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7
34
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Basic Properties of Circles (II)
7A
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7.1 Tangents to a Circle and their Properties
Key Concepts and Formulae
1. If PQ is the tangent to the circle at T, then PQ ⊥ OT.
[Abbreviation: tangent ⊥ radius]
2. Tests for tangency
If PQ ⊥ OT, then PQ is the tangent to the circle at T.
[Abbreviation: converse of tangent ⊥ radius]
3. If two tangents, TP and TQ, are drawn to acircle from an external point T, then
(a) TP = TQ,
(b) ∠ POT = ∠ QOT,
(c) ∠ PTO = ∠ QTO.
[Abbreviation: tangent properties]
O
P QT
O
P QT
O
P
Q
T
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7 Basic Properties of Circles (II)
In the following figures, AB is the tangent to the circle at P. Find x. (1 – 4)
1.
Solution
Q OP OC
OPC
OPC
x
=∠ = ∠
∠ =
== −=
−
( )
( )
( ) ( )
( )
( ) ( )
2
( )
∴ (base ∠ s, isos. ¡ )
(∠ sum of ¡ )
(tangent ⊥ radius)
2.
Solution
Q
OP OC
OPC OCP
OPC
CPB OPC
x CPB
=
∠ = ∠
∠ =
= °
∠ = ° − ∠
= ° − °
= °
= ∠
= °
° − °
180 60
2
60
90
90 60
30
30
(radius)
∴ (base ∠ s, isos. ¡ )
(∠ sum of ¡ )
(tangent ⊥ radius)
(base ∠ s, isos. ¡ )
x
60o
O
A BP
C
x
105oO
A BP
C
radius
OCP
180° 105°
37.5°
90° 37.5°
52.5°
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Measures, Shape and SpaceMeasures, Shape and Space
3.
Solution
Q ∠ OPB = 90° (tangent ⊥ radius)
OC = OP = 9 cm (radius)
By Pythagoras’ theorem,
OP PB OB
x
x
x
x
x
2 2 2
2 2 2
2
2
9 12 9
81 144 9
225 9
9 15
6
+ =
+ = +
+ = +
= +
+ =
=
( )
( )
( )
or
9 15
24
− = −
= −
x
xor (rejected)
x
45oO
A BC
P
4.
Solution
Q
OP OC
OPC OCP
OPC
x
=
∠ = ∠
∠ =
= °
= ° − °
= °
° − °180 45
2
67 5
90 67 5
22 5
.
.
.
(radius)
∴ (base ∠ s, isos. ¡ )
(∠ sum of ¡ )
(tangent ⊥ radius)
x cm9 cm
12 cm
O
A BP
C
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7 Basic Properties of Circles (II)
In the following figures, TA and TB are tangents to the circle at A and B respectively. Find x. (5 – 8)
5.
Solution
∠ TAB = ( ) − ( ) (adj. ∠ s on st. line)
= ( )
TA = ( ) (tangent properties)
∠ TAB = ∠ ( ) (base ∠ s, isos. ¡ )
= ( )
x = ( ) − ( ) − ( ) (∠ sum of ¡ )
= ( )
6.
x
110o
A
B
T
Solution
∠ OBT = 90° (tangent ⊥ radius)
∠ BOT = 180° − 90° − 25° (∠ sum of ¡ )
= 65°
∴ x = 65° (tangent properties)
x
25o
O
A
B
T
180° 110°
70°
TB
TBA
70°
180° 70° 70°
40°
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Measures, Shape and SpaceMeasures, Shape and Space
7.
Solution
∠ = °
∠ = ° − ° − °
= °
∠ = °
∠ = ° − ° − °
= °
=
= °
°
OAT
AOT
BOT
AOB
x
90
180 90 25
65
65
360 65 65
230
115
230
2
reflex
(tangent ⊥ radius)
(∠ sum of ¡ )
(tangent properties)
(∠ s at a pt.)
