Lesson Notes By Weeks and Term - Senior Secondary School 2







  • New General Mathematics SSS2 by M.F. Macraeetal.
  • Essential Mathematics SSS2 by A.J.S. Oluwasanmi. 





    -Definition of Properties of a  Circle.

    -Problems on Length of Arc and Chords.

    -Perimeter and Area of Sector and Segments of a Circle.


Definition of Properties of a Circle    radius

Parts (properties) of a circle are:

  1. Centre
  2. Circumference
  3. Arc

4    Radius

  1. Chord
  2.         Diameter diameter
  3.         Segment segment
  4.         Sector

1.Circumference:This is the curved outer boundary of a circle.

2.Arc:  Arc is a part/portion of the circumference of a circle

3.Major and Minor Arc: The chord which is not a diameter divides the circumference 

into two arc of diff sizes: a major and a minor arc.

4.Radius: this is any straight line joining the  centre to the circumference of a circle.

5.Diameter: A diameter is a chord which passes through the centre and divides the circle into  3 equal parts.

6.Chord: A chord of a circle is a line segment joining the centre is a line its 


7.Sector: This is the region between two radii and the circumference.

8.Segment: it is the region between a chord and the circumference.

9.Major and minor Segment: The chord also divides the circle into  two segments of difference sizes: major and minor segments



Draw a circle, locate and label all its properties in it


Arcs and Chord

Circumference of a circle (Perimeter) = 2π r

Lenght of Arc =      Ɵ      x 2πr


Perimeter of a sector = 2r +  Ɵ  × 2πr


Where  π= 22/7


Example 1

A chord of a circle is 12cm long the radius r of the circle is 10cm calculate the distance of the mid-point of the chord to the center.

O  is the center

AB =  12cm

AO = radius = 10cm

M = mid –point of AB

∆  AMO


|OA|2 = |OM|2 + |AM|2 (Pythagoras Theorem)

102= OM|2 + 62

|OM|2 = 102 - 62

|OM|2 = 100 -36

|OM|2 = 64

|OM =    64 = 8cm


| OM| = 8cm

The mid-point of the chord is 8cm from the centre of the circle


Example 2 

A chord of length 24cm is 13cm from the centre. . Calculate the radius of the circle radius of the circle.



∆ OAC is a right angled triangle 

|OA|2 = AC2 + C02

|OA|2 = 122 + 132

|OA|2 = 144 + 169

|OA|2 = 313

|OA|= 313 = 17.69                                                                                                                           

OA = 17.7cm


Example 3

Calculate the length of the minor arc /AB/ in example 2 above

Length of arc   = Ɵ  ×    2πr


Π= 22/7

         ÆŸ = 




Adj      13

Tan< AOC = 0.9231

Tan-1  (0.9231) =< AOC = 42.70



 Length of arc AB =  Ɵ     x 2πr


                                =   85.40×  2 × 22  ×  17.69cm

                                      3600               7




1)A  chord of a circle is 9cm long if its distance from the centre of the circle is 5cm, calculate.

i.The radius 

ii.The length of the minor arc.

2) What angle does an arc 5.5cm in length subtend at the centre of a circle diameter



Perimeter and Area of Sector and Segments of a Circle

Area of sector =   Ɵנ πr2


Area of segment = Area of sector – Area of the included triangle.

Perimeter of sector = 2r + length of  arc

Perimeter of segment = length of chord + length of arc.



The arc of a circle radius 7cm subtends an angle of 1350 at the centre.


i    the area of the sector

ii    The perimeter of the sector

Area  =Ɵ      x   π r2


         = 57. 75 cm


Perimeter = 2r + length of arc

But     Length of arc =   ÆŸ     × 2πr


                                 =1350×  2 × 22 × 7

3600             7                    =16. 5cm


Perimeter = 2(7) + 16.5

        = 14cm + 16.5cm



The angle of a sector of a circle radius 17.5cm is 600. AB is a chord. Find  

  1. Area of the sector
  2. Perimeter of the sector
  3. Area of the minor segment
  4. Perimeter of the minor segment



Theorem and Proofs Relating to Angles in a Plane.


Theorem  I.

Theorem: A straight line drawn from the centre of a circle to bisect a chord, which is not diameter is at right angle to the chord.

Given: a circle with centre O and Chord AB.

OM Such that  AM   =  MB

To prove: < AMO = 0

Construction: Join OA and AB 


    OA = OB    (radii)

    AM = MB    (given)S

            OM   =    OM

    AMO=       BMO   (SSS)


but    0

1800    =   900



Example I: The radius of a circle is 10cm and the length of a chord of the circle is 16cm. Calculate the distance of the chord from the centre of the circle.

