6 Basic Properties of Circles (I)...
Transcript of 6 Basic Properties of Circles (I)...
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NON-FOUNDATIONBasic Properties of Circles (I)
6A
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6.2 Chords of a Circle
Key Concepts and Formulae
1. If ON ⊥ AB, then AN = BN.
[Abbreviation: line from centre ⊥ chord bisects chord]O
A BN
2. If AN = BN, then ON ⊥ AB.
[Abbreviation: line joining centre to mid-pt. of chord⊥ chord]
O
A BN
O
C DN
A
BM
O
C DN
A
BM
3. If OM ⊥ AB, ON ⊥ CD and AB = CD, thenOM = ON.
[Abbreviation: equal chords, equidistant from centre]
4. If OM ⊥ AB, ON ⊥ CD and OM = ON, thenAB = CD.
[Abbreviation: chords equidistant from centre areequal]
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Measures, Shape and SpaceMeasures, Shape and Space
In this exercise, unless otherwise specified, O is the centre of a circle.
1. In the figure, N is the mid-point of the chord AB and ∠ = °NOB 55 .Find ∠ OAN.
Solution
ON ⊥ ( ) (line joining centre to mid-pt. of chord ⊥ chord)
OA = ( ) (radius)
∠ OAN = ∠ ( ) (base ∠ s, isos. ¡ )
But ∠ OBN = 180° − ( ) − ( ) (∠ sum of ¡ )
= ( )
∴ ∠ OAN = ( )
2. In the figure, OC = 5 cm, ON = 3 cm and ON ⊥ CD. Find CD.
Solution
CN = ND (line from centre ⊥ chord bisects chord)
By Pythagoras’ theorem,
CN
ND
CD
= −
=
=
=
5 3
4
4
8
2 2 cm
cm
cm
cm
∴
∴
O
A
B
N
55o
5 cm
3 cm
O
NC D
AB
OB
OBN
90° 55°
35°
35°
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6 Basic Properties of Circles (I)
3. In the figure, AB = 8 cm, OM = ON = 2 cm. Find
(a) the chord CD,
(b) OD.
(Give your answers in surd form if necessary.)
Solution
(a) CD = AB (chords equidistant from centre are equal)
∴ CD = 8 cm
(b) DN = NC (line from centre ⊥ chord bisects chord)
∴
DN =
=
8
2
4
cm
cm
By Pythagoras’ theorem,
OD ON ND= +
= +
=
=
2 2
2 22 4
20
2 5
cm
cm
cm
4. In the figure, OM = ON, OM ⊥ AB, ON ⊥ CD, ∠ MBO = 30°and OB = 5 cm. Find
(a) OM,
(b) CD.
Solution
(a)
OM = °
=
5 30
2 5
sin
.
cm
cm
O
A
B
N
C
D
M2 cm
8 cm
2 cm
5 cm
30o
O
A
B
N
C
DM
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Measures, Shape and SpaceMeasures, Shape and Space
O
A BN
5 cm
5. In the figure, AB = 24 cm, ON = 5 cm and ON ⊥ AB. Find theradius of the circle.
Solution
AN NB
AB
=
=
=2
12 cm
(line from centre ⊥ chord bisects chord)
By Pythagoras’ theorem,
OB ON NB= +
= +
=
2 2
2 25 12
13
cm
cm
∴ The radius of the circle is 13 cm.
(b)
MB = °
=
5 30
5 3
2
cos cm
cm
AM MB= (line from centre ⊥ chord bisects chord)
∴
AB = ×
=
2
5 3
5 3
2 cm
cm
CD AB= (chords equidistant from centre are equal)
∴ CD = 5 3 cm
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6 Basic Properties of Circles (I)
