Lesson Notes By Weeks and Term - Senior Secondary School 1
Introduction of circle and its properties
Introduction of circle and its properties
SUBJECT: MATHEMATICS
CLASS: SS 1
DATE:
TERM: 2nd TERM
REFERENCE BOOK
WEEK NINE
TOPIC: Introduction of circle and its properties
(a) Introduction of circle and its properties
Parts of a circle: The figure below shows a circle and its parts.
The centre is the point at the middle of a circle. The circumference is the curved outer boundary of the circle. An arc is a curved part of the circumference. A radius is any straight line joining the centre to the circumference. The plural of radius is radii. A chord is any straight line joining two points on the circumference. A diameter is a straight line which divides the circle into two equal parts or a diameter is any chord which goes through the centre of the circle.
Region of a circle
The figure below shows a circle and its different regions.
A sector is the region between two radii and the circumference. A semi-circle is a region between a diameter and the
Semi – circle
Segment
circumference i.e half of the circle. A segment is the region between a chord and the circumference.
EVALUATION
Draw a circle and show the following parts on it. Two radii, a sector, a chord, a segment, a diameter, an arc; label each part and shade any regions.
(b) Calculation of length of arc and perimeter of a sector
Given a circle centre O with radius r. The circumference of the circle is 2Ðr. Therefore, in the figure below, the length, L, of arc XY is given as:
L = θ x 2Ðr
360o
Where θ is the angle subtended at the centre by arc XY and r is the radius of the circle.
Also,
The perimeter of Sector XOY = r + r + L
Where
L = length of arc XY
= θ X 2 Ðr
360
Then
Perimeter of
Sector XOY = r + r + L
= 2r + θ x 2 Ðr
360o
EXAMPLES
Solutions
360
When L = 28cm , θ = 240, r = ?
Then
L = θ x 2 Ðr
360o
28 = 24 x 2 x 22 x r
3600 7
Cross-multiply:
24 x 44 x r = 28 x 360 x 7
15
7 60
r = 28 x 360 x 7 cm
24x44
4 11
r = 49 x 15 cm
11
r = 735 cm
11
Also
When L = 35cm, r = 735 cm
11
θ = ?
Then
L = θ x 2 Ðr
360o
35 = θ x 2 x 22 x 735
360 7 11
Then,
Cross multiply
1 11
35 x 360 x 77 = θ
44 x 735
4 105 3
360 = θ = 300
12
Thus, when the length of the arc is 35cm, the angle subtended at the centre is 300
360o
= 2 x7 + 108 x 2 x 22 x 7
360 7 1
3
= 14 + 108 x 44cm
360 10
= 14 + 3 x 44 cm
10
= 14 + 132 cm
10
= 14 + 13.2 cm
= 27.2 cm
EVALUATION
7.
Length of chord and perimeter of a segment.
Consider a circle centre O with radius r
If OC is the perpendicular distance from O to chord AB and angle
AOB = 2 θ, then the length of chord AB can be found as follows:
In right-angled triangle OCA
AC = Sin θ
r
Cross multiply:
__
AC = r Sin θ
Since
AB = 2 x AC
AB = 2r Sin θ
Where
r = radius of the circle
θ =Semi Vertical angle of the sector i.e half of the angle subtended at the centre by arc AB.
Also
The perimeter of segment ACBD = Length of chord AB + length of arc ADB
= 2r Sin θ + θ x 2 Ðr
360o
Example
In a circle of radius 6 cm, a chord is drawn 3cm from the centre.
(a) Calculate the angle subtended by the chord at the centre of the circle.
(b) Find the length of the minor arc cut off by the chord
Solution
= AOB = 2 θ
Where
θ = Semi vertical angle of the sector.
Then
Cos θ = 3cm = 1
6cm 2
Cos θ = 0.5000
θ = Cos-1 0.5000
θ = 600
-: Required angle = 2 θ
= 2 x 600
= 1200
b Length of minor arc ADB =θ x 2 Ðr
1 2 3600
= 120 x 2 x 22 x 6cm
360 7
3
1
= 4 x 22cm
7
= 88cm = 12 4cm
7 7
= Length of + length of arc
Chord AB ADB
= 2r Sin θ + 1247cm
= 2 x 6 x sin600 + 1247cm
= 12 x Sin 600 + 1247cm
= 12 x 0.8660 + 12.5714cm
= 10.3920 + 12.5714cm
= 10.3920
12.5714
22.9634cm
= 22.96 cm to 2 places of decimal.
EVALUATION
READING ASSIGNMENT
NGM SS BK 2, pg. 31,Ex2a, Nos.2,3,5.
(c) Area of sectors and segments. Area of triangles
Area of sectors
Area of a sector of a circle is given by the formular;
Area of sector θ x πr 2
360o
where r = radius of the circle, θ = angle subtended at the centre by XY or angle of the sector
Examples
Solutions
d = 2r = 28
where d = diameter and r = radius
thus 2r = 28
2r = 28 = 14cm
Area of sector = θ x πr 2
360o
= 45 x 22 x ( 14 ) 2
360 7
= 1/8 x 22/7 x 14 x 14 cm
= 77cm2
Since the area of the whole circle PQR = 72cm2
Then
Area of sector = θ x πr2
360o
But πr2= Area of the whole circle PQR = 72cm2
:. Area of = 40 x 72cm2
sector POQ 360o
= 8cm2
Evaluation
complete the table below for areas of sectors of circles. make a rough sketch in each case.
Radius | Angle of sector | Area of sector |
a.14cm | - | 462cm2 |
b. -- | 140 | 99cm2 |
Area of segments
A segments of a circle is the area bounded by a chord and an arc of the circle.Considering the figure below, we have a major segment and a minor segment .
Given the diagram below:
Area of the shaded segment= Area of sector POQ – Area of triangle POQ
= θ
360o x πr2 - ½ r2 sin θ
Where
r = radius of the circle
θ = angle subtended by the sector at the centre
Π= a constant = 22/7
Examples
2.Calculate the area of the shaded parts in the figure below. All dimensions are in cm and all arcs are circular.
Solutions
1 Area of the given shaded segment =θ x πr2 - 1/2r2 Sin θ
360o
= 56/360 x 22/7 x ( 15 ) 2 - ½ x ( 15)2 sin 560
= 1/45 x 22 x 15 x 15 - ½ x 15 x 15 sin 56o
= 22 x 5 - ½ x 225 x sin 560
= 110 - ½ x 225 x 0.8290
= 110 - 225 x 0. 4145
= 110 – 93.2625cm2
= 110 – 93.2625cm2
= 16.7375cm2
= 16.7cm2 to 3 s. f
2)
The arc in the given figure is part of a circle as shown in the figure above. Thus area of given shaded segment = Area of sector – area of triangle
= θ x πr2 – ½ r2 sin θ
360
= 90/ 360 x 22/7 x ( 14) 2 – ½ x (14) 2 sin 90o
=¼ x 22/7 x 14 x 14 – ½ x 14 x 14 x 1
= 11 x 14 – 14 x 7 cm2
= 154 - 98cm2
= 56cm2
EVALUATION
Calculate the area of the shaded parts in the figure below. All dimensions are in cm and all arcs are circular.
GENERAL EVALUATION
READING ASSIGNMENT
NGM SS BK1 Pages 134-139 Ex 12d Nos 6 and 9 139
WEEKEND ASSIGNMENT
( π = 22/7 )
THEORY
(Take 3.14)
A 21cm ________ 22cm
B ____ 108o 132cm
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