MATEMATIK TINGKATAN 2

Chapter 3 Algebraic Formulae CHAPTER 3

WALKING
THROUGH TIME
Al-Khwarizmi introduced negative and decimal
numbers. He also founded a mathematical
programme using a set of instructions to
complete a complex calculation.
For more information:

http://rimbunanilmu.my/mat_t2e/ms043

WHY STUDY THIS CHAPTER?
The algebraic formulae is applied by engineers,
statisticians, mathematicians and astronomers
in their respective jobs.

43

CHAPTER 3 Chapter 3 Algebraic Formulae

CREATIVE ACTIVITY

Aim: Identifying formula
Material: School Calendar
Steps:
1. Students carry out the activity in pairs.
2. Calculate the amount of money that can be saved from the following situations (assume

that the calculation starts from first to the last day of each month).

Situation 1
Badrul is a form 2 student who likes to save. On each school day, he receives RM5 as his
pocket money and spends RM4.50. What is the amount of Badrul’s savings in January?

Situation 2
Sedthu saves RM15 per month. If he receives RM10 as pocket money, calculate Sedthu’s
expenditure for one day in April.

3. State the method of calculating the savings.

From the situations above, write an equation for the total savings in relation to pocket money,
money spent and to the number of days using basic mathematical operations to get the total amount
of savings. Pocket money, money spent and number of days are variables. You can determine the
amount saved by changing the value of the variables.

3.1 Algebraic Formulae

Algebraic expression is a combination of two or more algebraic terms. The algebraic formulae
combines an algebraic expression using addition, subtraction, multiplication or division and is
written in the form of an equation.

3.1.1 Forming formula LEARNING
STANDARD

COGNITIVE STIMULATION Write a formula based on
Aim: Forming algebraic formulae a situation.
Material: Worksheets
Steps:

1 . Students carry out this activity in groups.
A cultural club will perform at a school-level cultural night. The table shows the number

of dancers according to the types of dance and race represented by an alphabet.

44

Chapter 3 Algebraic Formulae

Types of dance Malay Race Indian DO YOU KNOW CHAPTER 3
a Chinese 2a
Sumazau 2b 5b The Sumazau dance is
Kuda Kepang 2c 2c c known as the traditional
Lion b dance of the Kadazan
3a Dusun tribe in Sabah.
The Sumazau dance is
The alphabet a, b and c are known as variables. performed during the
Tadau Kaamatan festival
2. Write a formula for each of the following subject. and is celebrated every
(a) s, number of Chinese dancers. year in May.
(b) d, number of Kuda Kepang dancers.
(c) w, number of Indian and Malay dancers. http://www.jkkn.gov.my/
pemetaanbudaya/
Discussion:
(i) Difference in formula between the groups in your class.

(ii) Conclusion from the activity above.

The formula is expressed as s = 2c + b + 3a, d = 8b, w = 3a + 7b + 3c. TIPS
From the activity above, the formula is formed by the relationship
among a few variables. In the activity on the left,
s, d and w are subjects
EXAMPLE 1 of formula. They can
be written on the left or
Suzi sold two types of cakes of different prices. The chocolate cake right side.
sold at RM3 a slice. The cheese cake was sold at twice the price of the
chocolate cake. In conjunction with the opening of a new branch, she TIPS
gave 10% discount for all cakes. Determine a formula to calculate the
selling price of the cake, if m slices of chocolate cake and n slices of A variable in a formula can
cheese cake were sold. be represented by letters a
to z (in example 1, m and
Solution: n represents variables). z,
written on the left is known
Price of cheese cake = twice the price of the chocolate cake as subject of formula.
= 2 × RM3
= RM6 THINK SMART!

Selling price, z = �(number of chocolate cake × price) + Is this equation called a
(number of cheese cake × price)� × discount formula?

= �(m × RM3) + (n × RM6)� × (100% − 10%) (i) a × (b + c) = (a × b) + (a × c)
= (RM3m + RM6n) × 90%
= (3m + 6n) × 0.9 (ii) p + q = b
Let, z = Selling price a a a
m = number of chocolate cake
n = number of cheese cake

The algebraic formula is z = 0.9 (3m + 6n)
= 2.7m + 5.4n

Discuss.

45

Chapter 3 Algebraic Formulae

3.1.2 Changing the subject of formula LEARNING
STANDARD
The subject of a formula can be a variable for the algebraic formula
and the variable can be the subject of an algebraic formula. Change the subject of
formula of an algebraic
equation.

CHAPTER 3 bb TIPS

a Coefficient for the subject
of formula must be 1.
Perimeter, P for an equilateral triangle can be expressed in a and b.
Hence, P = a + 2b FLASHBACK

The subject of formula of the equation above can be changed as shown. 1×p =p
−1 × p = −p
(i) a = P – 2b (ii) b = P − a
2

EXAMPLE 2 0×p =0
State m, as the subject of formula.
1 ×p = p
3 3

(a) q = m + p (b) b = 2s – m − 1 ×p =− p
3 3
(c) a = 25m (d) t = m––3 n
TIPS
Solution:

(a) m + p = q (b) 2s – m = b Subject of formula should
m + p – p = q – p 2s –2s – m = b – 2s be on the left side of the
Then, m = q – p – m = b – 2s equation.

m in terms of p and q 1 × (– m) = 1 (b – 2s)
–1 –1

m = –b + 2s

m in terms of b and s Then, m = 2s – b

(c) a = 5 (d) m– n = t FLASHBACK
2m –3
You have learned to solve
a × 2m = 5 × 2m m–n × –31 = t × (–3) linear equation using the
2m 1 –3 following three methods:
1 1 (a) Trial and improvement
(b) Application of equality
2am = 5 m – n = – 3t concept
(c) Back track
1 2am = 5 m – n + n = – 3t + n
2a 2a m = – 3t + n
1
5 Then, m = n – 3t
Then, m = 2a m in terms of a
m in terms of n and t

46

EXAMPLE 3 Chapter 3 Algebraic Formulae

State p, as the subject of formula. TIPS
(�a 2)2 = a2
(a) q = ­ p (b) s = p2 �a 2 = a

(c) w = p (d) t = 1 CHAPTER 3
3 p2
ATTENTION

Solution: �x = x 1
(a) p = q 2

( p )2 = (q)2 Both sides of (b) p2 = s 1
p = q2 the equation are
squared p2 = s (�x )2 = (x 2)2
p = s
= x 1 × 2
2

= x

(c) p = w (d) t = 1 ATTENTION
3 p2

p 2 t × p2 =1 1 × p2 Reciprocals
3 p2
� � = w2 1 = a, x= 1
x a
1
tp2 = 1
p = w2 Squares
3 p2 = 1
t (�x )2 = a2, x = a2
p
13 × 3 = w 2 ×3 1 Square root
t
1 p = √x2 = √a, x = ±√a

p = 3w 2

3.1.3 Determining the value of variable LSETAANRNDAINRGD

The value of a subject of a formula can be obtained when all Determine the value of
variable values are given. On the other hand, the value of a a variable when the
variable can be obtained when the value of subject of the formula value of another variable
and variable is given. is given.

EXAMPLE 4 FLASHBACK

Given w = 7t – 5u, calculate the following. −a + a = 0
(a) value w when t = 3 and u = –2 −a − a = −2a
(b) value t when w = 15 and u = 4 −a × a = −a2
(− a ) × (− a ) = a2
Solution: −a ÷ a = −1
(−a) ÷ (−a) = 1
(a) Substitute t = 3 and u = –2 into the formula.
w = 7(3) − 5(−2) 47
= 21 + 10
= 31

Chapter 3 Algebraic Formulae

(b) Substitute w = 15 and u = 4 into the formula. FLASHBACK

7t – 5u = w Algebraic formulae

7t – 5 (4) = 15 Variables
Variables is a quantity where
7t = 15 + 20 the value is not known or can be
changed.
t = 35
7 Constant
CHAPTER 3 t = 5 Constant is a quantity where the
value is fixed.
EXAMPLE 5
Algebraic formulae
Given m = 1 (p – q)2, calculate the value q if m = 16 and p = 3. Algebraic formulae involve equations
4 that connect a few variables.