2 cm 4 cm
O
A
BM
6. In the figure, OB = 4 cm, OM = 2 cm and OM ⊥ AB. Find thelength of AB.
Solution
7. In the figure, OM = ON, AM = MC, BN = NC and ∠ ACB = 60°.Find ∠ MON.
Solution
OM ⊥ AC (line joining centre to mid-pt. of chord ⊥ chord)
∴ ∠ OMC = 90°
ON ⊥ BC (line joining centre to mid-pt. of chord ⊥ chord)
∴ ∠ ONC = 90°
∠ MON = 360° − 90° − 90° − 60°
= 120°
60oO
A
C
N
B
M
By Pythagoras’ theorem,
MB OB MO= −
= −
=
2 2
2 24 2
2 3
cm
cm
AM = MB (line from centre ⊥ chord bisects chord)
∴
AB = ×
=
(2 2 3
4 3
) cm
cm
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Measures, Shape and SpaceMeasures, Shape and Space
8. In the figure, BN = NC and ∠ BAN = 30°. Find ∠ OCN.
Solution
AN ⊥ BC (line joining mid-pt. of chord ⊥ chord)
∴ ∠ ANB = 90°
∠ ABN = 360° − 30° − 90° (∠ sum of ¡ )
= 60°
OA = OB (radius)
∠ OBA = 30° (base ∠ s, isos. ¡ )
OB = OC (radius)
∠ OBN = ∠ OCN (base ∠ s, isos. ¡ )
Q ∠ OBN = ∠ ABN − ∠ OBA
= 60° − 30°
= 30°
∠ OCN = 30°
A B
C
D
M
9. In the figure, CD is the common chord of two identicalcircles of centres A and B. M is the mid-point of CD. IfAM = (x + 1)2 cm and MB = (4x + 4) cm, find thevalue of x.
Solution
AM ⊥ CD (line joining centre to mid-pt. of chord ⊥ chord)
BM ⊥ CD (line joining centre to mid-pt. of chord ⊥ chord)
AM = MB (equal chord, equidistant from centre)
(x + 1)2 = 4x + 4
x 2 + 2x + 1 = 4x + 4
x 2 − 2x − 3 = 0
(x − 3)(x + 1) = 0
x = 3 or x = −1 (rejected)
30o
O
A
B N C
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6 Basic Properties of Circles (I)
10. In the figure, AB = 48 cm, BC = 56 cm and OC = 82 cm.Find
(a) OM,
(b) the area of ¡ OAC.
Solution
(a) AM = MB (line from centre ⊥ chord bisects chord)
∴
MB =
=
48
2
24
cm
cm
By Pythagoras’ theorem,
OM OC MC= −
= − +
=
=
2 2
224 56
18
82 cm
324 cm
cm
2 ( )
(b) The area of ¡ OAC
=
=
1
2
9
( )( )
(
AC OM
= 48 + 56)(18) cm
36 cm
1
22
2
48 cm 56 cm
82 cm
O
A BM
C
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Measures, Shape and SpaceMeasures, Shape and Space
11. In the figure, the radius of the circle is 13 cm. AB and CD are perpendicularchords intersecting at P. If AB = CD = 24 cm, find the length of BC.
Solution
Let M and N be the mid-points of AB and CD respectively.
OM ⊥ AB (line joining centre to mid-pt. of chord ⊥ chord)
∴
MB = 24
2
1
cm
= 2 cm
By Pythagoras’ theorem,
OM OB MB= −
= −
=
2 2
23 121 cm
5 cm
2
ON = OM (equal chords, equidistant from centre)
∴ ON = 5 cm
∴ PB = PC
= (12 + 5) cm
= 17 cm
BC PB PC= +
=
2 2
17 2 cm
O
A
N
M B
C
D
P
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6 Basic Properties of Circles (I)
12. In the figure, ABCD is a rectangle with vertices lying on a circle ofradius 15 cm.
(a) Prove that AOC is a straight line.
(b) If AB = 24 cm, find the area of ABCD.
Solution
(a) Join QN and PM as shown in the figure.
DA = CB property of rectangle
OQ = ON equal chords, equidistant from centre
DC = AB property of rectangle
OP = OM equal chords, equidistant from centre
Let OQ = ON = a and OP = OM = b.
