The document discusses Pythagoras' theorem, which states that for any right-angled triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides. It provides examples of using the theorem to calculate missing sides of triangles. It also notes that some triangles have integer sides, called Pythagorean triples, and lists some examples. The document suggests some ways questions could become more difficult, such as using multiple triangles or requiring algebraic manipulation.

1. Sketch this right-angled triangle in your book in the centre of a
new page. Work out the length of the longest side using a ruler.
4cm
3cm
STARTER

2. 3cm
3cm
4cm
4cm
5cm
5cm
Area = ?
Area = ?
Area = ?
Now turn each side of the
triangle into a square.
Can you notice anything
about the relationship of
the three areas?
32 + 42 = 52
STARTER

3. For any right-angled triangle
with hypotenuse 𝑐.
𝑎2
+ 𝑏2
= 𝑐2
𝑐
𝑏
𝑎
Hypotenuse
(the longest side)
! Write this down
Pythagoras’ Theorem
Bro Note: notice that it’s the longest side
that’s on it’s own on one side of the
equation. The (squared) shorter sides are
the ones that are added.

4. 3
4
𝑥
Step 1: Determine the
hypotenuse.
Step 2: Form an equation
32
+ 42
= 𝑥2
Step 3: Solve the equation to
find the unknown side.
𝑥2
= 9 + 16 = 25
𝑥 = 25 =5
The hypotenuse
appears on its own.
Example
Reveal >

5. 13
5
𝑥 Step 1: Determine the
hypotenuse.
Step 2: Form an equation
𝑥2
+ 52
= 132
Step 3: Solve the equation to
find the unknown side.
𝑥2
= 169 − 25 = 144
𝑥 = 144 =12
The hypotenuse
appears on its own.
Further Example
Reveal >

6. 11
𝑥
6 Step 1: Determine the
hypotenuse.
Step 2: Form an equation
𝑥2
+ 62
= 112
Step 3: Solve the equation to
find the unknown side.
𝑥2
= 121 − 36 = 85
𝑥 = 85 =9.22 ( 2 d.p)
The hypotenuse
appears on its own.
Further Example
Reveal >

7. 𝑥 = 85
= 9.22 (2 d.p)
A value written as the
square root of a number is
known as a surd.
Sometimes it’s better to
leave your answer in surd
form (we’ll see why later)
rather than as a decimal.
When we found
areas/circumferences of circles,
we often left our answer in terms
of 𝜋 so that it was exact.
Similarly, answers in surd form
are exact whereas decimal form
answers have to be rounded, and
are thus not exact.
Surd or decimal?

8. 6
8
𝑥
42
55
𝑥
𝑥
6
4
“To learn secret way of ninja,
find 𝑥 you must.”
1 1
𝑥
𝑥
10
12
1
2 3
4
5
Answer: 𝒙 = 𝟏𝟎
Answer: 𝒙 = 𝟒𝟕𝟖𝟗
Answer: 𝒙 = 𝟐
Answer: 𝒙 = 𝟒𝟒
Answer: 𝒙 = 𝟐𝟎
Test Your Understanding

9. We’ve so far written out the equation 𝑎2 + 𝑏2 = 𝑐2, filled in our information,
and rearranged to find the missing side. But it’s helpful to be able to do it in our
heads sometimes!
If you’re looking for the hypotenuse  Square root the sum of the squares
If you’re looking for another side  Square root the difference of the squares
3
5
ℎ
ℎ = 32 + 52 = 34
𝑥
4
7
𝑥 = 72 − 42 = 33
Pythagoras Mental Arithmetic

10. Pythagoras Game!
Everyone stand up. Each of you will be asked, one at a time, and in your head, to find
the missing side of the right-angled triangle. Answer must be in exact form.
If you get it wrong, you sit down, and the person who last sat down has the
opportunity to ‘steal’, where they will be able to stand up again if they correct the
answer.
3
5
𝟒
7
2
𝟓𝟑
Test Run:
(Note to teacher: You don’t need to specifically click on the green boxes. The next
answer will be removed by a mouse/right-arrow press anywhere)