Solution: Subject of formula
Subject of formula is a dependant
m ×4 = 1 (p – q)2 ×4 variable expressed in terms of an
4 independent variable of a formula.
The subject of formula always
4m = (p – q)2 has coefficient 1. The algebraic
formulae involves
�4m = √(p – q)2 Square root both sides of the equation (a) One of the basic mathematical

p – q = �4m operations.
(b) Squares and square root
– q = �4m – p (c) A combination of basic and

(– q) × 1 = �√4m – p� × 1 Multiply both sides of square operations or square root.
–1 –1
1 TIPS
q = − √4m + p the equation by –1
Alternative method
q = p – √4m Substitute m = 16 and p = 3

q = 3 – √4(16) Replace m = 16 and p = 3 16 = 1 (3 − q ) 2
4
q = 3 – 8
64 = (3 − q ) 2
q = –5
√64 = (3 − q )
8 = 3 − q
q = 3 − 8
q = −5

3.1.4 Solving problems LEARNING
STANDARD
EXAMPLE 6
Solving problems involving
formulae.

The price of a fried chicken at a school canteen is twice the price of a bun. With RM5, Azman
bought two buns and a piece of chicken. The balance of RM1 is saved. If Azman has RM12 and
decides to buy the same number of buns, how many pieces of fried chickens will he be able to buy?

48

Chapter 3 Algebraic Formulae

Solution:

Understanding the problem Planning the strategy

Number of fried chicken that can be Determine the price of a bun.
bought by Azman for RM12. (a) Represent the price of bun and chicken
with x.
Price of bun = RM x CHAPTER 3
Price of chicken = RM2 x

(b) The total price of bun + The total price of chicken + RM1 = Total expenditure

2(RM x ) + RM2x + RM1 = RM5

2x + 2x + 1 = RM5

4x + 1 = 5

x = 5−1
4

=1

Thus, the price of a bun is RM1 and the price of a piece of chicken is RM2.

Conclusion Implementing the strategy

Azman gets to buy 5 pieces (a) Represent the number of fried chicken with y.
of fried chicken.
(b) Total price of bun + Total price of chicken = RM12

(RM1 × 2) + (RM2 × y) = RM12

2 + 2y = 12

y = 12 – 2
2

=5

SELF PRACTICE 3.1

1. Express the letters in the brackets as subject of formula.

(a) z = m − qp [ m ] (b) v = u + 2 [u]

(c) 3y = 7w [ x ] (d) 3a = 5 4 b [ b ]
x +

(e) 5q = 3 − 5 [ u ] (f ) 2w = −4 + 5 [ v ]
u v

(g) 2a = √3b + 5 [ b ] (h) (−5t)2 = 25w2 [ w ]
36

(i) (−3m)2 = 4p − 8 [m] (j) √(9r2) = 4s − 7 [r]

2. The price of a shirt is RM35, while the cost of a pair of trousers is RM45. A discount of 15% is
given on the price of a shirt, while a discount of 10% is given on the price of a pair of trousers.
Write the formula for the total expenditure, z, if Syamsul wants to buy x shirts and y trousers.

49

Chapter 3 Algebraic Formulae

3. Solve the following.

(a) Given c = 4d + 8, calculate (b) Given 4p = 18 − 5q, calculate

(i) value c when d = 2 (i) value p when q = 2

(ii) value d when c = 10 (ii) value q when p = 2

(c) Given 1 m = 2 n + 8, calculate (d) Given √4m = n2 − 5 , calculate
3 3 2
CHAPTER 3
(i) value m when n = −15 (i) value n when m = 4

(ii) value n when m = 30 (ii) value m when n = 2

( e) Given 3u = 4r + s, calculate (f) Given 3 p = 2 q − 1 r , calculate
5 3 4

(i) value u when r = 5 and s = −2 (i) value p when q = 3 and r = 8

(ii) value r when u = 3 and s = 3 (ii) value q when r = −12 and p = 10
1
(iii) value s when u = 2 and r = 2 (iii) value r when p = −15 and q = −15

1 3 1
(g) Given √3a = 9b − 4 c, calculate (h) Given 1 2 s= 5 t2 + 3 u 2, calculate

(i) value a when b = 1 and c = 1 (i) value s when t = −5 and u = 3
3 2

(ii) value b when c = 3 and a = 12 (ii) value t when u = −6 and s = 28
(iii) value c when a = 3 and b = 3
(iii) value u when s = 4 and t = 5
6 6

4. Write the algebraic formula based on the following situations.
(a) The total price RMz that needs to be paid by a buyer who bought x workbook and y geometry
set. The workbook and the geometry set each costs RM5.90 and RM3.60 respectively.

(b) In a class party, a teacher buys p carton of canned drinks to be distributed to the q
students. From the total number of canned drinks, seven cans were distributed to the
subject teachers. If a carton contains 24 cans of drinks, calculate the number of cans
received by each student, b in terms of p and q.

(c) Shoe A is sold at RM35 a pair, while shoe B costs RM76 a pair. Beautiful Shoe Shop offers
a 15% discount on purchases of two pairs of shoes. Mei Ling buys m pairs of shoe A and n
pairs of shoe B. Calculate the price payable, P in terms of m and n.

(d) A car is able to travel as far as 10 km with a litre of petrol. Express the cost, RMx of the
petrol that needs to be filled for s km if a litre of petrol costs RMt.

GENERATING EXCELLENCE

1. Write the algebraic formula from the following situation.
(a) A represents area, x represents the length of a square. Write the formula that relates A to x.

50

Chapter 3 Algebraic Formulae

(b) The rental fee of a sepak takraw court is RM5 for the first hour. Payment for the next hour
is RM3. Write the formula that relates the amount of payment p, and the hours used, h.

(c) Acceleration, a is defined as the difference between the final velocity, v2 and initial velocity
v1 divided by time, t. Write the relationship between a, v2, v1 and t.

2. Express the letters in the brackets as a subject of formula. CHAPTER 3

(a) m = – 3 q + p [q] (b) x = – p – w [w]

(c) 2e = 4 g + 3h [g] (d) 3 m – 6p = 3 q [q]
4 4
3
(e) w = 3v 2 1 [v] (f) 2m = 4 n2 [n]

(g) 3w = (v + 1) 2 [v] (h) 45 f = k −5 7 [k]
2

3. Calculate the following value.

(a) Given w = x + y , calculate the value (b) Given 6 b = c −d 2, calculate the
1 + x value 9

(i) w, when x = 2 and y = − 8 (i) b, when c = 20 and d = 2

(ii) x, when w = 20 and y = 5 (ii) c, when b = 1 and d = 2
9
(iii) y, when w = 5 and x = 6

(iii) d, when b = 1 and c = 90
2

(c) Given −2p = (q + q1)), calculate the value (d) Given 4 s 2 = � 3t – 4u 2 calculate
(r + the value 5
�,

(i) p, when q = 3 and r = 3q

(ii) q, when p = 3 and r = 2q (i) s, when t = s − 1 and u = 2 s
1
(iii) r, when p = − 3 and q = 2p (ii) t, when s = −5 u and u = 3
1
(iii) u, when s = 3 t and t = 2 − u

4. The salary of fast food store branch manager is 3 times more than a part time employee salary,
RMx per day. Working hours for part time employees are half of the manager’s working time, y
within a month. If they work 26 days in a month, write the formula for the difference in salary,
RMz between the two workers in terms of x and y.

5. Julia takes 40 seconds to walk as far as 50 metres. Write a formula to help Julia calculate the
duration of the trip, t in minutes from her home to the school that is s kilometre away.

6. The area of the trapezium below is 36 cm2. If x + y = 11 cm, calculate the value of x and y.
x cm

4 cm
2y cm

51

Chapter 3 Algebraic Formulae

CHAPTER SUMMARY
Algebraic Formulae

CHAPTER 3 Algebraic formulae connect the Subject of formula is represented by a letter.
algebraic expressions through addition,
subtraction, multiplication and division Subject of formula can be changed
in an equation form. according to the value of the variable.

