CHAPTER

1
CIRCULAR
MEASURE

SIA
AY
AL
M
AN
IK
ID
ND
PE
AN
RI

What will be learnt?

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Radian
EN

Arc Length of a Circle

Area of Sector of a Circle
Application of Circular Measures
M
KE

List of Learning
Standards

bit.ly/2QDBAxI
In the 21st century, technology and Info Corner
innovation are evolving at a very rapid
pace. Innovatively designed buildings
can increase the prestige of a country. Euclid (325-265 BC) was a Greek
An architect can design very unique mathematician from Alexandria. He is well
and beautiful buildings with special known for his work ‘The Elements’, a study in
software together with his or her creative the field of geometry.
and innovative abilities. How can the Geometrical mathematics is concerned
buildings be structurally sound and yet with sizes, shapes and relative positions in
retain their dynamic designs? What does diagrams and space characteristics.

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an architect need to know to design a
major segment of a circular building like
the one shown in the picture?

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For more info:

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M
bit.ly/35KqImk

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Significance of the Chapter
IK An air traffic controller uses his skills in
reading and interpreting radar at the air
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traffic control centre to guide planes to
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fly safely without any collision in the air,

which may result in injury and death.
Odometer in a vehicle records the total
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mileage covered from the beginning

to the end of the journey by using
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the circumference of the tyre and the

number of rotations of the tyre.
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TE

Key words
EN
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Radian Radian
Degree Darjah
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Centre of circle Pusat bulatan

Radius Jejari
Segment Tembereng
Sector Sektor
Video on round Perimeter Perimeter
building architecture Arc length Panjang lengkok
Area of sector Luas sektor

bit.ly/35E1wh1
1
 1.1 Radian

The diagram on the right shows two sectors marked on a

10 cm 18
dartboard with radii 10 cm and 20 cm and their respective arc 20 cm
lengths of 10 cm and 20 cm. Since each arc length is the same 10 cm 10 cm
length as its radius, the angle subtended at the centre of the circle 1 rad
is defined as 1 radian. That is, the size of the angle subtended by 6
20 cm
both arcs at the centre of the circle should be the same.

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What can you say about the measurement of the angle of
1 radian?

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Relating angle measurement in radians and degrees

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Information Corner
In circular measures, the normal unit used to measure angles is
in degrees. However, in some mathematical disciplines, circular

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• “Rad” stands for “Radian”.
measures in degrees are less suitable. Therefore, a new unit • 1 rad can be written as 1r
or 1c.
called the radian is introduced to measure the size of an angle.

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The activity below will explain the definition of one radian and at the same time relates
angles measured in degrees to those measured in radians.
IK
1
ID
Discovery Activity Group STEM CT
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Aim: To explain the definition of one radian and then relate angles measured
in radians to angles in degrees
Steps:
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1. Scan the QR code on the right or visit the link below it. bit.ly/2R1JvEe
2. Each group is required to do each of the following activities by recording
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the angle subtended at the centre of the circle.

Drag slider a such that the length of the arc, s is the same length as the radius
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of the circle, r.
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Drag slider a such that the length of the arc, s is twice the length of the radius
of the circle, r.
EN

Drag slider a such that the length of the arc, s is three times the length of the
radius of the circle, r.
M
KE

Drag slider a such that the length of the arc, s is the length of the semicircle.

Drag slider a such that the arc length, s is the length of the circumference of
the circle.
3. Based on the results obtained, define an angle of 1 radian. Then, relate radians to degrees
for the angle subtended at the centre of the circle.
4. From this relation, estimate an angle of 1 radian in degrees and an angle of 1° in radians.
Discuss your answer.
2 1.1.1
 Circular Measure

From the Discovery Activity 1, the definition of one radian is HISTORY GALLERY R
as follows: PTE

CHA
B
One radian is the measure of an r r
angle subtended at the centre of a 1 rad
circle by an arc whose length is the O r A
same as the radius of the circle.
Gottfried Wilhelm Leibniz

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was a brilliant German
mathematician who
In general, for a circle with centre O and radius r units: introduced a method to

AY
calculate the value of
If the arc length AB = r, then ˙AOB = 1 radian. π = 3.142 without using a
If the arc length AB = 2r, then ˙AOB = 2 radians.

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circle. He also proved that
π can be obtained by using
If the arc length AB = 3r, then ˙AOB = 3 radians. 4

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If the arc length AB = π r, then ˙AOB = π radians. the following formula.
π =1– 1 + 1 – 1
If the arc length AB = 2π r, then ˙AOB = 2π radians. 4 3 5 7

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+ 1 – 1 +…
Note that when the arc length AB is 2π r, it means that OA 9 11

has made a complete rotation or OA has rotated through 360°.

IK
Hence, we can relate radians to degrees as follows.
ID
2π rad = 360°
DISCUSSION
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π rad = 180° 1 radian is smaller than 60°.

What are the advantages
PE

of using angles in radians

Hence, when π = 3.142, compared to angles in
1 rad = 180° ≈ 57.29° degrees? Discuss.
π
AN

and 1° = π ≈ 0.01746 rad

180°
RI

1
TE

Example Calculator Literate

Convert each of the following angles into degrees.
EN

[Use π = 3.142] To find the solution for

(a) 2 π rad
Example 1(b) using a
(b) 2.25 rad
5
M

scientific calculator.
1. Press
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Solution
(a) π rad = 180° (b) π rad = 180° 2. Press
2 π rad = 2 π × 180° 2.25 rad = 2.25 × 180°
5 5 π π 3. The screen will display
2 = 2.25 × 180°
= × 180° 3.142
5
= 72° = 128° 54

1.1.1 3
Example 2 Excellent Tip
(a) Convert 40° and 150° into radians, in terms of π.
Special angles:
(b) Convert 110° 30 and 320° into radians.
Angle in Angle in
[Use π = 3.142] degree radian
Solution 0° 0
π
(a) 180° = π rad (b) 180° = π rad 30° 6
40° = 40° × π π
110° 30 = 110° 30 × π
36°
180° 180° 5

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= 2 π rad = 110° 30 × 3.142 45°
π
9 180° 4
π
150° = 150° × π = 1.929 rad

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60°
180° 320° = 320° × π
3
π
5 180° 90°
= π rad

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2
6
= 320° × 3.142 180° π
180°