(∠ at centre twice ∠ at ¡ ce)
x
25oO
A
B
C
T
x55o
O
A
B
CT
Solution
Join AO and BO.
∠ = ∠ = °
∠ = ° − ° − ° − °
= °
∠ =
= °
= ° − °
= °
°
TAO TBO
AOB
ACB
x
90
360 90 90 55
125
62 5
180 62 5
117 5
125
2
.
.
.
(tangent ⊥ radius)
(∠ at centre twice ∠ at ¡ ce)
(opp. ∠ s, cyclic quad.)
8.
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7 Basic Properties of Circles (II)
9. In the figure, AB, BC and CA touch the circle at P, Qand R respectively. If AB = 11 cm, AR = 2.8 cmand RC = 4.8 cm, find BC.
Solution
RC = QC (tangent properties)
∴ QC = 4.8 cm
AR = AP (tangent properties)
∴ AP = 2.8 cm
PB = AB − AP
= (11 − 2.8) cm
= 8.2 cm
PB = BQ (tangent properties)
∴ BQ = 8.2 cm
Q BC = BQ + QC
= (8.2 + 4.8) cm
= 13 cm
A
B Q C
RP2.8 cm
4.8 cm
11 cm
10. In the figure, the sides of equilateral triangle ABC are tangentsto the circle at P, Q and R. Find ∠ AOP.
Solution
∠ CAB = 60° (prop. of equil. ¡ )
∠ RAO = ∠ OAP (tangent properties)
∴ ∠ OAP =
60
2
° (tangent properties)
= 30°
Join OP.
∠ OPA = 90° (tangent ⊥ radius)
∠ AOP = 180° − ∠ OPA − ∠ OAP (∠ sum of ¡ )
= 180° − 90° − 30°
= 60°
O
A B
R
C
Q
P
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Measures, Shape and SpaceMeasures, Shape and Space
11. In the figure, AB is the tangent to the circle at T and DOCB is astraight line. Prove that ∠ ODT = ∠ CTB.
Solution
∠ DTC = 90° ∠ in the semi-circle
∠ OTB = 90° tangent ⊥ radius
∠ OTD + ∠ OTC = ∠ CTB + ∠ OTC = 90°
Q OT = OD radius
∠ OTD = ∠ ODT base ∠ s, isos. ¡
∴ ∠ ODT + ∠ OTC = ∠ CTB + ∠ OTC
∴ ∠ ODT = ∠ CTB
12. In the figure, TA and TB are tangents to the circle at A andB respectively. If ∠ AQB = 115°, find
(a) reflex ∠ AOB,
(b) ∠ ATO.
Solution
(a) reflex ∠ AOB = 2 × 115° (∠ at centre twice ∠ at ¡ ce)
= 230°
(b) ∠ AOB = 360° − 230° (∠ s at a pt.)
= 130°
Join OT.
∠ AOT =
130
2
°(tangent properties)
= 65°
∠ OAT = 90° (tangent ⊥ radius)
∠ ATO = 180° − ∠ OAT − ∠ AOT (∠ sum of ¡ )
= 180° − 90° − 65°
= 25°
O
A B
D
C
T
O
A
B
T
Q115o
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7 Basic Properties of Circles (II)
13. In the figure, AB is the tangent to the circle at T. OB cuts thecircle at C. If CT = CB, find ∠ TOC.
Solution
Let ∠ CBT = a.
Q CT = CB (given)
∠ CTB = ∠ CBT = a (base ∠ s, isos. ¡ )
∠ OCT = a + a = 2a (ext. ∠ of ¡ )
OC = OT (radius)
∠ OCT = ∠ OTC = 2a (base ∠ s, isos. ¡ )
∠ OTB = 90° (tangent ⊥ radius)
∠ OTC + ∠ CTB = 2a + a = 90°
∴ a = 30°
∠ TOC = 180° − ∠ OTC − ∠ OCT (∠ sum of ¡ )
= 180° − 2a − 2a
= 180° − 4a
= 180° − 120°
= 60°
14. In the figure, EF and AC are two parallel tangents to the circleat F and A respectively. BE is another tangent to the circle atD. If ∠ OBE = 60° and ∠ FEO = 30°,
(a) find ∠ EBC,
(b) prove that ¡ OBD ~ ¡ EBO.