1. y = 3x – 5 w = – 6 – 8t

2. w = 6 – 7v
v
t = –6–w
1 8
3. L = 2 th

4. A = πr2

A variable in the subject of formula can be obtained Solving problem involves
when the value of the other variables is given. changing the subject of a
formula, combination of basic
Example: 2v mathematical operations, square
–v + and square root.
Given Q = u , calculate value u,

if v = 2, Q = 4

Thus, u = 3

SELF REFLECTION
At the end of the chapter, I am able to:

1. Write a formula based on a situation.

2. Change the subject of formula for an algebraic equation.
3. Determine the value of a variable when the value of another variable

is given.
4. Solve problems involving formulae.

52

Chapter 3 Algebraic Formulae CHAPTER 3

MINI PROJECT

Title: Counting board
Materials: Manila card, used card board, coloured paper, glue and scissors
Steps:
1. Create a counting board to calculate the price that needs to be paid by the student to

purchase three items.
2. Example of the things that needs to be purchased are pen, mineral water, and note book.
3. Price of the pen, mineral water and note book is determined by the students according to

the current price.

Items

Number a b c
Price a × RM b × RM c × RM

Total (i) (ii) (iii)
Overall total (iii)
+ +
(i) (ii)

Example of a counting board

53

Chapter 4 Polygon

BABCHAPTER 54 WHAT WILL YOU LEARN? In our daily life, there is a combination of

4.1 Regular Polygons polygons around us especially in the designs
4 of buildings. The combinations of polygons
produce interesting and diverse forms of
4.2 Interior Angles and Exterior Angles art. This geometric pattern can be seen at
of Polygons Tanjung Bungah Floating Mosque, Penang
whereby it has a unique combination of
local and Western Asian architecture.

WORD LINK

• Polygon • Poligon
• Regular polygon • Poligon sekata
• Irregular polygon • Poligon tak sekata
• Axis of symmetry • Paksi simetri
• Side • Sisi
• Interior angle • Sudut pedalaman
• Exterior angle • Sudut peluaran
• Supplementary angle • Sudut penggenap
• Origami • Origami

54
BAB 4

Chapter 4 Polygon

WALKING CHAPTER 54BAB
THROUGH TIME
4
Polygon refers to the words ‘poly’ which means
many and ‘gon’ which means angle. Polygon
is named by the number of sides. For larger
polygons, mathematicians name the polygon
according to the number of sides for example
17-gon.

For more information:

http://rimbunanilmu.my/mat_t2e/ms055

WHY STUDY THIS CHAPTER?
Creating logos, murals on school walls and
creating symmetry on drawing.
In the field of technology, knowledge of
polygon is used in building architecture,
roofing, interior designing, fabric design and
more.
Careers involved in this field are surveyors,
technicians, engineers, architects, graphic
designers and many others.

55

Chapter 4 Polygon

CHAPTER 54 CREATIVE ACTIVITY QR

Aim: Producing a pentagon using paper folding technique PS
(origami)
Materials: Square paper and scissors Diagram A
Steps: QTR
1. Fold the square paper into two sections.
2. Label each vertex with PQRS. PU S
3. Fold P towards line QR. Press the fold down. Open the fold. Diagram B
4. Fold Q towards line PS. Press the fold down. Open the fold.
5. There should be fold marks shaped X in the middle. Label it QTR

as X. PUS
6. Bring S to the centre marked X. Press the fold down. Diagram C
7. Bring the vertex that touches X and fold it back so that this
QR CODE
side rests on the furthest side.
8. Take P and fold it to line TU. Fold this shape back to get Scan the QR Code or visit
http://rimbunanilmu.my/
Diagram D. mat_t2e/ms056 to view the
9. Now, cut off the top as shown in Diagram D. tutorial video of the pentagon
10. Open the folds. State the shape of the origami. shape origami.

Diagram D

4.1 Regular Polygon LEARNING
STANDARD
4.1.1 Geometric properties of regular polygon
Describe the geometric
A regular polygon is a polygon that has sides with equal length properties of a regular
and interior angles of the same size. polygons using various
representations.
Identifying regular polygon
DO YOU KNOW
COGNITIVE STIMULATION
Origami originated from
Aim: Exploring geometrical characteristics of regular polygon Japan that means
Materials: Ruler and compasses ‘ori’ = art, ‘gami’ = paper
J
BE F FLASHBACK

IK A polygon is a closed form
on a plane that is bounded
A CD GH L by three or more straight
56 lines as the sides.

Chapter 4 Polygon

Steps:
1. Measure the length of the side and interior angle for all the polygons.

2. Complete the table below.

Triangle ABC Square DEFG Pentagon HIJKL

Length of Measurement Length of Measurement Length of Measurement
side of angle side of angle side of angle

AB ∠CAB DE ∠GDE HI ∠HIJ

BC ∠ABC EF ∠DEF IJ ∠IJK

CA ∠BCA FG ∠EFG JK ∠JKL

GD ∠FGD KL ∠KLH CHAPTER 54

LH ∠LHI

Conclusion: Conclusion: Conclusion:

Discussion: DO YOU KNOW
Your findings from the activity above.
Determining types of polygon.
Regular polygons are polygons for which all sides are equal and all A polygon can have 3 or more sides.
interior angles are of the same size. Regular polygons have congruent
interior angles. Irregular polygons are polygons with irregular sides. Regular polygon
All sides are equal. All interior
angle are of the same size.

EXAMPLE 1 3 sides 4 sides 5 sides
triangle square pentagon
Based on the diagram, which one is a regular polygon and which is
an irregular polygon? 6 sides 7 sides 8 sides
(a) (b) (c) hexagon heptagon octagon

Irregular polygon
Not all sides are equal in length.

3 sides 4 sides 5 sides
triangle quadrilateral pentagon

(d) (e) (f) 6 sides 7 sides 8 sides
hexagon heptagon octagon

Concave polygon
Has at least one angle

that is more than 180°.

Solution: (b) Irregular polygon Convex polygon
(a) Irregular polygon (d) Regular polygon No interior angle more
(c) Regular polygon (f) Irregular polygon than 180°.
(e) Irregular polygon
Complex polygon
Has lines that
intersects in the
polygon.

Non polygon

Circle Shape that Shape Three-
has a curved that dimensional
object
line is not

closed

57

Chapter 4 Polygon

Determining axis of symmetry
COGNITIVE STIMULATION

Aim: Describing the axis of symmetry of a regular polygon QR CODE
Materials: Dynamic geometry software and scissors
Steps: Scan the QR Code or visit
1. Open the file MS058A and print the worksheet. http://rimbunanilmu.my/
2. Divide the class into two groups. mat_t2e/ms058a to print
3. The first group cuts the regular polygons and second group the worksheet.
cuts the irregular polygons.
CHAPTER 54 4. By folding the polygon, determine the axis of symmetry for
all the regular polygons and irregular polygons.
5. Complete the table below.

Number of sides Number of axis of symmetry

Regular polygons

Irregular polygons

Discussion: QR CODE
(i) What is the relationship between the number of sides with the
Scan the QR Code or visit
number of axis of symmetry? http://rimbunanilmu.my/
(ii) Conclusion from the findings of the first group and second mat_t2e/ms058b to find
out names of the
group. multi-sided polygon.

The number of axis of symmetry for a regular polygon is equals to
the number of sides of the polygon.

For irregular polygons the number of axis of symmetry should be
explored using the folding method.

58

Chapter 4 Polygon

4.1.2 Constructing a regular polygon LEARNING CHAPTER 54
STANDARD
Regular polygon can be constructed by using various methods.
Explore the activity below. Construct regular
polygons using various
COGNITIVE STIMULATION methods and explain the
rationales for the steps of
Aim: Creating regular polygon construction.
Materials: Dynamic geometry software, paper and scissors
Steps: QR CODE
1. Open the file MS059A.
2. Click on the polygon instructions and choose regular polygon. Scan the QR Code or
3. Click any points on the Cartesian plane. visit http://rimbunanilmu.
4. Click any second point. my/mat_t2e/ms059a for
5. On the window of the regular polygon, at the vertices enter the cognitive stimulation.

number of edges that has to be built. For example, pentagon has
five vertices.
6. Repeat the same steps for regular hexagon, regular heptagon,
regular octagon and regular nonagon.
7. Print out the shapes.
8. Paste your work in your book.

Discussion:
Your findings from the activity above.

