TOPIC:
Mensuration
Content:
Definition and meaning of an arc sector and segment of a circle.
Length of arc of circle
Perimeter of sectors and segments
Areas of sectors and segments
Areas of sectors of a circle
Area of segments of a circle
DEFINITION AND MEANING, LENGTH OF ARCS OF CIRCLE
AN ARC: an arc of a circle is a part of the circumference of the circle.
Hence, an arc is a length or a distance along the circumference of a circle. It is never an area.
A SECTOR: a sector is a part or a fraction of a circle bounded by an arc and two radii.
Hence, an arc is a length whereas the sector covers an area of a circle.
A SEGMENT
The segment of a circle is the part cut off from the circle by a chord. A chord is the line segment AB.
https://youtu.be/viM7a4IUp9Y
LENGHT OF ARCS OF CIRCLES
If 5 sectors are cut off from 5 different circles and the lengths of the arcs l, radii r and angles θ measured and compared.
Then
i/2π = θ/360 or L = θ/360 x 2 πr = 2 πr θ/360
hence, in a circle of radius r, the length l of an arc that subtends angle θ at the centre is given by
L = θ/360 x 2 πr
Example 1:
Find the length of an arc of a circle of radius 5.6cm which subtends an angle of 600 at the centre of the circle (Take π = 22/7)
Solution
Length of arc = θ/360 x 2 πr
Given θ = 600, r = 5.6cm, π = 22/7
Substituting into the formula,
Length of arc AB = 60/360 x 2 x 22/7 x 5.6
= 1/6 x 2 x 22/7 x 5.6
= 1/6 x 2 x 22/7 x 0.8
= 17.6/3
= 5.8667cm
= 5.87cm (2 decimal places)
Example 2;
What angle does an arc 6.6cm in length subtends at the centre of a circle of radius 14cm. Use π = 22/7)
Solution
Length of arc xy = θ/360 x 2 πr
6.6 = θ/360 x 2 x 22/7 x 14
θx 2 x 22 x 14 = 6.6 x 360 x 7
θ = (6.6 x 360 x 7)/(2 x 22 x 14)
= (33 x 18)/11x2
= 3 x 9
= 270
θ = 270 , the angles subtend by the arc
Example 3:
An arc of length 12.57cm subtends an angle of 600 at the centre of a circle. Find
The radius of the circle
The diameter of the circle.
Solution
Arc= θ/360 x 2 πr
12.57 = 60/360 x 2 x 22/7 x r/1
12.57 x 360 x 7 = 60 x 2 x 22 x r
r = (12.57 x 360 x 7)/(60 x 2 x 22)
r = (12.57 x 6 x 7)/44
r = 527.94/44
r = 11.99
r = 12cm
Example 4:
An arc of a circle of diameter 28m subtends an angle of 1080 at the centre of the circle. Find the length of the major arc.
Solution
Minor arc angle = 1080
Major arc angle = 3600 – 1080
= 2520.
Arc = = θ/360 x 2 πr
= 252/360 x 2/1 x 22/7 x 14/1
= (252 x 44)/180
= 11088/180
= 61.6m
The length of major arc = 61.6m
https://youtu.be/C9z3FXS7nlo
Evaluation
Find the radius of a circle which subtends an angle of 1200 at the centre of the circle and is of length 2.8cm (π = 22/7)
In terms of π, what is the length of an arc of a circle of radius 31/2m?
https://youtu.be/SwrlKx52L5o
PERIMETER OF SECTORS AND SEGMENTS
The word perimeter simply means the distance round an object. So the perimeter of a sector of a circle is the distance round the circle.
hence the perimeter of a sector AOB is the sum of two radii (2r) and length of arc l, where r is radius and l = length of arc.
Perimeter of sector AOB = 2r + Lunits
Example
Find the perimeter of the sector of radius 3.5cm which subtends an angle of (i) 450 (ii)3150.
Solution
Length of arc = θ/360 x 2πr
Here θ = 450, r = 3.5cm, π = 22/7
Length of arc = 45/360 x 2 x 22/7 x 3.5
= 1/8 x 2/1x 22/7 x 7/2
= 22/8
= 2.75cm
The perimeter of the sector is 2r + l, here r = radius which is 3.5cm and l = 2.75cm.