11. Pythagoras Game!
3
4
𝟕
3
1
𝟏𝟎
4
6
𝟓𝟐
3
4
𝟕
7
2
𝟒𝟓
1
2
𝟑
2
3
𝟏𝟑
5
6
𝟏𝟏
8
2
𝟔𝟖
2
2
𝟖
5
𝟓𝟎
8
10
𝟔
12
5
𝟏𝟑
9
3
𝟕𝟐
4
11
𝟏𝟎𝟓
5

12. Pythagoras Game!
1
3
𝟖
2
5
𝟐𝟗
2
9
𝟕𝟕
4
3
𝟓
2
7
𝟓𝟑
4
𝟑𝟐
7
9
𝟑𝟐
10
5
𝟏𝟐𝟓
6
2
𝟑𝟐
3
10
𝟗𝟏
4
1
7
𝟒𝟖
6
5
𝟔𝟏
3
8
𝟕𝟑
3
6
𝟐𝟕
12
10
𝟒𝟒

13. Exercise 1
Find the side marked with the letter
(you do not need to copy the diagrams).
4.5
7
1.8
3.6
𝑚
𝑡 125
98
𝑧
𝑠
19
23
𝑝
5.1
6.2
9
7
𝑥
𝑦
2.2
1.4
a
b c
d e f
g
Solutions: (to 3sf)
(a) 8.32 (b) 3.12 (c) 77.6
(d) 29.8 (e) 5.66 (f) 2.61
(g) 8.03
1
2 To rescue a cat I put a ladder of length 10m
against a tree, with the foot of the ladder 2.5m
away from the tree. How high up the tree is the
cat? 9.68m
Alice and Bob want to get from one corner of
this rectangular field to the other. Alice walks
round the edge of the field. Bob cuts right
across. How much further did Alice walk?
The length of the shortest diagonal of an
octagon is 1. What is the length of the
longest diagonal?
240𝑚
90𝑚
Start
Finish
1
Four unit squares are placed edge to edge as
shown. What is the length of the line 𝑃𝑄?
Solution: 𝟏𝟑
80m
3
4
N
Solution: 𝟐
(if this were a proof
you’d need to justify
why it’s right-angled)
1
1
2

14. Starter
21
20
29
You may have noticed last lesson that
sometimes all three sides of the right-angled
triangle were integers.
These are known as Pythagorean triples.
For example: The sides could be 20, 21 and
29, as 202 + 212 = 292 and thus satisfy
Pythagoras’ Theorem.
How many Pythagorean triples can you find?
(3, 4, 5) (5, 12, 13) (8, 15, 17) (7, 24, 25)
(20, 21, 29) (12, 35, 37) (9, 40, 41) (28, 45, 53)
(11, 60, 61) (16, 63, 65) (33, 56, 65) (48, 55, 73)
(13, 84, 85) (36, 77, 85) (39, 80, 89) (65, 72, 97)
Note that you could also have any multiple of any of these triples as the triangles could be
scaled in size. So for example (3, 4, 5) could become (6,8,10) and so on.
A final note is that if you changed the powers from 2 to 3, or any higher number, then there would never be any solutions.
This is known as Fermat’s Last Theorem, which was unproven for hundreds of years before being proven in 1995.

15. Harder Questions
There’s a variety of ways in which Pythagoras questions could get harder:
6
3
𝑥
4
Multiple triangles
chained together.
A B Adding lines to form
right-angled triangles
that weren’t
originally there.
Area?
2
2 2
3
𝑥 7
9
C Requiring algebraic
manipulation.
𝑥 − 1
12
2𝑥 + 1

16. A :: Multiple Triangles
6
3
𝑥
4
What should we do first?
Find the central length
using the right triangle.
𝒚 = 𝟔𝟐 − 𝟑𝟐 = 𝟐𝟕
𝒚
Then what?
Now we can find 𝒙 using
the left triangle.
𝒙 = 𝟐𝟕 + 𝟏𝟔 = 𝟒𝟑
Notice that 27
2
= 27. This is why it’s often important
to leave your answers in surd form.

