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Geometry of the Circle - Chords and Angles
Geometry of the Circle
Chords and Angles
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Chords
and
Angles
CHORDS
AND
ANGLES
The Circle is a basic shape and so it can be found almost anywhere. This section will introduce you to
some properties that will make finding lines and angles inside circles easier.
Answer these questions, before working through the chapter.
I used to think:
What is a "Theorem" in Geometry?
What does the word 'subtend' mean?
What does it mean to say a quadrilateral is 'cyclic'?
Answer these questions after you have worked through the chapter.
But now I think:
What is a "Theorem" in Geometry?
What does the word 'subtend' mean?
What does it mean to say a quadrilateral is 'cyclic'?
What do I know now that I didn’t know before?
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Chords and Angles
Basics
Circle Terms
Here is a reminder of some terms which relate to a circle.
diam
eter
minor
segment
secant
centre
radius
chord
sector
arc
Point of
contact
tangent
Complicated Language
These are more terms relating to circles. The terms may sound complicated but they have simple meanings.
C
C
O
A
A
B
Chord AB subtends +ACB
at the circumference.
This means +ACB is
"standing" on chord AB and
touches the circumference.
2
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B
"Concentric circles" are
circles with the same centre
and different radii.
Arc AB subtends +ACB at
the circumference and
+AOB at the centre.
They're just circles inside
each other.
This means +ACB and
+AOB are "standing"
on arc AB.
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Chords and Angles
Questions
Basics
1. Label a to h in the following diagram according to where it lies in relation to the circle.
h
a
a
f
e
d
b
g
c
a
b
c
d
e
f
g
h
2. Complete the sentences according to the following diagram by filling in the blanks.
C
a
Chord AB subtends + _________ at the circumference.
b
Chord BC subtends + _________ at the circumference.
c
Chord _____ subtends + AOC at the centre.
d
Chord BC subtends + BOC at the _____________________.
e
Chord _____ subtends + ABC at the ___________________.
O
A
B
3. What is the difference between a secant and a chord?
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Using Chords with Circles
Chords can be used to solve problems within circles, depending on their length and position. Here are some
theorems explaining how chords are used.
Theorem 1a: Equal chords subtend equal angles at the centre
A
C
Given: AB = CD
To Prove: +AOB = +COD
O
B
D
Proof
In TAOB and TCOD
AB = CD
AO = CO
BO = DO
... TAOB / TCOD
... + AOB = + COD
Proof
(Given)
(Equal radii)
(Equal radii)
(SSS)
(Corresponding angles of congruent triangles)
This means that the angles (at the centre) standing on equal chords are equal to each other.
Theorem 1b (Converse of 1a): Equal angles at the centre subtend equal chords
A
Given: + AOB = + COD
To Prove: AB = CD
C
O
B
D
Proof
In TAOB / COD
+ AOB = + COD
AO = CO
BO = DO
... TAOB / TCOD
... AB = CD
Proof
(Given)
(Equal radii)
(Equal radii)
(SAS)
(Corresponding sides of congruent triangles)
This means that if the angles (at the centre) are equal to each other, then the chords they are standing on are equal.
"Converse" means "inverse".
Find the length of chord PQ given O is the centre of the circle
P
O is the centre
+ POQ = + GOH = 40c
... PQ = GH
Q
40c
O
40c
8 cm
H
G
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(Given)
(Given)
(Chords that subtend equal angles at the
centre are equal)
... PQ = 8 cm
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Theorem 1b
Chords and Angles
Knowing More
Theorem 2a: A perpendicular line from the centre to a chord bisects the chord
Given: + ABO = + CBO = 90c, O is the centre
To Prove: AB = BC
O
A
C
B
Proof
In TABO and TCBO
+ ABO = + CBO = 90c
AO = CO
BO is common
... TAOB / TCBO
... AB = BC
Proof
(Given)
(Equal radii)
(RHS)
(Congruent triangles; TAOB / TCOD )
This means that a perpendicular line drawn from the centre to the chord, cuts the chord in half.