COGNITIVE STIMULATION QR R
PDiagram A S
Aim: Producing a regular octagon using origami
Materials: Dynamic geometry software, printer, square shaped Q

coloured paper and scissors
Steps:
1. Open the file MS059B to watch the tutorial on making an

octagon shaped origami.

2. Fold the paper into two parts. Open the fold. QR

3. Fold the diagonal part of the paper into two parts. PDiagram BS
4. Take the centre point of the folded line and bring it close to the

diagonal line through the centre point. PT S
5. Cut away the extra paper. Q R Diagram C

6. Open the fold, then an octagon is produced. QR CODE

Discussion: PS Scan the QR Code or visit
Your findings from the activity above. Diagram D http://rimbunanilmu.my/

mat_t2e/ms059b to watch

the tutorial on creating an

COGNITIVE STIMULATION octagon shaped origami.

Aim: Building a regular polygon using geometry tools
Materials: Pencil, ruler, A4 paper and compasses

59

Chapter 4 Polygon

Activity 1:
Construct an equilateral triangle with the sides 5 cm

C

A 5 cm B AB AB AB

(a) Construct a line (b) Construct a curve with (c) Construct a curve with (d) Draw a line from A
segment AB with the radius 5 cm from radius 5 cm from point to C and B to C. An
length 5 cm. point A. B so that it intersects equilateral triangle
with the first curve. is created.
The intersection point
CHAPTER 54 Activity 2: is labelled C.
Construct a square with sides 4 cm

D
DC

A 4 cm B A B A B AB

(a) Construct a line (b) Construct a (c) Construct a curve with a (d) Construct two curves

segment AB with the perpendicular line with distance 4 cm from point with a distance of

length 4 cm. AB through point A. A so that it intersects 4 cm from B and

with the perpendicular D so that both the

line. The intersection curves intersect. The

point is labelled D. intersection point is

Activity 3: labelled C.

Construct a regular hexagon with sides 3.5 cm

B BC BC

A A A DA D

4 cm FE FE

(a) Construct a circle with (b) Construct a curve with (c) Construct a curve with (d) Draw line AB, BC, CD,

the radius of 3.5 cm. radius 3.5 cm from A the distance 3.5 cm from DE, EF and FA to form

Mark one point at the and label it B. B and label it C. Repeat a regular hexagon.

circumference and the steps until F.

label it A. QR CODE

Discussion: TIPS Scan the QR Code or
Your findings from the activity above. visit http://rimbunanilmu.

Regular polygons can my/mat_t2e/ms060 to

From all the activities that have been also be constructed by construct regular polygon
done, the most accurate method in dividing the angles at using geometry tools.
constructing regular polygons is by the centre of a circle
according to the number
using dynamic geometry software.
of sides.

60

Chapter 4 Polygon CHAPTER 54

SELF PRACTICE 4.1

1. Determine whether each polygon is a regular polygon or irregular polygon.
(a) (b) 120 ° 60° (c)


(d) (e) (f)

(g) (h) (i)

2. Trace the following diagrams. Determine the number of axis of symmetry for each diagram.
(a) (b) (c) (d)

3. Complete the table below with the characteristics of the polygons.

Regular polygon Polygon name Number of Number of Number of axis
sides vertices of symmetry

4. Construct the following regular polygons by using a ruler and compasses.
(a) Equilateral triangle with side length 3.4 cm.
(b) Square with sides 3.6 cm.
(c) Regular hexagon with sides 4 cm.
(d) Regular heptagon with sides 4.2 cm.
(e) Regular octagon with sides 4.5 cm

61

Chapter 4 Polygon

5. Draw the following regular polygon by dividing the vertices at the centre equally.
(a) Regular pentagon (b) Regular hexagon

CHAPTER 54 4.2 Interior Angles and Exterior Angles of DO YOU KNOW
Polygons
Exterior angle +
x Interior angle = 180°.
a
Exterior angle
yb c Interior 115°
z angle
65° 180°

Interior angle is an angle Exterior angle is an angle FLASHBACK
that is shaped by two that is formed when one side
of the polygon is extended. Total sum of interior angles
adjacent sides of a polygon. It is the supplementary of a triangle is 180°.
angle to the interior angle of
the adjacent side. a a + b + c = 180°
bc
Angle a, b, and c are Angle x, y, and z are
interior angles. exterior angles. LSETAANRNDAINRGD

4.2.1 Total sum of an interior angle Derive the formula for the
sum of interior angles of a
There is a relationship between the number of vertices of a polygon polygon.
with the sum of the interior angle. Explore the activity below:
QR CODE
COGNITIVE STIMULATION
Scan the QR Code or visit
Aim: Exploring the number of triangle in a polygon http://rimbunanilmu.my/
Materials: Paper and protractor mat_t2e/ms062 to obtain
Steps: the worksheets on polygon
1. Open the file MS062 to obtain information about polygon shapes.

shapes.
2. Print the riangle, square, pentagon, hexagon, heptagon,

octagon and nonagon.

62

Chapter 4 Polygon

3. Connect the edges of each polygon to form a triangle in the polygon as shown below.

1
1 1 13 2
2 3
2 4

4. Complete the table below.

Polygon Number of sides (n) Number of Total sum of interior angles
triangles
Triangle 3 1 × 180° = 180° CHAPTER 54
Square 4 1 2 × 180° = 360°
Pentagon 2
Hexagon
Heptagon
Octagon
Nonagon
Decagon

Discussion:

(i) What is the relationship between the number of sides, n with the number of triangles?

(ii) What is the relationship between the number of sides in a THINK SMART
triangle with the total sum of interior angles?
Pentagon can be divided
5. Total sum of interior angles of a polygon into 3 triangles. State the
= number of triangles × 180° number of interior angles of

= × 180° a pentagon.

the nth term

Total sum of interior angles of a polygon = (n – 2) × 180°.

DO YOU KNOW

EXAMPLE 2 Number Polygon Name
of sides
State the number of triangles formed for each of the following dodecagon
polygon. 12 tridecagon
13 tetradecagon
(a) 13 sided polygon (b) 18 sided polygon 14 pentadecagon
15 hexadecagon
Solution: 16 heptadecagon
17 octadecagon
(a) Number of triangles = 13 − 2 18 enneadecagon
19 icosagon
= 11 20

(b) Number of triangles = 18 − 2
= 16

63

Chapter 4 Polygon

EXAMPLE 3

CHAPTER 54 Calculate the value x for the following. (b) 60°
(a)
130°
100°
x x

130° (b) Total sum of interior angles,
= (n − 2) × 180°
60° = (4 − 2) × 180°
= 360°
Solution: x + 130° + 60° + 90° = 360°
(a) Total sum of interior angles,
= (n − 2) × 180° x + 280° = 360°
= (5 − 2) × 180° x = 360° − 280°
= 540° x = 80°
x + 100° + 130° + 60° + 90° = 540°

x + 380° = 540°
x = 540° − 380°
x = 160°

4.2.2 Total sum of exterior angles of polygons LEARNING
STANDARD
COGNITIVE STIMULATION
Make and verify conjectures
Aim: Exploring the total sum of exterior angle about the sum of exterior
Material: Dynamic geometry software angles of a polygon.

Polygon n The output angle enhancement QR CODE

Conjecture Verify (Yes / No) Scan the QR Code or visit
http://rimbunanilmu.my/
mat_t2e/ms064 to obtain
the worksheet.

Steps: TIPS
1. Open the file MS064 and print the file.
2. Do a conjecture for each of the polygon in the space provided in A conjecture is a proposition
or a theorem that looks
worksheet. right. Conjecture decisions
3. Open the file MS064 to view total sum of exterior angle. are not formally proven. The
4. Explore each polygon that is available. conjecture allows students
5. Drag the slider dilate to change the size of the polygon sides to speculate based on a
mathematical situation.
that is being displayed. For example, if we add 2
6. State the total sum of the exterior angles of a polygon. positive numbers then the
result is always greater than
Discussion: the number.
The sum of exterior angles of polygon.

The total sum of exterior angles of a polygon is 360°.

