1. 11 Pythagoras’ Theorem Case Study
11.1 Pythagoras’ Theorem and Its Proofs
11.2 Applications of Pythagoras’ Theorem
11.3 Converse of Pythagoras’ Theorem and
Its Applications
11.4 Surds and Irrational Numbers
Chapter Summary

2. Case Study
We have to know the height of the slide and the horizontal distance between the top and the bottom of the slide.
How can we find the length of the slide?
In the figure, y is the length of the slide, h is the height of the slide and x is the horizontal distance between the top and the bottom of the slide.
By using Pythagoras’ theorem, we know that y2  h2  x2.
As a result, we can find the length of the slide once we know the values of x and h.

3. 11.1 Pythagoras’ Theorem and Its Proofs
A. Pythagoras’ Theorem
The figure shows a right-angled triangle ABC, where C  90.
AB is the longest side of the right-angled triangle.
It is called the hypotenuse (the side opposite to the right angle) of the triangle.
Moreover, the side opposite to an angle is usually named by its corresponding small letter:
 a : opposite side BC of A
 b : opposite side AC of B
 c : opposite side AB of C

4. 11.1 Pythagoras’ Theorem and Its Proofs
A. Pythagoras’ Theorem
An important relationship among the 3 sides of a right-angled triangle:  
In a right-angled triangle, the sum of the squares of the 2 shorter sides is equal to the square of the hypotenuse. That is, in DABC, if C  90, then a2  b2  c2.
(Reference: Pyth. theorem)
The above result is known as the Pythagoras’ theorem.  
We are going to discuss some methods in proving the Pythagoras’ theorem in the next section.

5. Example 11.1T 11.1 Pythagoras’ Theorem and Its Proofs Solution:
A. Pythagoras’ Theorem
Example 11.1T
In DXYZ, Y  90, XY  16 and XZ  34. Find the value of a.
Solution:
In DXYZ,
XY 2  YZ 2  XZ 2 (Pyth. theorem)
162  a2  342
a2  1156  256
 900
∴ a 
 30

6. Example 11.2T 11.1 Pythagoras’ Theorem and Its Proofs Solution:
A. Pythagoras’ Theorem
Example 11.2T
In DXYZ, X  90, XY  2 and YZ  3. Find the value of f. (Give the answer in surd form.)
Solution:
In DXYZ,
XY 2  XZ 2  YZ 2 (Pyth. theorem)
22  f 2  32
f 2  9  4
 5
∴ f 

7. Example 11.3T 11.1 Pythagoras’ Theorem and Its Proofs Solution:
A. Pythagoras’ Theorem
Example 11.3T
In the figure, DXYZ is a right-angled triangle with Y  90, XY  cm, WZ  8 cm and XZ  11 cm.
(a) Find YW and XW.
(b) Find the perimeter of DXWZ.
Solution:
(a) In DXYZ,
XY 2  YZ 2  XZ 2 (Pyth. theorem)
∴ YZ  cm
 cm
 10 cm
∴ YW  (10  8) cm
 2 cm

8. Example 11.3T 11.1 Pythagoras’ Theorem and Its Proofs Solution:
A. Pythagoras’ Theorem
Example 11.3T
In the figure, DXYZ is a right-angled triangle with Y  90, XY  cm, WZ  8 cm and XZ  11 cm.
(a) Find YW and XW.
(b) Find the perimeter of DXWZ.
Solution:
(a) In DWXY,
YW 2  XY 2  XW 2 (Pyth. theorem)
∴ XW  cm
 cm
 5 cm
(b) Perimeter of DXWZ  XW  WZ  XZ
 (5  8  11) cm
 24 cm

9. 11.1 Pythagoras’ Theorem and Its Proofs
B. Different Proofs of Pythagoras’ Theorem
There is evidence that the ancient Chinese and Indians applied the concept of the Pythagoras’ theorem in early time.
However, Pythagoras was the first to discover the theorem by geometric proof.
Some methods in proving the Pythagoras’ theorem:
Pythagoras probably proved the theorem in another way. Please refer to Enrichment Mathematics, p.205 of Book 2B for details.
 Proved in Ancient Greece
 By Zhao Shuang in Ancient China
 By James Garfield in the United States