Theorem 2b (Converse of 2a): A line drawn from the centre to a chord's midpoint is perpendicular to the chord
Given: AB = BC; O is the centre
To Prove: OB = AC
O
A
C
B
Proof:
In TABO and TCBO
AB = BC
AO = CO
BO is common
... TAOB / TCOB
+ ABO = + CBO
Proof
(Given)
(Equal radii)
But + ABO + + CBO = 180c
... + ABO = + CBO = 90c
... OB = AC
(SSS)
(Corresponding angles of
congruent triangles)
(Supplementary angles)
This means that a line drawn from the centre, perpendicular to the chord, cuts the chord in half. Also, if a line is
perpendicular to a chord and bisects it ­— it has to pass through the centre.
In the diagram below O is the centre, OD is 50 mm and OE is 30 mm. Find the length of DE and DF
DE2 = OD2 OE2
= 502 302
DE = 40 mm
D
O
E
F
O is the centre
OE = DF
... DE = EF
... DE = EF = 40 mm
(Pythagorean Theorem)
(Given)
(Given)
(Perpendicular from centre to chord bisects chord)
Theorem 2a
DF = DE + EF = 40 + 40
... DF = 80 mm
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Theorem 3a: Equal chords are equidistant from the centre
Given: OE = DF; OB = AC; O is the centre; DF = AC
To Prove: OE = OB
F
E
D
Proof:
DE = 12 DF
AB = 12 AC
O
A
C
B
But AC = DF
` DE = AB
Proof
(Perpendicular from centre bisects chord)
(Perpendicular from centre bisects chord)
(Given)
In TODE and TOAB
DE = AB
+OED = +OBA = 90c
OD = OA
` TODE / TOAB
` OE = OB
Theorem 2a
(Proved above)
(OB = AC and OE = DF )
(Radii)
(RHS)
(Corresponding sides in congruent T's )
This means that if two chords are equal, then they are the same distance from the centre.
Theorem 3b (converse to 3a): Chords which are equidistant to the centre are equal
Given: OE = DF; OB = AC; O is the centre; OE = OB
To Prove: DF = AC
F
E
Proof:
In TODE and TOAB
OE = OB
+OED = +OBA = 90c
OD = OA
` TODE / TOAB
D
O
A
C
B
Proof
(Given)
(Given)
(Radii)
(RHS)
` DE = AB
but DE = 12 DF
(Corresponding sides in congruent T's )
and AB = 12 AC
(Perpendicular from centre bisects chord)
`
1
2
DF =
1
2
(Perpendicular from centre bisects chord)
AC
Theorem 2a
` DF = AC
This means that if two chords are the same distance from the centre, then the chords are equal.
O is the centre of the circle. Find the length of PR in the diagram below if OQ = ON = 15 cm and MN = 14 cm
R
M
14
O
Q
N
O is the centre
MN = NS = 14 cm
` MS = MN + NS = 28 cm
(Given)
(Perpendicular from centre bisects chord)
Theorem 2a
15
S
P
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But PR = MS
... PR = 28 cm
(Chords equidistant from the centre are equal)
Theorem 3b
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Chords and Angles
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Here are some examples using all the above theorems:
If O is the centre of the circle, find the length of BD if AB = 200 mm and OC = 60 mm
O is the centre
... AB is a diameter
... OB is a radius
` OB = 200 mm ' 2 = 100 mm
D
C
B
60 mm
BC2 = OB2 - OC2
` BC = 1002 - 602 = 80 mm
O
A
OC = BD
BC = CD = 80 mm
(Given)
(Pythagorean Theorem)
(Given)
(Perpendicular from centre bisects chord)
... BD = BC + CD
= 80 + 80 = 160 mm
Theorem 2a
Find the sizes of angles +A , +B and +C in the diagram below given EO = OD = OF and
+OFB = +ODA = +OEA = 90c
OE = OD
OE = AC and OD = AB
AB = AC
(Given)
D
Similarly: BC = AC
B
... AB = AC = BC
... TABC is equilateral
... + A = + B = + C = 60c
(Chords equidistant from the centre
are equal)
Theorem 2a
A
E
O
C
F
(Given)
(Chords equidistant from the centre
are equal)
(All sides are equal)
(Properties of equilateral triangle)
O is the centre in the circle below. Find the length of DE given and +AOC = +DOE = 50c and +OBA = 90c
D
O 50c
E
O is the centre
OB = AC
... AB = BC
... AC = AB + BC
= 10 + 10 = 20 cm
(Given)
(Given)
(Perpendicular from centre bisects chord)
50c
A
10 cm
B
C
+ AOC = + DOE = 50c
(Given)
... DE = AC
(Chords that subtend equal angles at
the centre are equal)
... DE = 20 cm
Theorem 1b
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Questions
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1. Find x in each of the following (all lengths in cm).
