of 4/3 to scale from the smaller triangle to the larger triangle.
Find the lengths of the missing sides.
Check that Pythagoras’ Theorem holds for both triangles.

26. Similar right-angled triangles
The following right-angled triangles are similar. 60 45 32 24 68 51
Teacher notes
Pupils can verify these calculations using their calculators.
Make sure that pupils can locate both the x2 key and the  key on their calculators. = =
= 4624
= 2601
= 682
= 512

27. Similar right-angled triangles
The following right-angled triangles are similar. 2
1.5
2.4
1.8 ?
2.5 ? 3
Teacher notes
Ask pupils if they can work out which Pythagorean triple both these triangles are based on.
For the first triangle, 2 and 1.5 are half of 4 and 3 respectively.
For the second triangle, 2.4 and 3 are 3/5 of 4 and 5 respectively.
These triangles are both fractional enlargements of a right-angled triangle with sides of length 3, 4 and 5.
Use this to find the lengths of the missing sides.
Alternatively, we could use a scale factors.
To scale from the smaller triangle to the larger triangle we scale from 2 to 2.4. In other words we would multiply by 2.4/2 or 6/5 or 1.2.
To scale from the larger triangle to the smaller triangle we scale from 2.4 to 2. In other words we would multiply by 2/2.4 or 5/6 or divide by 1.2.
Find the lengths of the missing sides.
Check that Pythagoras’ Theorem holds for both triangles.

28. Similar right-angled triangles
The following right-angled triangles are similar. 2
1.5
2.4
1.8 ?
2.5 3
Teacher notes
Conclude by explaining that most Pythagoras problems on the non-calculator paper will involve triangles based on Pythagorean triples. This is because these triangles have side lengths that can be written exactly (the side lengths are rational numbers). Right-angled triangles that are not based on Pythagorean triples have at least one side length that cannot be written as a terminating or recurring decimal, that is a side length that is irrational. Such lengths can only by written exactly as surds (unresolved roots). = =
= 6.25
= 9
= 2.52
= 32

30. Finding the length of the hypotenuse
Use Pythagoras’ Theorem to calculate the length of side a. c
5 cm
12 cm
Using Pythagoras’ Theorem,
c2 =
Teacher notes
Talk through the calculation to find the length of the hypotenuse.
Stress that if c2 = 169, to find a we need to find the square root of 169.
If pupils do not know the square root of 169, ensure that they are able to locate and use the  key on their calculators. Encourage pupils to learn at least the first 15 square numbers.
Pupils should also be encouraged to look at the triangle to make sure that their answer seems about right compared to the other two given sides.
c2 =
c2 = 169
c = 169
c = 13 cm

31. Finding the length of the hypotenuse
Use Pythagoras’ Theorem to calculate the length of side PR. P
0.7 m Q
2.4 m R
Using Pythagoras’ Theorem
PR2 = PQ2 + QR2
Substituting the values we have been given,
PR2 =
Teacher notes
This example uses the letters at each vertex to label the sides.
Stress to pupils that they must start by stating Pythagoras’ Theorem for the given triangle and show every step in their calculation.
They should also draw a diagram if one has not been given.
Again encourage pupils to think about whether the answer seems sensible given the lengths of the other two sides.
PR2 =
PR2 = 6.25
PR = 6.25
PR = 2.5 m

32. Finding the length of the hypotenuse
Use Pythagoras’ Theorem to calculate the length of side x to 2 decimal places.
6 cm x
3 cm
Using Pythagoras’ Theorem
x2 =
x2 =
x2 = 45
x = 45
x = 6.71 cm

34. Calculating the shorter sides

35. Find the correct equation
Teacher notes
One of the most common errors when solving problems involving Pythagoras’ Theorem is failing to identify which side is the hypotenuse before writing the equation of the relationship between the sides.
For each of the examples in the activity, ask pupils to identify the hypotenuse before selecting the correct equation.
Encourage pupils to go through the process of identifying the hypotenuse and writing down Pythagoras’ Theorem before attempting to solve any problem involving the lengths of the sides of right-angled triangles.

36. Complete this table