Solution
(a) ∠ OED = ∠ FEO = 30° (tangent properties)
∠ EBC = ∠ FED (alt. ∠ s, FE // AC)
= ∠ FEO + ∠ OED
= 30° + 30°
= 60°
O
A B
C
T
60o
30o
O
A B
F
C
D
E
(Solution continues on the next page.)
42
Measures, Shape and SpaceMeasures, Shape and Space
Solution
(b) Join OD.
∠ EOB = 180° − ∠ OED − ∠ EBO ∠ sum of ¡
= 180° − 30° − 60°
= 90°
∠ ODB = 90° tangent ⊥ radius
∴ ∠ EOB = ∠ ODB = 90°
∠ OBD = ∠ EBO = 60° common angle
∴ ¡ OBD ~ ¡ EBO A.A.A.
15. In the figure, the circle PQRS is inscribed in the quadrilateral MBCN.M and N are mid-points of AB and AC respectively. If MN // BC,find the perimeter of ¡ ABC.
Solution
MN =1
2BC (mid-pt. theorem)
= 9 cm
MP = MQ (tangent properties)
NP = NS (tangent properties)
BQ = BR (tangent properties)
CR = CS (tangent properties)
∴ MB + NC
= MQ + QB + NS + SC
= MP + BR + PN + RC
= (MP + PN ) + (BR + RC)
= (9 + 18) cm
= 27 cm
Q M and N are the mid-points of AB and AC respectively.
AB + AC + BC = (2MB + 2NC + 18) cm
= [2(MB + NC) + 18] cm
= (2 × 27 + 18) cm
= 72 cm
∴ The perimeter of ¡ ABC = 72 cm.
18 cm
A
B
N
C
Q
P
R
S
M
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43
NON-FOUNDATIONBasic Properties of Circles (II)
7B
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7.2 Angles in the Alternate Segment
Key Concepts and Formulae
1. A tangent-chord angle of a circle is equal to an angle in the alternate segment.
(a) (b)
∠ ATQ = ∠ ABT ∠ ATP = ∠ ACT
[Abbreviation: ∠ in alt. segment]
2. If ∠ ATQ = ∠ ABT, then PQ is the tangent to thecircle at T.
[Abbreviation: converse of ∠ in alt. segment]
A
B
PT
Q
A
C
PT
Q
A
B
PT
Q
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Measures, Shape and SpaceMeasures, Shape and Space
In the following figures, PQ is the tangent to the circle at T. Find the unknown x. (1 – 6)
1.
Solution
∠ TBA = ( ) (∠ in alt. segment)
x = 2 × ( ) ( )
= ( )
2.
25o
x
O
A
B
P QT
x
25o
O
A
B
TP Q
Solution
∠ OTP = 90° (tangent ⊥ radius)
∠ BAT = ∠ BTP (∠ in alt. segment)
= 90° + 25°
= 115°
∠ BTA = ∠ ABT = x (base ∠ s, isos. ¡ )
∠ BTA + ∠ ABT + ∠ BAT = 180° (∠ sum of ¡ )
x + x + 115° = 180°
x = 32 5. °
25°
25°
50°
∠ at centre twice ∠ at ¡ ce
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7 Basic Properties of Circles (II)
3.
Solution
∠ BTQ = 180° − 136° (adj. ∠ s on st. line)
= 44°
x = ∠ BTQ (∠ in alt. segment)
= 44°
4.
x
136o
A
B
P QT
x
62o
OA
B
P
Q
T
Solution
∠ ABT = 62° (∠ in alt. segment)
∠ ATB = 90° (∠ in the semi-circle)
∠ BAT = 180° − ∠ ABT − ∠ ATB (∠ sum of ¡ )
= 180° − 62° − 90°
= 28°
∠ BAT + ∠ AQT = ∠ ATP (ext. ∠ of ¡ )
28° + x = 62°
x = 34°
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Measures, Shape and SpaceMeasures, Shape and Space
5.
