1. 15/11/2018 Starter L.O. To be able to
Find the missing side of a right-angled triangle, given the other two sides
Solve problems using Pythagoras’ theorem, such as finding the diagonal of a rectangle
Solve problems using Pythagoras’ theorem in 3 dimensions
Key Words
hypotenuse, right angle, square, square root, theorem
Starter

2. FINDING THE HYPOTENUSE
15/11/2018
CONTENTS – Click to go to…
INTRODUCTION
FINDING THE HYPOTENUSE
FINDING ANY SIDE
PROBLEM SOLVING
PYTHAGORAS IN 3D
REVISION

3. 15/11/2018
INTRODUCTION

4. SQUARES AND SQUARE ROOTS
15/11/2018
SQUARES AND SQUARE ROOTS
On your mini whiteboards… Quick Questions!

5. 15/11/2018
Evaluate : 72

6. 15/11/2018 x
Square
Area = 49 m2

7. 15/11/2018
5 m
Square
Area = ?

8. 15/11/2018
Evaluate : 81

9. 15/11/2018 x
Square
Area = 144 m2

10. 15/11/2018
10 m
Square
Area = ?

11. 15/11/2018
Evaluate : 12

12. 15/11/2018
Evaluate :
400

13. 15/11/2018
Evaluate : 251/2

14. 15/11/2018
Evaluate : 1

15. Identifying the hypotenuse
15/11/2018
Identifying the hypotenuse
A right-angled triangle contains a right angle
The side opposite the right angle is always the LONGEST
It has a special name - HYPOTENUSE
hypotenuse

16. 15/11/2018
Use this exercise to identify the hypotenuse in right-angled triangles in various orientations.

17. Which side is the hypotenuse?
15/11/2018
Which side is the hypotenuse? f e a b d c g h i

18. 15/11/2018
Discovery task!
Make accurate constructions of the three triangles below
MEASURE accurately the missing length on each one. Copy and fill in this table…
Can you spot the pattern that links the last three columns?
8cm
6cm
12cm
5cm
4cm
3cm 𝒂 𝒃 𝒄
𝒂 𝟐
𝒃 𝟐
𝒄 𝟐 3 4 9 5 12 6 8

19. 15/11/2018 Pythagoras proof activity From your worksheet…
Colour in square 5 one colour, then cut it off
Colour in the square with pieces 1-4 a different colour, hen cut it off, also cutting along the dotted lines to get four shapes from that square
Try to arrange the five pieces you have so that they fit exactly into the larger bold square. Stick them down.
Write a sentence that describes what you have discovered from this activity
“I have shown that…_____________________
_____________________________________
_____________________________________”

20. 15/11/2018
Pythagoras’ Theorem
a + b = h 2 a b h

21. 15/11/2018
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.

22. c b a X X X a2 X c2 X X b2 X X + = a2 b2 c2

23. 15/11/2018
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 red 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 green right-angled triangles.
The area of the four green right-angled triangles is 4 × ½ab = 2ab.
c2 = (a + b)2 – 2ab
Expanding,
c2 = a2 + 2ab + b2 – 2ab
c2 = a2 + b2

24. 15/11/2018
Pythagorean trees! – Do you think you could draw one?!

25. FINDING THE HYPOTENUSE
15/11/2018
FINDING THE HYPOTENUSE

26. a + b = h h a b 11/15/2018 Pythagoras’ Theorem – ALGEBRAIC METHOD
Learn the formula!
Substitute the two values that you know into the formula
Solve the equation to find the unknown value a b h
a + b = h 2

27. Example – Algebraic method
15/11/2018
Example – Algebraic method
h 2 = a b 2
5 cm
12 cm x 1
h 2 = a b 2
4.6 cm
9.8 cm x 2

28. Pythagoras – QUICK METHOD
15/11/2018
Pythagoras – QUICK METHOD
Decide if the side you are LOOKING FOR is the HYPOTENUSE, or one of the two SHORTER SIDES
Use one of these quick and easy methods to work out what you need
Looking for the LONGEST SIDE?
(the hypotenuse!)
Square it
Add it
Square root it
Looking for a SHORTER SIDE?
(not the hypotenuse!)
Square it
Add it
Square root it