64

Chapter 4 Polygon

EXAMPLE 4

(a) Calculate the value x for the (b) In the diagram below, ABCDE is a regular pentagon.
diagram below. BCF and EDF are straight lines. Calculate the value x.
A
x
EB

120° 160° DC
x
Solution:
360° F
(a) Total sum of exterior angles = 360° (b) ∠FCD = 5 CHAPTER 54
x + 160° + 120° = 360° TIPS
x + 280° = 360° 72°
x = 360° − 280° = 180° − 72° − 72° Exterior angle of a = 360°
x = 80° x = 36° regular polygon n
=
Interior angle
= 180° − exterior angle

4.2.3 Total sum of exterior angles of polygons LEARNING
STANDARD

EXAMPLE 5 Determine the values of
Calculate the value of the interior angle for a regular hexagon. interior angles, exterior
Solution: angles and the number of
sides of a polygon.

Number of sides of regular hexagon, n = 6 TIPS

Total sum of interior angles = (n − 2) × 180° Interior angle of regular
polygon
= (6 − 2) × 180°
(n − 2) × 180°
= 4 × 180° =n

= 720°

Interior angle = = 7T2o60ta°l sNuummobferinotfersiiodreasngles

= 120°

EXAMPLE 6 30° 30° + b

Calculate the value of b for the diagram on the right. 60°
15°
Solution:
45° b
360° = (30° + b + b + 50° + 45° + 15° + 60° + 30°) 50°
360° = 230° + 2b

2b = 360° − 230°
2b = 130°
b = 65°

65

Chapter 4 Polygon

EXAMPLE 7 ATTENTION

Calculate the value of the exterior angle of a regular octagon. POLYGON

Solution: 34 5 6

triangle square pentagon hexagon

Number of sides of a regular octagon, n = 8 Total sum of interior angles
(n − 2) × 180°
Total sum of exterior angles = 360°
360°
Exterior angle = 8 Number of side
4 × 180° = 540°

= 45° Interior angle

CHAPTER 54 EXAMPLE 8 Total sum of
interior angles
Calculate the number of sides of the following regular polygon
Number of sides
or

180° − exterior angle

when given the value of interior angle. Exterior angle

(a) 108° (b) 144° 360°
Number of sides
Solution:
or
180° − interior angle

(a) Exterior angle = 180° − 108° (b) Exterior angle = 180° − 144°

= 72° = 36°
360° 360°
Number of sides, n = exterior angle Number of sides, n = exterior angle

360° 360°
n = 72° n = 36°


n = 5 n = 10

4.2.4 Solving problems LEARNING
STANDARD
EXAMPLE 9
Solve problems involving
The diagram on the right is a regular hexagon that is polygons.
enlarged from the design of a football.
(a) Calculate angle y. Q
(b) Calculate the difference between y and (x + z).
PR
Solution: xz
y
U S

T

Understanding the problem Planning the strategy Implementing the
Calculating angle y using formula strategy
(n − 2) × 180° (a) y = (6 − 2) × 180° (b) Difference between
6 y and (x + z)
n = 120° − (30° + 30°)
y = 120° = 60°
Angle x is in the equilateral triangle
∠UPQ = ∠TSR = y (b) x = 180° − 120° Conclusion
2
180° − UPQ
2 x = 30° (a) y = 120°

z = 30°(alternate angle) (b) y − (x + z) = 60°

66

Chapter 4 Polygon

SELF PRACTICE 4.2

1. State the number of triangles that can be found in the polygon below and calculate the total sum
of the exterior angles.

Polygon Number of triangles in the polygon Total sum of exterior angles
Pentagon

Hexagon

Heptagon CHAPTER 54

Octagon

Nonagon

2. Name all the interior angles and exterior angles for each of the following polygons.

(a) h g e f (b) i h
a d c

bc e bg
d
ja
f

Interior angles: Interior angles:

Exterior angles: Exterior angles:

3. Calculate the value x for each of the following diagrams. 75°

(a) 80° (b)

x 85°

130° 100°

x

(c) x (d) 50°
x
70° 76°
50°

112° 60°

67

Chapter 4 Polygon

4. For each of the diagram below, calculate the value p, q and r.

(a) (b) r
100° p p 112°

q 45° 60°
r
80°
q

5. Calculate the value a + b + c.

(a) c (b)
a
CHAPTER 54
60° b b 80°
80° c
a

(c) a c (d) c
b 98° b
85°
b a 65°

6. Determine the number of sides for a polygon if the total sum of exterior angles is

(a) 900° (b) 1 080° (c) 1 260°

7. Zaidi has a vegetable garden that is shaped like a regular polygon. The dotted lines is the axis
of symmetry of his garden.

(a) What is the actual shape of Zaidi’s garden?
(b) Calculate the value y.

y

8. The diagram shows two swimming pools at a sports centre in the shape of a regular octagon
and pentagon. What is the value of angle x?

x

GENERATING EXCELLENCE

1. Construct the following polygons using compasses and a ruler.
(a) Equilateral triangle ABC with sides 4 cm.
(b) Square PQRS with sides 3 cm.

68

Chapter 4 Polygon

2. Calculate the value p,q and r in the following.

(a) (b) r (c) 105° p
q p 140° q

45° p 85° r 75° q
40°
r 85°

135°

3. Calculate the value x for the following.

(a) x (b) (c) 60°
x 100°

120° x CHAPTER 54

110° 130°

2x

4. Calculate the number of sides for the following.

(a) 45° (b) 36°

(c) 140° (d) 150° 150°

140°

5. (a) Calculate the value x + y in the (b) The diagram shows a logo in the shape
diagram below. of a regular pentagon. FED is a straight
line. Calculate the value x + y.
150° 65° B
x
AC
y

y x
FE D

(c) In the diagram below, HIJKL is a pentagon. KJM is a straight line.
Calculate the value a + b + c + d.

H

La b cI

d 65°
K JM

69

Chapter 4 Polygon

6. Azreen wants to draw a logo for the Peers Counselling Club at her school. She chooses
to draw a regular hexagon with the radius 4 cm. Help Azreen draw her logo using a ruler,
protractor and compasses.

7. Total sum of all interior angles of a regular polygon is 2 700°. State the number of sides of
this polygon.

8. In the diagram below, calculate the value p + q.

60° p

CHAPTER 54 98° q
92°

70° 80°

9. In the diagram below, ABCDEFGH is a regular octagon and EFKLM is a regular pentagon.
Calculate ∠CBM.
AB

H L C
67°
K
G M
D

FE

10. The exterior angle of a regular polygon is 2h and the interior angle of the same polygon is 7h.
(a) Calculate the value of h.
(b) Calculate the value of the interior angle and exterior angle.
(c) Calculate the number of sides of the polygon and name the polygon.

11. The diagram below shows 4 regular pentagons and a square. Calculate the value x.

x
70

Chapter 4 Polygon

12. Bahar wants to construct a polygon that has an interior angle of 300°. Can Bahar construct the
polygon? Justify your answer.

13. The diagram below shows a partial design that has been formed from combining two tiles.
There are two types of tiles. They are tile A and tile B that are regular polygons. Calculate the
number of sides of tile A.

Tiles A

TiBles CHAPTER 54

Tiles A Tiles A

TiBles
Tiles A

14. Devaa is a graphic design student at a local university. Help Devaa calculate the value of x to
construct a photo frame that has the characteristics of combined polygons.

x

15. Calculate the value of x.

x
71

Chapter 4 Polygon

CHAPTER SUMMARY

The number of axis of symmetry
in a regular polygon with n side
and n axis of symmetry.

CHAPTER 54 Regular polygon Irregular polygon

• Interior angle Exterior angle of a polygon is a
(n − 2) × 180° complement to an interior angle of the
= n Total sum of interior polygon.
angles
• Exterior angle = (n − 2) × 180° Exterior angle + Interior angle = 180°

= 360° Total sum of exterior angle = 360°
n

Total sum of exterior angles
= 360°

Regular polygon is a polygon where all its Irregular polygon is a polygon where
sides have the same length and all interior not all sides are of the same length.
angles are equal. Examples

Exterior Interior angle
angle

= 360° = (3 − 2) × 180°
3 3

= 360° = (4 − 2) × 180°
4 4

= 360° = (5 − 2) × 180°
5 5

72

Chapter 4 Polygon

SELF REFLECTION

At the end of this chapter, I will be able to: CHAPTER 54

1. Describe the geometric properties of regular polygons using various
representations.

2. Construct regular polygons using various methods and explain the rationales
for the steps of construction.

3. Derive the formula for the sum of interior angles of a polygon.

4. Make and verify a conjecture about the sum of exterior angles of a
polygon.

5. Determine the values of interior angles, exterior angles and the
number of sides of a polygon.

6. Solve problems involving polygons.

MINI PROJECT

You are a food stall owner. Create your business logo using the combination of two or three
polygons. You can use the dynamic geometry software, geometry tools or origami to design
your logo. Present the rationale for the choice of your logo in the class.

