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Trigger Activity

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

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Trigger Activity

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  1. Trigger Activity

  2. Topic of discussion: Pythagoras’ Theorem

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

  4. Proof of Pythagoras’Theorem Student Activity

  5. One more Proof & demonstration of Pythagoras’Theorem

  6. Watch this !

  7. Proof using Similar Triangles 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 : BC2 = AB2 + AC2(Pythagoras’ Theorem)

  8. Application of Pythagoras’ Theorem

  9. 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 ?

  10. For any right-angled triangle, c b c2 = a2 + b2 a Summary of Pythagoras’ Theorem

  11. Worksheet Practice

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

  13. P 24 R 25 Hypotenuse Q 252 = 242 + QR2 (Pythagoras’ Theorem) QR2 = 252 - 242 QR = QR = 7 Qn 2:Find the length of QR. Solution:

  14. a 12 5 Qn 3:Find the value of a. Solution : a2 = 52 + 122 (Pythagoras’ Theorem)

  15. 6 b 10 Qn 4:Find the value of b . Solution: 102 = 62 + b2 (Pythagoras’ Theorem)

  16. c 7 25 Qn 5 : Find the value of c . Solution: 252 = 72 + c2 (Pythagoras’ Theorem)

  17. 24 d 10 Qn 6 :Find the length of diagonal d . Solution: d2 = 102 + 242 (Pythagoras’ Theorem)

  18. 852 = e2 + 842 (Pythagoras’ Theorem) 85 84 e Qn 7 : Find the length of e . Solution:

  19. Applications of Pythagoras’ Theorem to Word Problems

  20. Example 1 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. 16km N 12km ?

  21. 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 distance between the starting point and the destination point of the car is 20 km

  22. Example 2 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

  23. 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

  24. 13 m 5 m Example 3 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

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