Circles
Circumferences of Circles
The circumference (C) and the diameter (d) of a circle are
related by
circumference (C)
C  πd
diameter (d)
O
radius (r)
Since d = 2r, where r denotes the radius of the circle,
we have
C  2π r
Do you know how to find the circumference
of the circle with diameter 18 cm?
Take  = 3.14,
circumference of the circle
C=d
= 3.14  18 cm
= 56.52 cm
Follow-up question 1
Complete the following table.
22 

.
 Take π 
7 

Radius
14 cm
Diameter
3.5 m
28 cm
7m
7 mm
14 mm

Circumference of the circle
88 cm

22 m
44 mm
d = 2r
C=d
Example 1
Find the circumference of a circle with diameter 15 m.
(Take  = 3.14.)
Solution
Circumference of the circle    15 m
 3.14  15 m
 47.1 m
Example 2
Find the radius of a circle with circumference 88 cm.
22 

.
 Take  
7 

Solution
Let r cm be the radius of the circle.
88  2r
 22 
88  2   r
 7 
r  14
∴ The radius of the circle is 14 cm.
Example 3
s of a semi-circle and a rectangle from
The figure consists of a semi-circle and a rectangle from which
mi-circle is cut from it.
another semi-circle is cut from it. Find the perimeter of the figure.
of the figure.(Give your answer correct
(Give your answer correct to 2 decimal places.)
s.)
Solution
1
   10 cm
2
 5 cm
Perimeter of the larger semi-circle 
Perimeter of the smaller semi-circle  1    6 cm
2
 3 cm
∴ Perimeter of the figure  (5  3  6  10) cm
 (8  16) cm
 41.13 cm (cor. to 2 d.p.)
As shown in the figur
As shown in the figure, Karen’s ant
Example 4
wheels of radii 35 cm
wheels of radii 35 cm and 15 cm re
wheel makes 18 revol
wheels of radii 35 cm and 15 cm respectively. If the larger wheel makes 18 revolutions, find
(a) the distance trav
distance travelled by the bicycle,
(a) the distance travelled by the b
wheel makes 18 revolutions, find
(b) the number of r
number of revolutions that the smaller wheel has
(b) the number of revolutions tha
(a) the distance travelled by the bicycle,
made.
(b) the number of revolutions that the smaller wheel has made.
22 
22 

 made.
.
Take


.

22 

7 
7

 Take   . 
Solution
7


22 

.
(a) The required
distance travelled by the bicycle
 Take
7  distance travelled by the
(b)bicycle
Circumference of the smaller wheel
 The required
(a)
Circumference
the smaller
(b) (b)
Circumference
of theofsmaller
wheelwheel
n in the figure, Karen’s antique bicycle has two
f radii 35 cm and 15 cm respectively. If the larger
As shown in the figure, Karen’s antique bicycle has two
akes 18 revolutions, find
 (2    35)  18 cm
 (2    35)  18 cm
22
222   35  18 cm
7  18 cm
 2   35
73960 cm
 3960 cm
(b) Circumference of the smaller wheel
 2    15 cm
 15 cm
 2  215 cm
22
 2  22 15
22cm
2   15 cm
 15
 2  2  15 cm
7 cm
7 7
22
 2   15660
cm
660 7660
cm (cor. to 3 sig. fig.)
to 3fig.)
sig. fig.)
to 3 sig.
(cor.
  cm cm
7 (cor.
7 7
660
∴to 3The
number
of revolutions that the smaller
 ∴cm∴
(cor.
sig.
fig.)
The
number
of
revolutions
that the
wheelwhee
The
number
of
revolutions
thatsmaller
the smaller
7
660 that the smaller wheel
∴ has
Themade
number
of
revolutions
660
660