29. 15/11/2018 Example – Quick method Square it… 4 2 = 16 1 +
3 cm
4 cm x 1
Square it… = 9 +
Add it… = 25
Square root it… 𝑥= 25
= 𝟓cm
Square it… =
4.6 cm
9.8 cm x 2
Square it… = +
Add it… =117.2
Square root it… 𝑥=
=10.825…
=𝟏𝟎.𝟖 cm (1dp)

30. 15/11/2018 b = 15.62cm a = 7.21cm d = 9.68cm c = 7.21cm f = 15.43cm
Task – Find the missing side, to 2d.p.
10 cm a b 1) 2)
b = 15.62cm
4 cm
6 cm
12 cm
a = 7.21cm
5.3 cm
5 cm c
8.1 cm 3) 4) d
d = 9.68cm
c = 7.21cm
7cm
10.1 cm
12 cm
8.5 cm
9.7 cm 5) 6)
f = 15.43cm e
e = 8.60cm f

31. 15/11/2018
Task – Find the missing side, to 2d.p.
13.04 m
16.12 cm 1) 2) 3) x x
9.85 m
8 cm
9 m
7 m x
11 m
4 m
14 cm 4) 5) 6)
6 m x x
9.76 m
16.46 mm
7.81 m
5 m x
8.2 m
5.3 m
14.7 mm
7.4 mm

32. 15/11/2018
FINDING ANY SIDE

33. a + b = h h a b 11/15/2018 Pythagoras’ Theorem – ALGEBRAIC METHOD
Learn the formula!
Substitute the two values that you know into the formula
Solve the equation to find the unknown value a b h
a + b = h 2

34. 15/11/2018
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.

35. Example – Algebraic method
15/11/2018
Example – Algebraic method
a 2 = h 2 − b 2
25 m
7 m
x m

36. Pythagoras – QUICK METHOD
11/15/2018
Pythagoras – QUICK METHOD
Decide if the side you are LOOKING FOR is the HYPOTENUSE, or one of the two SHORTER SIDES
Use one of these quick and easy methods to work out what you need
Looking for the LONGEST SIDE?
(the hypotenuse!)
Square it
Add it
Square root it
Looking for a SHORTER SIDE?
(not the hypotenuse!)
Square it
Add it
Square root it

37. 15/11/2018 Example – Quick method Square it… 15 2 =225 1 −
9cm
15cm 1
Square it… = −
Subtract it… = 144
Square root it… 𝑥= 144
= 𝟏𝟐cm
Square it… = 49 x
5cm
7cm 2
Square it… = 25 −
Subtract it… = 24
Square root it… 𝑥= 24
= …
=4.9 cm (1dp)

38. 1) 2) 4) 3) 6) 5) 15/11/2018 h = 11.26cm g = 9.17cm i = 4.45cm
Task – Find the missing side, to 2d.p.
5.2 cm
10 cm
12.4 cm 1) 2)
h = 11.26cm
4 cm g h
g = 9.17cm 4) j 3)
14.5 cm i
8.4 cm
i = 4.45cm
j = 23.86cm
25.3 cm
13.8cm
2.9 cm 6)
23 cm 5) l k
k = 9.15cm
l = 19.26cm
9.6 cm
30 cm

39. 15/11/2018 Task – Find the missing sides, to 2.d.p 12.69 m 1) 2) 3) x
8 mm
9.75 m
24 mm
22.63 mm x
7 m 4) 5) 6)
14 m
52 mm
18 m
28 mm x x
20 m
17.29 m
5 m
14.28 m
43.82 mm x

40. 15/11/2018 Task – Find the missing sides, to 2.d.p 5.3 1) 2) 3) 10.4434 x x
14.32
27.50 9 6 x 13 20 4) 5) 6) x
1.8
18.25
4.48 14
9.80 x 23 x 14 10
4.1

41. 15/11/2018

42. Exam question – Spot the errors!
15/11/2018
Exam question – Spot the errors!
a) FIND PQ =
=586𝑐𝑚
So PQ is 586cm
b) FIND THE AREA OF PQR
15 × 19 2
= 𝑐𝑚

43. PYTHAGORAS CATCH PHRASE!
11/15/2018
PYTHAGORAS CATCH PHRASE!
On your mini whiteboards… Choose a question to answer. Get it right to reveal part of the picture and guess the catchphrase!