Since ( COA is a right angled triangle, using Pythagoras theorem


x2 = 102 – 82

x2  =  100 – 64

x2 =   36

x  =  √ 36  =  6cm 

Example 2:

The distance of a chord of a circle of radius 5cm from the centre of the circle is 4cm. Calculate the distance of the length of the chord.



                         Chord AB  =  |AC|  +  |CB|

                                | AC|  = | CB|

< AOC is a right angled triangle            O

                               Using Pythagoras:                    5cm

                                  |AC|2  =  52  -  42

=  25 – 16  =  9                  A            C              B

                                   |AC|2  =  9cm

                                    |AC|  =     √9     = 3cm

                                  |AB| = 3 + 3 = 6cm

                               Length of chord AB  =  6cm



Two parallel chords lie on opposite side of the centre of a circle of radius 13cm, their lengths are 10cm and 24cm, what is the distance between the chords?


Theorem 2

The angle that an arc of a circle subtends at the centre is twice that which it subtends at any point on the remaining part of the circumference. 

Given: a circle APB with centre O 

To prove:< AOB = 2 x

Construction:  Join PO and produce to any point Q

Proof :

OA  =   OP            ( radii)

x1  =  x2       (base angle of isosceles triangle)

1  +  x2  (exterior angle of  ∆ AOP)

2  (x1 = x2)

Similarly,  2

In fig.8.20 (a)  

                                        = 2x2 + 2y2

                                        = 2(x2  +  y2)

                           But,       < APB = x2 + y2


1.Find the value  of the lettered angle.  


q =  2 × 410      (angle at the centre= 2 × angle at circumference)

q= 840


2.Find  the lettered angles

x = 2 x 1190 = 2380 (angle at centre = 2x angle at circumference)

y = 3600 – x  (angle at a point)

y  =  3600 – 2380  = 1220

    z   =    y    =    1220=  610  (angle at centre = 2 x angle at circumference)

              2             2

    (x  =  2380

y  =  1220

    z  =  610



1.Find the lettered angles  in the  diagrams below

(a)                                   (b)

Theorems and Proofs Relating to Angles on the Same Segments.

Angle in the Same Segments

Theorem:Angles  in  the  same  segment  of  a  circle  are  equal.





                        O          B


Given: P and Q are any points on the major arc of circle APQB.

To proof: APB = AQB

Construction: Join A and B to O, the centre of the Circle.


2 (same reason)

 2x1 = 2x2 =

x1 = x2 = ½ (AOB)

APB  =  x1

AQB  =  x2


Since P and Q are any points on the major arc, all angles in the major segment are equal to each other. The theorem is also true for angles in the minor segments i.e.




                                a =  b = c


a = b = 400 (angle in the same segments)

c = 320 (angle in the same segment) 



Find the lettered angles.

Theorem and Proof:

(The angle in a semi circle is a right angle)


Theorem: The angle in a semi circle is a right angle.

Given: AB is a diameter on a circle centre O. X is any point on the circumference on the circle.


To prove0

Proof:  AOB = 2 x 

    But < AOB = 1800 (angle on a straight line)

    180 = 2 (A X B)

    180=  A X B



Example: in the fig below: PQ is a diameter of a circle PMQN, centre O if 0, find QNM.

In ∆ PQM

0 (angle in a semi circle)

0 – (900 + 650) [sum of angle in a ∆ ]

0 – 1530 = 270

0 (angle in the same segment)


Example 2:

Find i and j.

0 (angle in a semi circle)

i  = 650 (angle on the same segment with PRS)

j = 900 – 650( angle in a semicircle)

j = 250.



1.In the fig. O is the centre of the circle,BOC is a diameter and 0 ,what is


Find  the  value  of  the  lettered  angles

  1.                                                                     2.

  1. In a rectangular tank is 76cm long, 50cm wide and 40cm high. How many litres of water can it hold?
  2. A 2160 sector of radius 5cm is bent to form a cone. Find the radius of the base of the cone and its vertical angle.



Essential Mathematics for SSS2, page135-136, numbers 1-5.


WEEKEND  ASSIGNMENT                                               :

ObjectiveFind  the  lettered  angles

1  (a) 500   (b) 400   (c) 900    (d) 1000

  1. (a) 650   (b) 1000   (c) 2600   (d) 500

3.(a) 550    (b) 1100   (c) 1650   (d) 600

4.Two parallel chords lie on opposite sides of the centre of a circle of radius 13cm.Their lengths are 10cm and 24cm.What is the distance between the chords?   

(a)15cm    (b)16cm    (C)17cm     (d)18cm

5.The distance of a chord of a circle, of radius 5cm from the centre of the circle is 4cm, calculate the length of the chord.  (a) 6cm   (b) 5cm   (c) 4cm    (d) 7cm



1.Find w, x, y, z.

2.There are two chords AB and CD in a circle. AB=10cm, CD=8cm and the radius of the circle is 12cm.What is the distance of each chord from the centre of the circle? 



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