3960

 3960
 
 3960
has made
has made
7
7 7
660
has made  3960   42
 42  42
Areas of Circles
The formula of the area of circle can be deduced in the
following way:
1.
2.
3.
4.
Prepare a circle.
Divide it into 16 equal parts.
Cut the circle into 2 parts as shown.
Rearrange them to form the following figure.
Which kind of quadrilateral does
the figure look like? parallelogram
Assuming the radius of the circle is r, estimate the height and
the base of the figure obtained in terms of r.
r
height
base
Height  r
∴
1
Base   2 r =  r
2
The
height  base
The approximate
approximat earea
area  height
base
 r  πr
 πr 2
The figure looks like
a parallelogram.
If A and r denote the area and the radius of a circle
respectively, then
A  πr
2
r
For a circle of radius 28 cm,
22
take  =
,
7
22
its area =
 282 cm2
7
= 2464 cm2
A=r2
Follow-up question 2
Complete the following table. (Take  = 3.14.)
(Give your answers correct to 2 decimal places if necessary.)
Radius
Diameter
Area of the circle
4 mm
8 mm 
50.27 mm2
5 cm
10 cm
78.54 cm2
3.57 m
7.14 m

40 m2
d = 2r
A =  r2
Example 5
(a) the
Find
theofarea
of a with
circleradius
with radius
Find
area
a circle
6 cm. 6 cm.
m. (b) Find
(b) the
Find
the radius
of a with
circlearea
with40area
cm2. (Give
radius
of a circle
cm240
. (Give
m2. your
(Giveanswers
your answers
to 3 decimal
correctcorrect
to 3 decimal
places.)places.)
(a)
Solution
(a) Area of the circle    62 cm2
 113.097 cm2
(cor. to 3 d.p.)
(b) Let r cm be the radius of the circle.
40  r 2
r 
2
r
40

40

 3.568 (cor. to 3 d.p.)
∴ The radius of the circle is 3.568 cm.
Example 6
e, two
semi-circles
are cut
from a square
In the
figure, two
semi-circles
are ABCD.
cut from a square ABCD.
adedFind
areathe
correct
to 2 area
decimal
places.
shaded
correct
to 2 decimal places.
Solution
Radius of the two semi-circles  4  2 cm  2 cm
1
∴ Area of the two semi-circles  (  22 )   2 cm 2  4 cm 2
2
Area of the square ABCD  (4  4)2 cm2  64 cm2
∴ Area of the shaded area  area of the square ABCD  area of the two semi-circles.
 (64  4 ) cm2
 51.43 cm2
(cor. to 2 d.p.)
Example 7
sistsThe
of 3figure
semi-circles
having
same
consists
of 3the
semi-circles
having the same
meters 6 cm, 8 cm and 10 cm respectively.
centre with diameters 6 cm, 8 cm and 10 cm respectively.
f the shaded region in terms of .
Find the area of the shaded region in terms of .
Solution
2
1
 10 
Area of the semi-circle with diameter 10 cm       cm 2
2
2
 12.5 cm 2
2
1
8
Area of the semi-circle with diameter 8 cm       cm 2
2
2
 8 cm 2
2
1
6
Area of the semi-circle with diameter 6 cm       cm 2
2
2
 4.5 cm 2
∴ Area of the shaded region  (12.5  8  4.5 ) cm 2
 9 cm2
Example 8
The area of a semi-circle is 50 cm2. Find the perimeter of the
semi-circle, correct to 2 decimal places.
Solution
Let r cm be the radius of the semi-circle.
1 2
r  50
2
r 2  100
r
∴
100

 100 1
100 

 cm
  2
The perimeter of the semi-circle   2

2
 

 29.01 cm (cor. to 2 d.p.)
Arcs and Sectors
Arcs
B
The figure shows a circle with centre O.
C
If A and B are two points on the
circumference of a circle,
O

then curve AB is called an arc,
which is denoted by ‘arc AB’ or ‘ AB’.
A
D
angle at the centre
AOB is the angle subtended at the centre by the arc AB.
It can be simply called the angle at the centre.
To avoid confusion,


the shorter arc ACB is denoted by ACB
and the longer arc ADB is denoted by ADB .
Lengths of Arcs
B
For a circle with centre O and radius r,
circumference = 2 r.

