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Introduction

Introduction. Students’ activity Topic of discussion: Pythagoras’ Theorem Historical background Proof of Pythagoras’ Theorem Typical examples Classwork 1 (worksheets given) Applications of Pythagoras Theorem to long questions Classwork 2 Homework Summary. Students’ Activity. 0. 0. 1.

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Introduction

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  1. Introduction • Students’ activity • Topic of discussion: Pythagoras’ Theorem • Historical background • Proof of Pythagoras’ Theorem • Typical examples • Classwork 1 (worksheets given) • Applications of Pythagoras Theorem to long questions • Classwork 2 • Homework • Summary

  2. Students’ Activity

  3. 0 0 1 2 1 3 4 2 5 3 0 1 2 3 4 5 4 5 Now take out a square paper and a ruler. 1. cut a triangle with base 4 cm and height 3 cm 5 cm 3 cm 2. measure the length of the hypotenuse 4 cm

  4. a2 = b2 = c2 = 9 16 25 To guess the relationship among the hypotenuse and the two sides c = 5 a = 3 b = 4 So, we guess that a2 + b2 = c2

  5. Topics of discussion: Pythagoras’ Theorem

  6. Hypotenuse - it is the side opposite to the right angle Pythagoras’ Theorem a c b For any right-angled triangle, c is the length of the hypotenuse, a and b are the length of the other 2 sides, then c2 = a2+ b2

  7. Historical Background

  8. Pythagoras’ Theorem 畢氏定理 Pythagoras 畢達哥拉斯(~580-500 B.C.) He was a Greek philosopher responsible for important developments in mathematics, astronomy and the theory of music.

  9. 百牛定理

  10. Proof of Pythagoras’ Theorem

  11. a b a b b a a b Consider a square PQRS with sides a + b c c c c Now, the square is cut into - 4 congruent right-angled triangles and - 1 smaller square with sides c

  12. a + b a b A B P Q a c b c a + b c b c a R S C a D b = 4 + c2 Area of square ABCD Area of square PQRS = (a + b) 2 = a 2 + 2ab + b 2 2ab + c 2 a2 + b2 = c2

  13. Typical Examples

  14. Hypotenuse Example 1. Find the length of AC. A 16 B C 12 Solution : AC2 = 122 + 162 (Pythagoras’ Theorem) AC2 = 144 + 256 AC2 = 400 AC = 20

  15. P 24 R 25 Hypotenuse Q Example 2. Find the length of QR. Solution : 252 = 242 + QR2 (Pythagoras’ Theorem) QR2 = 625 - 576 QR2 = 49 QR = 7

  16. Classwork 1 (Worksheets Given)

  17. a2 = 52 + 122 (Pythagoras’ Theorem) a 12 5 1. Find the value of a. Solution:

  18. 6 102 = 62 + b2 (Pythagoras’ Theorem) b 10 2. Find the value of b . Solution:

  19. 252 = 72 + c2 (Pythagoras’ Theorem) c 7 25 3. Find the value of c . Solution:

  20. d2 = 102 + 242 (Pythagoras’ Theorem) 24 d 10 4. Find the length of diagonal d . Solution:

  21. 852 = e2 + 842 (Pythagoras’ Theorem) 85 84 e 5. Find the length of e . Solution:

  22. Applications of Pythagoras’ Theorem to Long Questions

  23. Application of Pythagoras’ Theorem A car travels 16 km from east to west. Then it turns left and travels a further 12 km.Find the displacement between the starting point and the destination point of the car. 16km N 12km ?

  24. 16 km B A 12 km C Solution : In the figure, AB = 16 BC = 12 AC2 = AB2 + BC2 (Pythagoras’ Theorem) AC2 = 162 + 122 AC2 = 400 AC = 20 The displacement between the starting point and the destination point of the car is 20 km

  25. Peter, who is 1.2 m tall, is flying a kite at a distance of 160 m from a tree. He has released a string of 200 m long and the kite is vertically above the tree. Find the height of the kite above the ground. 200 m ? 1.2 m 160 m

  26. A 200 m C B 1.2 m 160 m Solution : In the figure, consider the right-angled triangle ABC. AB = 200 BC = 160 AB2 = AC2 + BC2 (Pythagoras’ Theorem) 2002 = AC2 + 1602 AC2 = 14400 AC = 120 So, the height of the kite above the ground = AC + Peter’s height = 120 + 1.2 = 121.2 m

  27. Classwork 2

  28. 13 m 5 m The height of a tree is 5 m. The distance between the top of it and the tip of its shadow is 13 m. Find the length of the shadow L. Solution: 132 = 52 + L2 (Pythagoras’ Theorem) L2 = 132 - 52 L2 = 144 L = 12 L

  29. Summary

  30. For any right-angled triangle, c b a Summary of Pythagoras’ Theorem

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