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3π
= 5.586 rad 270° 2
360° 2π

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Self-Exercise 1.1 IK
1. Convert each of the following angles into degrees. [Use π = 3.142]
ID
(a) π rad (b) 3 π rad (c) 0.5 rad (d) 1.04 rad
8 4
ND

2. Convert each of the following angles into radians, in terms of π.

(a) 18° (b) 120° (c) 225° (d) 300°
PE

1.1
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Formative Exercise Quiz bit.ly/2QGcIWr

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1. Convert each of the following angles into degrees. [Use π = 3.142]

(a) 7 π rad (b) 1 1 π rad
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(c) 2 rad (d) 4.8 rad

12 3
EN

2. Convert each of the following angles into radians. Give answers correct to three decimal
places. [Use π = 3.142]
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(a) 76° (b) 139° (c) 202.5° (d) 320° 10

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3. In each of the following diagrams, POQ is a sector of a circle with centre O. Convert each
of the angles POQ into radians. [Use π = 3.142]
(a) (b) (c) (d) P
Q P
Q
O
O
118° 150.5°
73° P 220°
O Q
O
P Q

4 1.1.1
 Circular Measure

1.2 Arc Length of a Circle PTE

R

CHA
The diagram on the right shows a little girl on a swing. The
swing sweeps through 1.7 radians and makes an arc of a circle. 2.5 m
What is the arc length made by the little girl on that swing?
What formula can be used to solve this problem?

Determining the arc length, radius and the

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angle subtended at the centre of a circle

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Discovery Activity 2 Group 21st cl STEM CT

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Aim: To derive the formula for the arc length of a circle with centre O

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Steps:
1. Scan the QR code on the right or visit the link below it. ggbm.at/haatecxq

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2. Drag the point A or B along the circumference of the circle to change the
arc length AB.
IK
3. Note the arc length AB and the angle AOB in degrees subtended at the centre of the circle
when the point A or B changes.
ID
Minor arc length AB
4. What do you observe concerning the value of the ratios and
ND

Circumference
Angle AOB
? Are the ratios the same?
360°
PE

5. Drag the slider L to vary the size of the circle. Are the two ratios from step 4 above still
the same?
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6. Then, derive a formula to determine the minor arc length of a circle.

7. Record all the results from the members of your group on a piece of paper.
RI

8. Each group presents their findings to the class and finally come up with a conclusion
concerning this activity.
TE

From Discovery Activity 2, it is found that the arc length of a circle is proportional to the angle
EN

subtended at the centre of the circle.

Minor arc length AB Circumference
M

= B
∠AOB 360°
KE

Minor arc length AB r

= 2π r
q 360° θ
O A
r
2π r
Minor arc length AB = ×q
360°

where q is the angle in degrees subtended at the centre of the circle, O whose radius is r units.

1.2.1 5
However, if ˙AOB is measured in radians,
Information Corner
Minor arc length AB
= Circumference B
q 2π The symbol q is read as
s
s = 2π r r “téta”, which is the eighth
q 2π θ letter in the Greek alphabet
2π r ×q O r A and it is often used to
s=
2π represent an angle.

s = rq

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In general, DISCUSSION
s = rq From the definition of

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radian, can you derive the
where s is the arc length of the circle with radius r units and formula s = rq ?
q radian is the angle subtended by the arc at the centre of the

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circle, O.

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Example 3

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Find the arc length, s for each of the following sectors POQ with centre O.
[Use π = 3.142] IK
(a) (b) (c)
s
ID
P P
s
s
ND

5 cm 6 cm 2– π rad O 10 cm
Q
Q 3
140°
O
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O 0.9 rad Q P

AN

Solution
(a) Arc length, s = rq (b) Arc length, s = rq
RI

s = 5 × 0.9 s=6× 2π
s = 4.5 cm 3
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s = 4π
s = 4(3.142)
EN

s = 12.57 cm
(c) Ref lex angle POQ in radians
M

= (360° – 140°) × π Recall

180°
KE

= 220° × 3.142 The angle size of a reflex

180° angle is 180° , q , 360°.
= 3.84 rad
Arc length, s = rq
s = 10 × 3.84 θ
s = 38.4 cm

6 1.2.1
 Circular Measure

Example 4 Recall R
PTE
The diagram on the right
1

CHA
B 1.4 cm Major Major
shows a part of a circle with C
sector arc
centre O and a radius of r cm. 2.6 cm Minor
Given that ˙AOB = 1.3 rad sector
and the arc lengths AB and 1.3 rad O Minor
BC are 2.6 cm and 1.4 cm A
r cm O arc
respectively, calculate
(a) the value of r, Chord

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Segment
(b) ˙BOC, in radians.
Solution

AY
(a) For sector AOB, (b) For sector BOC,
s = 2.6 cm and QR Access

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s = 1.4 cm and r = 2 cm.
q = 1.3 rad. Hence, s = rq
Thus, s = rq
q= s

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Recognising a circle
r= s r
q 1.4

AN
q=
r = 2.6 2
1.3 q = 0.7 rad
r = 2 cm Thus, ˙BOC = 0.7 rad. IK bit.ly/37Tju0u
ID
Self-Exercise 1.2
ND

1. Find the arc length MN, in cm, for each of the following sectors MON with centre O.
[Use π = 3.142]
PE

(a) (b) (c) (d) M

M N
AN

M
5 cm
12 cm 2 rad O 5– π rad O M
6 10 cm
RI

O 2.45 rad
1.1 rad 8 cm
O P
TE

N N N

EN

2. The diagram on the right shows a circle with centre O.

E
Given that the major arc length EF is 25 cm and
M

∠EOF = 1.284 rad, find 25 cm 1.284 rad

(a) the radius, in cm, of the circle, O
KE

(b) the minor arc length EF, in cm. F

[Use π = 3.142]
3. The diagram on the right shows semicircle OPQR with a radius Q
of 5 cm. Given that the arc length QR is 5.7 cm, calculate 5.7 cm
(a) the value of q, in radians, θ
(b) the arc length PQ, in cm.
P 5 cm O R
[Use π = 3.142]
1.2.1 7
 Determining the perimeter of segment of a circle

The coloured region of the rim of the bicycle tyre with a

radius of 31 cm in the diagram consists of three identical
segments of a circle. The perimeter for one of the
segments is the sum of all its sides.
With the use of the arc length formula s = rq
and other suitable rules or formulae, can you find the
perimeter of any one of the segments?