Let AB be an arc on the circle
and  be the angle subtended at the centre
by the arc.
Let’s find the lengths of arc AB for
different values of ...
r
O

A
(a)
A
 = 180

Length of AB
1 
180
 circumference
=
of the circumference
2 
360
1
=  r  2 r
2
180
O
B
(b)
 = 90

Length of AB
190
 circumference
=
of the circumference
4 
360
1
= r 2 r
2
4
(c)
A
 = 45
90
A

Length of AB
145
 circumference
=
of the circumference
8 
360
1
= r 2 r
48
B
O
45
O
B
(d)
 = 1
A

Length of AB
1
1
ofcircumference
=
the circumference
360
1
 r2 r

=
360
180
O
What can you conclude
from the above results?
The ratio of the arc length to the
circumference of the circle =

360
.
B
For a circle with radius r and angle 
subtended at the centre by an arc,
B
r
O
Length of arc
Angle subtended at the centre
=angle subtended
Arc length
atangle
the centre
Round
Circumference

Circumfere nce
Length of arc
θ

2π r
360
round angle
Therefore,
θ
2π r
length of arc =
360

A
Follow-up question 3
In each of the following figures, O is the centre of the circle.
Find the length of the arc AB in terms of .
1.
2.
A
10 cm
240
72
O
9m
O
B
A
B
Solution
1.

72
AB 
 2π  10 cm
360
 4π cm
2.

240
AB 
 2π  9 m
360
 12π m
Example 9
e figure,
O is
the centre
circle.ofFind
length
of the
arc ABC
In the
figure,
O is of
thethe
centre
the the
circle.
Find
length of arc ABC
ct tocorrect
2 decimal
to 2places.
decimal places.
Solution
360  142
 2  6 cm
360
218

 2  6 cm
360
 22.83 cm (cor. to 2 d.p.)
The length of arc ABC 
Example 10

61
61
ABC
cm, ,find
cm
Thefigure
figureshows
showsaacircle
circlewith
withradius
radius15
15cm.
cm.IfIf ABC
find. .
The
66
61
cm , find .
ws a circle with radius 15 cm. If ABC 
6

Solution
∵
∴

ABC 
61
cm
6
360  
61
 2  15 
360
6
61 360
360   

6
30
360    122
  238
Example 11
In the figure, ∠ACB is the angle subtended at the centre C by arc AB.
BDC is a semi-circle with diameter BC. If AC  10 cm and ∠ACB  70,
find the perimeter of the figure correct to 3 significant figures.
ed at the centre C by arc AB.
Solution
AC  10 cm and ∠ACB  70,

BC  AC  10
cm (radii)
3 significant
figures.
1
   10 cm
2
 5 cm
70
AB 
 2  10 cm
360
35

cm
9
CDB 

Perimeter of the figure


 CDB  AB AC
35


  5 
 10  cm
9


 37.9 cm (cor. to 3 sig. fig.)
Areas of Sectors
A sector is the region enclosed by an arc and two radii
of a circle.

by AB , radii OA and OB.
A
In the figure, AOB is a sector enclosed
O
sector
B
AOB is called the angle of the sector. angle of the sector
For a sector with radius r and angle of the
sector ,
area of sector angle of the sector

area of circle
round angle
area of sector
θ

πr 2
360
Therefore,
area of sector =
θ
π r 2
360
A
r
O
θ
B
Refer to the following figure.
324
O
10 m
A
B
324
 π  10 2 m2
Area of sector OAB 
360
 90π m2
 area of sector
θ

πr 2
360
Follow-up question 4
Complete the following table.
(Give your answers in terms of π if necessary. )
Radius
Angle of the sector
Area of sector
10 mm
10 mm2
8 cm
36 
45 
3m
320 
8 m2

8 cm2
area of sector
θ

πr 2
360
Example 12
In the
figure,
AOB
COD
sectors.
Given
4 cm,
7 cm
In the
figure,
AOB
andand
COD
are are
sectors.
Given
thatthat
BDBD
 4 cm,
ODOD
 7cm
∠AOB
 130,
of the
shaded
region
ABDC
correct
andand
∠AOB
 130,
findfind
the the
areaarea
of the
shaded
region
ABDC
correct
to 2to 2
decimal
places.
decimal
places.
Solution
130
   (7  4) 2 cm 2
360
1573