OC = a b2 2+ and OA = a b2 2+
Q ON = PC and
PC = PD (line from centre ⊥ chord bisects chord)
∴ DC = 2a
Similarly, AD = 2b
By Pythagoras’ theorem,
AC b a
b
AO OC
= +
= +
= +
( ) ( )2 2
2
2 2
2a cm2
∴ AOC is a straight line.
(b) Consider ¡ AMO
AMAB=2
1= 2 cm
By Pythagoras’ theorem,
OM OA AM= −
= −
=
2 2
2 213 12
5
cm
cm
∴ CB = 10 cm
∴ The area of ABCD = (10 × 24) cm2 = 240 cm2
O
A B
CD
M
P
NQ
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Basic Properties of Circles (I)
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6.3 Angles in a Circle
Key Concepts and Formulae
1. ∠ = ∠AOB APB2
[Abbreviation: ∠ at centre twice ∠ at ¡ ce]
2. ∠ = °APB 90
[Abbreviation: ∠ in semi-circle]
3. x y=
[Abbreviation: ∠ s in the same segment]
O
AB
Q
P
OA B
P
x
y
A B
P
Q
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6 Basic Properties of Circles (I)
In this exercise, unless otherwise specified,O is the centre of a circle. Find the unknownsin the following figures. (1 – 12)
1.
x
70o
O
A
B
C
x
100o
O
A
B
C
Solution
∠ = ∠
=
ABC
x
1
2( )
( )
2.
Solution
reflex ∠ AOC = 2∠ ABC(∠ at centre twice ∠ at ¡ ce)
x = 2 × 100°
x = 200°
x
120oO
A
B
C3.
Solution
reflex ∠ AOC = 360° − 120° (∠ s at a pt.)
= 240°
reflex ∠ AOC = 2∠ ABC(∠ at centre twice ∠ at ¡ ce)
= 240°
240° = 2x
x = 120°
4.
x 30o
OA
B
C
Solution
∠ ABC = 90° (∠ in the semi-circle)
x + 90° + 30° = 180° (∠ sum of ¡ )
x = 60°
AOC
35°
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Measures, Shape and SpaceMeasures, Shape and Space
5.x60o
A
B
C
D
Solution
x = 60° (∠ s in the same segment)
x
70o
A
B
C
D6.
Solution
∠ ABC = 70° (base ∠ s, isos. ¡ )
∠ ACB = 180° − 70° − 70° (∠ sum of ¡ )
= 40°
x = 40° (∠ s in the same segment )
x
OA B
C7.
Solution
∠ CAB = x (base ∠ s, isos. ¡ )
∠ ACB = 90° (∠ in the semi-circle)
x + 90° + x = 180° (∠ sum of ¡ )
x = 45°
8.
x
30o
A B
E
C
D
Solution
∠ EBA = x (∠ s in the same segment)
∠ BEC + ∠ ECB = ∠ EBA (ext. ∠ of ¡ )
∴ 30° + 25° = x
∴ x = 55°
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6 Basic Properties of Circles (I)
9.
x
135o
O
A
B
C
D
Solution
∠ CBD = 180° − 135° (adj. ∠ s on st. line)
= 45°
∠ COD = 2∠ CBD(∠ at centre twice ∠ at ¡ ce)
x = 2 × 45°
= 90°
10.x
50o
110o
A
BE
C
D
Solution
∠ EDC = ∠ BAE(∠ s in the same segment)
= 50°
∠ CDE + ∠ DCE = ∠ CEB (ext. ∠ of ¡ )
50° + x = 110°
x = 60°
x
35o
OA B
C
D
x
60o
O
A
B
C
11.
Solution
∠ ACB = 90° (∠ in the semi-circle)
∠ CBA = 180° − 90° − 35° (∠ sum of ¡ )
= 55°
Consider ¡ BCD
x + 90° + 55° = 180° (∠ sum of ¡ )
x = 35°
12.
