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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. - PowerPoint PPT Presentation
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Pythagoras’ Theorem
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
c2 = a2+ b2
a
b
c
Historical
Background
Pythagoras’ Theorem
Pythagoras (~580-500 B.C.)He was a Greek philosopher responsible for important developments in mathematics, astronomy and the theory of music.
Proof of
Pythagoras’ Theorem
Consider a square PQRS with sides a + ba
a
a
a
b
b
b
b cc
cc
Now, the square is cut into
- 4 congruent right-angled triangles and
- 1 smaller square with sides c
a + b
a + b
A B
CD
Area of square ABCD
= (a + b) 2
b
b
a b
b
a
a
a
cc
cc
P Q
RS
Area of square PQRS
= 4 + c 2
ab
2
a 2 + 2ab + b 2 = 2ab + c 2
a2 + b2 = c2
Typical Examples
Example 1. Find the length of AC.
Hypotenuse
AC2 = 122 + 162 (Pythagoras’ Theorem)
AC2 = 144 + 256
AC2 = 400
AC = 20
A
CB
16
12Solution :
Example 2. Find the length of QR.
Hypotenuse
252 = 242 + QR2 (Pythagoras’ Theorem)
QR2 = 625 - 576
QR2 = 49
QR = 7
R
Q
P
25
24
Solution :
Further Questions
a2 = 52 + 122 (Pythagoras’ Theorem)
a
5 12
13
2 2
169
1. Find the value of a.
5
12a
Solution:
2. Find the value of b .
Solution:
102 = 62 + b2 (Pythagoras’ Theorem)
b
10 6
8
2 2
64
6
10
b
3. Find the value of c .
Solution:
252 = 72 + c2 (Pythagoras’ Theorem)
c
25 7
24
2 2
576
25
7 c
4. Find the length of diagonal d .
10
24 d
Solution:
d2 = 102 + 242 (Pythagoras’ Theorem)
d
10 24
26
2 2
676
5. Find the length of e .
e
84 85
Solution:
852 = e2 + 842 (Pythagoras’ Theorem)
e
85 84
13
2 2
169
Contextual Pythagoras’ 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
AB
C
Solution :
In the figure,AB = 16BC = 12
AC2 = AB2 + BC2 (Pythagoras’ Theorem)AC2 = 162 + 122
AC2 = 400AC = 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 theright-angled triangle ABC.
AB = 200BC = 160
AB2 = AC2 + BC2 (Pythagoras’ Theorem)2002 = AC2 + 1602
AC2 = 14400AC = 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
BC
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:132 = 52 + L2 (Pythagoras’ Theorem)L2 = 132 - 52
L2 = 144L = 12
Find the length of the shadow L.
5 m13 m
L
Summary
Summary ofSummary of
Pythagoras’ TheoremPythagoras’ Theorem
a
bc
c b2 2 = a 2
For any right-angled triangle,