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Example 5

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The diagram on the right shows a circle A
Alternative Method

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with centre O and a radius of 10 cm. To find the chord AC, draw a
The chord AC subtends an angle of 114° 114° perpendicular line, OD from
O to chord AC.

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at the centre of the circle. Calculate the O B In ∆ COD,
perimeter of the shaded segment ABC.
10 cm ˙COD = 114°

AN
[Use π = 3.142] 2
= 57
C
Solution sin ˙COD = CD
Since 180° = π rad, we have
IK OC
Hence, CD = OC sin ˙COD
114° = 114° × π
ID
= 10 sin 57°
180° = 8.3867 cm
= 1.990 rad
ND

Thus, AC = 2CD
= 2(8.3867)
Arc length ABC = rq = 16.77 cm
= 10 × 1.990
PE

= 19.90 cm
With cosine rule, the length of chord AC is
AN

AC 2 = 102 + 102 – 2(10)(10) cos 114°

Flash Quiz
AC = ! 200 – 200 cos 114°
RI

Can the length of AC be

= 16.77 cm
TE

obtained using sine rule,

Thus, the perimeter of the shaded segment ABC = 19.90 + 16.77 a = b = c ?
sin A sin B sin C
= 36.67 cm
EN
M

Self-Exercise 1.3
KE

1. For each of the following circles with centre O, find the perimeter, in cm, of the shaded
segment ABC. [Use π = 3.142]
(a) (b) (c) (d)
B B
C A
2.5 rad π
– rad
A 3 O
C
O 120° 9 cm C
A 6 cm O B O 8 cm
10 cm m
A 15 c
C
B
8 1.2.2
 Circular Measure

2. The diagram on the right shows a sector with centre O and a P R

PTE
radius of 7 cm. Given that the arc length PQ is 14 cm, find
1
14 cm

CHA
(a) the angle q, in degrees,
7 cm
(b) the perimeter of the shaded segment, in cm.
θ
O
Q

Solving problems involving arc lengths

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With the knowledge and skills of converting angles from degrees to radians and vice versa, as
well as the arc length formula, s = rq and other suitable rules, we can solve many problems in

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our daily lives involving arc length of a circle.

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Example 6 MATHEMATICAL APPLICATIONS
P

M
Q
The diagram on the right shows the region for the shot put event
drawn on a school field. The region is made up of two sectors from

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two circles, AOB and POQ, both with centre O. Given that 8m
˙AOB = ˙POQ = 50°, OA = 2 m and AP = 8 m, calculate the A B
perimeter of the coloured region ABQP, in m. [Use π = 3.142]
IK 2m
O
ID
Solution
ND

1 . Understanding the problem 2 . Planning the strategy

The shot put region consists of two Convert 50° into radians and use the
PE

sectors AOB and POQ from two formula s = rq to find the arc lengths
circles, both with centre O. AB and PQ.
AN

The sector AOB has a radius of 2 m, The perimeter of the shaded region
AP = 8 m and ˙AOB = ˙POQ = 50°. ABQP can be obtained by adding all
the sides enclosing it.
RI
TE

3 . Implementing the strategy

EN

180° = π rad
50° = 50° × 3.142
M

180°
= 0.873 rad
KE

Arc length AB, s = rq Thus, the perimeter of the shaded

s = 2(0.873) region ABQP
s = 1.746 m = arc length AB + BQ + arc length PQ + AP
Arc length PQ, s = rq = 1.746 + 8 + 8.73 + 8
s = 10(0.873) = 26.48 m
s = 8.73 m

1.2.2 1.2.3 9
 4 . Check and reflect
Arc length AB = 50° (2)(3.142)(2) Thus, the perimeter of the shaded
360° region ABQP
= 1.746 m = arc length AB + BQ
+ arc length PQ + AP
Arc length PQ = 50° (2)(3.142)(10)
360° = 1.746 + 8 + 8.73 + 8
= 8.73 m = 26.48 m

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Self-Exercise 1.4

AY
1. In each of the following diagrams, calculate the perimeter, in cm, of the shaded region.

AL
(a) (b) (c)
C C

M
A O
5 cm
3 cm
A 10 cm B

AN
4 cm 110°
O D C
O B D 3 cm B 1 cm A 0.5 rad
IK
2. The city of Washington in United States of America and the city of Lima in Peru lie on the
ID
same longitude but are on latitudes 38.88° N and 12.04° S respectively. Given that the earth
is a sphere with a radius of 6 371 km, estimate the distance, in km, between the two cities.
ND

3. The diagram on the right shows a part of a running

O
track which is semicircular in shape. Fazura wants to 25 m
PE

pass the baton to Jamilah, who is waiting at 85° from 85°

her. How far must Fazura run in order to pass the baton
to Jamilah?
AN

Fazura Jamilah
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4. The diagram on the right shows a window which

consists of a rectangle and a semicircle. The width
TE

and height of the rectangle are 70 cm and 100 cm

respectively. Find
EN

(a) the arc length of the semicircle of the window, 100 cm

in cm,
M

(b) the perimeter of the whole window, in cm.

70 cm
KE

5. The diagram shows the chain linking the front

25 cm
and back cranks of a bicycle. It is given that the
circumference of the front and back cranks are
50.8  cm and 30.5 cm respectively. Calculate the
length of the bicycle chain, in cm.
160°
25 cm 185°

10 1.2.3
 Circular Measure

Formative Exercise 1.2 Quiz

PTE
R

1
bit.ly/39W9p4V

CHA
1. The diagram on the right shows a circle with centre O. The
minor arc length RS is 15 cm and the angle of the major R
sector ROS is 275°. Find
15 cm O 275°
(a) the angle subtended by the minor sector ROS, in radians,
(b) the radius of the circle, in cm. S

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2. The diagram on the right shows sector UOV with centre O. U
Given that the arc length UV is 5 cm and the perimeter of

AY
5 cm
sector UOV is 18 cm, find the value of q, in radians.