10. 11.1 Pythagoras’ Theorem and Its Proofs
B. Different Proofs of Pythagoras’ Theorem
(a) Proved in Ancient Greece
Each figure has 4 identical right-angled triangles (in purple colour)
∵ The areas of the 2 figures are equal.
∴ a2  b2  c2

11. 11.1 Pythagoras’ Theorem and Its Proofs
B. Different Proofs of Pythagoras’ Theorem
(b) By Zhao Shuang in Ancient China
In ancient China, mathematicians proposed the Gougu theorem（勾股定理）to express the relationship among the 3 sides of a right-angled triangle, where the base and the height of the triangle were named as Gou（勾）and Gu（股）respectively.
In about 350 AD, Zhao Shuang used the Yuan-Dao（弦圖）to prove the theorem in the Chou Pei Suan Ching《周脾算經》.

12.  4   11.1 Pythagoras’ Theorem and Its Proofs
B. Different Proofs of Pythagoras’ Theorem
(b) By Zhao Shuang in Ancient China
A Yuan-Dao consists of 4 identical right-angled triangles with sides a, b, c and a square with side (b  a).
 4  
∵ Area of ABCD  4  Area of DADH  Area of EFGH
∴ c2  4  ab  (b  a)2
 2ab  b2  2ab  a2
 a2  b2

13. 11.1 Pythagoras’ Theorem and Its Proofs
B. Different Proofs of Pythagoras’ Theorem
(c) By James Garfield in the United States
The 20th president of the United States, James Garfield (1831  1881) developed his own proof in The Journal of Education (Volume 3 issue 161) in 1876:
In the figure, trapezium ABCD is formed by 2 congruent right-angled triangles ABE and ECD, and an isosceles right-angled triangle AED.
∵ Area of trapezium  2  Area of DABE  Area of DAED
∴ (a  b)2  2  ab  c2
(a  b)2  2ab  c2
a2  b2  c2

14. 11.2 Applications of Pythagoras’ Theorem
In our daily lives, we often come across problems that can be modelled as right-angled triangles.
Pythagoras’ theorem can be used to solve problems involving right-angled triangles.

15. Example 11.4T 11.2 Applications of Pythagoras’ Theorem Solution:
A ladder of length 3.2 m leans against a vertical wall. The distance of its top from the ground is the same as the distance of its foot from the wall. Find the distance of the bottom of the ladder from the wall. (Give the answer correct to 1 decimal place.)
Solution:
Let x m be the distance of bottom of the ladder from the wall.
x 2  x 2  3.22 (Pyth. theorem)
2x2  10.24
x2  5.12
∴ x 
 2.3 (cor. to 1 d. p.)
∴ The distance of the bottom of the ladder from the wall is 2.3 m.

16. Example 11.5T 11.2 Applications of Pythagoras’ Theorem Solution:
The figure shows a trapezium with AD // BC, A  B  90. If AB  12 cm, AD  15 cm and BC  20 cm, find the perimeter of the trapezium. E
Solution:
As shown in the figure, construct a line DE such that DE  BC.
∴ DE  12 cm and EC  5 cm
In DCDE,
CE 2  DE 2  CD 2 (Pyth. theorem)
∴ CD  cm
 13 cm
∴ Perimeter of the trapezium  AB  BC  CD  DA
 (12  20  13  15) cm
 60 cm

17. Example 11.6T 11.2 Applications of Pythagoras’ Theorem Solution:
Peter and Lily left school at 4:00 p.m. Peter walked east at a speed of 2.4 m/s to reach the library at 4:15 p.m. Lily walked north at a speed of 2.25 m/s to reach the bookstore at 4:12 p.m.
(a) How far did each of them walk?
(b) Find the distance between the library and the bookstore.
Solution:
(a) Distance travelled by Peter
 (2.4  15  60) m
Distance travelled by Lily
 (2.25  12  60) m
 2160 m
 1620 m
(b) As shown in the figure,
AB 2  AC 2  BC 2 (Pyth. theorem)
∴ BC  m
 2700 m
∴ The distance between the library and the bookstore is 2700 m.

