Chapter 10:

CIRCLES

Sub CHAPTER:

10.1 PARTS OF A CIRCLE

What is the locus of a moving object such that its distance from a fixed point is always

constant?

Is the ratio of the circumference to the diameter of the ‘ear’ greater

than or less than the same ratio for the ‘face’ ?

diameterncecircumfere

diameterncecircumfere

Object Circumference Diameterdiameter

ncecircumfere

Class Activity

Apparatus:

For any circle,

is a constant

Usually is taken to be approximately

equal to 3.14 (2 d.p.) or

3.142 (3 d.p.) or

diameterncecircumfere

Greek letter

722

Introduced by William Jones

in 1706

= 3.14159265358979323846264338327950288419716939937510

. . .

71322850559064483

58209749445923078164

9074660328231568084128976071124352843082689980268260

In 2004, it took more than 500 hours for a supercomputer to generate pi to 1.3511 trillion decimal places

pi

• current world record

holder for the memorization

of .

• in October 2006, he took 16

hours to recite to 100 000

d.p.

3.142

3.142

πdiameter

ncecircumfere

Formula FOR circumference

nceCircumfere

circumference = d or

circumference = 2r

nceCircumfere diameterπ

radius) (2

×

Find the circumference of a

circle, given its:

a) diameter

b) radius

circumference

Example 1:

Find the circumference of a circle with diameter:

Solution:

circumference = d

14

= 44 mm

7

22

[ ]7

22(a) 14 mm (b) 0.7 cm ( = 3.142 )

Solution:

circumference = d

= 2.199 cm

(3 d.p.)

14 mm 0.7 cm

2

7

22

=

14

0.7 3.142

= 0.73.142

[ ]7

22

Example 2:

Find the circumference of a circle with radius:

(a) 7 cm

0.5 mm7 cm

Solution:

circumference = 2r

= 2 x x 7

= 44 cm

Solution:

circumference = 2r

= 2(3.142)( 0.5)

= 3.142 mm

(3 d.p.)7

22

(b) 0.5 mm ( = 3.142)

Example:Find the perimeter of the diagram , in cm. (Take π = 3.142 )

70 m Length of the semicircle =

=

= 109.97

Perimeter of the diagram = 70 + 109.97 = 179.97 m

3.142 x 702

2

d

Find the :

a) diameterb) radius

given the circumference of a circle.

Example:(a) Given that the circumference of a circle is 66 cm. Find the

diameter of the circle.

Solution:

circumference = d d =

= 66

= 66 x

= 21 cm

7

22

22

7

Example:(b) Given that the circumference of a circle is 20 cm. Find the radius of the circle.

Solution:

circumference = 2 r

r =

=

= 10 cm

2

20

3

[ ]7

22

circumference

circumference

2

360centre at Angle

nceCircumferearc of Length

Arc of a circle

360centre at Angle

r 2arc of Length

Find the arc length, given the

a) angle at the centre b) radius

of a circle.

r2360

arc an of length The

360centre at Angle

r 2arc of Length

1. State the angle subtended by the arc

55o

O55o

345o

O

345o

O

140o

140o

2. State the angle subtended by the arc

= 320o

40o

OO

310o

= 50o

O

= 270o

360o 40o 360o 310o 360o 90o

77

222

360

Example 1:

Find the length of arc XY. [ Take ]

r 2 XYarc 360

7

22

O

X

Y7 cm

= 8.80 cm

72o 72o

(2 d. p.)

97

222

360

Example 2:

Find the length of arc AB. [ Take ]

r 2 arc AB 360

7

22

O

A

B9 cm

60o

= 47.14 cm

300o

(2 d. p.)

Perimeter = arc length + radius + radius

cm 86 212144

21 cm120o

Example 3:

Find the perimeter [ Take ]7

22

21 21 21

7

222

360

120 )(

r r r2360

( )O

1

3 1

3

r r r2360

( )

O

A

BC

D

Textbook, Pg. 59 No. 6:

Given : OA = 7 cm , AB = 2 cm

Perimeter = arc AD + arc BC + AB + CD

2 2 9 722

2 36052

7

7

222

360

52 )( )(= 6.356 + 8.171 + 2 + 2

= 18.53 cm (2d.p.)

52o

ETextbook, Pg. 59 No. 7:

Perimeter = arc AE + AD + DC + CB + BE

= 32.267 + 15.4 + 15.4 + 15.4 + 15.4

= 93.87 cm (2d.p.)

BA

D C

120o

15.4

cm

15.4 15.4 15.4 15.4 15.4

7

222

360

120( )1

3

Find the angle, given the

a) arc length

b) radius

of a circle.

360 r

lengtharc centre the at angle

2

360centre at Angle

r 2arc of Length

[ ]7

22

7 cm

2 mm

Example 1: Find the angle subtended at the centre

O

7 cm

11 cm

θ =

360

r lengtharc 2

7 7

222

11 360o

=

11 180o

22= = 90o

2

180

Find the radius, given the

a) arc length

b) angle at the centre

of a circle.

360 lengtharc r

2

360r 2arc of Length

360 lengtharc r 2

360 lengtharc r 2

Example 1: Find the radius (use = 3.142)

6.284 m

O

40o

r =

360

lengtharc

2

= 9 m

6.284 360o

=2 3.142 40o

2r360

sector of area

360centre at Angle

circle of Areasector of Area

Area of a Sector

360r sector of Area2

Find the area of a sector,

given the

a) radius

a) angle at the centre

O

O

[ ]7

22(a) radius = 7 cm (b) diameter = 8.4 m ( = 3.142)

Example 1: Find the area of the shaded region, given :

156o

Solution:

area = x x 7 x 7

= 115.5 cm2

360

2707

22

Solution:

radius = 8.4 2 = 4.2 m

area = x 3.142 x 4.2 x 4.2

= 24.02 m2

360

156

Example 2 : Calculate the area of the shaded region

O

14cm

Solution:

Area of shaded region

= Area of sector Area of triangle

=

= 154 98 cm2

= 56 cm2

14147

22

360

90

[ ]7

22

141421

Textbook, Pg. 72 No. 6

40o

Solution :

Area of shaded region

= area of larger sector area of smaller sector = = 50.286 8.730

= 41.56 cm2

12 cm

5 cm

5

57

22

360

4012

127

22

360

40

Find the angle, given the area of the sector

360r sector of Area2

360 r sector

2 ofarea

Example 1: Find the angle subtended at the centre

7 cm

78.57 cm2O

10

cm

( )use = 3.142

θ = 360o

r2

area of sector

=78.57

3.142 10 10 360o

= 90o

Find the radius, given the area of the sector and the angle at the centre

360 sector of area

r

360r sector of Area2

360 sector of

area

360 sector) of (area r 2

r 2

Example 1: Find the radius. [ ]7

22

O

180o

19.25 m2

( )

angle 360 area r

m 3.5 12.25

7

22 180 360 19.25r

7

3960 6930

3960

7 6930