AL
θ
O V

M
3. The diagram on the right shows sector EOF of a circle with E
centre O. Given that OG = 4 cm and OE = 5 cm, find

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5 cm
(a) the value of q, in radians,
(b) the perimeter of the shaded region, in cm. θ
IK O 4 cm G F

4. The diagram on the right shows two sectors, OPQ and

ID
R
ORS, with centre O and radii 2h cm and 3h cm respectively. P
Given that ˙POQ = 0.5 radian and the perimeter of the
ND

2h
shaded region PQSR is 18 cm, find
(a) the value of h, in cm, 0.5 rad
O
PE

(b) the difference in length, in cm, between the arc lengths Q S

of RS and PQ. 3h
AN

5. The diagram on the right shows a part of a circle with M

centre O and a radius of 10 cm. Tangents to the circle
RI

10 cm
at point M and point N meet at P and ˙MON = 51°.
TE

Calculate O 51° P
(a) arc length MN, in cm,
EN

(b) the perimeter of the shaded region, in cm.

N
M

6. A wall clock has a pendulum with a length of 36 cm. If it swings through an angle of 21°,
KE

find the total distance covered by the pendulum in one complete oscillation, in cm.
7. The diagram on the right shows the measurement of a car
tyre. What is the distance travelled, in m, if it makes 14 cm
(a) 50 complete oscillations?
(b) 1 000 complete oscillations? 38 cm
[Use π = 3.142]
14 cm

11
 1.3 Area of Sector of a Circle

A pizza with a radius of 10 cm is cut into 10 equal pieces. Can

you estimate the surface area of each piece?
What formula can be used to solve this problem?

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Determining the area of sector, radius and the angle subtended at the
centre of a circle

AY
The area of a sector of a circle is the region bounded by the arc length and the two radii. The
following discovery activity shows how to derive the formula for the area of a sector of a circle

AL
by using the dynamic GeoGebra geometry software.

M
Discovery Activity Group 21st cl STEM CT

AN
Aim: To derive the formula for the area of a sector of a circle with centre O
Steps:
IK
1. Scan the QR code on the right or visit the link below it. ggbm.at/rdpf3rx9
2. Drag the point A or B along the circumference to change the area of
ID
the minor sector AOB.
ND

3. Pay attention to the area of the sector AOB and the angle AOB in degrees subtended at the
centre of the circle when the point A or B changes.
4. What are your observations on the values of the ratios Area of minor sector AOB and
PE

Area of the circle

Angle AOB
? Are the values of the two ratios the same?
360°
AN

5. Drag the slider L to change the size of the circle. Are the two above ratios still the same?
6. Subsequently, derive the formula for the area of a minor sector of a circle. Record all the
RI

values from the members of your group on a piece of paper.

TE

7. Each group presents their findings to the class and subsequently draws a conclusion from
this activity.
EN

8. Members from other groups can give feedback on the presentations given.
M

From Discovery Activity 3, we found that:

Area of minor sector AOB = Area of the circle B
KE

∠AOB 360° r
Area of minor sector AOB = π r 2 O θ
q 360°
r
Area of minor sector AOB = π r × q
2

360° A

where q is the angle in degrees subtended at the centre of the circle, O whose radius is r units.

12 1.3.1
 Circular Measure

However, if ˙AOB = q is measured in radians, R

PTE
Area of minor sector AOB = Area of the circle QR Access
1

CHA
q 2π B
A = πr 2
r Alternative method to
q 2π derive the formula of area
O θ A of a sector of a circle,
π
A= r ×q
2
A = 1 r 2q.
2π r 2
1
A = r 2q A
2

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In general,
A = 1 r 2q
bit.ly/39YqDOT
2

AY
where A is the area of a sector of the circle with radius r units and

AL
q radian is the angle subtended by the sector at the centre O of the circle.

M
Example 7

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Find the area of sector, A for each sector MON with centre O. [Use π = 3.142]
(a) (b) (c)
M
M
IK
O M
ID
2.2 rad
12 cm 8 cm O 124°
1.7 rad
ND

N 10 cm
O N N

PE

Solution
(a) Area of the sector, A = 1 r 2q (b) Area of the sector, A = 1 r 2q
AN

2 2
A = 1 (12)2(1.7) A = 1 (8)2(2.2)
2 2
RI

1
A = (14 4)(1.7) A = 1 (6 4)(2.2)
2 2
TE

A = 122.4 cm2 A = 70.40 cm2

EN

(c) Ref lex angle MON in radians

= (360° – 124°) × π
Information Corner
180°
M

= 236° × 3.142 Area of a sector, A is A = 1 r 2q,

180°
KE

2
where q is the angle in
= 4.12 rad radians. Since s = rq,
Area of the sector, A = 1 r 2q we obtained:
2 A = 1 r(rq)
A = 1 (10)2(4.12)
2

2 1
A = rs
A = 1 (100)(4.12)
2

2
A = 206 cm2

1.3.1 13
Example 8
P
The diagram on the right shows a sector POQ which subtends an r cm
angle of q radians and has a radius of r cm. Given that the area of
the sector POQ is 35 cm2, find θ O
(a) the value of r if q = 0.7 rad,
(b) the value of q if the radius is 11 cm. Q

Solution
(a) Area of sector POQ = 35 cm2 (b) Area of sector POQ = 35 cm2

SIA
1 r 2q = 35 1 r 2q = 35
2 2
1 1 (11)2q = 35

AY
r (0.7) = 35
2
2 2
r 2 = 35 × 2 1 (121)q = 35

AL
0.7 2
r 2 = 100
q = 35 × 2

M
r = ! 100 121
r = 10 cm q = 0.5785 rad

Self-Exercise 1.5
AN
IK
1. For each of the following sectors of circles AOB with centre O, determine the area, in cm2.
ID
[Use π = 3.142]
(a) (b) (c) (d) A
ND

A 5– π rad
3
O
PE

1.1 rad 10 cm A O 135°

2.15 rad O
6 cm 5 cm
O 20 cm
AN

A B B B B
RI

2. A sector of a circle has a radius of 5 cm and a perimeter of 16 cm. Find the area of the
TE

sector, in cm2.
3. The diagram on the right shows a major sector EOF with E
EN

centre O, a radius of r cm and an area of 195 cm2. Calculate

(a) the value of r, in cm,
M

O r cm
(b) the major arc length EF, in cm,
(c) the perimeter of the major sector EOF, in cm. 3.9 rad F
KE

4. The diagram on the right shows a sector VOW with centre O

and a radius of 10 cm. Given that the area of the sector is O 10 cm
V
60 cm2, calculate θ
(a) the value of q, in radians,
(b) the arc length VW, in cm,
(c) the perimeter of sector VOW, in cm. W
14 1.3.1
 Circular Measure

Determining the area of segment of a circle R

PTE

CHA
The diagram on the right shows a circular piece of a table cloth with
centre O with an inscribed hexagon pattern. The laces sewn around the
hexagon form segments on the table cloth. What information is needed
to find the area of each lace? O
By using the formula of a sector, A = 1 r 2q and other suitable
2
formulae, this problem can be solved easily and fast.