Solution
∠ BTP = x (base ∠ s, isos. ¡ )
∠ ABT = ∠ BPT + ∠ BTP (ext. ∠ of ¡ )
= x + x
= 2x
∠ ATQ = ∠ ABT (∠ in alt. segment)
= 2x
∠ BTP + 93° + ∠ ATQ = 180° (adj. ∠ s on st. line )
93° + 3x = 180°
x = 29°
6.
x93o
A
B
P QT
x
70o
40o
A
B
Q
C
TP
Solution
∠ TCA = 40° (∠ in alt. segment)
∠ CAT = 70° (∠ in alt. segment)
∠ BAC = x (base ∠ s, isos. ¡ )
∠ BCT + ∠ BAT = 180° (opp. ∠ s, cyclic quad.)
40° + 70° + x + x = 180°
x = 35°
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7 Basic Properties of Circles (II)
7. In the figure, TA and TB are tangents to the circle at A and¡_ ¡_
B respectively. If ∠ BAC = 40° and BC : AC = 4 : 7,
find ∠ ATB.
Solution
∠∠
= =ABC
BAC
7
4(arcs prop. to ∠ s at ¡ ce)
∴ ∠ ABC = 70°
∠ ACB = 180° − ∠ BAC − ∠ ABC (∠ sum of ¡ )
= 180° − 40° − 70°
= 70°
∠ TAB = ∠ ACB (∠ in alt. segment)
= 70°
∠ TBA = ∠ ACB (∠ in alt. segment)
= 70°
∠ ATB = 180° − ∠ TAB − ∠ TBA (∠ sum of ¡ )
= 180° − 70° − 70°
= 40°
40o
A
B
C
T
8. In the figure, PQ and RS are tangents to the circle at A and C
respectively. If ∠ ABC = 65°, find a + b.
Solution
Join BD.
∠ ADB = a (∠ in alt. segment)
∠ BDC = b (∠ in alt. segment)
∴ ∠ ADC = ∠ ADB + ∠ BDC
= a + b
∠ ADC + ∠ ABC = 180° (opp. ∠ s, cyclic quad.)
a + b + 65° = 180°
a + b = 115°
a
b
65o
A
B
Q
C
D
P
S
R
¡_AC¡_BC
48
Measures, Shape and SpaceMeasures, Shape and Space
9. In the figure, RS is the tangent to the circle at A. PQ cuts the circle
at B and C. If PQ // RS and ∠ CAB = 60°, prove that ABC is an
equilateral triangle.
Solution
∠ CAR = ∠ CBA ∠ in alt. segment
∠ BAS = ∠ BCA ∠ in alt. segment
∠ CBA = ∠ BAS alt. ∠ s, PQ // RS
∴ ∠ CBA = ∠ BCA
Q ∠ CBA + ∠ BCA + ∠ CAB = 180° ∠ sum of ¡
2∠ CBA + 60° = 180°
∠ CBA = 60°
∴ ∠ BCA = 60°
∴ ABC is an equilateral triangle.
60o
A
BQ
CP
SR
50o
A
B
QO
P
S
R
10. In the figure, AB is the tangent to the circle at P. QS is a diameter
of the circle and ∠ PQS = 50°. Find
(a) ∠ QRP, (b) ∠ QPA.
