1 / 24

Pythagoras’ Theorem

Pythagoras’ Theorem. 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. c 2 = a 2 + b 2. Historical Background. Pythagoras’ Theorem.

wilmet
Download Presentation

Pythagoras’ Theorem

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Pythagoras’ Theorem

  2. 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

  3. Historical Background

  4. Pythagoras’ Theorem Pythagoras (~580-500 B.C.) He was a Greek philosopher responsible for important developments in mathematics, astronomy and the theory of music.

  5. Proof of Pythagoras’ Theorem

  6. 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

  7. 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

  8. Typical Examples

  9. 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

  10. 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

  11. Further Questions

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

  13. 6 102 = 62 + b2 (Pythagoras’ Theorem) b 10 2. Find the value of b . Solution:

  14. 252 = 72 + c2 (Pythagoras’ Theorem) c 7 25 3. Find the value of c . Solution:

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

  16. 852 = e2 + 842 (Pythagoras’ Theorem) 85 84 e 5. Find the length of e . Solution:

  17. Contextual Pythagoras’ Questions

  18. 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 ?

  19. 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

  20. 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

  21. 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

  22. 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

  23. Summary

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

More Related