Solution
∠ OAB = x (base ∠ s, isos. ¡ )
∠ AOB = 180° − x − x (∠ sum of ¡ )
= 180° − 2x
∠ AOB = 2∠ ACB
180° − 2x = 2 × 60°
x = 30°
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Measures, Shape and SpaceMeasures, Shape and Space
13. In the figure, AB is a diameter of the circle. If AB is producedto C such that DA = DC and ∠ ABD = 60°, find ∠ ADC.
Solution
∠ ADB = 90° (∠ in the semi-circle)
∠ DAB = 180° − 60° − 90° (∠ sum of ¡ )
= 30°
∠ DCA = 30° (base ∠ s, isos. ¡ )
∠ BDC = 60° − 30° (ext. ∠ of ¡ )
= 30°
∴ ∠ ADC = ∠ ADB + ∠ BDC
= 90° + 30°
= 120°
60o
AB
C
D
14. In the figure, AB intersects OC at K. If ∠ ACK = 70° and∠ BOC = 110°, find ∠ KBO.
Solution
∠ CAB = 1
2× 110° (∠ at centre twice ∠ at ¡ ce)
= 55°
∠ CKB = 70° + 55° (ext. ∠ of ¡ )
= 125°
∠ KBO + ∠ BOK = ∠ CKB (ext. ∠ of ¡ )
∠ KBO = 125° − 110°
= 15°
70o
110oOA
B
K
C
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6 Basic Properties of Circles (I)
30oA
BC
D15. In the figure, diameter AB is produced to C. If BC = BD and∠ BCD = 30°, find ∠ DAB.
Solution
∠ BDC = 30° (base ∠ s, isos. ¡ )
∠ DBA = 30° + 30° (ext. ∠ of ¡ )
= 60°
∠ ADB = 90° (∠ in the semi-circle)
∠ DAB = 180° − ∠ ADB − ∠ DBA (∠ sum of ¡ )
= 180° − 90° − 60°
= 30°
6
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NON-FOUNDATION
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Date :
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Basic Properties of Circles (I)
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6.4 Relationships among Arcs, Chords and Angles
Key Concepts and Formulae
1. If ∠ AOB = ∠ COD, then
(a)¡_ ¡_AB = CD
[Abbreviation: equal ∠ s, equal arcs]
(b) AB = CD
[Abbreviation: equal ∠ s, equal chords]
2.¡_ ¡_
If AB = CD, then ∠ AOB = ∠ COD.
[Abbreviation: equal arcs, equal ∠ s]
3. If AB = CD, then ∠ AOB = ∠ COD.
[Abbreviation: equal chords, equal ∠ s]
O
A
B C
D
O
A
B C
D
O
A
B C
D
O
A
B C
D
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6 Basic Properties of Circles (I)
Key Concepts and Formulae
5.¡_ ¡_
If AB = CD, then AB = CD.
[Abbreviation: equal arcs, equal chords]
6.¡_ ¡_AB : CD = x : y
[Abbreviation: arcs prop. to ∠ s at centre]
7.¡_ ¡_AB : CD = m : n
[Abbreviation: arcs prop. to ∠ s at ¡ ce]
A
BC
D
A
BC
D
x
y
O
AB
C
D
m
n
A
B
Q
C
D
P
4.¡_ ¡_
If AB = CD then AB = CD.
[Abbreviation: equal chords, equal arcs]
18
Measures, Shape and SpaceMeasures, Shape and Space
In this exercise, unless otherwise specified, Ois the centre of a circle. Find the unknowns inthe following figures (1 – 4)
1.
Solution
= x
x
30°=
=
( )
( )
(arcs prop. to ∠ s at ¡ ce)
4
630
20
× °
= °
Solution
x
y
x
y
7 5
25
125
125
3
7 5
1 5
50
.
.
.
=
=
=
= °°
(arcs prop. to ∠ s at centre)
(arcs prop. to ∠ s at centre)
x cm
y
25o
125o
3 cm
7.5 cm
OA
B C
DE
6 cm
x30o
4 cm Solution
∠ ACB = 180° − 60° − 70° (∠ sum of ¡ )
= 50°
x
x
12
50
60
10
=
=
°°
(arcs prop. to ∠ s at ¡ ce)
4.
x cm
60o 70o
12 cm
AB
C
x
12 cm 6 cm
OA
B
C
3.