44. The diagram shows a trapezium ABCD . Calculate the length of AB.
6.71cm
The diagram shows a trapezium ABCD . Calculate the length of AB.
10cm
7cm
8cm A D C B
5cm
This rectangle has a diagonal of 13cm and a width of 12cm. Calculate it’s height. j
12cm
13cm 9m c
12.73m
miles
Cardiff is 130 miles west of London. Hull is 200 miles north east of Cardiff. How far is Hull from London?
130 miles
200 miles
Cardiff
London
Hull 3m 8m d
8.54m
23cm
16cm f
16.52cm
13.23cm
15cm a
20cm
12cm
8cm b
8.94cm
6.93m k 4m 8m

45. 15/11/2018
BAD HAIR DAY!

46. 15/11/2018
PROBLEM SOLVING

47. 15/11/2018
Problem solving with Pythagoras
Draw yourself a diagram which includes a right-angled triangle
Identify which side is the hypotenuse
Use Pythagoras in the normal way to find the missing side

48. 15/11/2018
Change the length and the position of the ladder to generate various problems.

49. 15/11/2018
Use Pythagoras’ Theorem to calculate the distance of the aeroplane from the starting point.

50. 15/11/2018 Examples 1 2 d 6 cm 9.3 cm 7.8 cm 4.3 cm x cm
Find the diagonal of the rectangle
6 cm
9.3 cm 1 d
A rectangle has a width of 4.3 cm and a diagonal of 7.8 cm. Find its perimeter. 2
7.8 cm
4.3 cm
x cm
Perimeter = 2( ) = 21.6 cm

51. 15/11/2018
Example
A boat sails due East from a Harbour (H), to a marker buoy (B), 15 miles away. At B the boat turns due South and sails for 6.4 miles to a Lighthouse (L). It then returns to harbour. Make a sketch of the journey. What is the total distance travelled by the boat? H B L
15 miles
6.4 miles
Total distance travelled = = 37.7 miles

52. 15/11/2018
Example
A 12 ft ladder rests against the side of a house. The top of the ladder is 9.5 ft from the floor. How far is the base of the ladder from the house?
12 ft
9.5 ft L

53. PROBLEM SOLVING WITH PYTHAGORAS
11/15/2018
PROBLEM SOLVING WITH PYTHAGORAS
On your mini whiteboards… Quick Questions!

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60. Finding the distance between two points
15/11/2018
Finding the distance between two points 1 2 3 4 5 6 –1 –2 –3 –4 –5 –6 y x
A(–4, 5)
The points A(–4, 5), B(–4, –2), and C (6,–2) form a right angled triangle.
7 units
What is length of the line AB?
B(–4, –2)
10 units
C (6, –2)
AB = 5 – –2
BC = 6 – –4
= 5 + 2
= 6 + 4
= 7 units
= 10 units

61. Finding the distance between two points
15/11/2018
Finding the distance between two points 1 2 3 4 5 6 –1 –2 –3 –4 –5 –6 y x
A(–4, 5)
The points A(–4, 5), B(–4, –2), and C (6,–2) form a right angled triangle.
12.21 units (to 2 d.p.)
7 units
What is length of the line AB?
B(–4, –2)
10 units
C (6, –2)
Using Pythagoras,
AC2 = AB2 + AC2 = =
= 149
AC = 149
= units (to 2 d.p.)

62. Finding the distance between two points
15/11/2018
Finding the distance between two points
We can use Pythagoras’ Theorem to find the distance between any two points on a co-ordinate grid.
Find the distance between the points A(7, 6) and B(3, –2). 1 2 3 4 5 6 7 –1 –2 –3 x y
A(7, 6)
We can add a third point at C to form a right-angled triangle.
AB2 = AC2 + BC2 = =
B(3, –2)
C (7, –2)
= 80
AC = 80
= 8.94 units (to 2 d.p.)