SIA
9

AY
Example

For each of the following given sectors POQ with centre O, find the area of the

AL
segment PRQ, in cm2.
[Use π = 3.142]

M
(a) (b) Q
Q

AN
R 3.5 cm

O 4 cm R
2.2 rad IK
ID
O 6 cm P P
Alternative Method
ND

Solution Q
(a) 2.2 rad = 2.2 × 180°
3.142
PE

S
= 126° 2
63°1'
Area of sector POQ = 1 r 2q
O
AN

6 cm P
2
= 1 (6)2(2.2)
In ∆ POQ,

2
RI

∠POS = 126° 2
= 39.60 cm2 2
TE

= 63° 1
Area of ∆ POQ = 1 (OP)(OQ) sin ˙POQ sin 63° 1 = PS
2 6
EN

= 1 (6)(6) sin 126° 2 PS = 6 × sin 63° 1

2 = 5.3468 cm
= 14.56 cm2
M

PQ = 2PS
Area of the segment PRQ = 39.60 – 14.56 = 2 × 5.3468
KE

= 25.04 cm2 = 10.6936 cm

Q
QS OS = ! 62 – 5.34682
(b) In ∆ QOP, sin ˙QOS = = 2.7224 cm
OQ 3.5 cm
2 cm
2 Therefore, area of ∆ POQ
=
3.5 O S = 1 × PQ × OS
2
˙QOS = 34° 51 = 1 × 10.6936 × 2.7224
2
= 14.56 cm2
P

1.3.2 15
 Hence, ˙POQ = (2 × 34° 51) × π
180° Recall
= 69° 42 × 3.142
180° C
= 1.217 rad
Area of sector POQ = 1 r 2q b a
2
= 1 (3.5)2(1.217) A c B
2
= 7.454 cm2 (a) Area of ∆ ABC

SIA
In ∆ POQ, the semiperimeter, s = 3.5 + 3.5 + 4
= 1 ab sin C
2
2
s = 5.5 cm = 1 ac sin B

AY
2
Area of ∆ POQ = ! s(s – p)(s – q)(s – o) = 1 bc sin A
2

AL
= ! 5.5(5.5 – 3.5)(5.5 – 3.5)(5.5 – 4) (b) Formula to find area of
triangle by using
= ! 5.5(2)(2)(1.5)

M
Heron’s formula:
= ! 33 Area of ∆ ABC
= 5.745 cm2 = ! s(s – a)(s – b)(s – c),

AN
Area of the segment PRQ = 7.454 – 5.745 where s = a + b + c is
2
= 1.709 cm2 the semiperimeter.
IK
ID
ND

Self-Exercise 1.6
1. For each of the following sectors AOB with centre O, find the area of the segment ACB.
PE

[Use π = 3.142]
(a) (b) (c) (d) A
AN

C A
A C C
5 cm
A B 9 cm
2– π rad C 58° O
RI

cm

7 cm 1.5 rad 3
15

O
TE

O B
O 10 cm B B
EN

2. The diagram on the right shows sector MON of a circle with M

centre O and a radius of 3 cm. Given that the minor arc length 3 cm
MN is 5 cm, find
M

O 5 cm
(a) ˙MON, in degrees,
KE

(b) the area of the shaded segment, in cm2. N

H
3. The diagram on the right shows sector HOK of a circle with
centre O and a radius of 4 cm. The length of chord HK is the
same as the length of the radius of the circle. Calculate K
4 cm O
(a) ˙HOK, in radians,
(b) the area of the shaded segment, in cm2.

16 1.3.2
 Circular Measure

Solving problems involving areas of sectors R

PTE

CHA
The knowledge and skills in using the area of a sector formula, A = 1 r 2q or other suitable
2
formulae can help us to solve many daily problems involving areas of sectors.

Example 10 MATHEMATICAL APPLICATIONS

The diagram on the right shows a paper fan fully spread

SIA
out. The region PQNM is covered by paper. Given that
OP = 15 cm, OM : MP = 2 : 3 and ∠POQ = 120°,

AY
P 120°
calculate the area covered by the paper, in cm2. Q
M N

AL
O
Solution

M
1 . Understanding the problem 2 . Planning the strategy

AN
PQNM is the region covered with Find the length of OM by using the ratio
paper when the paper fan is opened IK OM : MP = 2 : 3.
up completely. Convert 120° into radians and use the
Given OP = 15 cm, OM : MP = 2 : 3 formula A = 1 r 2q to find the area of
ID
and ∠POQ = 120°. 2
Find the area, in cm2, of the region the sector POQ and the area of the
ND

covered by the paper. sector MON.

Subtract the area of the sector MON
from the area of the sector POQ to
PE

obtain the area covered by the paper.