Perimeter = 2r + length of arc
= 2r + 2.75cm
= (2 x 3.5) + 2.75cm
= 7.0 + 2.75cm
= 9.75cm
(ii) θ = 3150, π = 22/7 , r = 3.5cm
Length of arc = 315/360 x 2 x 22/7 x 3.5cm
= 63/72 x 2 x 22/7 x 7/2 cm
= 7/8 x 2 x 22/7 x 7/2 cm
= 77/4 = 19.25cm
Perimeter = 2r + 19.25cm
= (2 x 3.5) + 19.25
= 7 + 19.25
= 26.25cm
https://youtu.be/_y4E1cbga4Y
EVALUATION
Calculate the perimeter of a sector of a circle of radius 14cm, where the sector angle is 600. Take π = 22/7
Page 163, 15, 16, 23, 27
PERIMETER OF SEGMENT
The perimeter of a segment = length of arc + length of chord.
To find the length of chord AB, we bisect <AOB
Example 2:
The perimeter of a sector is 61.43cm. If the angle subtended by the sector at the centre is 1200. Find the radius of the sector.
Solution
Perimeter of a sector = Arc + 2r
61.43 = = θ/360 x 2πr + 2r
61.43 = = 120/360 x 2 x 22/7 x r + 2r
61.43 = = 1/3 x 2 x 22/7 x r + 2r
Multiply through by L.c.m = 21
21 x 61.43 = (211/3)x 2 x 22r)/ 3 x 7+ 2r)
1290.03 = 2 x 22r + 21 + 2r
1290.03 = 44r + 42r
1290.03 = 86r
86r = 1290.03
r = 1290.03/86
r = 15.00cm
https://youtu.be/3j8p1HW9GkI
Example 3
The perimeter of a sector is 61.43cm. If the radium of the sector is 15cm. Find the angle it subtends at the centre.
Solution
Perimeter of a sector = Arc + 2r
61.43 = θ/360 x 2 πr + 2r
61.43 = θ/360 x 2 x 22/7 x 15 + 2 x 15
61.43 = 11θ/(6 x 7) + 30
61.42 – 30 = 11θ/42
31.43/1 = 11θ/42
11θ = 42 x 31.43
θ = (42 x 31.43)/11
θ = 1200
https://youtu.be/vVAl1jyL8X0
https://youtu.be/umkOuw25nX0
EVALUATION
The angle of a sector of a circle radius 12.5cm is 680. Calculate the perimeter of the sector.
A rope of length 18m is used to form a sector of a circle of radius 2.5m on a school playing field. What is the size of the angle of the sector? Correct to the nearest degree?
The perimeter of a sector is 75.43cm. If the angle – subtended by the sector at the centre is 1350. Find the radius of the sector.
An arc of length 21.34cm subtends an angle 1010 at the centre of a circle. Find the diameter of the circle.
Using trigonometric ratios
AD/r = Sin θ/2
AD = r Sin θ/2
But, AD = DB
Hence
AB chord AB = AD + DB
= r Sin θ/2 + r Sin θ/2
= 2r Sin θ/2 units
Length of Chord = 2r Sin θ/2units
Thus,
Perimeter of segment = θ/360 x 2πr + 2r Sin θ/2
= θ/180 x πr + 2r Sin θ/2
Example 3:
AB is a chord of a circle with centre o and radius 4cm, AOB = 1200. Calculate the perimeter of the minor segment (π = 22/7).
Solution
Chord AB = rSinθ/2, r = 4cm, θ = 1200.
θ/2 = 120/2 = 600.
So chord AB = 2 x 4 Sin 600
= 8 x sin 600
= 8 x(√3)/2
= 4√3cm
= 6.92cm
Length of arc AB = 120/360 x 2 x 4 x 22/7
= 1/3 x 8 x 22/7
= 176/21
= 8.38cm (2 d.p)
Perimeter of minor segment = length of arc + chord
= 8.38 + 6.92cm
= 15.30cm
https://youtu.be/no64hhheEjs
EVALUATION
A sector subtends an angle of 840 at the centre of a circle of radius 5.6cm. calculate the perimeter of the sector to the nearest cm.