17. Test Your Understanding
4
6
12
𝑥
𝑦 = 42 + 62 = 52
𝑥 = 52 + 144 = 14

18. B :: Adding Lines
𝑥
7
4
1
𝑦 = 72 + 42 = 65
𝑥 = 65 − 12 = 8
Sometimes the line(s) you add to form right angled
triangle(s) are fairly obvious…
𝑦

19. Quickfire Heights!
Reminder of Pro Bro Tip: The height of an equilateral triangle is
3 times half the side length.
2
Height = 3
Area = 3
4
Height = 2 3
Area = 4 3
2 3
Height = 3
Area = 3 3
1
Height =
3
2
Area =
3
4

20. Test Your Understanding
4
6
Find the height of this
isosceles triangle.
Solution: 𝟕
Medium Difficulty
Harder Difficulty
An equilateral triangle is cut out of a square of
side 2 cm, as shown. What area of the square
remains?
Solution: 𝟒 − 𝟑

21. C :: Algebraic Triangles
4𝑎
3𝑎
15 3𝑎 2
+ 4𝑎 2
= 152
9𝑎2 + 16𝑎2 = 225
25𝑎2
= 225
𝑎2 = 9
𝒂 = 𝟑
(You will likely encounter more interesting algebraic
Pythagoras problems next year once you cover
expanding two brackets)

22. Exercise 2 (exercises on provided sheet)
12
4
5
𝑥
Give answers in exact form unless
specified.
𝑥 = 185
1
𝑦
2
3
5
𝑦 = 13
1
2

23. Exercise 2 (exercises on provided sheet)
Two snowmen are back to
back, facing in opposite
directions. They each walk
3km forward, turn left and
then work a further 4km.
How far are the snowmen
from each other?
Solution: 10km
2
1
3
1
𝑧
𝑧 = 8 − 3
4 6
5
6
6
6
(a) What is the height of
this equilateral triangle?
Solution: 𝟐𝟕 or 𝟑 𝟑
(b) The area?
Solution: 𝟗 𝟑 𝒐𝒓 𝟑 𝟐𝟕
7
25
25
48
Find the area of this
isosceles triangle.
Solution: 168
Find the height of this
isosceles triangle.
Solution: 𝟒 𝟐𝟎 = 𝟖 𝟓
6
6
8
8

24. Exercise 2 (exercises on provided sheet)
27
𝑥
2𝑥
Determine 𝑥.
𝒙𝟐 + 𝟐𝟕 = 𝟒𝒙𝟐
𝟑𝒙𝟐 = 𝟐𝟕
𝒙𝟐
= 𝟗
𝒙 = 𝟑
9

25. Exercise 2 (exercises on provided sheet)
N5
𝑥
9
7
𝑥
Determine 𝑥.
Solution: We can find the (square of
the) central length in two different
ways:
𝒙𝟐 + 𝟒𝟗 = 𝟖𝟏 − 𝒙𝟐
𝟐𝒙𝟐
= 𝟑𝟐
𝒙𝟐
= 𝟏𝟔
𝒙 = 𝟒
N6
𝑥
3
27
3𝑥
Determine 𝑥 (to 2dp).
Solution: Using similar strategy to
the previous question:
𝟐𝟕𝟐
− 𝟑𝒙 𝟐
= 𝒙𝟐
− 𝟑𝟐
𝟕𝟐𝟗 − 𝟗𝒙𝟐 = 𝒙𝟐 − 𝟗
𝟏𝟎𝒙𝟐
= 𝟕𝟑𝟖
𝒙𝟐
=
𝟕𝟑𝟖
𝟏𝟎
𝒙 = 𝟐. 𝟕𝟐