Pythagoras Theorem Graphs

1. How can we apply
Pythagoras’ Theorem
in graphs and other
triangles?
Learning Objective

2. • To revise all parts of the theorem
• To calculate the length of one of the sides
• To devise a way to use the Theorem in 3D
shapes
Learning Outcomes

3. To find c use: 𝑐2
= ℎ2
− 𝑏²
To find b use: 𝑏2
= ℎ2
− 𝑐²
To find the hypotenuse, we use: ℎ² = 𝑏² + 𝑐²
The Theorem only works on a
_________________:

4. To find c use: 𝑐2
= ℎ2
− 𝑏²
To find b use: 𝑏2
= ℎ2
− 𝑐²
To find the hypotenuse, we use: ℎ² = 𝑏² + 𝑐²
The Theorem only works on a
_________________:

5. To find c use: 𝑐2
= ℎ2
− 𝑏²
To find b use: 𝑏2
= ℎ2
− 𝑐²
To find the hypotenuse, we use: ℎ² = 𝑏² + 𝑐²
The Theorem only works on a
_________________:

6. To find c use: 𝑐2
= ℎ2
− 𝑏²
To find b use: 𝑏2
= ℎ2
− 𝑐²
To find the hypotenuse, we use: ℎ² = 𝑏² + 𝑐²
The Theorem only works on a
_________________:

7. To find c use: 𝑐2
= ℎ2
− 𝑏²
To find b use: 𝑏2
= ℎ2
− 𝑐²
To find the hypotenuse, we use: ℎ² = 𝑏² + 𝑐²
The Theorem only works on a right
angle triangle:

8. Learning Outcomes
• To revise all parts of the theorem
• To calculate the length of one of the sides
• To devise a way to use the Theorem in 3D
shapes

9. Learning Outcomes
• To revise all parts of the theorem
• To calculate the length of one of the sides
• To devise a way to use the Theorem in 3D
shapes

10. Edexel Pythagoras Question

11. Edexel Pythagoras Question
We use Pythagoras’ Theorem to calculate
the distance between Norwich and York.

12. Edexel Pythagoras Question
We use Pythagoras’ Theorem to calculate
the distance between Norwich and York.
The Hypotenuse is the distance between
them:
ℎ2 =?2 +?2

13. Edexel Pythagoras Question
We use Pythagoras’ Theorem to calculate
the distance between Norwich and York.
The Hypotenuse is the distance between
them:
ℎ2 =?2 +?2
b
c

14. Edexel Pythagoras Question
We use Pythagoras’ Theorem to calculate
the distance between Norwich and York.
The Hypotenuse is the distance between
them:
ℎ2 = 𝑏2 + 𝑐2
b
c

15. Edexel Pythagoras Question
We use Pythagoras’ Theorem to calculate
the distance between Norwich and York.
The Hypotenuse is the distance between
them:
ℎ2 = 𝑏2 + 𝑐2
ℎ2 = 1682 + 1572
b
c

16. Edexel Pythagoras Question
We use Pythagoras’ Theorem to calculate
the distance between Norwich and York.
The Hypotenuse is the distance between
them:
ℎ2 = 𝑏2 + 𝑐2
ℎ2 = 1682 + 1572
ℎ2
= 52873
b
c

17. Edexel Pythagoras Question
We use Pythagoras’ Theorem to calculate
the distance between Norwich and York.
The Hypotenuse is the distance between
them:
ℎ2 = 𝑏2 + 𝑐2
ℎ2 = 1682 + 1572
ℎ2
= 52873
ℎ = 52873
b
c

18. Edexel Pythagoras Question
We use Pythagoras’ Theorem to calculate
the distance between Norwich and York.
The Hypotenuse is the distance between
them:
ℎ2 = 𝑏2 + 𝑐2
ℎ2 = 1682 + 1572
ℎ2
= 52873
ℎ = 52873
ℎ = 229.94𝑘𝑚 (𝑡. 2. 𝑑. 𝑝)
b
c