AN

3 . Implementing the strategy

RI

OM = 2 × OP Area of sector POQ, A = 1 r 2q

TE

5 2
= 2 × 15 1
A = (15)2(2.0947)
5
EN

2
= 6 cm A = 235.65 cm2
π Area of sector MON, A = 1 r 2q
M

q in radians = 120° ×
180° 2
1
KE

= 120° × 3.142 A = (6)2(2.0947)

2
180°
= 2.0947 rad A = 37.70 cm2
Thus, the area covered by the paper
= 235.65 – 37.70
= 197.95 cm2

1.3.3 17
 4 . Check and reflect
Excellent Tip
Area of sector POQ, A = 120° × 3.142 × 152
360° A
A = 235.65 cm2 r
Area of sector MON, A = 120° × 3.142 × 62 θ
A
360° O B
A = 37.70 cm2
If the angle q is in degrees,
Thus, the area covered by the paper

SIA
then the area of the sector
= 235.65 – 37.70 of a circle, A = q × π r 2.
360°
= 197.95 cm2

AY
AL
Self-Exercise 1.7

M
1. The diagram on the right shows a semicircular garden

AN
R
SRT with centre O and a radius of 12 m. The region 16 m
PQR covered by grass is a sector of circle with
IK
centre Q and radius 16 m. The light brown coloured 14 m
patch is fenced and planted with flowers. Given that the
ID
arc length PR is 14 m, find S P O Q T
(a) the length of the fence, in m, used to fence around
ND

12 m
the flowers,
(b) the area, in m2, planted with flowers.
PE

2. The diagram on the right shows the cross-section

of a water pipe with the internal radius of 12 cm. 12 cm O
AN

Water flows through it to a height of h cm and the

horizontal width of the water, EF is 18 cm. Calculate
h cm E 18 cm
(a) the value of h, F
RI

(b) the cross-section area covered by water, in cm2.

TE

3. The diagram on the right shows two discs with radii

EN

11 cm and 7 cm touching each other at R. The discs A

are on a straight line PDCQ. R
M

B
(a) Calculate ˙BAD, in degrees. 11 cm
7 cm
(b) Subsequently, find the shaded area, in cm2.
KE

P Q
D C

4. The diagram on the right shows a wall clock showing the

time 10:10 in the morning. Given that the minute hand is
8 cm, find
(a) the area swept through by the minute hand when the
time shown is 10:30 in the morning, in cm2,
(b) the angle, in radians, if the area swept through by the
minute hand is 80 cm2.
18 1.3.3
 Circular Measure

Formative Exercise 1.3 Quiz

PTE
R

1
bit.ly/2NdT3uH

CHA
1. The diagram on the right shows sector AOB with centre O and B
another sector PAQ with centre A. It is given that OB = 6 cm,
OP = AP, ˙PAQ = 0.5 rad and the arc length AB is 4.2 cm. 6 cm 4.2 cm
Calculate Q
(a) the value of q, in radians, θ
(b) the area of the shaded region, in cm2. O A

SIA
P
0.5 rad

2. The diagram on the right shows sector VOW with centre O V

AY
and a radius of 5 cm. Given that OW = OV = VW, find
(a) the value of q, in radians,

AL
(b) the area of the shaded segment VW, in cm2.
θ

M
O 5 cm W
3. A cone has a base with a radius of 3 cm and a Q

AN
height of 4 cm. When it is opened up, it forms
sector POQ as shown on the right. Given that 4 cm
O
˙POQ = q radian, find IK P θ
(a) the value of q,
3 cm
ID
(b) the area of sector POQ, in cm2.
ND

K
4. The diagram on the right shows a circle with centre O and a 4 cm
radius of 4 cm. It is given that the minor arc length KL is 7 cm.
O θ 7 cm
(a) State the value of q, in radians.
PE

(b) Find the area of the major sector KOL, in cm2.

L
AN

5. In the diagram on the right, O is the centre of the circle with

radius 9 cm. The minor arc AB subtends an angle of 140° at the A 9 cm
centre O and the tangents at A and B meet at C. Calculate O
RI

(a) AC, in cm, 140°

TE

(b) the area of the kite shaped OACB, in cm , 2

B
(c) the area of the minor sector OAB, in cm2,
EN

(d) the area of the shaded region, in cm2. C

6. The diagram on the right shows a circular ventilation window Q
M

in a hall. PQR is a major arc of a circle with centre S. The

lines OP and OR are tangents to that circle. The other four S
KE

panels are identical in size to OPQR. O is the centre of P R

ventilation window that touches the arc PQR at Q. It is given 6 cm 60°
that OS = 6 cm and ˙OSR = 60°. O
(a) Show that RS = 3 cm.
(b) Calculate the area of the panel OPQR, in cm2.
(c) The window has a rotational symmetry at O to the T
nth order; find the value of n and the area labelled T
between two panels, in cm2.
19
 1.4 Application of Circular Measures

Study the following two situations in daily lives.

A rainbow is an optical phenomenon which displays a

spectrum of colours in a circular arc. A rainbow appears
when the sunlight hits the water droplets and it usually
appears after a rainfall. The rainbow shown in the photo

SIA
is an arc of a circle. With the formula that you have
learned and the help of the latest technology, can you
determine the length of this arc?

AY
AL
M
The cross-section of a train tunnel is usually in the

AN
form of a major arc of a circle. How do we find the
arc length and the area of this cross-section tunnel?
IK
ID

The ability to apply the formulae from circular measures, that is, the arc length, s = rq and
ND

the area of a sector, A = 1 r 2q, where q is the angle in radians and other related formulae, can
2
help to solve the problems mentioned above.
PE

Solving problems involving circular measures

AN

The following example shows how the formula in circular measures and other related formulae
are used to solve problems related to the cross-section of a train tunnel in the form of a major
RI

segment of a circle.
TE

Example 11
EN

The diagram on the right shows a major segment ABC of B

a circular train tunnel with centre O, radius of 4 m and
M

˙AOC = 1.8 rad.

[Use π = 3.142]
KE

(a) Show that AC is 6.266 m.

(b) Find the length of major arc ABC, in m. O
(c) Find the area of the cross-section of the train
4m
tunnel, in m2. 1.8 rad

A C

20 1.4.1
 Circular Measure

Solution O R
PTE
(a) 1.8 rad = 1.8 × 180°
1

CHA
4m 1.8 rad 4m
3.142
= 103° 7
By using the cosine rule, A C
AC 2 = OA2 + OC 2 – 2(OA)(OC) cos ˙AOC
= 42 + 42 – 2(4)(4) cos 103° 7
AC = ! 42 + 42 – 2(4)(4) cos 103° 7
= ! 39.2619

SIA
= 6.266 m
B
(b) Ref lex angle AOC = 2π − 1.8

AY
= 4.484 rad
Length of major arc ABC = rq 4.484 rad

AL
= 4 × 4.484
4m O
= 17.94 m

M
(c) By using the area of a triangle formula: A C
Area ∆ AOC = 1 × OA × OC × sin ˙AOC