a
b
10
x
x
10
c
d
x
42c
42c
x
5
25c
2. A circle with diameter 60 cm has a chord MN 18 cm from the centre.
a
Draw a rough sketch of the circle and chord in the box provided.
b
How long is MN?
8
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Chords and Angles
Questions
Knowing More
3. The circle below has a diameter of 400 units. OB = 120 units and AP = 160 units.
Q
S
a
Use the Pythagorean Theorem to find the length of OA
b
Find the length of RS
O
B
A
120
160
R
P
4. Find the length of chord AB.
C A
47c
B
11
86c
O 88c
E
D
12
F
5. The circle below has diameter 50 cm. ED = 30 cm and FG = 40 cm. Find the distance between the two chords.
(Hint: Sketch in the distances between the chords and the centre)
G
E
O
D
F
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Chords and Angles
Questions
Knowing More
B
A
E
C
D
6. A and C are the centres of the above circles (which have the same radius). Use the above diagram to answer the following questions.
a
Show that BE = ED.
b
On the diagram construct lines AB, BC, CD and DA.
c
Prove that BD = AC .
10
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Chords and Angles
Questions
Knowing More
d
Now show that TBEC / TDEC and TAEB / TAED
e
Show that AE = EC.
f
Show that ABCD is a rhombus.
g
Complete this sentence:
When a line is drawn between the centres of different circles through a common chord,
then this line and the chord _________ each other at an angle of _____.
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Chords and Angles
Using Our Knowledge
Using Angles with Circles
Here are some theorems showing how to find and use angles appearing inside circles.
Theorem 4: An angle subtended at the centre is twice the angle subtended at the circumference
standing on the same arc
In these 3 diagrams, the same arc (AB) subtends an angle at the centre ( +AOB ) and at the circumference ( +ACB )
Diagram b
Diagram a
Diagram c
Proof
C
M
O
O
M
A
B
A
B
O
M
C
B
A
C
Given: O is the centre of the circle
To Prove: +AOB = 2 # +ACB
Proof
Construct line CM which passes through the centre O
In all 3 diagrams
OC = OA
` +OCA = +OAC
but +AOM = +OCA + +OAC
` +AOM = 2+OCA
similarly +BOM = 2+BCO
In Diagram a
+AOB = +AOM + +BOM
` +AOB = 2+OCA + 2+BCO
= 2 (+OCA + +BCO)
= 2+ACB
(Equal radii)
(Angles opposite equal sides)
(Exterior angle of a triangle)
In Diagram b
reflex +AOB = +AOM + +BOM
` reflex +AOB = 2+OCA + 2+BCO
= 2 (+OCA + +BCO)
= 2+ACB
In Diagram c
+AOB = +BOM - +AOM
` +AOB = 2+BCO - 2+OCA
= 2 (+BCO - +OCA)
= 2+ACB
Find the sizes of the angles labeled x
a
b
x
12
35c
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x = 1 # angle at centre
x = 2 # angle at circumference
= 2 # 35c
= 70c
=
40c
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x
2
1
2
# 40c
= 20c
Chords and Angles
Using Our Knowledge
Theorem 5: The angle in semicircle is a right angle
C
O is the centre
B
A
O
Proof
+AOB = 180c
(Straight line)
+ACB = 1 +AOB
(Angle at centre is twice angle at
circumference on same arc)
2
180c
+ACB = 1 # 180c = 90c
2
Theorem 4
This means that an angle standing on the diameter of a circle is 90c. Also, if an angle on the circumference is
90c then it must be standing on the diameter.
