3 tangents to circles - … · 3 tangents to circles 61 3 tangents to circles activity activity 3.1...

3 Tangents to Circles

61


Activity

Activity 3.1 (p. 3.4)1. 90PMO (line joining centre to mid-pt. of chord

chord)

2.

22

22

xr

AMOAOM

(Pyth. theorem)

3. (a) Yes(b) AM = 0; OM = r(c) Yes

Classwork

Classwork (p. 3.5)1. 90x (tangent radius)

48

9042

y

y (tangent radius)

2. Refer to the figure.

CBTATB (alt. s, TA // CB) 25x

65

9025

90

90

y

y

yx

ATQ (tangent radius)

3. OT = OA (radii)

30x

OATOTA (base s, isos. )

60

9030

90

90

y

y

yx

OTQ (tangent radius)

Classwork (p. 3.14)1. QTOPTO (tangent properties)

25x

cm6

y

TPTQ (tangent properties)

2. QOTPOT (tangent properties)

QOTx

72

144

144

x

xx

QOTPOT

72QOT

108

18072

180

y

y

QOTROQ (adj. s on st. line)

3. TQ = TP (tangent properties)x = cm7

PQTPTQ 60TPQ (prop. of equil. )

60y

Classwork (p. 3.23)1.

70x

ABTATP ( in alt. segment)

45y

BATBTQ ( in alt. segment)

2.

50x

BTQBAT ( in alt. segment)

74y

ATPABT ( in alt. segment)

3.

103

1803542

180

TBA

TBA

TBAATBBAT ( sum of )

103x

TBAATQ ( in alt. segment)

4.x

PTBBAT

( in alt. segment)

35

702

180110

180

x

x

xx

TBAATBBAT ( sum of )

Quick Practice

Quick Practice 3.1 (p. 3.5)

552

1102

1BODOAD ( at centre twice at ce)

90OBC (tangent radius)In ABC,

35

1809055

180

ACB

ACB

OBCOADACB

NSS Mathematics in Action 5A Full Solutions

62


(a)

Construct XY as the common tangent to the circles at C.XY O1C tangent radiusXY O2C tangent radius

O1CO2 is a straight line.i.e. C lies on the straight line O1O2.

(b)

Construct O2D such that O2D O1A.DABO2 is a rectangle.

cm42

BODA

cm5

cm)49(11

DAAODO

cm13

cm)49(2121

COCOOO

Consider right-angled triangle O1DO2.

cm12

cm513 222

22

21

221

DO

DODOOO (Pyth. theorem)

cm122

DOAB

Quick Practice 3.3 (p. 3.7)AB is a diameter of the circle.

90ADB

cm16

cm12)1010( 22

22

BDABAD Pyth. theorem

90

90180

180 ADBCDB adj. s on st. line

cm9

cm1215 22

22

BDBCCD Pyth. theorem

cm25

cm)916(

CDADAC

2

222

cm625

cm25

AC

2

22222

cm625

cm)1520(

BCAB

AC2 = AB2 + BC2

CB AB converse of Pyth. theoremBC is the tangent to the circle converse of tangent at B. radius

Quick Practice 3.4 (p. 3.8)AC = AB given

30

ABCACB base s, isos.

60

3030

ACBABCOAC ext. of

OA = OC radii

60

OACOCA base s, isos.

90

3060

ACBOCAOCB

BC is the tangent to thecircle at C. converse of tangent radius

Quick Practice 3.5 (p. 3.15)(a)

30

OBCABO (tangent properties)

25

OCAOCB (tangent properties)

70

180225230

180

BAC

BAC

BACACBABC ( sum of )

35

702

12

1BAC

CAOBAO (tangent properties)

(b)

115

1803035

180

AOB

AOB

AOBOBABAO ( sum of )

Quick Practice 3.6 (p. 3.16)Let BX = x cm, then AX = (9 – x) cm.

cm)9( x

AXAZ

(tangent properties)

cmx

BXBY


CY = (10 – x) cm

cm)10( x

CYCZ


AZ + CZ = 11 cm

4

11219

11)10()9(

x

x

xx

cm4BX


63


Join OC. 90ABC tangent radius

ODA = 90 tangent radius

60

9030180

180 ABCOADBCD sum of

30

602

12

1BCD

OCBOCD tangent properties

30OCDOAD 90ODCODA proved

OD = OD common sideAOD COD AAS

AD = CD corr. sides, sD is the mid-point of AC.