AN
2 B
1
= × 4 × 4 × sin 103° 7

2
= 7.791 m2
IK
4.484 rad
ID
Area of the major sector ABC = 1 r 2q O
2 4m
ND

1.8 rad
= 1 × 42 × 4.484
2 A C
= 35.87 m2
PE

Thus, the cross-section area of the train tunnel is 7.791 + 35.87 = 43.66 m2
AN

Self-Exercise 1.8
RI

1. The diagram on the right shows a moon-shaped kite whose O

line of symmetry is OS. AQB is an arc of a sector from 20 cm
TE

a circle with centre O and a radius of 20 cm. APBR is a P

A B
16 cm
EN

semicircle with centre P and a radius of 16 cm. TRU is also

an arc from a circle with centre S and a radius of 12 cm. Q
Given that the arc length of TRU is 21 cm, calculate
M

(a) ˙AOB and ˙TSU, in radians, R

KE

T U
(b) the perimeter of the kite, in cm,
12 cm
(c) the area of the kite, in cm2.
S

2. In the diagram on the right are three identical 20 cent coins

with the same radii and touching each other. If the blue
coloured region has an area of 12.842 mm2, find the radius of
each coin, in mm.

1.4.1 21
Formative Exercise 1.4 Quiz bit.ly/2FzIlu7

1. A cylindrical cake has a radius and a height of 11 cm and

8 cm respectively. The diagram on the right shows a uniform Q
cross-section of a slice of a cake in the form of a sector POQ P
being cut out from the cylindrical cake with centre O and 11 cm
8 cm
a radius of 11 cm. It is given that ˙POQ = 40°.
O
(a) Calculate
(i) the perimeter of sector POQ, in cm,

SIA
(ii) the area of sector POQ, in cm2,
(iii) the volume of the piece of cake that has been cut out, in cm3.
(b) If the mass of a slice of the cake that has been cut out is 150 g, calculate the mass of the

AY
whole cake, in grams.

AL
2. The diagram on the right shows the plan of a swimming 12 m
A B
pool with a uniform depth of 1.5 m. ABCD is a rectangle

M
with the length of 12 m and the width of 8 m. AED and
BEC are two sectors from a circle with centre E. Calculate

AN
8m
(a) the perimeter of the floor of the swimming pool, in m, E
(b) the area of the floor of the swimming pool, in m2,
IK
(c) the volume of the water needed to fill the swimming
D C
pool, in m3.
ID

3. The diagram on the right shows the cross-section area of R

ND

10 cm P Q
a tree trunk with a radius of 46 cm floating on the water.
The points P and Q lie on the surface of the water while θ
PE

46 cm
the highest point R is 10 cm above the surface of the
water. Calculate O
(a) the value of q, in radians,
AN

(b) the arc length PRQ, in cm,

(c) the cross-section area that is above the water, in cm2.
RI

4. The diagram on the right shows the logo of an ice cream

TE

company. The logo is made up of three identical sectors

EN

AOB, COD and EOF from a circle with centre O and a

radius of 30 cm. It is given that ˙AOB = ˙COD A B
= ˙EOF = 60°.
M

(a) Calculate 30 cm
KE

(i) the arc length of AB, in cm, F C

(ii) the area of sector COD, in cm2, O
(iii) the perimeter of segment EF, in cm,
(iv) the area of segment EF, in cm2.
(b) The logo is casted in cement. If the thickness is D
E
uniform and is 5 cm, find the amount of cement
needed, in cm3, to make the logo.
(c) If the cost of cement is RM0.50 per cm3, find the total cost, in RM, to make the logo.
22
 Circular Measure

REFLECTION CORNER PTE

R

CHA
CIRCULAR MEASURE

Convert radians into Arc length Area of a sector

SIA
degrees and vice versa of a circle of a circle

AY
A A

AL
r r
× 180° O θ C s O θ A C

M
π
Radians Degrees B B

AN
× π Arc length, s = rq Area of sector, A = 1 r 2q
180° 2
Perimeter of segment ABC
IK Area of segment ABC
= s + AB = A – area of ∆  AOB
ID
ND

Applications
PE
AN
RI

Journal Writing
TE
EN

1. Are you more inclined to measure an angle of a circle in degrees or radians? Give
justification and rationale for your answers.
M

2. Visit the website to obtain the radius, in m, for the following six Ferris wheels:
KE

(a) Eye on Malaysia (b) Wiener Riesenrad, Vienna (c) The London Eye
(d) Tianjin Eye, China (e) High Roller, Las Vegas (f) The Singapore Flyer
If the coordinates of the centre of each Ferris wheel is (0, 0), determine
(i) the circumference of each Ferris wheel, in m,
(ii) the area, in m2, covered by each Ferris wheel in one complete oscillation,
(iii) the equation for each Ferris wheel.

23
 Summative Exercise
1. The diagram on the right shows sector KOL from a circle K
with centre O and a radius of 10 cm. Given that the area of 10 cm
the sector is 60 cm2, calculate PL 2
θ O
(a) the value of q, in radians,
(b) the perimeter of sector KOL, in cm.
L

SIA
2. The diagram on the right shows sector AOB from a circle A
with centre O. Given that AD = DO = OC = CB = 3 cm,

AY
find PL 2 D
(a) the perimeter of the shaded region, in cm,

AL
2 rad
(b) the area of the shaded region, in cm2.
O C B

M
AN
3. The diagram on the right shows sectors POQ and ROS R
with the same centre O. Given that OP = 4 cm, the ratio P
OP : OR = 2 : 3 and the area of the shaded region is 10.8 cm2,
IK 4 cm
find PL 3 O θ
ID
(a) the value of q, in radians,
(b) the perimeter of the shaded region, in cm.
ND

Q
S

4. The diagram on the right shows sector MON from a circle with M
PE

an angle of q radian and a radius of r cm. It is given that the

perimeter of the sector is 18 cm and its area is 8 cm2. PL 3 N r cm
AN

(a) Form a pair of simultaneous equations containing r and q. θ

(b) Subsequently, find the values of r and q.
RI

5. The diagram on the right shows a square ABCD with a side

TE

A P B
of 4 cm. PQ is an arc from a circle with centre C whose
radius is 5 cm. Find PL 3
EN

Q
5 cm
(a) ˙PCQ, in degrees,
(b) the perimeter of the shaded region APQ, in cm,
M

(c) the area of the shaded region APQ, in cm2.