Theorem 6: Angles subtended on the circumference by the same arc (in the same segment) are equal
D
C
Given: O is the centre of the circle
To Prove: +ACB = +ADB
Proof:
Construct radii AO and BO
+ACB = 12 +AOB
O
B
A
+ADB = 12 +AOB
` +ACB = +ADB
Proof
(Angle at centre is twice angle at
circumference on same arc)
(Angle at centre is twice angle at
circumference on same arc)
Imagine there is a chord joining A and B. The angles standing on the same arc are equal if they are on the same
side of the imaginary chord. The angles standing on AB at the circumference below AB, would NOT be equal to
the angles meeting at the circumference above. If angles are on the same side of the imaginary chord then they
are "in the same segment".
Find the values of x and y in the following diagram
60c
x
(Angles in same segment on same arc)
x = 60c
y = 120c
(Angles in same segment on same arc)
Theorem 6
120c
y
Notice x and y are not equal to each other even though they are
standing on the same arc. This is because x is above the imaginary
chord, and y is below the imaginary chord.
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Using Our Knowledge
Here are some examples using the above theorems.
O is the centre of the circle below. Find the sizes of the angles labeled x, y and z
P
y = 2+PQR
` y = 2 # 60c = 120c
S z
y
O
60c
x
Q
OP = OR
` +PRO = +RPO = x
` x + x + 120c = 180c
` 2x = 60c
` x = 30c
(Angle at centre is twice angle at
circumference on same arc)
(Equal radii)
(Angles opposite equal sides)
(Sum of angles in a triangle)
reflex +POR = 360c - y = 240c (Angles around a point)
R
z = 1 # reflex +POR (Angle at centre is twice angle at
`z=
2
1
2
circumference on same arc)
# 240c = 120c
O is the centre of the circle below. Show that + DEO = + EGF
D
G
+EGF = 12 +EOF
(Angle at centre is twice angle at
circumference on same arc)
and +EDF = 12 +EOF
(Angle at centre is twice angle at
circumference on same arc)
O
OD = OE
` +ODE = +DEO
but +ODE = +EGF
E
F
` +DEO = +EGF
(Equal radii)
(Angles opposite equal sides)
(Angles in same segment on same arc)
(Both equal + ODE)
In the following diagram O is the centre of the circle and + MPR = 90c. Show that + NPR = + NQP
M
+MQP = 90c
N
(Angle in a semicircle)
` +NQP = 90c - +MQN
now +NPR = 90c - +MNP
Q
O
but +MQN = +MPN
(Angles in same segment on same arc)
` +NPR = 90c - +MQN
` +NPR = +NQP
P
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R
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Chords and Angles
Questions
Using Our Knowledge
1. Write only the sizes of x and y (no reasons necessary) in each of the following diagrams:
a
b
x=
x
y
x=
y 272c
y=
30c
x
y=
52c
c
d
y
x=
x=
x
y
x
y=
y=
35c
41c
2. Find the size of + ACB.
C
A
20c
O
B
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Chords and Angles
Questions
Using Our Knowledge
3. Find + QOR in the diagram below.
R
Q
P
30c
O
4. O is the centre of the circle below and LN is the diameter.
a
Find + LOM.
b
Find + LNM.
N
O
K
x
L
c
Find + LMN.
d
Let + KLO = x. Find + KOL in terms of x.
e
Prove + LKO + + KNO = 90c.
16
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60c
M
Chords and Angles
Questions
Using Our Knowledge
A
E
B
F
O
x
C
x
D
5. In the diagram above O is the centre of the circle. Let + ODC = + ACB = x.
a
Show that AB = BC.
b
Use + CAB to find + BOC in terms of x.
c
Find + BCD.
d
Find + OCF in terms of x.
e
Find + DEC in terms of x and prove that DE || CA.