cm 2
36
130
Area of sector COD 
   7 2 cm 2
360
637

cm 2
36
Area of sector AOB 
∴
Area of the shaded region   1573  637  cm 2
 36
 26 cm 2
36 
 81.68 cm 2
(cor. to 2 d.p.)
Example 13
tthe
theradius
radiusof
ofaapizza
pizzaisis12
12cm.
cm.As
Asshown
shownininthe
thefigure,
figure,
It is given that the radius
of
a
pizza
is
12
cm.
As shown in
22
he
e pizza
pizzawith
witharea
area24
24cm
cm isiseaten.
eaten.Find
Findthe
theangle
angleofof
the figure, one slice of the pizza with area 24 cm2 is eaten.
gsector
sectorof
ofthe
thepizza.
pizza.
Find the angle  of the remaining sector of the pizza.
Solution
∵
∴
Area of the sector  24 cm2
360  
   122  24
360
360
360    24 
144
360    60
  300
Example 14
2 120.
and ∠AOB
cm2 AOB
AOB
sectorthe
area
the In
theoffigure,
areaisof80sector
is 80 cm
and ∠AOB  120.
figures.
3 significant
correct
the sector,
us ofFind
the radius
of thetosector,
correct to
3 significant figures.
Solution
Let r cm be the radius of the sector.
∵ Area of the sector  80 cm2
360  120
 r 2  80
∴
360
240
 r 2  80
360
80 360

r
 240
 6.18 (cor. to 3 sig. fig.)
∴ The base radius of the sector is 6.18 cm.
Cylinders
A cylinder is a solid with uniform cross-section and its two
bases are circles.
The solids below are all cylinders.
Volume of Cylinders
We have learnt that
volume of a prism = base area × height
Similarly,
volume of a cylinder = base area × height
For a cylinder of base radius r,
height h and volume V,
then
r
h
V =  r 2h
base
Refer to the following figure.
5 cm
2 cm
Volume of the cylinder
=   22  5 cm3
 V = r2h
= 20 cm3
= 62.83 cm3
(cor. to 2 d.p.)
Follow-up question 5
The figure shows a cylinder of base radius r and height h.
Complete the following table.
(Give your answers in terms of π if necessary. )
Volume (cm3 )
r (cm)
h (cm)
4
7
112
2
10
40
5
11

h
r
275
V =  r2 h
Example 15
The figure shows a dumbbell which consists of three cylinders of length
10 cm each. The base radii of the left and the right cylinders are 4 cm and
s a dumbbell
which
consists
three cylinders
of the volume of the
the base radius
of the
middleof
cylinder
is 1 cm. Find
h. The
base radii
theanswer
left and
the right
dumbbell.
(Giveof
your
in terms
of .)cylinders
base radius of the middle cylinder is 1 cm. Find the
mbbell.
(Give your answer in terms of)
Solution
Volume of the dumbbell  [(  42  10)  2    12  10] cm3
 (320  10 ) cm3
 330 cm3
Example 16
900 cm3 of water is poured into a cylindrical tank and the depth of water is 12 cm. Find
the base radius of the tank, correct to 3 significant figures.
Solution
Let r cm be the base radius of the tank.
r 2  12  900
900
12
 4.89 (cor. to 3 sig. fig.)
The radius of the tank is 4.89 cm.
r
∴
Example 17
figure
showsthe
themetal
metal cup
inner
andand
outer
diameters
are 4 are
The
figure
shows
cup
whose
inner
diameters
TheThe
figure
shows
the
metal
cupwhose
whose
inner
andouter
outer
diameters
are44
cm and 6 cm respectively. If the height and the thickness of the base
cm
cmand
and66cm
cmrespectively.
respectively.IfIfthe
theheight
heightand
andthe
thethickness
thicknessof
ofthe
thebase
base
of the cup are 7 cm and 1 cm respectively, find the volume of metal
of
ofthe
thecup
cupare
are77cm
cmand
and11cm
cmrespectively,
respectively,find
findthe
thevolume
volumeof
ofmetal
metal
required to make the metal cup.(Give your answer in terms of .)
requiredtotomake
makethe
themetal
metalcup.(Give
cup.(Giveyour
youranswer
answerininterms
termsof
of.).)
required
Solution
2
  6 2

