2. Trigger Activity

3. Topic of discussion: Pythagoras’ Theorem

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

6. Proof of Pythagoras’Theorem Student Activity

8. One more Proof & demonstration of Pythagoras’Theorem

9. Watch this !

10. In the right angled triangle ABC, can you spot two other triangles that are similar to it ? By comparing the ratios of the corresponding lengths of the 2 similar triangles, we can lead to the proof that : BC 2 = AB 2 + AC 2 (Pythagoras’ Theorem) Proof using Similar Triangles

11. Application of Pythagoras’ Theorem

12. Locked Out & Breaking In You’re locked out of your house and the only open window is on the second floor, 4 metres above the ground. You need to borrow a ladder from your neighbour. There’s a bush along the edge of the house, so you’ll have to place the ladder 3 metres from the house. What length of ladder do you need to reach the window ?

14. Summary of Pythagoras’ Theorem a b c For any right-angled triangle, c 2 = a 2 + b 2

15. Worksheet Practice

16. Qn 1 : Find the length of AC. Hypotenuse A CB 16 12 Solution : AC 2 = 12 2 + 16 2 (Pythagoras’ Theorem) AC= AC = 20

17. Qn 2 : Find the length of QR. Hypotenuse R Q P 25 24 Solution : 25 2 = 24 2 + QR 2 (Pythagoras’ Theorem) QR 2 = 25 2 - 24 2 QR= QR= 7

18. a 2 = 5 2 + 12 2 (Pythagoras’ Theorem) Qn 3 : Find the value of a. 5 12 a Solution :

19. Qn 4 : Find the value of b. Solution: 10 2 = 6 2 + b 2 (Pythagoras’ Theorem) 6 10 b

20. Qn 5 : Find the value of c. Solution: 25 2 = 7 2 + c 2 (Pythagoras’ Theorem) 25 7 c

21. Qn 6 : Find the length of diagonal d. 10 24 d Solution: d 2 = 10 2 + 24 2 (Pythagoras’ Theorem)

22. Qn 7 : Find the length of e. e 84 85 Solution: 85 2 = e 2 + 84 2 (Pythagoras’ Theorem)

23. Applications of Pythagoras’ Theorem to Word Problems

24. 16km 12km A car travels 16 km from east to west. Then it turns left and travels a further 12 km. Find the distance between the starting point and the destination point of the car. N ? Example 1

25. 16 km 12 km A B C Solution : In the figure, AB = 16 BC = 12 AC 2 = AB 2 + BC 2 (Pythagoras’ Theorem) AC 2 = 16 2 + 12 2 AC 2 = 400 AC = 20 The distance between the starting point and the destination point of the car is 20 km

26. 160 m 200 m 1.2 m ? 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. Example 2

27. Solution : In the figure, consider the right-angled triangle ABC. AB = 200 BC = 160 AB 2 = AC 2 + BC 2 (Pythagoras’ Theorem) 200 2 = AC 2 + 160 2 AC 2 = 14400 AC = 120 So, the height of the kite above the ground = AC + Peter’s height = 120 + 1.2 = 121.2 m 160 m 200 m 1.2 m A B C

28. The height of a tree is 5 m. The distance between the top of it and the tip of its shadow is 13 m. Solution: 13 2 = 5 2 + L 2 (Pythagoras’ Theorem) L 2 = 13 2 - 5 2 L 2 = 144 L = 12 Find the length of the shadow L. 5 m 13 m L Example 3