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Brief History Finding the shorter side A Pythagorean Puzzle Pythagoras’ Theorem Using Pythagoras’ Theorem Menu Further examples.

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Presentation on theme: "Brief History Finding the shorter side A Pythagorean Puzzle Pythagoras’ Theorem Using Pythagoras’ Theorem Menu Further examples."— Presentation transcript:

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3 Brief History Finding the shorter side A Pythagorean Puzzle Pythagoras’ Theorem Using Pythagoras’ Theorem Menu Further examples

4 Pythagoras was a Greek philosopher and religious leader. He was responsible for many important developments in maths, astronomy, and music. Pythagoras (~ B.C.)

5 His students formed a secret society called the Pythagoreans. As well as studying maths, they were a political and religious organisation. Members could be identified by a five pointed star they wore on their clothes. The Secret Brotherhood

6 They had to follow some unusual rules. They were not allowed to wear wool, drink wine or pick up anything they had dropped! Eating beans was also strictly forbidden! The Secret Brotherhood

7 A right angled triangle A Pythagorean Puzzle Ask for the worksheet and try this for yourself! © R Glen 2001

8 Draw a square on each side. A Pythagorean Puzzle © R Glen 2001

9 x y z Measure the length of each side A Pythagorean Puzzle © R Glen 2001

10 Work out the area of each square. A Pythagorean Puzzle x z y x² y² z² © R Glen 2001

11 A Pythagorean Puzzle x² y² z² © R Glen 2001

12 A Pythagorean Puzzle © R Glen 2001

13 1  A Pythagorean Puzzle © R Glen 2001

14 1 2  A Pythagorean Puzzle © R Glen 2001

15 1 2  A Pythagorean Puzzle © R Glen 2001

16 1 2 3  A Pythagorean Puzzle © R Glen 2001

17 1 2 3  A Pythagorean Puzzle © R Glen 2001

18  A Pythagorean Puzzle © R Glen 2001

19 A Pythagorean Puzzle © R Glen 2001

20 A Pythagorean Puzzle © R Glen 2001

21 What does this tell you about the areas of the three squares? The red square and the yellow square together cover the green square exactly. The square on the longest side is equal in area to the sum of the squares on the other two sides. A Pythagorean Puzzle © R Glen 2001

22 Put the pieces back where they came from. A Pythagorean Puzzle © R Glen 2001

23 A Pythagorean Puzzle Put the pieces back where they came from. © R Glen 2001

24 A Pythagorean Puzzle Put the pieces back where they came from. © R Glen 2001

25 A Pythagorean Puzzle Put the pieces back where they came from. © R Glen 2001

26 A Pythagorean Puzzle Put the pieces back where they came from. © R Glen 2001

27 A Pythagorean Puzzle Put the pieces back where they came from. © R Glen 2001

28 This is called Pythagoras’ Theorem. A Pythagorean Puzzle x² y² z² x²=y²+z² © R Glen 2001

29 It only works with right-angled triangles. hypotenuse The longest side, which is always opposite the right-angle, has a special name: This is the name of Pythagoras’ most famous discovery. Pythagoras’ Theorem

30 x z y x²=y²+z² Pythagoras’ Theorem

31 x y x x y y z z z x y z x²=y²+z²

32 1m 8m Using Pythagoras’ Theorem What is the length of the slope?

33 1m 8m x z= y= x²=y²+ z² x²=1²+ 8² x²= x²=65 ? Using Pythagoras’ Theorem

34 How do we find x? We need to use the square root button on the calculator. It looks like this √ Press x²=65 √, Enter 65 = So x= √ 65 = 8.1 m (1 d.p.) Using Pythagoras’ Theorem

35 Example 1 x 12cm 9cm y z x²=y²+ z² x²=12²+ 9² x²= x²= 225 x = √ 225= 15cm

36 x 6m 4m s y z x²=y²+ z² s²=4²+ 6² s²= s²= 52 s = √ 52 =7.2m (1 d.p.) Example 2

37 What’s in the box? 24cm 7cm 25 cm 7m 5m 8.6 m to 1 dp Problem 1 Problem 2

38 7m 5m h x y z x²=y²+ z² 7²=h²+ 5² 49=h² + 25 ? Finding the shorter side

39 49 = h² + 25 We need to get h² on its own. Remember, change side, change sign! Finding the shorter side = h² h²= 24 h = √ 24 = 4.9 m (1 d.p.)

40 169 = w² + 36 x w 6m 13m y z x²= y²+ z² 13²= w²+ 6² 169 – 36 = w² w = √ 133 = 11.5m (1 d.p.) w²= 133 Example = w² + 36 Change side, change sign!

41 x z x²= y²+ z² 11²= 9²+ PQ² 121 = 81 + PQ² 121 – 81 = PQ² PQ = √ 40 = 6.3cm (1 d.p.) PQ²= 40 y 9cm P 11cm R Q Example 2 81 Change side, change sign!

42 What’s in the box 2? 9cm A 11cm C B 6.3 cm to 1 dp 9m 4.5m 7.8m to 1 dp Problem 1 Problem 2

43 x y z x²=y²+ z² r²=5²+ 7² r²= r²= 74 r = √ 74 =8.6m (1 d.p.) 14m 5m r r 7m Example 1 ½ of 14 ?

44 x y z 23cm 38cm p 23cm x²= y²+ z² 38²= y²+ 23² 1444 = y² – 529 = y² y = √ 915=30.2 y²= 915 So p =2 x 30.2 = 60.4cm Example Change side, change sign!

45 What’s in the box 3? 20m 8m r r = 12.8m to 1 dp 30cm 42cm p p =58.8m to 1 dp Problem 1 Problem 2


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