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Pythagoras Theorem Graphs

This document discusses how to apply Pythagoras' theorem to calculate distances and lengths in triangles, both right-angled and non-right angled. It provides examples of using the theorem to solve examination questions involving finding lengths, areas, perimeters, volumes, and calculating how much water a tank can hold.

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How can we apply Pythagoras’ Theorem in graphs and other triangles? Learning Objective
• 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
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 _________________:
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 _________________:
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 _________________:
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 _________________:
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:
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
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
Edexel Pythagoras Question
Edexel Pythagoras Question We use Pythagoras’ Theorem to calculate the distance between Norwich and York.
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
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
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
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
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
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
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
𝑏2 = ℎ2 − 𝑐2 𝑥
𝑏2 = ℎ2 − 𝑐2 𝑥2 = 62 − 22 𝑥
𝑏2 = ℎ2 − 𝑐2 𝑥2 = 62 − 22 𝑥2 = 36 − 4 𝑥
𝑏2 = ℎ2 − 𝑐2 𝑥2 = 62 − 22 𝑥2 = 36 − 4 𝑥2 = 34 𝑥
𝑏2 = ℎ2 − 𝑐2 𝑥2 = 62 − 22 𝑥2 = 36 − 4 𝑥2 = 34 𝑥 = 34 𝑥
𝑏2 = ℎ2 − 𝑐2 𝑥2 = 62 − 22 𝑥2 = 36 − 4 𝑥2 = 34 𝑥 = 34 𝑥 = 5.65685 … 𝑥
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
Pythagoras – Non right angled triangle 𝑥
𝑥² + 2² = 4² Pythagoras – Non right angled triangle 𝑥
𝑥² + 2² = 4² 𝑥² = 4² - 2² Pythagoras – Non right angled triangle 𝑥
𝑥² + 2² = 4² 𝑥² = 4² - 2² 𝑥² = 16 - 4 Pythagoras – Non right angled triangle 𝑥
𝑥² + 2² = 4² 𝑥² = 4² - 2² 𝑥² = 16 – 4 𝑥 = 12 Pythagoras – Non right angled triangle 𝑥
𝑥² + 2² = 4² 𝑥² = 4² - 2² 𝑥² = 16 – 4 𝑥 = 12 𝑥 = 3.5cm (1.d.p) Pythagoras – Non right angled triangle 𝑥
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
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
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
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
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
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
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
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
Examination Questions
• V = Area of the cross section x depth Examination Questions
• V = Area of the cross section x depth. • We need to find the area of the cross section. Examination Questions
• V = Area of the cross section x depth. • We need to find the area of the cross section. • It’s a triangle!!! Examination Questions
• 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
• 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
• 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
• Volume We now need to put the answer into the Volume formula. Examination Questions
• Volume • V = ½(4.5x2.8) x 3.5 We now need to put the answer in to the Volume formula. Examination Questions
• 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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
We can also use Pythagoras’ Theorem to calculate the length of a diagonal line on a graph. Pythagoras on a Graph
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
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
P1: (1,3) P2: (7,8) Vertical = y2 – y1 Horizontal = x2 – x1 Pythagoras on a Graph
P1: (1,3) P2: (7,8) Vertical = 8 – 3 Horizontal = 7 – 1 Pythagoras on a Graph
P1: (1,3) P2: (7,8) Vertical = 8 – 3 Horizontal = 7 – 1 Pythagoras on a Graph
P1: (1,3) P2: (7,8) Vertical = 8 – 3 = 5 Horizontal = 7 – 1 = 6 Pythagoras on a Graph
P1: (1,3) P2: (7,8) Vertical = 8 – 3 = 5 Horizontal = 7 – 1 = 6 Now we can use Pythagoras’ Theorem. Pythagoras on a Graph
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
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
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
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
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
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
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
Pythagoras on a Graph
P1 = (3,2) P2 = (9,7) Distance = 9 − 3 2 + (7 − 2)² = = Pythagoras on a Graph
P1 = (3,2) P2 = (9,7) Distance = 9 − 3 2 + (7 − 2)² = 36 + 25 = Pythagoras on a Graph
P1 = (3,2) P2 = (9,7) Distance = 9 − 3 2 + (7 − 2)² = 61 = Pythagoras on a Graph
P1 = (3,2) P2 = (9,7) Distance = 9 − 3 2 + (7 − 2)² = 61 = 7.81cm (2.d.p) Pythagoras on a Graph
Pythagoras on a Graph
P1 = (4,-4) P2 = (-5,3) Distance = = = Pythagoras on a Graph
P1 = (4,-4) P2 = (-5,3) Distance = −5 − 4 2 + (3 − −4)² = = Pythagoras on a Graph
P1 = (4,-4) P2 = (-5,3) Distance = −5 − 4 2 + (3 − −4)² = −9 ² + 7² = Pythagoras on a Graph
P1 = (4,-4) P2 = (-5,3) Distance = −5 − 4 2 + (3 − −4)² = 130 = Pythagoras on a Graph
P1 = (4,-4) P2 = (-5,3) Distance = −5 − 4 2 + (3 − −4)² = 130 = 11.40cm (2.d.p) Pythagoras on a Graph
P1 = P2 = Distance = = = Pythagoras on a Graph
P1 = (4,-4) P2 = (2,5) Distance = 2 − 4 2 + (5 − −4)² = = Pythagoras on a Graph
P1 = (4,-4) P2 = (2,5) Distance = 2 − 4 2 + (5 − −4)² = 4 + 81 = Pythagoras on a Graph
P1 = (4,-4) P2 = (2,5) Distance = 2 − 4 2 + (5 − −4)² = 85 = Pythagoras on a Graph
Pythagoras on a Graph P1 = (4,-4) P2 = (2,5) Distance = 2 − 4 2 + (5 − −4)² = 85 = 9.21cm (2.d.p)
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
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