Solution
(a) ∠ QPS = 90° (∠ in the semi-circle)
∠ PQS + ∠ QPS + ∠ QSP = 180° (∠ sum of ¡ )
50° + 90° + ∠ QSP = 180°
∠ QSP = 40°
∠ QRP = ∠ QSP (∠ s in the same segment)
= 40°
(b) ∠ QPA = ∠ QSP (∠ in alt. segment)
= 40°
49
7 Basic Properties of Circles (II)
11. In the figure, PQ and PS are tangents to the circle at C and Arespectively. If PS // CB and ∠ BAS = 52°, find ∠ QPS.
Solution
∠ CBA = ∠ BAS = 52° (alt. ∠ s, CB // PS)
∠ CAP = ∠ CBA (∠ in alt. segment)
= 52°
∠ PCA = ∠ CBA (∠ in alt. segment)
= 52°
∠ QPS = 180° − ∠ PCA − ∠ PAC (∠ sum of ¡ )
= 180° − 52° − 52°
= 76°
52o
A
B
Q
C
P S
12. In the figure, PQ and PS are tangents to the circle at C and Arespectively. If ∠ CBA = 116°, find ∠ CPA.
Solution
Join CA.
∠ BAP = ∠ BCA (∠ in alt. segment)
∠ BCP = ∠ BAC (∠ in alt. segment)
∠ CBA + ∠ BCA + ∠ BAC = 180° (∠ sum of ¡ )
116° + ∠ BCA + ∠ BAC = 180°
∠ BCA + ∠ BAC = 64°
∠ CPA + ∠ PCA + ∠ PAC = 180° (∠ sum of ¡ )
∠ CPA + (∠ BCP + ∠ BCA) + (∠ BAP + ∠ BAC) = 180°
∠ CPA + (∠ BCP + ∠ BAC) + (∠ BAP + ∠ BCA) = 180°
∠ CPA + 2∠ BAC + 2∠ BCA = 180°
∠ CPA = 180° − 2(∠ BAC + ∠ BCA)
= 180° − 2 × 64°
= 52°
116o
A
B
Q
C
P
S
50
Measures, Shape and SpaceMeasures, Shape and Space
13. In the figure, PQ is the tangent to the circle at A. CB is a diameterof the circle, ∠ CDA = 110° and DC = DA. Find ∠ BAQ.
Solution
Join AC.
∠ CDA + ∠ CBA = 180° (opp. ∠ s, cyclic quad.)
110° + ∠ CBA = 180°
∠ CBA = 70°
∠ CAB = 90° (∠ in the semi-circle)
∠ ACB + ∠ CBA + ∠ BAC = 180°
∠ ACB + 70° + 90° = 180°
∠ ACB = 20°
∠ BAQ = ∠ ACB (∠ in alt. segment)
= 20°
110o
AB
Q
C
P
O
D
14. In the figure, PQ and QR are tangents to the smallercircle at A and the larger circle at C respectively. AEC isa straight line. If ∠ BAP = 2x and ∠ RCB = 3x, find x.
Solution
Join BE.
∠ AEB = ∠ PAB (∠ in alt. segment)
= 2x
∠ BDC = ∠ BCR (∠ in alt. segment)
= 3x
∠ BDC = ∠ BEC (∠ in the same segment)
∴ ∠ BEC = 3x
∠ AEB + ∠ BEC = 180° (adj. ∠ s on st. line)
2x + 3x = 180°
x = 36°
A
B
E R
Q
CD
P
2x
3x
51
7 Basic Properties of Circles (II)
15. In the figure, PQ is the tangent to the circle at B and AC is adiameter of the circle. AOCQ is a straight line and CB = CQ.
(a) Prove that ¡ ABQ is an isosceles triangle.
(b) Prove that ¡ BCQ ~ ¡ ABQ.
(c) Find ∠ CQB.
Solution
(a) Let ∠ CQB = a.
CB = CQ given
∠ CBQ = ∠ CQB base ∠ s, isos. ¡
= a
∠ CAB = ∠ CBQ ∠ in alt. segment
= a
∴ ∠ CAB = ∠ CQB
∴ AB = BQ base ∠ s equal
∴ ¡ ABQ is an isosceles triangle.
(b) ∠ CQB = ∠ BQA common angle
∠ CBQ = ∠ BAQ ∠ in alt. segment
∴ ¡ BCQ ~ ¡ ABQ A.A.A.