63. 15/11/2018 Task – Coordinate cards Choose any TWO CARDS
Sketch them and join with a line. This is your hypotenuse.
Sketch in the horizontal and vertical sides to make a right angled triangle.
Use Pythagoras to find the length of the hypotenuse
Do as many as you can
Difficulty Levels
Green
( , )
coordinates in the first quadrant
POSITIVE ONLY
Orange
( , ) 
coordinates in the other three quadrants
POSITIVE AND NEGATIVE
Red
( , )  
coordinates with decimal numbers
+ AND - INCLUDING DECIMALS
Blank
  ( , )
make your own!
AS DIFFICULT AS YOU LIKE

64. Finding the distance between two points - Algebraically
15/11/2018
Finding the distance between two points - Algebraically
What is the distance between two points with coordinates A(xA, yA) and B(xB, yB)?
The horizontal distance between the points is
xB – xA
The vertical distance between the points is
yB – yA
Using Pythagoras’ Theorem the square of the distance between the points A(xA, yA) and B(xB, yB) is
(xB – xA)2 + (yB – yA)2
The distance between the points A(xA, yA) and B(xB, yB) is
(xB – xA)2 + (yB – yA)2

65. 15/11/2018
Challenge problem!
Use Pythagoras’ theorem to find the length and width of the rectangle and hence find its perimeter and area in surd form.
Perimeter = √10 + 2√10 +
√10 + 2√10 1 3
√10 6 2
2√10
= 6√10 units
Area = √10 × 2√10
= 2 × √10 × √10
= 2 × 10
= 20 units2

66. Find the diagonals of the kite
15/11/2018
Find the diagonals of the kite
5 cm
12 cm
6 cm
5 cm
x cm
y cm

67. 15/11/2018 Real life problem B A W
130 miles
170 miles B
An aircraft leaves RAF Waddington (W) and flies on a bearing of NW for 130 miles and lands at a another airfield (A). It then takes off and flies 170 miles on a bearing of NE to a Navigation Beacon (B). From (B) it returns directly to Waddington. Make a sketch of the flight. How far has the aircraft flown?

68. 15/11/2018
PYTHAGORAS IN 3D

69. Example – Find the distance BG
15/11/2018
Example – Find the distance BG
Find FG first
FG2 =
FG = ( )
FG = 13 cm A B E F C D G H
3 cm
5 cm
12 cm
Use FG to find BG
BG2 =
FG = ( )
FG = 13.3 cm
13 cm

70. Example – Find the distance BG
15/11/2018
Example – Find the distance BG
Find EC first
EC2 =
EC = ( )
EC = 10.67cm A B D C E F
3.1 cm
9.2 cm
5.4 cm
Use EC to find BG
BE2 =
BE = ( )
BE = 11.1 cm (1 dp)
10.67cm

71. 15/11/2018
Problem solving
My pencil case is in the shape of a cuboid.
It measures 8cm, by 7cm by 5cm.
I want to put my new pencil into the pencil case. The pencil measures 12cm. Does it fit?

72. 15/11/2018
REVISION

73. 15/11/2018 Right Angled Triangle Topics (printable version)
Area of the triangle
Questions involve a BASE and PERPENDICULAR HEIGHT
It will ask you for the area of the triangle
Use AREA =
Pythagoras
Involves __________ sides, ____ given and ____ missing
If you want the LONGEST side (hypotenuse): square it, square it, ADD it, then square-root it
If you want one of the SHORTER sides: square it, square it, SUBTRACT it, then square-root it
Trigonometry
Involves ____ sides and ____ angle
____ of these things will be given and ____ will be missing
Use SOH___TOA
HYPOTENUSE! O S H × 𝜃 A C H × 𝜃 O T A × 𝜃

74. 15/11/2018
Task – Complete the review worksheet of Pythagoras questions
Click to reveal solutions…
…and click again to reveal Qs 7-10…