Solution
∠ ACB = 90° (∠ in the semi-circle)
∠ CBA = 180° − 90° − x (∠ sum of ¡ )
= 90° − x
12
6
90
30
=
= °
° − x
x
x
(arcs prop. to ∠ s at ¡ ce)
2.
4
6
19
6 Basic Properties of Circles (I)
5.¡_ ¡_
In the figure, if AB : BC = 3 : 4, ∠ AOB = 45°, OB and AC intersectat K, find
(a) ∠ BOC, (b) ∠ CAB, (c) ∠ OKC.
Solution
(a)¡_ ¡_AB : BC = ∠ AOB : ∠ BOC (arcs prop. to ∠ s at ¡ ce)
3
4=
45°
∠ BOC
∴ ∠ BOC = 60°
(b) ∠ CAB = 1
2∠ BOC (∠ at centre twice ∠ at ¡ ce)
= 1
2× 60°
= 30°
(c) ∠ AOC = ∠ AOB + ∠ BOC
= 45° + 60°
= 105°
OA = OC (radius)
∠ OAC = ∠ OCA (base ∠ s, isos. ¡ )
= 180 105
2
° − °(∠ sum of ¡ )
= 37.5°
∠ OKC = 45° + 37.5° (ext. ∠ of ¡ )
= 82 5. °
45o
O
A B
C
K
20
Measures, Shape and SpaceMeasures, Shape and Space
6. In the figure, ABC and AED are straight lines. If AD = AC,¡_ ¡_
find EB : BC.
Solution
∠ BDC = 110° − 65° (ext. ∠ of ¡ )
= 45°
Q ∠ ADC = 65° (base ∠ s, isos. ¡ )
∴ ∠ EDB = 65° − 45°
= 20°¡_ ¡_
∴ EB : BC = 20 : 45
= 4 : 9
E
65o110o
AB C
D
7.¡_ ¡_
In the figure, AB is a diameter of the circle and BC = CD.¡_ ¡_
If ∠ CAB = 15°, find BD : DA.
Solution
Q¡_BC
¡_= CD
∴ ∠ DAC = 15° (arcs prop. to ∠ s at ¡ ce)
Join OD.
∠ ADO = 30° (base ∠ s, isos. ¡ )
∠ DOB = 30° + 30° (ext. ∠ of ¡ )
= 60°
∠ DOA = 180° − ∠ DOB (adj. ∠ s on st. line)
= 180° − 60°
= 120°
∴¡_ ¡_BD : DA = 60 : 120 (arcs prop. to ∠ s at centre)
= 1: 2
15o OA B
C
D
21
6 Basic Properties of Circles (I)
8.¡_ ¡_
In the figure, ∠ PQS = 30°, ∠ QRS = 75° and SR = 2PS. If PR and
QS intersect at K, find ∠ PKS.
Solution¡_
Q SR¡_
= 2PS
∴ ∠ SQR = 2 × 30° (arcs prop. to ∠ s at ¡ ce)
= 60°
∠ SRP = 30° (∠ s in the same segment)
∴ ∠ KRQ = 75° − 30°
= 45°
∠ QKR = 180° − 60° − 45° (∠ sum of ¡ )
= 75°
∴ ∠ PKS = 75° (vert. opp. ∠ s)
30o
75o
K
PS
Q
R
9. In the figure, AB is a diameter of the circle. If BC = CD,prove that OC // AD.
Solution
Let ∠ DAC = a.
∠ CAB = a arcs prop. to ∠ s at ¡ ce
OA = OC radius
∠ ACO = ∠ CAO base ∠ s, isos. ¡
= a
∴ OC // AD alt. ∠ s equal
O
A
B
C
D
22
Measures, Shape and SpaceMeasures, Shape and Space
Solution
(a) AE = AE common side
BE = ED line joining centre ⊥ chord bisects chord
∠ AEB = 180° − 90° adj. ∠ s on st. line
= 90°
∴ ∠ AEB = ∠ AED
∴ ¡ ABE ≅ ¡ ADE S.A.S.
10. In the figure, diameter AC of the circle is perpendicular to chord BD.
(a) Prove that ¡ ABE ≅ ¡ ADE.