20. 𝑏2 = ℎ2 − 𝑐2
𝑥

21. 𝑏2 = ℎ2 − 𝑐2
𝑥2 = 62 − 22
𝑥

22. 𝑏2 = ℎ2 − 𝑐2
𝑥2 = 62 − 22
𝑥2 = 36 − 4
𝑥

23. 𝑏2 = ℎ2 − 𝑐2
𝑥2 = 62 − 22
𝑥2 = 36 − 4
𝑥2 = 34
𝑥

24. 𝑏2 = ℎ2 − 𝑐2
𝑥2 = 62 − 22
𝑥2 = 36 − 4
𝑥2 = 34
𝑥 = 34
𝑥

25. 𝑏2 = ℎ2 − 𝑐2
𝑥2 = 62 − 22
𝑥2 = 36 − 4
𝑥2 = 34
𝑥 = 34
𝑥 = 5.65685 …
𝑥

26. 𝑏2 = ℎ2 − 𝑐2
𝑥2 = 62 − 22
𝑥2 = 36 − 4
𝑥2 = 34
𝑥 = 34
𝑥 = 5.65685 …
The perpendicular height of the triangle is 5.7cm to 1 decimal place.
𝑥

27. Pythagoras – Non right angled triangle
𝑥

28. 𝑥² + 2² = 4²
Pythagoras – Non right angled triangle
𝑥

29. 𝑥² + 2² = 4²
𝑥² = 4² - 2²
Pythagoras – Non right angled triangle
𝑥

30. 𝑥² + 2² = 4²
𝑥² = 4² - 2²
𝑥² = 16 - 4
Pythagoras – Non right angled triangle
𝑥

31. 𝑥² + 2² = 4²
𝑥² = 4² - 2²
𝑥² = 16 – 4
𝑥 = 12
Pythagoras – Non right angled triangle
𝑥

32. 𝑥² + 2² = 4²
𝑥² = 4² - 2²
𝑥² = 16 – 4
𝑥 = 12
𝑥 = 3.5cm (1.d.p)
Pythagoras – Non right angled triangle
𝑥

33. a) Calculate the length of AC
giving your answer correct
to 3 significant figures.
b) Calculate the area of
triangle ABC.
c) Calculate the perimeter of triangle ABC giving your answer correct to
1 decimal place.
Examination Questions

34. a) Calculate the length of AC
giving your answer correct
to 3 significant figures.
b) Calculate the area of
triangle ABC.
c) Calculate the perimeter of triangle ABC giving your answer correct to
1 decimal place.
a) AC² = 4² + 6² = 52
Examination Questions

35. a) Calculate the length of AC
giving your answer correct
to 3 significant figures.
b) Calculate the area of
triangle ABC.
c) Calculate the perimeter of triangle ABC giving your answer correct to
1 decimal place.
a) AC² = 4² + 6² = 52
AC = √52 = 7.211102551
Examination Questions

36. a) Calculate the length of AC
giving your answer correct
to 3 significant figures.
b) Calculate the area of
triangle ABC.
c) Calculate the perimeter of triangle ABC giving your answer correct to
1 decimal place.
a) AC² = 4² + 6² = 52
AC = 52 = 7.211102551
AC = 7.21 cm
Examination Questions

37. a) Calculate the length of AC
giving your answer correct
to 3 significant figures.
b) Calculate the area of
triangle ABC.
c) Calculate the perimeter of triangle ABC giving your answer correct to
1 decimal place.
a) AC² = 4² + 6² = 52
AC = 52 = 7.211102551
AC = 7.21 cm
b) Area = ½ x 4 x 6
Examination Questions

38. a) Calculate the length of AC
giving your answer correct
to 3 significant figures.
b) Calculate the area of
triangle ABC.
c) Calculate the perimeter of triangle ABC giving your answer correct to
1 decimal place.
a) AC² = 4² + 6² = 52
AC = 52 = 7.211102551
AC = 7.21 cm
b) Area = ½ x 4 x 6
= 12 cm²
Examination Questions

39. a) Calculate the length of AC
giving your answer correct
to 3 significant figures.
b) Calculate the area of
triangle ABC.
c) Calculate the perimeter of triangle ABC giving your answer correct to
1 decimal place.
a) AC² = 4² + 6² = 52
AC = 52 = 7.211102551
AC = 7.21 cm
b) Area = ½ x 4 x 6
= 12 cm²
c) P = 4 + 6 + 7.211102551
Examination Questions

40. a) Calculate the length of AC
giving your answer correct
to 3 significant figures.
b) Calculate the area of
triangle ABC.
c) Calculate the perimeter of triangle ABC giving your answer correct to
1 decimal place.
a) AC² = 4² + 6² = 52
AC = 52 = 7.211102551
AC = 7.21 cm
b) Area = ½ x 4 x 6
= 12 cm²
c) P = 4 + 6 + 7.211102551
= 17.2 cm
Examination Questions

41. Examination Questions

42. • V = Area of the cross
section x depth
Examination Questions

43. • V = Area of the cross
section x depth.
• We need to find the area
of the cross section.
Examination Questions

