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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

35

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)


(∠ sum of ¡ )


(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°

36

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)


(∠ sum of ¡ )


x cm9 cm

12 cm

O

A BP

C

37


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°

38


7.

Solution

∠ = °

∠ = ° − ° − °

= °

∠ = °

∠ = ° − ° − °

= °

=

= °

°

OAT

AOT

BOT

AOB

x

90

180 90 25

65

65

360 65 65

230

115

230

2

reflex


(∠ 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

.

.

.


(∠ at centre twice ∠ at ¡ ce)

(opp. ∠ s, cyclic quad.)

8.

39


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

40


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

41


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


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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7.2 Angles in the Alternate Segment


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

44


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

45


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°

46


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°

47


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


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


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


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


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

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7.3 Concyclic Points


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


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


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°


(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


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°


62o

16o36o

98o

A

DC

B

P

56


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


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


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


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


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


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


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


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

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