Circles I

Math CHAPTER
Form 2
10

5.1 Properties Of Circles

5.1.1 Parts of the Circle

5.3 Circumference and Area of a
Circle

Edited from teaching aids
Prepared By:
Norisah bt Mustaffa
KOSPINT 2004

5.1.1 Know the Parts Of Circles
The concept of circle.

The circle is the locus of a moving point which
is at constant distance from a fixed point, o.

point

Locus of o
point
centre

Parts of a Circle. radius

• Circumference centre
• Diameter
• Radius

The perimeter of
the circle is called
the circumference.

diameter

Parts of a Circle. • Chord

The line PR is a chord. • Arc
- Minor arc
- Major arc

Minor arc P

PQR is the minor arc. Q o S
PSR is the major arc. R
Major arc

Parts of a Circle. • Segment
- Minor segment
minor segment - Major segment

Q

Minor segment

PR

minor segment o

Major segment

S

Parts of a Circle. • Sector
- Minor sector
- Major sector

A

minor sector o C
major sector
Minor sector
Major sector

B

Circumference and Area of a Circle

Circumference of a Circle

 is read as “pi” .

 = Circumference (C) Diameter = 2 x radius (r)
= 2r
Diameter (d)
Where  = 3.142 or 22
C = d 7

radius

= 2r diameter

i.e. Circumference = Diameter x 
= 2 x radius x 

Solving Problem Involving Circumference of a Circle.

1. Calculate the 2. If the circumference of
circumference of a circle a circle is 22 cm, find
which has a diameter of the length of its radius.
14 cm.
Solution:
Radius = 14 ÷
2 2r = 22
= 7 cm
2x 22 x r = 22
Solution: 71
r = 22 x 7
Circumference = 2r 2 x 22 1

= 2 x 22 x 7 r =7
7 2

= 44 cm

= 3.5 cm

Solving Problem Involving Circumference of a Circle.

3. The diagram shows a circle

with centre, O. Find the
value of x if circumference Solution:

of circle is 16 cm. 2r = 16

X = d = 2r 8

r = 16

2 1

x r = 8 cm

o x=2x8

= 16 cm

Solving Problem Involving Circumference of a Circle.

4. The diagram shows two Solution:
circles in a rectangle. The
length of the radius of each Perimeter of rectangle +
circle is 7 cm. Find the perimeter of two circles.
perimeter of the shaded
part. = 2(28 + 14) + 2(2 x 22 x 7)
7

= 2(42) + 88

7cm 7cm 7cm = 84 + 88
28 cm
7cm 14 cm = 172 cm

Find the
length and
width of the
rectangle

first.

Arc of a Circle

Length of Arc of a Circle.

The length of arc is proportional to the angle at the
centre of the circle.

Length of arc = Angle at the centre

Circumference 360º

Length of arc AB, or B

=  x 2r 
360º

A

Solving Problem Involving Length of Arc of a Circle.

1. Find the length of arc PQ in
the figure below.

( Taken  = 22)
7

Solution:

Length of arc PQ,

o 14 cm = º x 2r

P 360º
60º
12
Q
= 60º x 2 x 22 x 14

360º 6 71

= 14.67 cm

Solving Problem Involving Angle at the Center of a Circle.

2. Find the value of  in the Solution:
figure below. Given that the
length of arc AB = 4.4 cm. º = length of arc
( Taken  = 22) 360º circumference
7
 = 360º x 4.4
o 2 x 22 x 6.3
7
6.3cm
º 0.1
AB
 = 360 x 4.4 x 7 1
2 x1 22 x 6.3 0.9

= 360º
9

= 40º

Solving Problem Involving Length of Radius of a Circle.

Solution:

3. Find the value of r in the º = length of arc
figure below. Given that the 360º circumference
length of arc XY = 11 cm.

( Taken  = 22) 45º = 11
7 360º 2r

8

2r = 11 x 360

Y 45 1

11 cm r r = 88

o 2 x 22

45º 7
º
2
X
= 88 x 7

2 x 22 1

= 14 cm

Area of a Circle

Finding Area of a Circle.

•The area of a circle is the region enclosed by

the circumference of the circle.

r2 = Area

Area of circle = r2 So, r = Area


Example:

Find the area of the circle below. Solution:

Area = r2

o = 22 x 7 x 7
7 22
3.5 cm
= 38.5 cm2

Solving Problem Involving Area of a Circle.

Find the area of the following Solution:
annulus given that OP = 5 Area of annulus,
cm and OQ = 6 cm.
( Taken  = 3.142) = R2 - r2
= (R2 – r2)
Let say OQ = 6cm=R = (62 – 52)
and OP = 5cm= r = (36 – 25)

Q = 3.142 x 11

P
O 5 cm

= 34.56 cm2

Solving Problem Involving Radius and Diameter of a circle.

Find the radius and diameter of Solution:
a circle given that its area is
616 cm2. r = Area


r = 616 x 7 ÷ 22

= 196

r = 14 cm

d = 2 x 14 cm

d = 28 cm

Finding The Area Of A Circle Given

The Circumference And Vice Versa.

Find the area of a circle Solution:
given that its
circumference is 176 cm. 2r = 176

r = 176

2

= 88 x 7
22

= 28 cm

Area = 22 x 28 x 28
7

= 2 464cm2

Area of Sector of a Circle
•The area of a sector of a circle is proportional
to the angle at the centre.

Area of sector = Angle at the center

Area of circle 360o

Example:

Find the area of minor sector OPQ in

the circle below. Solution:

o Area = r2

60º Q = 60º x 22 x 21 x 21
21 cm 360º 7

P = 231 cm2

Angle at the Centre of a Circle

Find the value of y in the figure below,

given that the area of minor sector OAB

= 154 cm2. (Taken  = 22) Solution:
7

y = area of sector
A B 360º area of circle

yº y = 154 x 360º
14 cm 22 x 14 x 14
7
0

= 154 x 360º
616

= 90º

Length of Radius of a Circle

Find the radius of the circle Solution:
below, given that the area of
minor sector OAB = 4 5cm2. 60º = area of sector
360º area of circle
7
(Taken  = 22) r2= 360º x 33

7 60º 7

o rB r2 = 198

60º 7

r2 = 198 x 7
7 22

A r=9

r = 3 cm

Solving Problem Involving Area of Sectors and Area of Circles.

The figure shows a semicircle Solution:
with AD as diameter and a square Area of the square,
ABCD. Find the area of the figure. = 14 x 14
= 196 cm2
D 14 cm C

Area of the semicircle

= 1 x 22 x 7 x 7
27

= 77 cm2
A B Total area = 196 + 77

= 273 cm2