4
Volume of metal required       7       (7  1) cm3
2
  2 

 (63  24 ) cm3
 39 cm3
Example 18
igure,
flowswater
from flows
a pipefrom
into aa pipe
cylindrical
tank of basetank of base
In water
the figure,
into a cylindrical
0.8 mradius
and height
In the
the tank is half
filledis half filled
0.8 m 1.2
and m.
height
1.2beginning,
m. In the beginning,
the tank
ater.with
If thewater.
waterIfflows
from flows
the pipe
at athe
constant
of 50 rate of 50
the water
from
pipe at rate
a constant
find cm
the3/s,
time
taken
filltaken
up thetotank
in the
minutes.
find
the to
time
fill up
tank in minutes.
Solution
The unfilled volume of the tank    0.82  1.2 
1 3
m
2
 0.384 m3
 384 000 cm3
unfilled volume of the tank
The time taken to fill up the tank 
rate of water flow
384 000 cm3

50 cm3 / s
 7680 s
 128 minutes
Total Surface Areas of Cylinders
The lateral face of a cylinder is a curved surface.
How can we find the curved surface area?
1. Prepare a cylindrical can that has a piece of wrapper.
2. Cut the wrapper vertically as shown.
3. Spread out the wrapper.
A
D
B
C
What
Whatisisthe
therelationship
relationshipbetween
betweenAD
ABand
the
and
circumference
the height of of
thethe
can?
base of the can?
= height of the of
can
ADAB
= circumference
the base of the can
A
D
B
C
Then, what is the curved surface area of the can?
Curved surface area of the can
= area of rectangle ABCD
= AD × AB
= the circumference of the base of the can
× the height of the can
For a cylinder of base radius r and height h,
curved surface area of a cylinder = 2 rh
The total surface area of the cylinder can be found by adding
the areas of its two bases to its curved surface area:
base area = r 2
r
base area = r 2
curved surface area
= 2rh
total surface area of a cylinder = 2 rh + 2 r 2
Refer to the following figure.
4 cm
3 cm
Total surface area of the cylinder
2

4
  
2
= π  4  3  2π     cm  Total surface area = 2 r h + 2 r 2
 2  

= 20 cm2
Follow-up question 6
The figure shows a cylinder of base radius r and height h.
Complete the following table.
(Give your answers in terms of π if necessary. )
h
r
r (cm)
h (cm)
1
3
5
7
2
4
Total surface area (cm 2 )

120 π
Total surface area
= 2 r h + 2 r 2
Example 19
(a)
(a)
(a) the
Find
thesurface
total surface
area
ofcylinder
the cylinder
infigure.
the figure.
Solution
Find
total
area
of
the
in
the
Find the total surface area of the cylinder in the figure.
(a) Total surface area of the cylinder
2

4
4  2

(a) Find
total
area ofarea
the of
cylinder
in
 the
6  2     cm
(a) the
Find
thesurface
total surface
the cylinder
inthefigure.
2figure.
(a) Find the total surface area of the cylinder in the figure.
2
 2  
 (24  8 ) cm 2
2

32

cm
(b) The solid as shown is half of the cylinder in (a). Find the total surface area of
 101 cm 2 (cor. to the nearest cm 2 )
the solid.
(b) The solid as shown is half of the cylinder in (a). Find the total s
(b) The solid as shown is half of the cylinder in (a). Find the total surfa
(b) The
solid
as
shown
is halfisofhalf
the of
cylinder
in (a). in
Find
total
area ofarea of
(b)the
The
solid
as shown
the cylinder
(a).the
Find
thesurface
total surface
solid.
(b) solid.
The solid as shown is half of the cylinder in (a). Find the total surface area of
the
the
solid.
the solid.
(b) Total surface area of the solid
the solid.
(Give your answers correct to the nearest cm2.)
1