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Thinking More
Cyclic Quadrilaterals
These are called "cyclic quads" for short. These are quadrilaterals that would allow a circle to pass through
all its vertices.
This is a cyclic quad.
This is NOT a cyclic quad.
This is NOT a cyclic quad.
All 4 vertices touch
the circumference.
One vertex lies outside
the circle.
One vertex does not touch
the circumference.
Cyclic quads have their own properties and theorems which can be used to solve problems.
Theorem 7: Opposite angles of a cyclic quad are supplementary
A
Given: O is the centre of the circle
To Prove: + C + + A = 180c and + B + + D = 180c
y
Proof
Proof:
Construct radii OB and OD
2x
O
D
2y
Let +C = x and +A = y
` reflex +BOD = 2x
and +BOD = 2y
B
but 2x + 2y = 360c
x
(Angle at centre is twice angle at
circumference on same arc)
(Angle at centre is twice angle at
circumference on same arc)
(Angles around a point)
` x + y = 180c
` +C + +A = 180c
` The opposite angles are supplementary.
Similarly, by constructing radii OA and CD it can be shown
+ B + + D = 180c
C
This also means that a quadrilateral is cyclic if its opposite angles add up to 180c
Find the sizes of angle m and n in the cyclic quadrilateral below
A
m + 80c = 180c
n
` m = 180c - 80c = 100c
Theorem 7
B 80c
115c
m
n + 115c = 180c
D
` n = 180c - 115c = 65c
C
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(Opposite angles of cyclic quad are
supplementary)
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(Opposite angles of cyclic quad are
supplementary)
Chords and Angles
Thinking More
Theorem 8: The exterior angle of a cyclic quad is equal to the interior opposite angle
B
Given: O is the centre of the circle
To Prove: + ABC = + CDE
C
Proof
Proof:
+ABC + +ADC = 180c
+CDE + +ADC = 180c
E
D
A
` +ABC = +CDE
(Opposite angles of cyclic quad
are supplementary)
(Angles on a straight line)
(Both angles are supplementary
with + ADC)
This also means that a quadrilateral is cyclic if the exterior angle is equal to its interior opposite angle.
Find the size of angle x and y
R
127c
Q
x
T
110c
S
x = 110c
(Exterior angle of cyclic quad equals
interior opposite angle)
y + 127c = 180c
(Opposite angles of cyclic quad are
supplementary)
` y = 180c - 127c
= 53c
y
P
Find the following angles if O is the centre of the circle
a
O
A
55c
D
Find + ACD
AD is the diameter
... + ACD = 90c
(O is the centre)
(Angle in semicircle is right angle)
E
b
Find + ADC
+ ADC + + ACD + + CAD = 180c (Interior angles of a triangle)
... + ADC = 90c + 55c = 180c
... + ADC = 35c
c
Find + ABC
+ CDE + + ADC = 180c
... + CDE + 35c = 180c
... + CDE = 145c
B
C
Now + ABC = + CDE
(Angles on a straight line)
(Exterior angle of cyclic quad
equals interior opposite angle)
... + ABC = 145c
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Chords and Angles
Questions
Thinking More
1. Find the sizes of the angle x and y in each of the following.
a
b
101c
136c
124c
y
77c
x
x
y
c
d
x
102c
61c
x
y
e
y
f
x
88c
111c
y
60c
132c
y
x
20
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Chords and Angles
Questions
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2. Answer the following questions about this diagram given AB = BC and AD = DC.
a
A
Show TABD / TCBD.
B
C
D
b
Find the sizes of + DAB and + DCB.
c
Is BD the diameter of the circle? Prove it.
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Chords and Angles
Questions
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3. Opposite angles of a cyclic quad are supplementary. Is trapezium ABCD a cyclic quad?
A
30c
B
D
40c
70c
C
4. Opposite angles of a cyclic quad are supplementary.
a
What do you know about the opposite angles of a parallelogram?
b
When are the opposite angles of a parallelogram supplementary?
c
Is it possible for a non-rectangular parallelogram to be a cyclic quadrilateral?