60

BCQBAC in alt. segment

BCA = CAR alt. s, BC // ARCAR = ABC in alt. segment

BCA = ABC

60

1202

18060

180

ABC

ABC

ABCABC

BACBCAABC sum of

60BACBCAABCABC is an equilateral triangle.

Quick Practice 3.9 (p. 3.25)(a) Consider ABD and BCD.

90ADB in semi-circle

90

90180

180 ADBBDC adj. s on st. line

BDCADB BAD = CBD in alt. segment

ABD ~ BCD AAA(b) ABD ~ BCD (proved in (a))

cm9

cm4

62

2

CD

BDAD

CD

BD

BD

AD (corr. sides, ~ s)

cm13

cm)49(

CDADAC

Quick Practice 3.10 (p. 3.26)(a) TBAACB ( in alt. segment)

TBABAC (alt. s, AC // TP)ACB = BAC

55

180702

180

ACB

ACB

ABCBACACB ( sum of )

(b) 55ACBTBA in alt. segment

55

1805570

180

TAB

TAB

TBAATBTAB sum of

TAB = ACBTA is the tangent to thecircle at A. converse of in alt.

segment

Further Practice

Further Practice (p. 3.8)1. Let r cm be the radius of the circle.

90ATO (tangent radius)

16

1288

144168

12)4(

)(

22

222

222

222

r

r

rrr

rr

OTATPAOP

OTATOA (Pyth. theorem)

The radius of the circle is 16 cm.

2. (a)

Let PA be the common tangent to the circles at A.AOPA 1 tangent radius

AOPA 2 tangent radius

AO1O2 is a straight line.(b) cm2cm)46(1221 AOAOOO

cm)102(orcm40

cm40

cm]6)46[(

1

2

222

22

221

21

BO

BOOOBO (Pyth. theorem)

3.

Join OD.


64

CD = AD given

30

CADDCA base s, isos.

OD = OA radii

30

OADODA base s, isos.

90

180)30(3030

180

ODC

ODC

CDAACDCAD sum of

CD is the tangent to the circle at D. converse oftangent radius

Further Practice (p. 3.17)1.

120

240360

reflex360 POQPOQ (s at a pt.)

POT = QOT = x (tangent properties)

60

021

x

xx

POQQOTPOT

OPT = 90 (tangent radius)

30

0819060

180

y

y

OTPOPTPOT ( sum of )

2.

cm3 BYBX (tangent properties)

cm7 CYCZ (tangent properties)

cm14

cm)721(

ZCACAZ

cm14 AZAX (tangent properties)

cm17

cm)314(

BXAXAB

3. AP = AQ tangent propertiesAPQ = AQP base s, isos.

APQ = RQP alt. s, AP // QRAQP = RQPPQ bisects AQR.

Further Practice (p. 3.27)1. BAC = ADC ( in alt. segment)

62ADC

CADBACBAD

39

782

18040)62(

180

ADC

ADC

ADCADC

ADBABDBAD ( sum of )

2. OB = OA radii

20

OABOBA base s, isos.

70

9020

90

90

PBA

PBA

OBAPBA

OBP tangent radius

70

PBABAC alt. s, AC // PQ

BAC = QCB = 70QC is the tangent to thecircle at C. converse of in

alt. segment

Exercise

Exercise 3A (p. 3.9)Level 11. PG = PE (given)

60

PEGPGE (base s, isos. )

60

PGEGPB (alt. s, AB // CD)

30

9060

90

x

x

OPB (tangent radius)

2. OP = OC (radii)OPC = OCP (base s, isos. )

252

130180

180

OPC

OCPOPCPOC ( sum of )

65

9025

90

x

x

OPB (tangent radius)

3. OP AB (tangent radius)OP2 = AO2 – AP2 (Pyth. theorem)

cm5

cm25

cm)24(7 22

OP

8

cm8

cm64

cm5)89( 22

222

x

PB

OPBOPB (Pyth. theorem)

4. OA = OB (radii)

34

OBAOAB (base s, isos. )

68

3434AOT (ext. of )

OAT = 90 (tangent radius)

22

1809068

180

ATO

ATO

ATOOATAOT ( sum of )


65

5. OT = OP (radii)

62

OPTOTP(base s, isos. )

90OTB (tangent radius)

28

6290PTB

34

6228

PBT

PBT

OPTPTBPBT (ext. of )

6.cm6

OAOB radii

2

22222

cm100

cm)68(

OBPB

2

22

222

cm100

cm10

cm)64(

PO

PB2 + OB2 = PO2

OB PB converse of Pyth. theoremPB is the tangent to thecircle at B. converse of tangent radius

7. AD = DC givenOA = OB radii

OD // BC mid-pt. theorem

90

AODOBC corr. s, OD // BC

BC is the tangent to the converse of tangent radiuscircle at B.

