Lesson Notes By Weeks and Term - Senior Secondary School 2
DEDUCTIVE PROOF OF CIRCLE GEOMETRY
DEDUCTIVE PROOF OF CIRCLE GEOMETRY
SUBJECT: MATHEMATICS
CLASS: SS 2
DATE:
TERM: 2nd TERM
REFERENCE BOOKS
WEEK EIGHT
TOPIC: DEDUCTIVE PROOF OF CIRCLE GEOMETRY
CONTENT
-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:
4 Radius
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
circumference.
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
Evaluation
Draw a circle, locate and label all its properties in it
Arcs and Chord
Circumference of a circle (Perimeter) = 2π r
Lenght of Arc = Æ x 2πr
3600
Perimeter of a sector = 2r + Æ × 2πr
3600
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
â 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.
Solution
â 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 = Æ × 2πr
3600
Π= 22/7
Æ = Æ= Given: Tan Adj 13 Tan< AOC = 0.9231 Tan-1 (0.9231) =< AOC = 42.70 Length of arc AB = Æ x 2πr 360 = 85.40× 2 × 22 × 17.69cm 3600 7 =26.38cm Evaluation 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 7cm. Perimeter and Area of Sector and Segments of a Circle Area of sector = Æ× πr2 360 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. Example The arc of a circle radius 7cm subtends an angle of 1350 at the centre. Calculate: i the area of the sector ii The perimeter of the sector Area =Æ x π r2 3600 = 57. 75 cm Perimeter = 2r + length of arc But Length of arc = Æ × 2πr 3600 =1350× 2 × 22 × 7 3600 7 =16. 5cm Perimeter = 2(7) + 16.5 = 14cm + 16.5cm =30.5cm Evaluation The angle of a sector of a circle radius 17.5cm is 600. AB is a chord. Find 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 = Construction: Join OA and AB Proof: OA = OB (radii) AM = MB (given)S OM = OM AMO= BMO (SSS) but 2 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 Solution 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. Solution 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 Evaluation 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) Similarly, In fig.8.20 (a) = 2x2 + 2y2 = 2(x2 + y2) But, < APB = x2 + y2 Examples: 1.Find the value of the lettered angle. Solution 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 Evaluation 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. Q P O B A 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. Proof: 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 Example a = b = 400 (angle in the same segments) c = 320 (angle in the same segment) Evaluation 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 prove: Proof: AOB = 2 x But < AOB = 1800 (angle on a straight line) 180 = 2 (A X B) 180= A X B 2
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