22
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Chords and Angles
Questions
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5. Use the information given below to answer the questions that follow.
Given (fill this on your diagram)
• + HED = 50c
• + GHI = 20c
• O is the centre of the larger circle
• GFOH is a cyclic quad in the smaller circle
a
H
D
Find + HOD.
O
I
G
F
E
b
Find + FGH.
c
Find + GFH.
d
Find + FHO.
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Chords and Angles
Visual Theorems
Visual Theorems
Here is a visual summary of all the theorems in this chapter.
Theorem 1a: Equal chords subtend equal angles
at the centre
Theorem 4: An angle subtended at the centre is
twice the angle subtended at the circumference by
the same arc.
x
2x
Theorem 1b: Equal angles at the centre subtend
equal chords
Theorem 5: The angle in a semicircle is a right angle
Theorem 2a: A perpendicular line from the centre
to a chord bisects the chord
Dia
me
ter
Theorem 6: Angles subtended by the same arc in
the same segment are equal
Theorem 2b: A line drawn from the centre to a
chord’s midpoint is perpendicular to the chord
Theorem 3a: Equal chords are equidistant to the centre
Theorem 7: Opposite angles of a cyclic quad are
supplementary
y
180c - x
x
180c -y
Theorem 3b: Chords which are equidistant to the
centre are equal
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K 15
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TOPIC
Theorem 8: The exterior angle of a cyclic quad is
equal to the interior opposite angle
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Chords and Angles
Answers
Basics:
1. a
c
Knowing More:
Sector
b
Chord
Arc
d
Tangent
6. b
A
e
Centre
f
Diameter
g
Secant
h
Point of contact
g
2. a
+ACB
b
+BAC
c
+AOC
d
Centre
e
Circumference
1. a
c
When a line is drawn between the centres
of different circles through a common
chord, then this line and the chord MEET
each other at an angle of 90c
x = 30c
x = 41c
y = 41c
Knowing More:
2. a
b
x = 88c
y = 44c
d
x = 70c
y = 145c
2. +ACB = 70c
x = 20 cm
b
x = 5 cm
x = 5 cm
d
x = 130c
3. +QOR = 60c
b
MN = 48 cm
4.
a
+LOM = 60c
30
b
+LNM = 30c
30
c
+LMN = 90c
d
+KOL = 180c - 2x
M
C
y = 52c
However, a chord is a line that is formed
between two points on the circumference.
c
E
Using Our Knowledge:
3. A secant is a line that cuts through a circle but
extends beyond the circumference of the circle.
1. a
B
18
N
3. a
OA = 120 units
4. a
AB = 11
b
RS = 320 units
5. OM = 20 cm
5. b
BOC = 2x
c
+BCD = 90c
d
+OCF = 90c - 2x
e
+DEC = 90c - 2x
ON = 15 cm
` distance between chords = 35 cm
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25
Chords and Angles
Answers
Thinking More:
1. a
x = 77c
b
y = 56c
y = 101c
c
x = 61c
d
x = 114c
f
b
x = 37c
y = 51c
y = 66c
2.
x = 102c
y = 78c
y = 29c
e
x = 44c
+DAB = 90c
+DCB = 90c
3. ABCD is a cyclic quadrilateral
4. a They are equal.
b
When a parallelogram is a square or
rectangle.
c
No, because opposite angles are not
supplementary, they are equal.
5. a +HOD = 100c
c
26
+GFH = 80c
K 15
SERIES
TOPIC
b
+FGH = 80c
d
+FHO = 30c
100% Geometry of the Circle – Chords and Angles
Mathletics 100%
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Chords and Angles
Notes
100% Geometry of the Circle – Chords and Angles
Mathletics 100%
© 3P Learning
K 15
SERIES
TOPIC
27
Chords and Angles
28
K 15
SERIES
TOPIC
Notes
100% Geometry of the Circle – Chords and Angles
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Geometry of the Circle - Chords and Angles
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