KE

D 4 cm C

6. The diagram on the right shows a quadrant with centre O and R

a radius of 10 cm. Q is on the arc of the quadrant such that
the arc lengths PQ and QR are in the ratio 2 : 3. Given that Q
˙POQ = q radian, find PL 3
(a) the value of q, θ
P
(b) the area of the shaded region, in cm2. 10 cm O

24
 Circular Measure

7. In the diagram on the right, PQRS is a semicircle with Q R R

PTE
centre O and a radius of r cm. Given that the arc lengths
1

CHA
of PQ, QR and RS are the same, calculate the area of the
shaded region, in cm2. Give the answer in terms of r.
[Use π = 3.142] PL 5 P
O r cm S

8. The diagram on the right shows a sector VOW from a circle

with centre O. The arc VW subtends an angle of 2 radians at V
centre O. The sector is folded to make a cone such that the

SIA
64 cm 2 rad
arc length VW is the circumference of the base of the cone. W
Find the height of the cone, in cm. PL 5 O

AY
9. The diagram on the right shows semicircle AOBP with O as P

AL
its centre and ∆ APB is a right-angled triangle at P. Given
that AB = 16 cm and ˙ABP = π radian, find PL 3

M
6 π
– rad
(a) the length of AP, in cm, 6
A B
(b) the area of ∆ ABP, in cm2,

AN
O
(c) the area of the shaded region, in cm2.
IK
10. In the diagram on the right, AOB is a semicircle with y
ID
centre D and AEB is an arc of a circle with centre C(7, 7).
y A
The equation of AB is x + = 1. Calculate PL 4 C (7, 7)
6 8
ND

(a) the area of ∆ ABC, D x– + –y = 1

(b) ˙ACB, in degrees, 6 8
PE

E
(c) the area of the shaded region, in units2.
x
O B
AN

11. The diagram on the right shows a semicircle ABCDE C

RI

with centre F and BGDF is a rhombus. It is given that G (5, 8)

the coordinates of E, F and G are (9, 6), (5, 6) and
TE

(5, 8) respectively and ˙BFD = q radian. Calculate PL 5 B θ D

E (9, 6)
EN

(a) the value of q, in radians, A

F (5, 6)
(b) the area of sector BFD, in units2,
(c) the area of the shaded region, in units2.
M

K
KE

12. The diagram on the right shows the sector of a circle JKLM
with centre M, and two other sectors, JAM and MBL with
centres A and B respectively. Given that the major angle JML M
is 3.8 radians, find PL 4
(a) the radius of the sector of a circle JKLM, in cm, J L
(b) the perimeter of the shaded region, in cm, 1 rad 1 rad
7 cm 7 cm
(c) the area of sector JAM, in cm2,
(d) the area of the shaded region, in cm2. A B

25
13. The diagram on the right shows a circle with Q
centre O and a radius of 2 cm inscribed in sector
A
PQR from a circle with centre P. The lines PQ and
2 cm
PR are tangents to the circle at point A and point B.
Calculate PL 4 P 60° O
(a) the arc length of QR, in cm,
(b) the area of the shaded region, in cm2. B
R

14. The diagram on the right shows the plan for a

SIA
A
garden. AOB is a sector of a circle with centre O
and a radius of 18 m and ACB is a semicircle with 18 m

AY
AB as its diameter. The sector AOB of the garden is
covered with grass while creepers are planted in the

AL
shaded region ACB. Given that the area covered by O θ C
grass is 243 m2, calculate PL 4

M
(a) the value of q, in radians,
(b) the length of the fence needed to enclose the

AN
creepers, in m, B
(c) the area planted with creepers, in m2. IK
15. Hilal ties four tins of drinks together by a string as shown
ID
in the diagram. The radius of each tin is 5.5 cm. Calculate
the length of the string used by Hilal, in cm. PL 5
ND
PE

16. A rectangular piece of aluminium measuring 200 cm by 110 cm is bent into a

AN

semicylinder as shown in the diagram. Two semicircles are used to seal up the two ends of
the semicylinder so that it becomes a container to hold water as shown below. PL 5
RI

200 cm 200 cm
TE

110 cm 118°
EN

110 cm P Q
M

The container is held horizontally and water is poured into the container. PQ represents
KE

the level of water in the container and O is the centre of the semicircle and
˙POQ = 118°.
(a) Show that the radius of the cylinder is about 35 cm, correct to the nearest cm.
(b) Calculate
(i) the area of sector POQ, in cm2,
(ii) the area of the shaded segment, in cm2,
(iii) the volume of water in the container, in litres.

26
 Circular Measure

17. The diagram on the right shows a uniform prism where D R

its cross-section is a sector of a circle with radius 3 cm. PTE

CHA
AOB and CED are identical cross-sections of the prism
40°
with points A, B, C and D lying on the curved surface of E C
the prism. Given that the height of the prism is 4 cm and
˙CED = 40°, find PL 4 4 cm
B
(a) the arc length AB, in cm,
(b) the area of sector AOB, in cm2,
O 3 cm A
(c) the volume of the prism, in cm3,
(d) the total surface area of the prism, in cm2.

SIA
18. The mathematics society of SMK Taman Pagoh Indah
organised a logo design competition for the society. The M

AY
diagram on the right shows a circular logo designed by S K
Wong made up of identical sectors from circles with

AL
radii 5 cm. Find PL 4 T I
(a) the perimeter of the coloured region of the logo, in cm, P

M
(b) the area of the coloured region of the logo, in cm2.

AN
MATHEMATICAL EXPLORATION
IK
ID
Mathematicians in the olden days suggested that the constant π is the ratio of the
ND

circumference of a circle to its diameter.

The information below shows the estimated value of π based on the opinion of four
well-known mathematicians.
PE

A Greek Ptolemy, a
AN

mathematician, Greco-Roman
Archimedes was able mathematician
to prove that showed that the
RI

estimated value of π
3 10 , π , 3 1 .
71 7 is 3.1416.
TE
EN
M

Euler, a Swiss
mathematician wrote Lambert, a German
KE

mathematician proved
that π = 1 + 12
2

6 1 that π is an
+ 12 + 12 + 12 + … irrational number.
2 3 4

In our modern age, computers can evaluate the value of π to ten million digits.
Use the dynamic Desmos geometry software to explore the value of π.

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