(c) ∠ BCA = ∠ CQB + ∠ CBQ (ext. ∠ of ¡ )
= a + a (by (a))
= 2a
∠ ABP = ∠ BCA (∠ in alt. segment)
= 2a
∠ ABC = 90° (∠ in the semi-circle)
∠ ABP + ∠ ABC + ∠ CBQ = 180° (adj. ∠ s on st. line)
2a + 90° + a = 180°
a = 180°
∴ ∠ CQB = 30°
O
A
BP
C
Q
7
52
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Date :
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Basic Properties of Circles (II)
7C
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7.3 Concyclic Points
Key Concepts and Formulae
Tests for concyclic points
1. If p = q, then A, B, Q and P are concyclic.
[Abbreviation: converse of ∠ s in the same segment] p
q
A B
Q
P
A
B C
D
A
B EC
D
p
q
2. If ∠ A + ∠ C = 180° or ∠ B + ∠ D = 180°,then A, B, C and D are concyclic.
[Abbreviation: opp. ∠ s supp.]
3. If q = p, then A, B, C and D are concyclic.
[Abbreviation: ext. ∠ = int. opp. ∠ ]
53
7 Basic Properties of Circles (II)
In the following figures, (1 – 4)
(a) prove that A, B, C and D are concyclic,
(b) find x.
1.
Solution
(a) ∠ DAB + ∠ DCB
= ( ) + ( ) + ( ) + ( )
= ( )
∴ A, B, C and D are concyclic. ( )
(b) ∠ BAC = ∠ BDC (∠ s in the same segment)
∴ x = 48°
2.
Solution
(a) ∠ DBC = 180° − ∠ BDC − ∠ BCD ∠ sum of ¡
= 180° − 40° − 110°
= 30°
Q ∠ DBC = ∠ DAC = 30°
∴ A, B, C and D are concyclic. converse of ∠ s in the same segment
x
61o
32o
48o
39oA
B C
D
x
110o
40o
30 o
A
B C
D
(Solution continues on the next page.)
39° 32°
opp. ∠ s supp.
48° 61°
180°
54
Measures, Shape and SpaceMeasures, Shape and Space
Solution
(b) ∠ BAC = ∠ BDC (∠ s in the same segment)
x = 40°
3.
x41o
85o
85o
51oA
B
EC D
F
4.
x
70o A
B
C D
E
Solution
(a) Q ∠ CAD = ∠ DBC = 90°
∴ A, B, C and D are concyclic. converse of ∠ s in the same segment
(b) ∠ BAC = 180° − ∠ BAE − ∠ CAD (adj. ∠ s on st. line)
= 180° − 70° − 90°
= 20°
∠ BDC = ∠ BAC (∠ s in the same segment)
x = 20°
Solution
(a) ∠ BFA = 85° vert. opp. ∠ s
∠ ABF = 180° − ∠ BFA − ∠ BAF ∠ sum of ¡
∠ BAT = 180° − 85° − 51°
= 44°
55
7 Basic Properties of Circles (II)
Solution
(a) ∠ ABF + ∠ FBC
= 44° + 41°
= 85°
= ∠ ADE
∴ A, B, C and D are concyclic. ext. ∠ = int. opp. ∠
(b) ∠ CAD = ∠ DBC
x = 41°
5. In the figure, the diagonals AC and BD of quadrilateral ABCDmeet at P. Given that ∠ ADB = 36°, ∠ BDC = 16°, ∠ APD = 62°and ∠ DAB = 98°, prove that A, B, C and D are concyclic.
Solution
∠ DAP = 180° − ∠ ADP − ∠ APD ∠ sum of ¡
= 180° − 36° − 62°
= 82°
∠ CAB = ∠ DAB − ∠ DAP
= 98° − 82°
= 16°
Q ∠ CAB = ∠ CDB = 16°
∴ A, B, C and D are concyclic. converse of ∠ s in the same segment
62o
16o36o
98o
A
DC
B
P
56
Measures, Shape and SpaceMeasures, Shape and Space
6. In the figure, PR and QS intersect at M. If PQ // SR, ∠ PQM = 55°and MP = MQ, prove that P, Q , R and S are concyclic.