(b)¡_ ¡_
If ∠ EAD = 30°, prove that AD = 2CD.
(b) ∠ ADE = 180° − 90° − 30° ∠ sum of ¡
= 60°
∠ ABE = 60° corr. ∠ s, ≅ ¡ s
= = =∠
∠ABE
EAD
60
30
2
1¡_ ¡_
∴ AD = 2CD
O
A
B
C
DE
¡_CD
¡_AD
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NON-FOUNDATIONBasic Properties of Circles (I)
6D
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6.5 Cyclic Quadrilaterals
Key Concepts and Formulae
1. ∠ A + ∠ C = 180° and ∠ B + ∠ D = 180°
[Abbreviation: opp. ∠ s, cyclic quad.]
2. ∠ DCE = ∠ DAB
[Abbreviation: ext. ∠ , cyclic quad.]
A
B
C
D
A
B
CE
D
In this exercise, unless otherwise specified, O is the centre of a circle. Find the unknowns in thefollowing figures (1 – 11)
1.
Solution
x + ( ) = 180° (opp. ∠ s, cyclic quad.)
x = ( )
y = ( ) (ext. ∠ , cyclic quad.)
x
y
85o
115o
85°
105°
115°
24
Measures, Shape and SpaceMeasures, Shape and Space
2.
Solution
∠ DAC = 25° (base ∠ , isos. ¡ )
∠ ADC = 180° − 25° − 25° (∠ sum of ¡ )
= 130°
∠ ADC + x = 180° (opp. ∠ s, cyclic quad.)
x = 180° − 130°
= 50°
x
25o
OA B
CD
3.
x
y
55o45o
A
B C
D
Solution
x = 45° (∠ in the same segment)
∠ ADC + ∠ ABC = 180° (opp. ∠ s, cyclic quad.)
x + 55° + y = 180°
45° + 55° + y = 180°
y = 80°
25
6 Basic Properties of Circles (I)
4.
x
D
C
B
A
Solution
∠ ADC = 60° (prop. of equil. ¡ )
∠ ADC + ∠ ABC = 180° (opp. ∠ s, cyclic quad.)
∴ ∠ ABC = 120°
∠ BCA = x (base ∠ s, isos. ¡ )
x + x + ∠ ABC = 180° (∠ sum of ¡ )
2x + 120° = 180°
x = 30°
5. In the figure, ABCDE is a pentagon inscribed in the circle. If ∠ EBC = 70°,∠ BAE = 120° and DE = DC, find ∠ BED.
Solution
∠ = ° − °
= °
∠ = ∠
∠ =
= °
∠ = ° − °
= °
∠ = ° − ° − °
= °
° − °
EDC
DEC DCE
DEC
ECB
BEC
180 70
110
35
180 120
60
180 70 60
50
180 110
2
Q ∠ = ∠ + ∠
∴ ∠ = ° + °
= °
BED BEC CED
BED 50 35
85
120o
70o
A
B C
D
E
(opp. ∠ s. cyclic quad.)
(base ∠ s, isos. ¡ )
(∠ sum of ¡ )
(opp. ∠ s, cyclic quad.)
(∠ sum of ¡ )
26
Measures, Shape and SpaceMeasures, Shape and Space
7. In the figure, ∠ ADB = 40°, CB = CD and AD // BC, find ∠ BAD.
6. In the figure, BCDF is a cyclic quadrilateral. ABC, AFD, EFB and
EDC are all straight lines. If ∠ FDE = 96° and ∠ BFD = 120°,find
(a) ∠ FED, (b) ∠ ABF.
Solution
(a)
∠ = ° − °
= °
FED 120 96
24
(ext. ∠ of ¡ )
(b) ∠ FDC = 180° − 96° (adj. ∠ s on st. line)
= 84°
∠ ABF = ∠ FDC (ext. ∠ , cyclic quad.)
= 84°
120o
96o
A
B
F
C DE
40oA
B C
D
(alt. ∠ s, AD // BC)
(base ∠ s, isos. ¡ )
(∠ sum of ¡ )
(opp. ∠ s, cyclic quad.)