Examples of logo

73

Chapter 5 Circles

AWNHDAAT WAKILAL NYOMUEMLEPAERLANJ?ARI The clockwise movement of the hand of a

5.1 Properties of Circles clock forms a circle as it completes a 360°
5.2 Symmetrical Properties of Chords rotation. This is called ‘kirkos’ in Greek,
5.3 Circumference and Area of a Circle which means to rotate and to become a curve.

CHAPTER 5

WORD LINK

• Circle • Bulatan
• Circumference • Lilitan
• Radius • Jejari
• Centre • Pusat
• Diameter • Diameter
• Chord • Perentas
• Segment • Tembereng
• Sector • Sektor
• Minor sector • Sektor minor
• Major sector • Sektor major
• Minor segment • Tembereng minor
• Major segment • Tembereng major
• Symmetry • Simetri

7744

Chapter 5 Circles CHAPTER 5

WALKING
THROUGH TIME
A circle is a curved pathway locus of a point
that is equidistant from a fixed point. This
fixed point is known as the centre and the
distance from this fixed point to the pathway
is called the radius. A circle is also a curve that
is joined which is known as circumference.
A mathematician named Euclid was the first
person to study circles. He is also known as the
‘Father of Geometry’ due to his research.

For more information:

http://rimbunanilmu.my/mat_t2e/ms075

WHY STUDY THIS CHAPTER?
The application of this chapter is in the field of
architecture, astronomy, design and astrology.

7755

Chapter 5 Circles

CREATIVE ACTIVITY

Aim: Getting to know circles
Materials: Coloured paper, glue, scissors, string and punch
Steps:
1. Students form groups.
2. Each group is required to draw circles of various sizes.

Examples are like the ones in the diagram on the right.
3. The circles will be used to decorate the class.
4. Write down the mathematical formulae of area of a rectangle, area of triangle, volume of

cube, volume of cuboid, Pythagoras theorem and so on in the circles.

CHAPTER 5 5.1 Properties of Circles

5.1.1 Getting to know the parts of the circle LEARNING
STANDARD
COGNITIVE STIMULATION
Aim: Knowing parts of a circle Recognise parts of a circle
Material: Dynamic geometry software and explain the properties
of a circle.

Steps:

1. Open the file MS076.

2. The perimeter of a circle is called .

3. Drag point A in the all directions.

(i) Point A is called the of the circle.

4. Drag point B around the circle.

(i) The line from the centre of the circle to any point on the circumference of the

circle is called .

5. Drag point C around the circle.

(i) The CC’ line that goes through the centre and touches the circumference is called .

6. Drag point D and the point E around the circle. QR CODE
(i) The line that connects two points on the circumference

is called . Scan the QR Code or visit
http://rimbunanilmu.my/
(ii) The region is called . mat_t2e/ms076 to explore
parts of the circle.
7. Drag points C and D.

(i) Name two lines generated. Lines AC and .

(ii) The region bounded by these two radius is called .

Discussion:
Make a conclusion on your explorations.

From the activity above, several parts of the circle have been identified as in the diagrams on the next page.
76

Chapter 5 Circles

Radius Circumference Centre
Perimeter of A fixed point where
A straight line from
the centre of the circle a circle. all points on the
circumference are
to any point on the equidistance from it.
circumference.

Diameter Minor Major CHAPTER 5
A straight line Segment Segment
that touches the
circumference Parts of a Segment
through the centre of Circle
The region enclosed
the circle.
by a chord

and an arc.

Major

Minor Sector

Sector

Sector

Chord The region enclosed

A straight line that joins Major by two radii
any two points on the
circumference. Minor Arc Arc and an arc.

Arc TIPS
Part of the
Diameter is the longest
circumference. chord in a circle.

EXAMPLE 1 TIPS

Identify the following parts of a circle. de A circle is a curved
pathway of a point that is
Solution: b, Diameter a bO c equidistant from a fixed
d, Circumference f point.
a, Chord f, Arc
c, Radius THINK SMART
e, Sector
Why are balls, globes
and marbles not
considered as circles?

77

Chapter 5 Circles

5.1.2 Constructing a circle LEARNING
STANDARD
COGNITIVE STIMULATION
Construct a circle and
Aim: Constructing a circle and parts of the circle based on the parts of the circle based
conditions given on the conditions given.
Materials: Compasses, protractor, ruler and pencil
Steps:

Conditions Steps Solution

(a) Construct a circle 1. Mark point O. 3 cm
O
with a radius of 3 cm 2. Using compasses measure 3
from the centre O.
cm on a ruler.

CHAPTER 5 3. Place the sharp point of the
compasses on point O and
draw a circle with the radius
of 3 cm.

(b) Construct a 1. Join points O and Q with a Step 1
diameter that passes straight line using a ruler.
through point Q in O
a circle with the 2. Extend the line until it Q
centre O. touches the circumference.
The extended straight line Step 2
that passes through Q and
centre O is the diameter. diameter
O

Q

(c) Construct two 1. Using compasses measure Step 1
chords of 3 cm in P
length from point P 3 cm on a ruler.
on a circle. O
2. Place the sharp point of the A
compasses on point P.

3. Draw the arc that cuts on the
circumference and label it as
point A.

78

Chapter 5 Circles

4. Join point P to point A that Step 2
has been marked on the P
circumference.
3 cm O
5. Thus, the line PA is a chord.

(d) Construct a sector 1. Draw a circle with the radius Step 1 A
A
with an angle of 60° OA 2 cm. 2 cm
Step 2 O
at the centre of a 2. Measure 60° with a protractor

circle with a radius as shown.

of 2 cm. 3. Draw the radius OB. AOB is

the sector of the circle. CHAPTER 5

O

Step 3
B

60°

AO

Discussion: TIPS
From the activity above, what parts of the circle has been
constructed? C

From the activities above, students are able to B
(a) construct a circle when the radius or diameter is given.
(b) construct a diameter through a given point in a circle. A Base
(c) construct a chord through a given point when the length of the Outer line Inner
scale Origin scale
chord is given.
(d) construct a sector when angle of the sector and the radius is To measure the ABC angle,
place the protractor's centre
given. point on the vertex of that
angle. Make sure the line with
the value 0 is located on the
AB line. Read the angle using
the external scale. Thus, the
angle for ABC is 120°.

79

Chapter 5 Circles

SELF PRACTICE 5.1 C

1. Name the following parts O D
(i) point O A
(ii) line AOC
(iii) dector AOB B
(iv) line OA
(v) arc AB
(vi) line BC
(vii) the shaded area BCD.

CHAPTER 5 2. Construct a circle with radius (b) 4.5 cm
(a) 3 cm (d) 6 cm
(c) 2.5 cm

3. Construct a diameter that passes through point Q for each of the circles with the centre O.

(a) (b)
Q

OO

Q

4. Construct the chord of a circle with radius and length given below.

Radius Length of Chord

(a) 3 cm 4 cm

(b) 4.5 cm 6.7 cm

5. Using a protractor, construct the sector AOB with O as the centre of the circle. The radius and
∠AOB as given below.

Radius ∠ AOB

(a) 3 cm 70°

(b) 3.6 cm 120°

80

Chapter 5 Circles

5.2 Symmetry and Chords LEARNING
STANDARD
5.2.1 Features in a circle
Verify and explain that
COGN ITIVE STIMULATION (i) diameter of a circle is

Aim: Verifying an axis of symmetry of
the circle;
(i) properties of the diameter of a circle. (ii) a radius that is
perpendicular to a
(ii) the relationship of a radius with chords. chord bisects the
chord and vice versa;
Material: Dynamic geometry software (iii) perpendicular bisectors
Steps: of two chords intersect
at the centre;
1. Open the file MS081. (iv) chords that are equal
in length produce arcs
2. Click on the Activity box. of the same length and
vice versa;
3. Drag point Q to points P, T, U, B1, V and Z. (v) chords that are
equal in length are
(i) Name the diameter of the circle. Lines . equidistant from the CHAPTER 5
centre of the circle and
(ii) Observe the value of angle at the centre when the vice versa.

diameter QQ’ is moved. Does it produce the same value? DO YOU KNOW

Are the resulting shapes similar? The circle has an infinite
number of axes of symmetry
(iii) If you fold the circle on the line QQ'‚ do the shapes because any straight
line that passes through
overlap each other perfectly? the centre is the axes of
symmetry of the circle.
(iv) The diameter of a circle is known as .