Solution
∠ QSR = 55° alt. ∠ s, PQ // SR
PM = QM given
∠ MPQ = ∠ MQP base ∠ s, isos. ¡
∠ MPQ = 55°
∴ ∠ QSR = ∠ MPQ = 55°
∴ P, Q, R and S are concyclic. converse of ∠ s in the same segment
7. In the figure, ABC is an equilateral triangle, ACD is an
isosceles triangle and ∠ DAC = 30°. Prove that A, B, C
and D are concyclic.
Solution
∠ ABC= 60° prop. of equil. ¡
∠ DCA = ∠ DAC base ∠ s, isos. ¡
= 30°
∠ ADC = 180° − ∠ DCA − ∠ DAC ∠ sum of ¡
= 180° − 30° − 30°
= 120°
∠ ADC + ∠ ABC = 120° + 60°
= 180°
∴ A, B, C and D are concyclic. opp. ∠ s supp.
55o
Q
R
P S
M
30oA
B
C
D
57
7 Basic Properties of Circles (II)
A B
CD8. In the figure, AC = BD, ∠ DBA = ∠ CAB. Show that A, B, Cand D are concyclic.
Solution
BA = AB common side
∠ DBA = ∠ CAB given
BD = AC given
∴ ¡ DBA ≅ ¡ CAB S.A.S.
∴ ∠ ADB = ∠ BCA corr. ∠ s, ≅ ¡ s
∴ A, B, C and D are concyclic. converse of ∠ s in the same segment
9. In the figure, the tangents to the circle at A and B meet at T.
(a) Prove that O, B, T and A are concyclic.
(b) Find ∠ AOB + ∠ ATB.
Solution
(a) ∠ OAT = 90° tangent ⊥ radius
∠ OBT = 90° tangent ⊥ radius
Q ∠ OAT + ∠ OBT = 180°
∴ O, B, T and A are concyclic. opp. ∠ s supp.
(b) ∠ AOB + ∠ ATB = 180° (opp. ∠ s, cyclic quad.)
O
A
B
T
58
Measures, Shape and SpaceMeasures, Shape and Space
10. The figure shows a semi-circle with diameter AB and centre O.AOD is a right-angled triangle.
(a) Show that ¡ AOD ~ ¡ ACB.
(b) Prove that D, O, B and C are concyclic.
Solution
(a) ∠ ACB = 90° ∠ in the semi-circle
∠ DAO = ∠ BAC common angle
∴ ¡ AOD ~ ¡ ACB A.A.A.
OA B
C
D
(b) ∠ DOB + ∠ DCB
= 90° + 90°
= 180°
∴ D, O, B and C are concyclic. opp. ∠ s supp.
7
Name :
Date :
Mark :
59
NON-FOUNDATION
7D
Basic Properties of Circles (II)
Multiple Choice Questions
1. In the figure, a circle is inscribed in¡ ABC. Find x.
3. In the figure, TB and TC are tangents tothe circle at B and C respectively,AC = BC. Find ∠ BTC.
A. 25°B. 26°C. 27°D. 31°
2. In the figure, AC is the tangent to thecircle at B and DEF is a straight line.Find x.
A. 38°B. 40°C. 42°D. 44°
A. 32°B. 34°C. 36°D. 38°
4. In the figure, PQ is the tangent to thecircle at A and PQ // BD. Find x.
A. 80°B. 83°C. 94°D. 99°
x
34o
31o
O
A
B C
x
36o
78o
A
B
F
G
C
DE
32o
A B
TC
x
47o
A
B
Q
C
D
P
A
C
C
A
60
Measures, Shape and SpaceMeasures, Shape and Space
5. In the figure, PA and PB are tangents tothe circle at A and B respectively. Finda + b.
8. In the figure, AD is the tangent to the circleat D. If AD = 12 cm and AB = 6 cm,find the radius of the circle.
A. 110° B. 112°C. 117° D. 121°
6. In the figure, find ∠ ABC.
A. 70° B. 72°C. 74° D. 80°
7. In the figure, a circle is inscribed in aright-angled triangle ABC. AB, BC andAC touch the circle at P, Q and Rrespectively. Find PQ.