8. ABC = 90 (tangent radius)

58

3290180BOD ( sum of )

29

582

12

1BODBAC ( at centre twice at ce)

61

2990180

)ofsum(180

ACB

BACABCACB

9.

34

682

12

1TOQTPQ ( at centre twice at ce)

OTB = 90 (tangent radius)

41

180)1590(34

180

TBQ

TBQ

TBQPTBTPB ( sum of )

10. ADB = 90 ( in semi-circle)

42

4890

ODBADBODA

OA = OD (radii)

42

ODAOAD (base s, isos. )

ABC = 90 (tangent radius)

48

1804290

180

ACB

ACB

OADACBABC ( sum of )

11. Let the radius of the circle be r cm.OTA = 90 (tangent radius)

5

1664144

)8(1222

222

222

r

rrr

rr

AOATOT (Pyth. theorem)

The radius of the circle is 5 cm.

12. AB is the tangent to the circle at P.Also, APQ = 90.

PQ passes through the centre of the circle.PQ is a diameter of the circle.

13.

25

6540

CED

CED

EDBCEDDCE ext. of

25

CEDAEO vert. opp. s

AO = EO radii

25

AEOEAO base s, isos.

90

6525180

180 EDBEAOOBD sum of

BC is the tangent to the converse of tangentcircle at B. radius

14.

54

4

BDA

BD

AOB

BOD

80

1809

4

BOD

BOD

OB = OD radiiODBOBD base s, isos.

502

80180

180

OBD

BODODBOBD sum of

BD = CD given

40

DCBDBC base s, isos.

90

4050

DBCOBDOBC

BC is the tangent to the circle at B. converse oftangent radius

arcs prop. to s atcentre


66

Level 215. (a) Reflex

232

128360POT (s at a pt.)

116

2322

1

reflex2

1POTPQT ( at centre twice at ce)

(b)

64

116180

180 PQTOTQ (int. s, QP // TO)

OT AB (tangent radius)

26

6490

90 OTQQTA

16. (a)

100

TCBPAT (ext. , cyclic quad.)

38

10042180

180 PATTPAPTA ( sum of )

(b) OT PT (tangent radius)

52

3890

90 PTAATO

OA = OT (radii)

52

)isos.s,(baseATOTAO

76

5252180

180

TOA

TOAATOTAO ( sum of )

17. (a) Consider ABC and ADB.ADB = 90 in semi-circleABC = 90 tangent radius

ADB = ABCBAC = DAB common angle

ABC ~ ADB AAA(b) ABC ~ ADB (proved in (a))

cm5

33

cm6cm10

cm6

AD

ADAB

AD

AC

AB(corr. sides, ~ s)

cm5

26

cm5

3310

ADACCD

18. (a) OBC = 90 tangent radiusODC = 90 tangent radius

180

9090ODCOBC

B, C, D and O are concyclic. opp. s supp.

(b)

BCDBOD

BODBCD

180

quad.)cyclics,(opp.180

BCD

BCD

BODBAD

2

190

)180(2

1

)at twicecentreat(2

1 ce

180)24(ABCBCDADCBAD

( sum of polygon)

110

552

1

360)9020()9015(2

190

BCD

BCD

BCDBCD

19.

BDAD givenAD = BD equal arcs, equal chordsDAB = DBA base s, isos.

ADB = 90 in semi-circle 180DABADBABD sum of

45

)90180(2

1

)180(2

1ADBABD

90

90180

180 ADBBDC adj. s on st. line

BD = CD givenDBC = DCB base s, isos.

180DCBBDCDBC sum of

45

)90180(2

1

)180(2

1BDCDBC

90

4545

DBCABDABC

BC is the tangent to the converse of tangentcircle at B. radius

20.

Join OP.Let OQB = .OQ = OP radiiOQB = OPB = base s, isos.OPA = 90 tangent radius

BPA = 90 Consider QOB.