44. • V = Area of the cross
section x depth.
• We need to find the area
of the cross section.
• It’s a triangle!!!
Examination Questions

45. • V = Area of the cross section x
depth.
• We need to find the area of the
cross section.
• It’s a triangle!!!
• Area = ½ (Base x Height)
• We need to calculate the
height
Examination Questions

46. • V = Area of the cross section x
depth.
• We need to find the area of the
cross section.
• It’s a triangle!!!
• Area = ½ (Base x Height)
• We need to calculate the
height
Height : we know the
longest side and the
base.
Examination Questions

47. • Pythagoras
• a² = b² + c²
• 5.3² = 2.8²+c²
• c² = 5.3²-2.8²
• = 20.25
• c = 20.25
• = 4.5
Height : we know the
longest side and the
base.
Examination Questions

48. • Volume
We now need to put
the answer into the
Volume formula.
Examination Questions

49. • Volume
• V = ½(4.5x2.8) x 3.5
We now need to put
the answer in to the
Volume formula.
Examination Questions

50. • Volume
• V = ½(4.5x2.8) x 3.5
• = ½(12.6) x 3.5
• = 6.3 x 3.5
• = 22.05cm³
We now need to put
the answer in to the
Volume formula.
Examination Questions

51. This is a representation of a water tank. For every 1cm³ there is 1ml. Calculate how
many litres of water would the tank hold.
Examination Questions

52. This is a representation of a water tank. For every
1cm³ there is 1ml. Calculate how many litres of water
would the tank hold.
First we need to find the area of the
cross section.
Examination Questions

53. This is a representation of a water tank. For every
1cm³ there is 1ml. Calculate how many litres of water
would the tank hold.
First we need to find the area of the
cross section.
That means finding the width and the
length.
Examination Questions

54. This is a representation of a water tank. For every
1cm³ there is 1ml. Calculate how many litres of water
would the tank hold.
First we need to find the area of the
cross section.
That means finding the width and the
length.
We don’t know the Length!!!!
Examination Questions

55. First we need to find the area of the
cross section.
That means finding the width and the
length.
We don’t know the Length!!!!
To find the width we use Pythagoras’
Theorem!
This is a representation of a water tank. For every
1cm³ there is 1ml. Calculate how many litres of water
would the tank hold.
Examination Questions

56. First we need to find the area of the
cross section.
That means finding the width and the
length.
We don’t know the Length!!!!
To find the width we use Pythagoras’
Theorem!
Which side do I need to find?
This is a representation of a water tank. For every
1cm³ there is 1ml. Calculate how many litres of water
would the tank hold.
Examination Questions

57. This is a representation of a water tank. For every
1cm³ there is 1ml. Calculate how many litres of water
would the tank hold.
First we need to find the area of the
cross section.
That means finding the width and the
length.
We don’t know the Length!!!!
To find the width we use Pythagoras’
Theorem!
Which side do I need to find?
ONE OF THE SHORTER SIDES
So we use:
Examination Questions

58. This is a representation of a water tank. For every
1cm³ there is 1ml. Calculate how many litres of water
would the tank hold.
First we need to find the area of the
cross section.
That means finding the width and the
length.
We don’t know the Length!!!!
To find the width we use Pythagoras’
Theorem!
Which side do I need to find?
ONE OF THE SHORTER SIDES
So we use:
b² = a² - c²
Examination Questions

59. This is a representation of a water tank. For every
1cm³ there is 1ml. Calculate how many litres of water
would the tank hold.
First we need to find the area of the
cross section.
That means finding the width and the
length.
We don’t know the Length!!!!
To find the width we use Pythagoras’
Theorem!
Which side do I need to find?
ONE OF THE SHORTER SIDES
So we use:
b² = a² - c²
= 24.7² - 16.3²
Examination Questions

60. This is a representation of a water tank. For every
1cm³ there is 1ml. Calculate how many litres of water
would the tank hold.
First we need to find the area of the
cross section.
That means finding the width and the
length.
We don’t know the Length!!!!
To find the width we use Pythagoras’
Theorem!
Which side do I need to find?
ONE OF THE SHORTER SIDES
So we use:
b² = a² - c²
= 24.7² - 16.3²
b = 18.56cm (2dp)
Examination Questions

61. This is a representation of a water tank. For every
1cm³ there is 1ml. Calculate how many litres of water
would the tank hold.
First we need to find the area of the
cross section.
That means finding the width and the
length.
We don’t know the Length!!!!
To find the width we use Pythagoras’
Theorem!
Which side do I need to find?
ONE OF THE SHORTER SIDES
So we use:
b² = a² - c²
= 24.7² - 16.3²
b = 18.56cm (2dp)
Now we use the Volume formula
Vol = Width x Length x Depth
= 18.56 x 16.3 x 53.1
Examination Questions