A. 7 cm B. 8 cm
C. 9 cm D. 11 cm
9. In the figure, O is the centre of twoconcentric circles. AB is the tangent to C1.If the radius of C1 and C2 are 5 cm and13 cm respectively, find AB.
A. 20 cm B. 22 cm
C. 24 cm D. 27 cm
10. In the figure, a circle is inscribed in¡ ABC. Given that AP = 2.5 cm,CQ = 1.5 cm and CA = CB, find theperimeter of ¡ ABC.
b
a
59o
A
BP
C
52o
52o
56o
A B
C
D
4 cm
3 cm
A
B
P
C
R
Q
12 cm
6 cmO
AB
C
D
C1 C2O
A B
A. 2 cm B. 3 cm
C. 2 cm D.5
2 cm
A. 5 cm B. 13 cm
C. 16 cm D. 18 cm
1.5 cm
2.5 cmA B
R
C
P
Q
C
C
A
D
B
B
61
7 Basic Properties of Circles (II)
11. In the figure, a circle is inscribed in aregular pentagon PQRST. TP, PQ and QRtouch the circle at A, B and C respectively.Find θ.
A. 96° B. 100°C. 108° D. 116°
12. In the figure, PA is the tangent to thecircle at A. Given that ∠ CAP = 28°,∠ CPA = 30° and BCP is a straight line,find ∠ BAC.
A. 58° B. 62°C. 75° D. 94°
13. In the figure, BF is a diameter of the circle.Find x.
A. 15° B. 19°C. 20° D. 22°
A
B
θ
Q
C
R
P
T
S
30o
28oA
B
C
P
x
38 o
A B
EF
C D
G
x
53o
AB
C
D
E
A. 14° B. 15°C. 16° D. 20°
15. In the figure, PQ is the tangent to the¡_ ¡_
circle at A and AD : DC = 3 : 1. Find∠ BAC.
63o
37oA
B
P
C
D
Q
A. 50° B. 59°C. 65° D. 70°
16. In the figure, the circle touches the sidesof quadrilateral ABCD at P, Q, R and S.If AB = a cm and CD = b cm. Theperimeter of the trapezium is
14. In the figure, ED is a diameter of the circle,AED is a straight line and AC is the tangentto the circle at B. Find x.
A. (ab − 3) cm.
B. (2b + a + 1) cm.
C. (2a + 2b) cm.
D. a(b + 1) cm.
A B
Q
C
D
P
S
R
B
D
C
C
B
C
62
Measures, Shape and SpaceMeasures, Shape and Space
A. 40 B. 50
C. 62 D. 65
18. In the figure, AB, AC and PR touch thecircle at B, C and Q respectively. IfAB = 15 cm, find the perimeter of ¡ APR.
A. 15 cm B. 30 cm
C. 45 cm D. 60 cm
19. In the figure, ABC is the tangent to thecircle at B. If AC // ED, find ∠ EBD.
A. 30°B. 31°C. 35°D. 40°
x o 50o
A
B
C
T
AB
R
C
P
Q
70o
AB
C
DE
30o
O
A
B
QC
P
17. In the figure, TA and TC are tangents tothe circle at A and C respectively,∠ ABC = x° and ∠ ATC = 50°, find x.
20. In the figure, O is the centre of thecircle. OA is parallel to BC. PQ is thetangent to the circle at C. If∠ BCP = 30°, find ∠ OAB.
A. 48°B. 52°C. 55°D. 60° D
B
D
D