18. 11.3 Converse of Pythagoras’ Theorem and Its Applications
In the previous section, we learnt the Pythagoras’ theorem.
In fact, the converse of Pythagoras’ theorem is also true and is stated below:
In a triangle, if the sum of the squares of the 2 shorter sides is equal to the square of the longest side, then the triangle is a right-angled triangle with the right angle opposite to the longest side.
That is, in DABC, if a2  b2  c2,
then C  90.
(Reference: converse of Pyth. theorem)

19. 11.3 Converse of Pythagoras’ Theorem and Its Applications
Example 11.7T
In the figure, XZ  25, YZ  30 and XW  20. W is the mid-point of YZ.  
(a) Prove that XWZ  90.  
(b) Prove that DXYZ is an isosceles triangle.
Solution:
(a) In DXWZ,
(b) In DXWY,
WZ  30  2
 15
WY  15
XWY  90
WX 2  WZ 2  152  202
WX 2  WY 2  XY 2 (Pyth. theorem)
 225  400
XY 
 625
 25
XZ 2  252
 XZ
 625
∴ DXYZ is an isosceles triangle.
∵ WX 2  WZ 2  XZ 2
∴ XWZ  90 (converse of Pyth. theorem)

20. 11.3 Converse of Pythagoras’ Theorem and Its Applications
Example 11.8T
The figure shows a piece of triangular paper. It is known that XW  YZ, XW  12 cm, YW  9 cm and WZ = 16 cm.  
(a) Find XY and XZ.  
(b) Prove that the paper is in the shape of
a right-angled triangle.
Solution:
(a) In DWXY,
In DWXZ,
WX 2  WY 2  XY 2 (Pyth. theorem)
WX 2  WZ 2  XZ 2 (Pyth. theorem)
XY  cm
XZ  cm
 15 cm
 20 cm
(b) In DXYZ,
XY 2  XZ 2  152  202
∵ XY 2  XZ 2  YZ 2
 625
∴ YXZ  90 (converse of Pyth. theorem)
YZ 2  (9  16)2
∴ The paper is in the shape of a right-angled triangle.
 625

21. 11.4 Surds and Irrational Numbers
A. Surds on a Number Line
In Book 1A Chapter 1, we learnt how to represent real numbers on a number line.
We can also represent surds on a number line.
For example, to represent on a number line, first construct a right-angled triangle OAB with OA  1 unit and AB  1 unit.
∴ OB  units
 units C
Using a pair of compasses,
draw an arc with centre O and radius OB to meet the number line at C.
∵ OC  OB (radii)
∴ OC  units

22. Example 11.9T 11.4 Surds and Irrational Numbers Solution:
A. Surds on a Number Line
Example 11.9T
Represent on a number line.
Solution:
Step 1: Consider the number 13 as the sum of 2 perfect squares,
i.e., 32  22  13.
Step 2: Construct a right-angled triangle with OA  3 units and AB  2 units.
Step 3: Draw an arc with centre O and radius OB to meet the number line at C. We have OC  C

23. Example 11.10T 11.4 Surds and Irrational Numbers Solution:
A. Surds on a Number Line
Example 11.10T
Represent on a number line.
Solution:
Step 1: Consider the number 11 as the sum of perfect squares,
i.e., 32  12  12  11.
Step 2: Construct a right-angled triangle with legs 3 units and 1 unit.
Then mark on the number line.
Step 3: Continue to construct a
right-angled triangle with legs
units and 1 unit. Then
mark on the number line.