62. This is a representation of a water tank. For every
1cm³ there is 1ml. Calculate how many litres of water
would the tank hold.
First we need to find the area of the
cross section.
That means finding the width and the
length.
We don’t know the Length!!!!
To find the width we use Pythagoras’
Theorem!
Which side do I need to find?
ONE OF THE SHORTER SIDES
So we use:
b² = a² - c²
= 24.7² - 16.3²
b = 18.56cm (2dp)
Now we use the Volume formula
Vol = Width x Length x Depth
= 18.56 x 16.3 x 53.1
= 16064.2cm³
Examination Questions

63. This is a representation of a water tank. For every
1cm³ there is 1ml. Calculate how many litres of water
would the tank hold.
First we need to find the area of the
cross section.
That means finding the width and the
length.
We don’t know the Length!!!!
To find the width we use Pythagoras’
Theorem!
Which side do I need to find?
ONE OF THE SHORTER SIDES
So we use:
b² = a² - c²
= 24.7² - 16.3²
b = 18.56cm (2dp)
Now we use the Volume formula
Vol = Width x Length x Depth
= 18.56 x 16.3 x 53.1
= 16064.2cm³
Not there yet!
Now we calculate the litres
1cm³ = 1ml,
1000 ml = 1L
Examination Questions

64. This is a representation of a water tank. For every
1cm³ there is 1ml. Calculate how many litres of water
would the tank hold.
First we need to find the area of the
cross section.
That means finding the width and the
length.
We don’t know the Length!!!!
To find the width we use Pythagoras’
Theorem!
Which side do I need to find?
ONE OF THE SHORTER SIDES
So we use:
b² = a² - c²
= 24.7² - 16.3²
b = 18.56cm (2dp)
Now we use the Volume formula
Vol = Width x Length x Depth
= 18.56 x 16.3 x 53.1
= 16064.2cm³
Not there yet!
Now we calculate the litres
1cm³ = 1ml,
1000 ml = 1L
Hence:
16064.2 ÷ 1000 =
Examination Questions

65. This is a representation of a water tank. For every
1cm³ there is 1ml. Calculate how many litres of water
would the tank hold.
First we need to find the area of the
cross section.
That means finding the width and the
length.
We don’t know the Length!!!!
To find the width we use Pythagoras’
Theorem!
Which side do I need to find?
ONE OF THE SHORTER SIDES
So we use:
b² = a² - c²
= 24.7² - 16.3²
b = 18.56cm (2dp)
Now we use the Volume formula
Vol = Width x Length x Depth
= 18.56 x 16.3 x 53.1
= 16064.2cm³
Not there yet!
Now we calculate the litres
1cm³ = 1ml,
1000 ml = 1L
Hence:
16064.2 ÷ 1000 = 16.06L (2dp)
Examination Questions

66. Learning Outcomes
• To revise all parts of the theorem
• To calculate the length of one of the sides
• To devise a way to use the Theorem in 3D
shapes

67. Learning Outcomes
• To revise all parts of the theorem
• To calculate the length of one of the sides
• To devise a way to use the Theorem in 3D
shapes

68. We can also use Pythagoras’
Theorem to calculate the
length of a diagonal line on
a graph.
Pythagoras on a Graph

69. We can also use Pythagoras’
Theorem to calculate the
length of a diagonal line on
a graph.
P1: (1,3)
P2: (7,8)
Pythagoras on a Graph

70. We can also use Pythagoras’
Theorem to calculate the
length of a diagonal line on
a graph.
P1: (1,3)
P2: (7,8)
Use the co-ordinates to find
the horizontal and vertical
difference between the two
points.
Pythagoras on a Graph

71. P1: (1,3)
P2: (7,8)
Vertical = y2 – y1
Horizontal = x2 – x1
Pythagoras on a Graph

72. P1: (1,3)
P2: (7,8)
Vertical = 8 – 3
Horizontal = 7 – 1
Pythagoras on a Graph

73. P1: (1,3)
P2: (7,8)
Vertical = 8 – 3
Horizontal = 7 – 1
Pythagoras on a Graph

74. P1: (1,3)
P2: (7,8)
Vertical = 8 – 3 = 5
Horizontal = 7 – 1 = 6
Pythagoras on a Graph

75. P1: (1,3)
P2: (7,8)
Vertical = 8 – 3 = 5
Horizontal = 7 – 1 = 6
Now we can use Pythagoras’
Theorem.
Pythagoras on a Graph