4. Click the Activity box for the next activity.

5. Drag the slider Drag Me until the end.

(i) Radius that bisects a chord is to the chord.

(ii) Radius that is perpendicular to the chord
the chord.

(iii) Equal chords produce arc.

QR CODE

Discussion: Scan the QR Code below
State the conclusions for all the activities above. or visit http://rimbunanilmu.
my/mat_t2e/ms081 for the
properties of symmetric
chord 1.

The diameter of a circle is the A radius which is
axis of symmetry of the circle. perpendicular to the chord
bisects the chord.

TIPS

The diameter is the chord
that passes through the
centre of the circle.

81

Chapter 5 Circles

COGNITIVE STIMULATION

Aim: Verifying
(i) properties of the bisector of the two chords.
(ii) properties of equal chords in a circle.

Material: Dynamic geometry software
Steps:
1. Open the file MS082.
2. Drag point A where, AB = CD.
3. Click on the box length of the perpendicular line from the centre of the circle.
4. Repeat steps 1 and 2 to get the others values.

CHAPTER 5

Discussion: .
(i) Where do lines OP and OQ meet?
(ii) Is the length of arc AGB and CID the same?
(iii) If the length of AB = CD, the distance of OP = the distance of
(iv) Is the distance of OP and OQ the same?

Perpendicular bisectors of two chords meet at the centre of the circle. QR CODE

O Scan the QR Code or visit
http://rimbunanilmu.my/
mat_t2e/ms082 for the
properties of symmetric
chord 2.

Equal chords or chords of the same length produce arc of the same length.

O

Equal chords are equidistant from the centre of the circle. THINK SMART

O How many axes of
symmetry are there in half
a circle?

82

Chapter 5 Circles

EXAMPLE 2 TIPS

A O

M AB
Two radii and a chord
PK O Q forms an isosceles
triangle.
N
B FLASHBACK

The diagram above shows a circle with centre O and the line MN Pythagoras theorem
is the chord. A
(a) Name the axes of symmetry of this circle.
(b) Given OK = 3 cm and NK = 4 cm, calculate length of ON. ac CHAPTER 5
(c) Name the angle that is equal to ∠ONK.

Solution:

(a) AOB and POQ K 3 cm O Bb C
4 cm
AB 2 + BC 2 = AC 2

N or
a2 + b2 = c2

�(16 + 9) ON = OM THINK SMART
�25
P

O

(c) ∠OMK MQ
EXAMPLE 3
O is the centre of the circle.
What is the relationship
between OP, OQ and OM?

The diagram on the right shows a circle with the radius OP that is perpendicular to the chord MN.

(a) Is the length MS equal to length of SN? Explain.

(b) If the radius of the circle is 10 cm and OS = 8 cm, calculate the length

of the chord MN. O

Solution: 10 cm 8 cm

(a) Yes, MS = SN (b) MS = �102 − 82 MS N
The radius OP which is MS = �100 − 64 P
perpendicular bisects MN.
MS = �36
MS = SN = 6

Therefore, MN = 12 cm.

83

Chapter 5 Circles

EXAMPLE 4

The diagram on the right shows two equal chords, RS and TU. M

POQ is a straight line passing through the centre of the circle O. R P S
Given OP = 5 cm and RS = 24 cm.

(a) Calculate the length of PR. O
(b) Are minor arcs RMS and TNU equal in length? Explain.

(c) Calculate the radius of the circle. T QU

Solution: N

(a) A radius that is perpendicular to the chord bisects the chord

into two equal lengths. Length of PR = 24 ÷ 2 cm = 12 cm.

(b) Yes, chords that are equal in length produce arc of the DO YOU KNOW
same length.

CHAPTER 5 (c) OR = �PR2 + OP2 Chords RS and TU are
= �122 + 52 equal in length

= �144 + 25 OR, OS, OT and OU are radii of a circle
= �169
= 13 cm Angle at the circumference
of a semicircle is 90°.

5.2.2 Centre and radius of a circle LEARNING
STANDARD
COGNITIVE STIMULATION
Determine the centre
Aim: Determining the centre and radius of the circle and radius of a circle by
Materials: Compasses, rulers, pencils and rounded material geometrical construction.
Steps:
1. Trace a circle on a piece of paper. PQ
2. Construct two chords, PQ and PR from point P.
3. Construct a perpendicular line for the chords PQ and PR.
4. The intersection point of two perpendicular lines is indicated by O.
5. Draw a line from O to the circumference and label it as OT.

Discussion: O T
(i) Properties of point O. R

(ii) Properties of line OT.

A perpendicular bisector for any chord will always intersect at the centre of the circle.

84

5.2.3 Solving problems Chapter 5 Circles

EXAMPLE 5 LEARNING
STANDARD
A blacksmith was asked to build a round-shaped window frame as
shown below. The window is 50 cm in diameter. Three iron rods, Solve problems involving
PR, US and QT that are not equal in length are used to support the symmetrical properties
window. Calculate the length of PR. of chords.

U

PQ O T 48 cm
R 31 cm

Solution: S

Understanding the problem Planning the strategy Implementing the strategy CHAPTER 5
Diameter of window = 50 cm �252 − 242
QT = 31 cm Radius = diameter �625 − 576
US = 48 cm 2 �49
Calculate length PR.
= 50 �252 − 242
2 �625 − 576
�49
= 25 cm

OT = �OU 2 − UT 2
OQ = QT − OT

PQ = �OP 2 − OQ 2
PR = PQ × 2

Conclusion
Therefore, PR is 14 cm.

SELF PRACTICE 5.2 M L
KN
1. In the diagram on the right, O is the centre of the circle. P
MNOP and KNL are straight lines. N OO
Given that MN = 8 cm and NP = 18 cm. Calculate the length of KL.
P
2. The diagram on the right shows a circle with the centre O. M
JKL and KOM are straight lines.
Given that JK = KL = 15 cm and radius of the circle is 25 cm. JO
Calculate the length of KOM.
K
L

85

Chapter 5 Circles

5.3 Circumference and Area of a Circle LEARNING
STANDARD
5.3.1 Relationship between circumference and
diameter Determine the relationship
between circumference
Circumference is the measurement around a circle. The diagram and diameter of a circle,
shows a round table that needs to be lined with skirting for a and hence define π and
wedding. What is the length of the skirting needed? derive the circumference
formula.
The length for the skirting can be
calculated using the formula that
involves π(pi).

CHAPTER 5 COGNITIVE STIMULATION

Aim: Determining the relationship between circumference and diameter
Materials: Stopwatch, pail, bicycle tyre, measuring tape, pencil or any circular material.
Steps:
1. Measure the circumference of the stopwatch, pail and bicycle tyre with the measuring

tape. Record the results in the table below.
2. Measure the diameter of the three items and record them in the table.
3. Complete the table below.

Material Circumference (cm) Diameter (cm) Circumference
1. Stopwatch Diameter

2. Pail

3. Bicycle tyre

Discussion:
(i) The relationship between diameter and circumference.

(ii) What is the ratio of the circumference to the diameter?

From the above activities, the ratio of circumference to diameter,

is π of a circle which is 3.142 or 22 . REMEMBER !
7
Diameter = 2 × radius
Circumference =π
Diameter

86

Chapter 5 Circles

The circumference of a circle is π multiplied by the diameter.

Circumference = π × diameter
= πd

The circumference of a circle can also be expressed as follows.
Circumference = π × 2 × radius
= 2πr

5.3.2 Formula for area of a circle LEARNING
STANDARD

COGNITIVE STIMULATION Derive the formula for

Aim: Expressing formula of a circle the area of a circle.