67

90

90180OBQ sum of

90

OBQABP vert. opp. s

APBABP ABAP sides opp. equal s

21. APQ = 90 (tangent radius)PO1O2Q is a straight line.

cm5

cm]13)9(2[11

QOPQPO

cm12

cm513 22

21

21

2

AP

POAOAP (Pyth. theorem)

PBAP (line from centre chord bisects chord)

cm24

cm122

2

AP

PBAPAB

22. (a) (i) 90OTA (tangent radius)AT = TB (line from centre

chord bisects chord)

cm8

cm162

12

1

ABAT

OT = OC = 15 cm (radii)OA2 = OT2 + AT2 (Pyth. theorem)

cm17

cm815 22

OA

The radius of the larger circle is 17 cm.(ii) Consider OAB and OCD.

OD

OB

OC

OA (radii)

CODAOB (common angle)OAB ~ OCD (ratio of 2 sides, inc. )

cm17

214

cm17

cm15

cm16

CD

CDOA

OC

AB

CD(corr. sides, ~ s)

(b) EF = ABEF and AB are equidistant from the centre O.

(equal chords, equidistant from centre)The mid-point of chord EF lies on the smallercircle.i.e. EF touches the smaller circle at the

mid-point of EF.EF is a tangent to the smaller circle.

23. OC = OD (radii)OCD = ODC = 66 (base s, isos. )

48

6666180

)ofsum(180

COD

ODCOCDCOD

48

CODFAB (corr. s, AB // OC)

OBA = 90 (tangent radius)BOC = OBA = 90 (alt. s, AB // OC)OC = OE (radii)OCE = OEC (base s, isos. )

45

902

OEC

OEC

BOCOECOCE (ext. of )

FAB BECA, B, F and E are not concyclic.

Exercise 3B (p.3.17)Level 11. TATB (tangent properties)

68

TABTBA (base s, isos. )

44

1806868

x

x ( sum of

2. BTOATO (tangent properties)

252

130180

180130

ATO

BTOATO (adj. s on st. line)

90OAT (tangent radius)

65

2590180x ( sum of

3. OAT 90 (tangent radius)

cm16

cm1220

cm)1220(

22

2222

222

AT

AT

ATOAOT

(Pyth. theorem)

cm16x (tangent properties)

4. AR AP 3 cm (tangent properties)

cm5

cm)38(

RC

cm5 RCQC (tangent properties)

cm2 BPBQ (tangent properties)

cm7

cm)52(

QCBQBC

5.

25

OCAOCB (tangent properties)

34

18025121

180

OBC

OBC

OCBBOCOBC ( sum of )

34

OBCOBA (tangent properties)

62

180252342

180

BAC

BAC

BCAABCBAC ( sum of )

OABOAC (tangent properties)


68

31

622

12

1BACOAB

6. Let the length of AR be x cm.

cm)8(

cm

x

APABBP

xARAP


cm)8( x

BPBQ


cm)2(

cm)]8(6[

x

x

BQBCCQ

cm)2(

x

CQCR (tangent properties)

cm10

cm68 22

22

BCABAC (Pyth. theorem)

6

122

)2(10

x

x

xx

CRARAC

cm6AR

7. TBTA (tangent properties)TBATAB (base s, isos. )

68

180244

180

TBA

TBA

TBATABATB ( sum of )

BMQ 25 68 (ext. of )

43BMQ

137

18043

180

AMQ

AMQ

BMQAMQ (adj. s on st. line)

8. Let O be the centre of the circle.

Join OC, OD and OQ.

1085

180)25(PCQ ( sum of polygon)

54

1082

12

1PCQ

OCPOCQ (tangent properties)

QDR PCQ 108

54

1082

12

1QDR

ODRODQ (tangent properties)

OCQ ODQOQC OQD 90 (tangent radius)

OQ OQ (common side)OCQ ODQ (AAS)

CQ = DQ (corr. sides, s)Q is the mid-point of CD.