76. P1: (1,3)
P2: (7,8)
Vertical = 8 – 3 = 5
Horizontal = 7 – 1 = 6
Now we can use Pythagoras’
Theorem.
P1P2² = 5² + 6²
Pythagoras on a Graph

77. P1: (1,3)
P2: (7,8)
Vertical = 8 – 3 = 5
Horizontal = 7 – 1 = 6
Now we can use Pythagoras’
Theorem.
P1P2² = 5² + 6²
P1P2 = 61 =
Pythagoras on a Graph

78. P1: (1,3)
P2: (7,8)
Vertical = 8 – 3 = 5
Horizontal = 7 – 1 = 6
Now we can use Pythagoras’
Theorem.
P1P2² = 5² + 6²
P1P2 = 61 = 7.81cm (2dp)
Pythagoras on a Graph

79. The distance between the points A(x1, y1) and B(x2, y2)
is ( ) ( )
x x y y
 
2 2
2 1 2 1
+
Pythagoras on a Graph

80. The distance between the points A(x1, y1) and B(x2, y2)
is ( ) ( )
x x y y
 
2 2
2 1 2 1
+
P1 = (4,7) P2 = (5,3)
Distance = 5 − 4 2 + (3 − 7)²
Pythagoras on a Graph

81. The distance between the points A(x1, y1) and B(x2, y2)
is ( ) ( )
x x y y
 
2 2
2 1 2 1
+
P1 = (4,7) P2 = (5,3)
Distance = 5 − 4 2 + (3 − 7)²
= 1 + 16
Pythagoras on a Graph

82. The distance between the points A(x1, y1) and B(x2, y2)
is ( ) ( )
x x y y
 
2 2
2 1 2 1
+
P1 = (4,7) P2 = (5,3)
Distance = 5 − 4 2 + (3 − 7)²
= 17
= 4.12cm (2.d.p)
Pythagoras on a Graph

83. Pythagoras on a Graph

84. P1 = (3,2) P2 = (9,7)
Distance = 9 − 3 2 + (7 − 2)²
=
=
Pythagoras on a Graph

85. P1 = (3,2) P2 = (9,7)
Distance = 9 − 3 2 + (7 − 2)²
= 36 + 25
=
Pythagoras on a Graph

86. P1 = (3,2) P2 = (9,7)
Distance = 9 − 3 2 + (7 − 2)²
= 61
=
Pythagoras on a Graph

87. P1 = (3,2) P2 = (9,7)
Distance = 9 − 3 2 + (7 − 2)²
= 61
= 7.81cm (2.d.p)
Pythagoras on a Graph

88. Pythagoras on a Graph

89. P1 = (4,-4) P2 = (-5,3)
Distance =
=
=
Pythagoras on a Graph

90. P1 = (4,-4) P2 = (-5,3)
Distance = −5 − 4 2 + (3 − −4)²
=
=
Pythagoras on a Graph

91. P1 = (4,-4) P2 = (-5,3)
Distance = −5 − 4 2 + (3 − −4)²
= −9 ² + 7²
=
Pythagoras on a Graph

92. P1 = (4,-4) P2 = (-5,3)
Distance = −5 − 4 2 + (3 − −4)²
= 130
=
Pythagoras on a Graph

93. P1 = (4,-4) P2 = (-5,3)
Distance = −5 − 4 2 + (3 − −4)²
= 130
= 11.40cm (2.d.p)
Pythagoras on a Graph

94. P1 = P2 =
Distance =
=
=
Pythagoras on a Graph

95. P1 = (4,-4) P2 = (2,5)
Distance = 2 − 4 2 + (5 − −4)²
=
=
Pythagoras on a Graph

96. P1 = (4,-4) P2 = (2,5)
Distance = 2 − 4 2 + (5 − −4)²
= 4 + 81
=
Pythagoras on a Graph

97. P1 = (4,-4) P2 = (2,5)
Distance = 2 − 4 2 + (5 − −4)²
= 85
=
Pythagoras on a Graph

98. Pythagoras on a Graph
P1 = (4,-4) P2 = (2,5)
Distance = 2 − 4 2 + (5 − −4)²
= 85
= 9.21cm (2.d.p)

99. Learning Outcomes
• To revise all parts of the theorem
• To calculate the length of one of the sides
• To devise a way to use the Theorem in 3D
shapes

100. Learning Outcomes
• To revise all parts of the theorem
• To calculate the length of one of the sides
• To devise a way to use the Theorem in 3D
shapes