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Pythagoras Theorem Graphs

  • 1.
    How can weapply Pythagoras’ Theorem in graphs and other triangles? Learning Objective
  • 2.
    • To reviseall 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 cuse: 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 cuse: 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 cuse: 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 cuse: 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 cuse: 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 • Torevise 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 • Torevise 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.
  • 11.
    Edexel Pythagoras Question Weuse Pythagoras’ Theorem to calculate the distance between Norwich and York.
  • 12.
    Edexel Pythagoras Question Weuse Pythagoras’ Theorem to calculate the distance between Norwich and York. The Hypotenuse is the distance between them: ℎ2 =?2 +?2
  • 13.
    Edexel Pythagoras Question Weuse 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 Weuse 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 Weuse 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 Weuse 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 Weuse 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 Weuse 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 – Nonright 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 thelength 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 thelength 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 thelength 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 thelength 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 thelength 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 thelength 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 thelength 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 thelength 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.
  • 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 nowneed 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 arepresentation 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 arepresentation 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 arepresentation 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 arepresentation 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 needto 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 needto 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 arepresentation 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 arepresentation 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 arepresentation 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 arepresentation 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 arepresentation 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 arepresentation 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 arepresentation 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 arepresentation 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 arepresentation 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 • Torevise 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 • Torevise 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 alsouse Pythagoras’ Theorem to calculate the length of a diagonal line on a graph. Pythagoras on a Graph
  • 69.
    We can alsouse 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 alsouse 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 betweenthe 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 betweenthe 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 betweenthe 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 betweenthe 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.
  • 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.
  • 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 aGraph P1 = (4,-4) P2 = (2,5) Distance = 2 − 4 2 + (5 − −4)² = 85 = 9.21cm (2.d.p)
  • 99.
    Learning Outcomes • Torevise 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 • Torevise 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

Editor's Notes

  • #2 An overview of the content of the lesson Must be in the form of a question where appropriate Students should be able to answer the question at the end - either fully, partly or in a way that demonstrates they understand what gaps in their knowledge they need to address Verbs such as to understand / to know / to gain confidence / to learn Ask students to give the question a go and point out that, at the end of the lesson, they should be able to answer fully
  • #3 Measureable outcomes that students can demonstrate and self-assess against Must be written using Bloom’s taxonomy verbs Verbs based on students ability and pitch of lesson It must be clear that students understand the outcomes before moving on Make an activity of this slide: Ask students to read this aloud Ask them to paraphrase Ask that they explain what they mean Ask what they already know related to these outcomes There may be as few as 2 outcomes, or max 4
  • #9 Revisit the first outcome and use the polling function to allow students to privately self-assess You may feel that the students do not need privacy to self-assess and in this instance, the chat box may be used Polling must be used until you can fully assess their confidence to use the chat box and express honesty If students self-assess as a 4/5, ensure that you are fully confident in their assessment Ask questions Ask for examples Students to ask each other questions If a few students self-assesses as a 3, but others as a 4/5, discretely ask the higher ones to give examples and to explain their achievement/understanding If all students are a 3 or below, do not move on. Move to a blank page at the end of the presentation and use as a whiteboard to further explain If students are ½, go back to the beginning Always ask students what the gaps are and help them to identify these in order to promote metacognition
  • #10 1. The outcome changes colour when achieved to the same colour as the objective to demonstrate the connection, progress and what happens next
  • #67 1. The outcome changes colour when achieved to the same colour as the objective to demonstrate the connection, progress and what happens next
  • #68 1. The outcome changes colour when achieved to the same colour as the objective to demonstrate the connection, progress and what happens next
  • #69 Explanation and use the writing pad.
  • #70 Explanation and use the writing pad.
  • #71 Explanation and use the writing pad.
  • #72 Explanation and use the writing pad.
  • #73 Explanation and use the writing pad.
  • #74 Explanation and use the writing pad.
  • #75 Explanation and use the writing pad.
  • #76 Explanation and use the writing pad.
  • #77 Explanation and use the writing pad.
  • #78 Explanation and use the writing pad.
  • #79 Explanation and use the writing pad.
  • #80 Explanation and use the writing pad.
  • #81 Explanation and use the writing pad.
  • #82 Explanation and use the writing pad.
  • #83 Explanation and use the writing pad.
  • #84 Explanation and use the writing pad.
  • #85 Explanation and use the writing pad.
  • #86 Explanation and use the writing pad.
  • #87 Explanation and use the writing pad.
  • #88 Explanation and use the writing pad.
  • #89 Explanation and use the writing pad.
  • #90 Explanation and use the writing pad.
  • #91 Explanation and use the writing pad.
  • #92 Explanation and use the writing pad.
  • #93 Explanation and use the writing pad.
  • #94 Explanation and use the writing pad.
  • #95 Explanation and use the writing pad.
  • #96 Explanation and use the writing pad.
  • #97 Explanation and use the writing pad.
  • #98 Explanation and use the writing pad.
  • #99 Explanation and use the writing pad.
  • #100 1. The outcome changes colour when achieved to the same colour as the objective to demonstrate the connection, progress and what happens next
  • #101 1. The outcome changes colour when achieved to the same colour as the objective to demonstrate the connection, progress and what happens next