9. (a)

25

502

12

1ATB

OTBOTA (tangent properties)

OAT 90 (tangent radius)

65

2590180

180 OTAOATAOC ( sum of )

(b) OC OA (radius)OCA OAC (base s, isos. )

5.57

180652

180

OCA

OCA

AOCOACOCA ( sum of )

5.32

5.5725

TAC

TAC

OCAOTATAC (ext. of )

10. OQB 90 (tangent radius)

37

1809053

180

OBQ

OBQ

OQBBOQOBQ ( sum of )

37

OBQOBR (tangent properties)

27

18037116

180

OAB

OAB

OBAAOBOAB ( sum of )

27

OAROAP (tangent properties)

54

2727

OAROAPPAR


69

11. PB QB (tangent properties)BPQ BQP (base s, isos. )

45

180902

180

BQP

BQP

PBQBQPBPQ ( sum of )

QC = RC (tangent properties)CQR = CRQ (base s, isos. )

28

1801242

180

CQR

CQR

QCRCQRCRQ ( sum of )

107

1802845

180

PQR

PQR

PQRCQRBQP (adj. s on st. line)

73

180107

180

PSR

PSR

PQRPSR (opp. s, cyclic quad.)

12. (a) OAT 90 tangent radiusOBT 90 tangent radius

OAT OBT 180O, B, T and A are concyclic. opp. s supp.

(b) O, B, T and A areconcyclic.OAB = OTB s in the same segmentOTB = OTA tangent propertiesOAB = OTA

Level 213.

Join OD.

AEDOEDOEA 2

1(tangent properties)

BCDOCDOCB 2

1(tangent properties)

180BCDAED (int. s, AE // BC)

90

18022

OCDOED

OCDOED

90

90180

180 OCDOEDEOC ( sum of )

14. BP BR (tangent properties)CR CQ (tangent properties)

BR CR 6 cmBP CQ 6 cm

CQ 6 cm – BP …… (1)

)2(......cm7cm9 CQBP

CQACABBP

AQAP


By substituting (1) into (2), we have

cm2

)cm6(cm7cm9

PB

PBPB

cm11cm)29( AP

15. AP AS, BP BQ, CQ CR and DR DS(tangent properties)

Perimeter of ABCD

cm40

cm)82122(

22

)()(

)()(

)()(

CDAB

CDABCDAB

DRCRAPBPCDAB

DSCQASBQCDAB

ASDSCDCQBQAB

DACDBCAB

16. Let O be the centre of the smaller circle and r cm be theradius of the smaller circle.

OZY 90 (tangent radius)OXC 90 (tangent radius)OYC 90 (tangent radius)

222

90

ORROOO

OOR

(Pyth. theorem)

(rejected)25or1

0)25)(1(

02526

8161025168

)4()49()4(

2

222

222

rr

rr

rr

rrrrrr

rrr

The radius of the smaller circle is 1 cm.

17. Let AOT = x.

x

AOTCOT

tangent properties

xBOA

COTAOTBOA

2180

180

adj. s on st. line

OB = OA radiiOABOBA base s, isos.

x

xOAB

BOAOABOBA

2

)2180(180

180 sum of

xAOTOAB BA // OT alt. s equal

18. BT AT tangent propertiesBT AT BAABT ATB TAB 60 prop. of equil.DAT 90 tangent radius

30

6090180

180 ATBDATCDB sum of

90CBA in semi-circle


70

180ABTCBACBD adj. s on st. line

30

6090180CBD

CDB CBD = 30CB CD sides opp. equal s

19. (a) AB BD tangent propertiesBD BC tangent properties

AB BC

(b) BAD BDA base s, isos.BDC BCD base s, isos.

90

90

180)(

180

ADC

BDCADB

DCBBDCADBDAB

DCAADCDAB sum of

20. PQ and RS are not parallel.PQ and RS must meet ata point C.

CA CB tangent propertiesCAB CBA = x base s, isos.CBA RBA 180 adj. s on st. line

x y 180

21. (a)

a

DADQ


b

BCQC


ba

QCDQDC

(b) PAD 90 tangent radiusQPA 90 given

QPA PAD = 180AD // PQ int. s supp.

Consider CRQ and CAD.CRQ CAD corr. s, PQ // ADCQR CDA corr. s, PQ // AD

CRQ ~ CAD AAA

ba

abQR

ba

b

a

QRCD

CQ

AD

QR

corr. sides, ~ s

(c) Consider APR and ABC.ABC 90 tangent radiusPAR BAC common angle

APR ~ ABC AAA

PR

BC

AR

AC corr. sides, ~ s

CRQ ~ CAD proved in (b)

QRba

abPR

ba

b

b

PRba

b

b

PRba

b

BC

PRa

ba

ab

AC

AR

DA

QR

AC

ARACDA

QR

AC

CR

1

1

1

1

corr. sides, ~ s

Exercise 3C (p. 3.27)Level 11. TB TA (given)

x

TABTBA

(base s, isos. )

x

TABBTQ

( in alt. segment)

ATB 111 x

69

111180

180)111(

180

x

xxx

ATBTBATAB ( sum of )

2. ATB 90 ( in semi-circle)

25

6590180

180 TBAATBTAB ( sum of )

25

TABBTP ( in alt. segment)

40

2565

x

x

TPBBTPTBA (ext. of )

3.