New General Mathematics for senior secondary school, Book 1 pages 154 exercise 12d, Nos 7 – 10
AREA OF SECTORS OF A CIRCLE
The area of a sector = θ/360 x area of circle
= θ/360 x πr2, where θ is the angle formed at the centre by the arc of the circle.
Example 1:
Find the area of the sector of a circle of radius 4.8cm which subtends an angle of 1350 at the centre.(Take π = 3.142)
Solution
The area of a sector = θ/360 x πr2
Here θ = 1350, r = 4.8cm, π = 3.142
The area of a sector = 135/360 x 3.142 x (4.8)2
= 135/360 x 3.142 x 4.8 x 4.8
= 27 x 3.142 x 0.4 x 0.8
= 27.14688cm2
Area of sector = 27.15cm2 (decimal places)
Example 2:
AB is an arc of a circle of length 9.2cm with centre 0 and the radius is 4.6cm. Find the area of the sector AOB.
Solution
Length of arc AB = θ/360 x 2πr
r = 4.6cm, length of arc = 9.2cm
Length of arc AB = θ/360 x 2 x π x 4.6
9.2 = θ/360 x 2 x π x 4.6
9.2 x 360 = θ x 2 x π x 4.6
θ = (9.2 x 360)/(2 π x 4.6)
θ = ( 360)/π
Area of sector AOB
= θ/360 x πr2
= θ/360 x πr2
= θ/360 x 1/360 x π x (4.6)2
= (4.6)2
= (4.6) x (4.6)
= 21.16cm2
https://youtu.be/7f9_U85_YX4
EVALUATION
Calculate the area of a sector of a circle which subtends an angle 450 at the centre of the circle, radius 14cm.
A sector of 800 is removed from a circle of radius 12cm. What area of the circle is left? Use 22/7.
New general mathematics for senior secondary school, Book 1 page 154 exercise 12d Nos 1 – 6.
SEGMENTS OF CIRCLES
Area of segments = area of sector – area of triangle
∴ Area of segment = θ/360 x πr2 - 1/2r2 Sin θ/2
= (r2/2) [πθ/180- Sin θ]
Example: The arc AB of a circle, radius 6.5cm, subtends an angle of 450 at the centre o. Find the area of the (i) Minor segment cut off by the chord AB (Take π = 22/7)
Solution
(i) Area of sector AOB = θ/360 x πr2
θ = 450, r = 6.5, π = 22/7
Area of sector AOB = 45/360 x 22/7 x (6.5)2cm2
= 1/8 x 22/7 x 6.5 x 6.5
= 1/8 x 22/7 x 42.25
= 16.598
= 16.60cm2 (2d.p)
(ii.) Area of segment = Area of sector – Area of ∆
Area of ∆ = 1/2 AO X OBSin450
= 1/2 X 0.5 x 6.5Sin450
= 42.25/2 x 1/√2
= (42.25√2)/4
= (42.25 x 1.414)/4
= 14.935375cm2
∴area of segment = 16.60 – 14.35375
= 1.66cm2
https://youtu.be/vVAl1jyL8X0
EVALUATION
An arc AB of a circle radius 4.8cm subtends an angle of 1580 at the centre O. Find
The area of the sector AOB
The area of the minor segment cut off by the chord AB (π = 3.142)
https://youtu.be/vYR9qicyWB4
GENERAL EVALUATION
What is the length of an arc which subtends an angle of 60° at the centre of a circle of radius 1/2m?
A chord AB of a circle radius 9.4cm is 12.8cm. find:
The angle subtended by the chord at the centre of the circle
The area of the minor segment cut off by this chord.
ASSIGNMENT
Solve questions from New General Maths for senior secondary schools 1 page 151. Ex 12c Qs 5 and 6.
Page 154, Ex 12d. Qs 5, 6, 7a, b and 11.
READING ASSIGNMENT
Read New General Maths for senior secondary schools 1 pages 150 – 155.
REFERENCE TEXTS
M. F. Macrae A. O e tal (2011). New General Mathematics for senior secondary school 1
Mrs Maria David – Osuagwuetal (2000). New General Mathematics for senior secondary schools.
Fundamental General Mathematics for senior secondary school by Idode G. O