Solution
∠ = °
∠ = °
∠ = ° − ∠ − ∠
= ° − ° − °
= °
∠ = ° − ∠
= ° − °
= °
DBC
BDC
BCD DBC BDC
BAD BCD
40
40
180
180 40 40
100
180
180 100
80
27
6 Basic Properties of Circles (I)
8.¡_ ¡_
In the figure, ABCD is a cyclic quadrilateral. If BC = 3BA and
∠ BDC = 60°, find
(a) ∠ ADB, (b) ∠ ABD.
Solution
(a)
∠∠
∠°
= =
∠ = °
ADB
BDC
ADB
ADB
60
1
3
20
(arcs prop. to ∠ s at ¡ ce)
∴
(b) ∠ BAD = 180° − ∠ BCD (opp. ∠ s, cyclic quad.)
= 180° − 50°
= 130°
∠ ABD = 180° − ∠ ADB − ∠ BAD (∠ sum of ¡ )
= 180° − 20° − 130°
= 30°
60o
50o
A
B
C
D
O
A
B C
D
9.¡_ ¡_ ¡_ ¡_
In the figure, AB : BC : CD : DA = 6 : 5 : 4 : 3.
(a) Find
(i) ∠ BCA, (ii) ∠ ACD.
(b) Hence, determine whether BA ⊥ AD.
Solution
(a) (i)¡_ ¡_ ¡_ ¡_AB + BC + CD + DA = circumference of the circle
∠ BOA + ∠ BOC + ∠ COD + ∠ DOA = 360°
∠ BCA + ∠ BAC + ∠ CAD + ∠ ACD = 180° (∠ at centre twice ∠ at ¡ ce)
But ∠ BCA : ∠ BAC : ∠ CAD : ∠ ACD¡_ ¡_ ¡_ ¡_
= AB : BC : CD : DA (arcs prop. to ∠ s at ¡ ce)
= 6 : 5 : 4 : 3
∴ ∠ BCA
=
= °+ + +
6
6 5 4 3
60
× 180°
28
Measures, Shape and SpaceMeasures, Shape and Space
A
B
C D
E
10. In the figure, AC = AD, prove that ∠ ABC = ∠ AED.
Solution
(ii) ∴ ∠ ACD
=
= °+ + +
3
6 5 4 3
30
× 180°
(b) ∠ BCD = ∠ BCA + ∠ ACD
= 60° + 30°
= 90°
∠ BAD = 180° − ∠ BCD (opp. ∠ s, cyclic quad.)
= 180° − 90°
= 90°
∴ BA ⊥ AD
∠ ACD = ∠ ADC base ∠ s, isos. ¡
∠ ABC = 180° − ∠ ADC opp. ∠ , cyclic quad.
∠ AED = 180° − ∠ ACD opp. ∠ , cyclic quad.
∴ ∠ ABC = ∠ AED
29
6 Basic Properties of Circles (I)
O
A BC
E
D
11. In the figure, ABC is a straight line, AO // BD and AO = AB.
(a) Show that ∠ AOD = 2∠ DBC.
(b) If ∠ CBD = 60°, prove that OABD is a parallelogram.
Solution
(a) ∠ DBC = ∠ AED ext. ∠ , cyclic quad.
∠ AOD = 2∠ AED ∠ at centre twice ∠ at ¡ ce
∴ ∠ AOD = 2∠ DBC
(b) ∠ AOD = 2 × 60°
= 120°
∠ OAB = ∠ CBD corr. ∠ s, AO // BD
∠ OAB + ∠ AOD = 60° + 120°
= 180°
∴ OD // AB int. ∠ s, supp.
Since OD // AB and AO // BD, OABD is a parallelogram.
6
30
NON-FOUNDATION
Name :
Date :
Mark :
6E
Basic Properties of Circles (I)
Multiple Choice Questions
In this exercise, unless otherwise specified,O is the centre of a circle.