24. 11.4 Surds and Irrational Numbers
B. First Crisis of Mathematics
At the time of Pythagoras, people believed that all things could be explained by numbers which are either integers or fractions.
The discovery of shocked the society and lead to the first crisis of mathematics.
After the Pythagoras’ theorem was proved, a follower of Pythagoras, Hippasus of Metapontum (about 500 BC) tried to demonstrate the length of the diagonal of a square with side 1 unit by applying the theorem.
However, he found that such length (the square root of 2) was neither an integer nor a fraction.
His discovery went against the belief of the Greeks, especially the Pythagoreans.
They did not accept the existence of irrational numbers.

25. 11.4 Surds and Irrational Numbers
B. First Crisis of Mathematics
Legend reveals that Hippasus was then caught and sentenced to death by drowning.
Although Hippasus was executed, Pythagoras admitted the existence of irrational numbers.
It was only until 2000 years later that the irrational numbers were defined using the concept of rational numbers.

26. Chapter Summary 11.1 Pythagoras’ Theorem
In a right-angled triangle ABC,
a2  b2  c2
(Reference: Pyth. theorem)
There are many different proofs of the Pythagoras’ theorem.
1. Proved in Ancient Greece
2. By Zhao Shuang in Ancient China
3. By James Garfield in the United States

27. 11.2 Applications of Pythagoras’ Theorem
Chapter Summary
11.2 Applications of Pythagoras’ Theorem
In our daily lives, we often come across problems that can be modelled as right-angled triangles.
Such problems can be solved by the Pythagoras’ theorem.

28. 11.3 Converse of Pythagoras’ Theorem and Its Applications
Chapter Summary
11.3 Converse of Pythagoras’ Theorem and Its Applications
In DABC,
if a2  b2  c2,
then C  90.
(Reference: converse of Pyth. theorem)

29. 11.4 Surds and Irrational Numbers
Chapter Summary
11.4 Surds and Irrational Numbers
1. We can represent surds on a number line.
2. The discovery of caused the first crisis of mathematics.

30. Follow-up 11.1 11.1 Pythagoras’ Theorem and Its Proofs Solution:
A. Pythagoras’ Theorem
Follow-up 11.1
In DPQR, P  90, PQ  60 and QR  65. Find the value of y.
Solution:
In DPQR,
PQ 2  PR 2  QR 2 (Pyth. theorem)
602  y2  652
y2  4225  3600
 625
∴ y 
 25

31. Follow-up 11.2 11.1 Pythagoras’ Theorem and Its Proofs Solution:
A. Pythagoras’ Theorem
Follow-up 11.2
In DPQR, R  90, PQ  4 and QR  Find the value of d.
Solution:
In DPQR,
PR 2  QR 2  PQ 2 (Pyth. theorem)
d 2  (2 )2  42
d 2  12  16
d 2  4
∴ d 
 2

32. Follow-up 11.3 11.1 Pythagoras’ Theorem and Its Proofs Solution:
A. Pythagoras’ Theorem
Follow-up 11.3
In the figure, DPQR is a right-angled triangle with R  90. QS : SR  2 : 1. PR  12 cm and PS  13 cm.
(a) Find SQ.
(b) Find the area of DPQR.
Solution:
(a) In DPRS,
PR 2  RS 2  PS 2 (Pyth. theorem)
∴ RS  cm
(b) Area of DPQR
 cm
 5 cm
∵ QS : SR  2 : 1
∴ SQ  10 cm

33. Follow-up 11.4 11.2 Applications of Pythagoras’ Theorem Solution:
A ladder of length 2 m leans against a vertical wall. Its foot is 1.6 m away from the wall. Find the vertical distance of the top of the ladder from the ground.
Solution:
Let x m be the vertical distance of the top of the ladder from the ground.
1.6 2  x 2  22 (Pyth. theorem)
x  
 1.2
∴ The vertical distance of the top of the ladder from the ground is 1.2 m.