37

CTQTBC ( in alt. segment)

75

BTPBCT ( in alt. segment)

37

TBCACB (alt. s, CA // TB)

31

180)3775()37(

180

x

x

ACTABT (opp. s, cyclic quad.)


71

4. TA TB (tangent properties)TAB TBA (base s, isos. )

62

180562

180

TAB

TAB


62

TABACB ( in alt. segment)

86

1806232

180

ABC

ABC

ABCACBBAC ( sum of )

5. (a)

30

150180

180

ADB

PDBADB (adj. s on st. line)

50

3080PTQ

QBDPTQADB (ext. of )

(b) TA TB (tangent properties)TAB TBA (base s, isos.

65

180502

)ofsum(180

TBA

TBA

ATBTBATAB

65

TBABCA ( in alt. segment)

6. (a) Consider BCT and CAT.

TACATCACT

TCBCTBCBT

TACTCB

ATCCTB

180

180

ofsum

ofsum

segmentalt.in

anglecommon

CBT ACTBCT CAT AAA

(b) Let the length of AB be x cm.BCT ~ CAT

17

258

200)8(8

8

210

210

8

x

x

x

x

AT

CT

CT

BT(corr. sides, ~ s)

The length of AB is 17 cm.

7. TP = TA (given)TPA = TAP (base s, isos. )

37

742

TAP

TAP

ATQTAPTPA (ext. of )

37

TAPBTP ( in alt. segment)

69

1807437

BTA

BTA (adj. s on st. line)

8. Let ACB x.

x

CTQBAT

x

BCACTQ

x

BCABACABT

x

BCABAC

BCBA

2

segment)alt.in(

)s,(alt.

)of(ext.

)isos.s,(base

(given)

PQ//AC

Consider ATB.

34

180278

180

x

xx

BATABTATB ( sum of )

34ACB

9.

Join BD.

66

PABADB ( in alt. segment)

32

MCBBDC ( in alt. segment)

98

3266

BDCADBADC

82

98180

180 ADCABC (opp. s, cyclic quad.)

10. (a) Consider ABC and BTC.

90

90180

90

BCT

ACB

TBCBAC

linest.onsadj.

circle-semiin

segmentalt.in

BCTACB

BTC

BCTTBC

ACBBACABC

180

180

ofsum

ofsum

ABC ~ BTC AAA

(b) 222 CBACAB (Pyth. theorem)

cm34

cm35 22

AB

ABC ~ BTC

BC

AC

BT

AB (corr. sides, ~


72

cm5

343orcm

5

306

cm3065

3

5cm34

TB

TB

TB

BC

AC

BT

AB

11.

ACBBAC

BCBA

CBEACB

isos.s,base

given

//s,alt.

DEAC

BAC CBEDE is the tangent to the converse of incircle at B. alt. segment

12.

BCTBTC

BCBT

TCATAB

TCTA

isos.s,base

given

.isoss,base

given

BTC TABPC is the tangent to the converse of incircle at T. alt. segment

13.

2

1

6

32

1

39

6

PT

PBPA

PT

TPABPT common angle

2

1

PT

PB

PA

PT

BPT ~ TPA ratio of 2 sides, inc. PTB PAT corr. s, ~ sTP is the tangent to the converse of incircle at T. alt. segment

14.

BCDBDC

BCBD

)isos.s,(base

(given)

Let BCD t.

t

BDCBCDABD

2 (ext. of )

t

ABDx

2 ( in alt. segment)

ty

tyt

BDCyx

3180

1802

180

(adj. s on st. line)

If 20t ,

40

)20(2x

120

60180

)20(3180y

If 30t ,

60

)30(2x

90

90180

)30(3180y

(or any other reasonable answers)

Level 215.

70

ABEEDB ( in alt. segment)

EBD EDC ( in alt. segment)Consider BCD.

30

180)70(50

180

EDC

EDCEDC

BDCBCDEBD ( sum of )

16.