Material: Dynamic geometry software CHAPTER 5
Steps:

1. Open the file MS087.

2. Drag the radius up to value 3, and drag the n until it reaches the value of 6. Take note of

the changes.

3. Repeat step 2 by changing the value of radius and n. Take note of the changes.

Discussion:

(i) The the sector of the circle is divided into the clearer is the rectangular shape

produced.

(ii) Height of rectangles = of the circle.

(iii) Rectangular base = of the circle. QR CODE

From the activity above, Scan the QR Code or visit
Area of circle = area of rectangle http://rimbunanilmu.my/
mat_t2e/ms087 to explore
= base × height area of a circle.

= 1 × circumference × height
2
1
= 2 × 2πr × r

= πr 2

Therefore, area of circle = πr 2

87

Chapter 5 Circles

5.3.3 Circumference, area of a circle, length of LEARNING
arc and area of sector STANDARD

To determine the circumference of a circle Determine the
EXAMPLE 6 circumference, area of a
circle, length of arc, area
of a sector and other
related measurements.

Calculate the circumference of a circle, if

(a) diameter, d = 14 cm (Use π= 22 ) (b) Radius, r = 21.3 cm (Use π = 3.142)
7
Solution:

(a) Radius = πd (b) Circumference = 2πr
22 = 2 × 3.142 × 21.3
= 7 × 14
= 133.85 cm
CHAPTER 5 = 44 cm

EXAMPLE 7

(a) Given the circumference of a circle is 88 cm, calculate the diameter of the circle in cm.
22
(Use π = 7 )

(b) Given the circumference of a circle is 36.8 cm, calculate the radius of the circle in cm and

round off the answer to two decimal places. (Use π = 3.142)

Solution:

(a) Circumference = πd (b) Circumference = 2πr

88 = 22 ×d 2πr = 36.8
7
2 × 3.142 × r = 36.8
7
d = 88 × 22 r = 36.8
6.284
d = 28 cm
r = 5.86 cm

To determine area of a circle REMEMBER !
EXAMPLE 8
diameter
22 2
7

22 10 22
72 7
88

Chapter 5 Circles

EXAMPLE 9

Given the area of a circle is 616 cm2, calculate the radius and diameter. (Use π = 22 )
7
Solution:

Area = πr2 Diameter = 2 × 14 THINK SMART
πr 2 = 616 = 28 cm

22 × r2 = 616
7
17 7 O
1 22 × 22 1 × r2 = 616 × 22 (a) Calculate the area of
71
7 quadrant if the radius
r2 = 616 × 22
is 7 cm.

r2 = 196 O CHAPTER 5

r = �196 (b) Calculate the area of
r = 14 cm the semi circle if the
radius is 7 cm.

EXAMPLE 10

Given the circumference is 66 cm, calculate the area of the circle. O
22
(Use π = 7 ) (c) Calculate the area of
the three quadrant if
Solution: the radius is 7 cm.

Circumference = 66 cm Area = πr 2

22 2πr = 66 = 22 × 10.52
7 7
2× × r = 66
7 = 346.5 cm2
44
r = 66 ×

r = 10.5 cm

EXAMPLE 11

Given area of circle is 75.46 cm2. Calculate the circumference. THINK SMART
22
(Use π = 7 ) 4 cm
O 8 cm
Solution: 4 cm

Area = πr 2 Circumference = 2πr The diagram shows two
circles in a bigger circle.
22 πr 2 = 75.46 = 2 × 22 × 4.9 Calculate the area of the
7 × r 2 = 75.46 7 shaded region.
75.46 × 7 = 30.8 cm
r2 = 22

r 2 = 24.01

r = �24.01

r = 4.9 cm

89

Chapter 5 Circles

Determining length of arc in a circle

The arc AB is part of the circumference of the circle. The length of arc is proportional to the angle

at the centre of the circle. A

Length of arc Angle at centre B

Circumference 360° 

Therefore, Length of arc  O
2πr 360° TIPS

EXAMPLE 12 The symbol  is read as

CHAPTER 5 The diagram below shows a circle with a radius of 14 cm and theta, a Greek letter used to
centred at O. Calculate the length of minor arc PQ which represent angle.
encloses 60° at the centre. Write your answer to two decimal places.
FLASHBACK
Solution: Q

Length of arc =  P 60° Acute angle
2πr 360° O 0° <  < 90°

Length of arc =  × 2πr  Obtuse angle
360° 90° <  < 180°

Length of arc = 60° × 2 × 22 × 14 Reflex angle
360° 7 180° <  < 360°

= 14.67 cm Right angle 90°

EXAMPLE 13

The diagram below shows a circle with a radius of 21 cm and centered at O.
∠ROS is 72°. Calculate the length of major arc RS.

Solution: DO YOU KNOW

Angle at centre = 360° − 72° O 1Radian Length of
= 288° 72° arc
r
Length of arc =  R S
2πr 360°

Length of arc =  × 2πr Angles can be measured
360° using radians. 1 radian
(1 rad) is the angle at the
Length of arc = 288° ×2× 22 × 21 centre of the circle when
360° 7 the length of the arc is
equal to the radius.
= 105.6 cm

90

Chapter 5 Circles

EXAMPLE 14

Given the length of the arc of a circle is 11 cm and the angle at the centre of the circle is 45°.
Calculate in cm the radius of the circle.

Solution:

360° = Length of arc 360° THINK SMART
2πr = 2πr arc 
R
Length of
× APB
14 cm 14 cm
2× 22 × r = 11 × 360° 14 cm D
7 45° CQ

r = 11 × 360° × 7 × 1 S
45° 22 2
ARC, APB, BSD and CQD CHAPTER 5
r = 27 720 are arcs of the circles
1 980 wheras AB, AC, BD and
CD are the diameter of the
r = 14 cm circles. Calculate the area
of the shaded region.

To determine area of a sector

The area of a sector is a region bounded by an arc and two radii. The area of the sector is proportional

to the area of the circle.

Angle at centre A
360° O
Area of sector =
Area of circle

Therefore,

Area of AOB  B
πr 2 360°
=

EXAMPLE 15

The diagram below shows a circle with centre O and radius 21 mm. Calculate the area of the minor
sector MON.

Solution: M

Area of sector =  100° O
πr 360° 21 mm
2
N
Area of sector MON = 100° × 22 × 212
360° 7

= 385 mm2

91

Chapter 5 Circles

EXAMPLE 16 THINK SMART

Given the area of a sector QOP is 18.48 cm2 and the radius is 12 cm. annulus

Calculate the value of . QP 6 cm
Solution: 8 cm O

Area of a sector =  12 cm  The shaded area is an
πr 2 360° annulus.Determine the
O formula to calculate the
area of the annulus.
 18.48
360° = LEARNING
22 × 122 STANDARD
7
Solve problems involving
= 18.48 × 360° circles.

22 × 12 × 12
7

CHAPTER 5  = 14.7°

5.3.4 Solving problems

EXAMPLE 17

Majlis Bandaraya Melaka Bersejarah intends to build a rectangular
recreational park with a length of 63 m and a width of 58 m. At every
corner of the park, a quadrant with radius of 7 m will be planted with
flowers. A circular shaped fish pond with a diameter of 28 m will be
built in the middle of the park.The remaining areas will be planted 58 m

with grass. Calculate the area covered with grass.(Use π = 22 )
7
63 m
Solution:

Understanding the problem Planning the strategy
Radius of quadrant = 7 m
Garden is rectangular. Recreational park area = length × width
Length = 63 m
Width = 58 m Flower area = 4 × 1 πr 2
4
Diameter of fish pond = 28 m
Calculate the area covered with The fish pond area = πr2
grass.
Area covered with grass
Conclusion = Recreational area − flower area − fish pond area
Thus, the area covered with
grass is Implementing the strategy
3 654 m2 − 154 m2 − 616 m2
= 2 884 m2 (i) Recreational park area = 58 × 63 (iii) Fish pond area
= 3 654 m2
92 = πr2

(ii) Flower area = 4 × 1 × πr2 = 22 × 142
4 7
= 616 m2
= 22 × 72
7

= 154 m2