34. Follow-up 11.5 11.2 Applications of Pythagoras’ Theorem Solution:
There are 2 vertical lampposts standing on a horizontal ground. One post is 6 m tall and the other is 14 m tall. If the distance
between the tops of the lampposts is 17 m, find
the horizontal distance between the 2 lampposts.
Solution:
As shown in the figure, construct a horizontal line such that the difference between the heights is 8 m.
In DABC,
AC 2  BC 2  AB 2 (Pyth. theorem)
∴ BC  m
 15 m
∴ The horizontal distance between the 2 lampposts is 15 m.

35. Follow-up 11.6 11.2 Applications of Pythagoras’ Theorem Solution:
Ships A and B left the port P at the same time. Ship A sailed due west and ship B sailed due south. After 3 hours, they reached ports S and T respectively. Suppose PT  72 km and ST  90 km.
(a) Find the speeds of ships A and B.
(b) After receiving a rescue message from ship A at ports, a helicopter
was sent out from port P at a speed of 81 km/h. How long did it
take to reach port S?
Solution:
(a) In DPST,
PS 2  PT 2  ST 2 (Pyth. theorem)
PS  km
 54 km
∴ Speed of Ship A  54 km  3 hours
 18 km/h
Speed of Ship B  72 km  3 hours
 24 km/h

36. Follow-up 11.6 11.2 Applications of Pythagoras’ Theorem Solution:
Ships A and B left the port P at the same time. Ship A sailed due west and ship B sailed due south. After 3 hours, they reached ports S and T respectively. Suppose PT  72 km and ST  90 km.
(a) Find the speeds of ships A and B.
(b) After receiving a rescue message from ship A at ports, a helicopter
was sent out from port P at a speed of 81 km/h. How long did it
take to reach port S?
Solution:
(b) PS  54 km
∴ Time required  (54  81) hours
 hours
 40 minutes

37. 11.3 Converse of Pythagoras’ Theorem and Its Applications
Follow-up 11.7
In the figure, PR  16, QR  17 and SQ  15. S is the mid-point of PR. Prove that QS is the perpendicular bisector of PR.  
Solution:
In DSQR,
RS  16  2
 8
QS 2  RS 2  152  82
 225  64
 289
QR 2  172
 289
∵ QS 2  RS 2  QR 2
∴ QSR  90 (converse of Pyth. theorem)
∴ QS is the perpendicular bisector of PR.

38. 11.3 Converse of Pythagoras’ Theorem and Its Applications
Follow-up 11.8
The figure shows a farmland PQRS. A partition QS is used to separate the chickens and the pigeons. It is known that PQ  m, QR  12 m, RS  5 m, SP  11 m and R  90.  
(a) Find the length of the partition QS.  
(b) Prove that DPQS is a right-angled triangle.
Solution:
(a) In DQRS,
QR 2  RS 2  QS 2 (Pyth. theorem)
QS  cm
 13 cm
(b) In DPQS,
PQ 2  PS 2  (4 )2  112
∵ PQ 2  PS 2  QS 2
 48  121
∴ QPS  90 (converse of Pyth. theorem)
 169
∴ DPQS is a right-angled triangle.
QS 2  169

39. Follow-up 11.9 11.4 Surds and Irrational Numbers Solution:
A. Surds on a Number Line
Follow-up 11.9
Represent on a number line.
Solution:
Step 1: Consider the number 58 as the sum of 2 perfect squares,
i.e., 72  32  58.
Step 2: Construct a right-angled triangle with OA  7 units and AB  3 units.
Step 3: Draw an arc with centre O and radius OB to meet the number line at C. We have OC  C

40. Follow-up 11.10 11.4 Surds and Irrational Numbers Solution:
A. Surds on a Number Line
Follow-up 11.10
Represent on a number line.
Solution:
Step 1: Consider the number 30 as the sum of perfect squares,
i.e., 52  22  12  30.
Step 2: Construct a right-angled triangle with legs 5 units and 2 units.
Then mark on the number line.
Step 3: Continue to construct a right-angled triangle with legs
units and 1 unit.
Then mark on the number line.