  6  4  32   cm 2
2

 (24  16 ) cm 2
 74 cm 2
(Give (Give
your answers
correctcorrect
to the to
nearest
cm22.) cm2.)
your answers
the nearest
(Give your
answers
correct
to the nearest
cm .)
(Give
your
answers
correct
to the
(cor. to the nearest cm 2 )
nearest2 cm2.)
(Give your answers correct to the nearest cm .)
Example 20
If the curved surface area and the height of a cylinder are 42 cm2
and 7 cm respectively, find its volume in terms of .
Solution
Let r cm be the base radius of the cylinder.
∵ Curved surface area  42 cm2
2r  7  42
42
14
3
∴ Volume of the cylinder    32  7 cm3
∴
r
 63 cm3
Extra Teaching Example
Example 2 (Extra)
shows
formed
two formed
circles with
a common
centre
Thea ring
figure
showsby
a ring
by two
circles with
a common centre
perimeter
theperimeter
ring.
O. Findofthe
of the ring.
.14.)(Take  = 3.14.)
Solution
Circumference of the inner circle  2(3.14) (6) cm
 37.68 cm
Circumference of the outer circle  2(3.14) (6  4) cm
 2(3.14) (10) cm
 62.8 m
∴ The perimeter of the ring
 37.68 cm  62.8 cm
 100.48 cm
Example 8 (Extra)
Infigure,
the figure,
ABCD
a square
andof
ADE
is a rsemi-circle
of radius
r
In athe
ABCD
is aissquare
and ADE
is
a semi-circle
of radius
r
CD is
square
and
ADE
aissemi-circle
radius
2
2
2area
The
of ABCDE
is 40
cm. The
is 40 cm
. cm .
ABCDE
iscm.
40area
cmof
. ABCDE
(a) (a)
Find r.Find r.
Find
the perimeter
of ABCDE.
your answers correct to 2
(b) (b)
the
perimeter
ABCDE.
(Give(Give
your answers
rimeter
ofFind
ABCDE.
(Give of
your
answers
correct
to 2 correct to 2
decimal
places.)
decimal
places.)
Solution
2 22 1 11 2 2 2
(a)
(
2
r
(
2
r
))  rr r40
4040
(
2
r
)
(a)
2
22
1 11r2 2 2 40
2r22 
4
44
r r   2rr 40
 40
22
1
2 
1 1   40
2 r  4 
r 2 4  2   40
r  4     40


2 
2

r
40
40
r  4  1 40

r 4  12 
1
 2.6796
42... 
2 to 2 d.p.)
22.68
準確至二位小數
)
.68 ((cor.
 2.68 (準確至二位小數 )
Solution
1
 2r ) cm
2
 (6r  r ) cm
 [6(2.6796)   (2.6796)] cm
(b) Perimeter of ABCDE  (3  2r 
 24.50 cm (cor. to 2 d.p.)
Example 10 (Extra)


The figure shows a circle with radius 18 cm. If ABC  18 cm, show
ABC
 18 cm, show
The figure
shows
circle18
with
18 cm.
 18If cm,
ure shows
a circle
witharadius
cm.radius
If ABC
show


that ABC is a semi-circle.
ABC is a semi-circle.
C is that
a semi-circle.
Solution
∵
∴

ABC  18 cm
AOC
 2  18  18
360
AOC  18 
∴

 180
ABC is a semi-circle.
360
36
Example 14 (Extra)
In the figure, the perimeter of sector AOB is 12 cm and ∠AOB =
22 

140. Find the radius and the area of sector AOB.  Take   .
7 

Solution
∠AOB
=be the radius of the sector.
Let r cm
∵ Perimeter of sector AOB  12 cm
140
∴
 2r  2r  12
360
22
 140

r
 2
 2   12
7
 360

r  2.7
22 
 .
7 
∴ The radius of the sector is 2.7 cm.
140
 r 2 cm 2
360
140 22

  2.7 2 cm 2
360 7
 8.91 cm 2
Area of sector 
Example 19 (Extra)
The
shows
a solid
formed
by drilling
a cylindrical
hole of base
Thefigure
figure
shows
a solid
formed
by drilling
a cylindrical
hole of base
radius
from
a rectangular
wooden
block block
of dimensions
20 cm  20 cm 
radius6 cm
6 cm
from
a rectangular
wooden
of dimensions
30
5 cm.
Find
the the
totaltotal
surface
area of
the of
solid.
30cm
cm 
5 cm.
Find
surface
area
the(Give
solid.your
(Give your
answer correct to 1 decimal place.)
answer correct to 1 decimal place.)
Solution
Total surface area of the solid
 (20  30  2  30  5  2  20  5  2  2  6  5  2  62 ) cm 2
 (1200  300  200  60  72 ) cm 2
 1662.3 cm 2
(cor. to 1 d.p.)