1. In the figure, POQ is a straight line,OC = 10 cm, AB = CD = 16 cm.Find PQ.
3. In the figure, BC is a diameter of thecircle and AB = AD = DC. Find∠ ABD.
16 cm
16 cm
O
A B
QC D
P
A. 11 cm B. 12 cm
C. 13 cm D. 14 cm
2. In the figure, AM = MB. Which of thefollowing is / are true?
O
A BM
I. OM ⊥ AB
II. AO ⊥ OB
III. ¡ OAM ≅ ¡ OBM
IV. ¡ OBM ≅ ¡ OBA
A. I
B. I and II
C. I and III
D. I, III and IV
A
B C
D
A. 27°B. 28°C. 29°D. 30°
4. In the figure, AC and BD intersect at K.Find x.
x
42o
45o
58o
A
B
C
D
K
A. 32°B. 35°C. 45°D. 50° B
C
DB
31
6 Basic Properties of Circles (I)
5. In the figure, find x. 8. In the figure, AB is a diameter of thecircle and AD // OC. Find x.
x
35o 23o
O
A
B
C
A. 115° B. 116°C. 117° D. 118°
6. In the figure, AB is a diameter of thecircle. Find x.
x
44o
A B
C
D
A. 19° B. 21°C. 23° D. 25°
7.¡_ ¡_
In the figure, AD = DC. Find x.
x
100o
45o
A
B
C
D
A. 65° B. 67°C. 69° D. 80°
x
35o
OA B
C
D
A. 100° B. 105°C. 107.5° D. 110°
9. In the figure, find x + y.
xO
A
B
C D
Ey
A. 215° B. 220°C. 225° D. 230°
10. In the figure, BD is a diameter of thecircle. Find ∠ DBC.
53o
A
B C
D
A. 30° B. 35°C. 36° D. 37°
C
C
A
B
C
D
32
Measures, Shape and SpaceMeasures, Shape and Space
11. In the figure, ABC and AED are straightlines. Find ∠ EDB.
36o48o
A
B
C
D
E
A. 11° B. 12°C. 18° D. 22°
12.¡_ ¡_ ¡_
In the figure, BCD : CDA : DAB =3 : 4 : 6. Find ∠ ADC.
A
B
C
D
A. 105° B. 110°C. 115° D. 121°
13. In the figure, AOB is a diameter of thecircle and ∠ DAO = °50 . If DC CB= ,find the value of x y+ .
xy
50o
OA B
CD
14. In the figure, COD is a diameter of thecircle and AB ⊥ CD. If AB = 8 cm andCM = 2 cm, find the radius of the circle.
O
A B
D
C
M
A. 2.5 cm B. 3 cm
C. 4 cm D. 5 cm
15. In the figure, chord ED and AB areproduced to meet at C. If ED = DB,AB = AE and ∠ DCB = 30°, find ∠ EAB.
30o
O
A
BC
D
E
A. 90° B. 98°C. 105° D. 112°
A. 50° B. 55°C. 58° D. 60°
16. In the figure, ∠ ABC = 105°, ∠ BCE = 70°and AB = BC, find x.
A. 100° B. 102.5°C. 105° D. 107.5°
x
70o
105o
A B
EC
D
B
D
C
B
D
D
33
6 Basic Properties of Circles (I)
17. In the figure, ABDF is inscribed in thecircle. AFE, BDE, FDC and ABC are thestraight lines. If ∠ E = 22° and ∠ C = 30°,find ∠ A.
30o
22o
F
A BC
D
E
A. b + c = 180°B. a + d = 180°C. a = x + y
D. d = b + c − 90°
19. In the figure, AB and BC are two chordsof the circle. If reflex ∠ COA = 285°,find x + y.
x y
a
b
c
d
AB
F
E
C
D
A. 30°B. 37.5°C. 52.5°D. 75°
20. In the figure, AB is a diameter of thecircle. Find ∠ DOC.
x
y
285o
O
A
B
C
125o
OA B
CD
K
A. 60°B. 64°C. 68°D. 78°
18. In the figure, which of the followingMUST be true?
A. 70°B. 75°C. 78°
D. 82°
B
A
B
C