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

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

a 2 = b 2 = c 2 = So, we guess that a 2 + b 2 = c 2 c = 5 a = 3 b = 4 To guess the relationship among the hypotenuse and the two sides

Topics of discussion: Pythagoras’ Theorem

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, then c 2 = a 2 + b 2 a b c

Historical Background

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

百牛定理

Proof of Pythagoras’ Theorem

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

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

Typical Examples

Example 1. Find the length of AC. Hypotenuse AC 2 = (Pythagoras’ Theorem) AC 2 = AC 2 = 400 AC = 20 A CB Solution :

Example 2. Find the length of QR. Hypotenuse 25 2 = QR 2 (Pythagoras’ Theorem) QR 2 = QR 2 = 49 QR= 7 R Q P Solution :

Classwork 1 (Worksheets Given)

a 2 = (Pythagoras’ Theorem) 1. Find the value of a a Solution:

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

3. Find the value of c. Solution: 25 2 = c 2 (Pythagoras’ Theorem) 25 7 c

4. Find the length of diagonal d d Solution: d 2 = (Pythagoras’ Theorem)

5. Find the length of e. e Solution: 85 2 = e (Pythagoras’ Theorem)

Applications of Pythagoras’ Theorem to Long Questions

16km 12km 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. N ? Application of Pythagoras’ Theorem

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 = AC 2 = 400 AC = 20 The displacement between the starting point and the destination point of the car is 20 km

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.

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

Classwork 2

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 = L 2 (Pythagoras’ Theorem) L 2 = L 2 = 144 L = 12 Find the length of the shadow L. 5 m 13 m L

Summary

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