35

145180

180 CTQCTP (adj. s on st. line)

35

CTPCAT ( in alt. segment)

60

3525

CTPTPCACT (ext. of )

85

3560180

180 CATACTCTA ( sum of )

95

18085

180

ABC

ABC

ABCCTA (opp. s, cyclic quad.)

17. Let ABC .

ABCCAP ( in alt. segment)

AC CP (given)CAP CPA (base s, isos. )

CPA BAC 90 ( in semi-circle)Consider BPA.

30

180)90(

180

PABBPAABP ( sum of )

30ABC

18. (a)

39

ACBABP ( in alt. segment)

36

39105180

180 ABPPABAPB ( sum of )

(b)

39

ABPADB ( in alt. segment)

39

ACBDAE (alt. s, AD // BC)

78

3939

DAEADBAEB (ext. of )


73

19.

53

CTQCBT ( in alt. segment)

60

DABBCT (ext. , cyclic quad.)

67

1805360

180

CTB

CTB

CTBCBTBCT ( sum of )

20.

Join OA.OT OC (radii)

40

OCTOTC (base s, isos. )

100

180 OCTOTCTOC ( sum of )

100

TOCAOT (equal chords, equal s)

502

1AOTABT ( at centre twice at ce)

50

ABTATP ( in alt. segment)

21.

Join AM, BM, AN, BN and AB.

AMBANBANB

AMB

AMB

ANB

2

1

2

(arcs prop. to s at ce)

60

1803

180

AMB

AMB

ANBAMB (opp. s, cyclic quad.)

60ABP

AMBABP ( in alt. segment)

22. (a) PD PC (tangent properties)PDC PCD (base s, isos. )

180DPCPDCPCD ( sum of )

652

501802

180 DPCPCD

65

PCDCBD ( in alt. segment)

36

BCNBDC ( in alt. segment)

AB AD (given)ABD ADB (base s, isos. )ABC ADC 180 (opp. s, cyclic quad.)

5.39

792

1803665)(

180)()(

ADB

ADB

ADBABD

BDCADBDBCABD

5.75

365.39

BDCADBADC

(b)

36

BCNBAC ( in alt. segment)

5.75

365.39

BAKABKAKD (ext. of )

23.

Join AB.TA TB (tangent properties)TAB TBA (base s, isos. )

55

180702

180

TBA

TBA


125

55180

180

QBA

TBAQBA (adj. s on st. line)

125

QBAACB ( in alt. segment)

24.

100

ATQABT ( in alt. segment)

25

10055180

180 ABTPBCCBT (adj. s on st. line)

25

CBTCAT (s in the same segment)

Consider APT.

33

100)25(42

100)(42

BAC

BAC

CATBAC

ATQPATAPT (ext. of )


74

25. (a) Consider AMD and BMP.

BMPAMD

PBMDAM

MBAM

BMPAMD

90

ASA

squareofproperties

given

sopp.vert.

(b) AMD BMPADPB (corr. sides, s)AB (properties of square)

BPA PAB (base s, isos. )ABP 90 (proved in (a))

45

180902

180

PAB

PAB

ABPPABBPA ( sum of )

(c) BCA 45 properties of squarePAB 45 proved in (b)

BCAPAB PA is the tangent to the converse of circle at A. in alt. segment

Revision Exercise 3 (p. 3.35)Level 11.

xTABTBA

TBTA

)isos.s,(base

)properties(tangent

61

180582

180

x

x


yCBFACB

TBAACB

61

)//s,(alt.

segment)alt.in(

TFAC

61y

2.

90

90

ODC

OAB

radius)tangent(

radius)tangent(

Sum of interior angles of pentagon

540

)25(180

132

1281009090540AOD

3. (a) 90BAP (tangent radius)

180

9090ABQBAP

AP // BQ (int. s supp.) 29CAP (alt. s, AP // BQ)

(b) Join OC.

61

2990

90 CAPBAC

BC = BA (tangent properties)

61

BACBCA (base s, isos. )

32

2961APC

BCACAPAPC (ext. of )

4.

80

BCQBAC ( in alt. segment)

64

1808036

180

x

x

TACBACPAB (adj. s on st. line)

64y

TACABC ( in alt. segment)

TCTA (tangent properties)TCATAC (base s, isos. )

52

6464180

180

z

TCATACz ( sum of )

5. Let OCA = a.

a

ACBABC

ACAB

aACB

OCB

aOCAOAC

OCOA

90

90

90

)△isos.s,(base

(given)

radius)(tangent

)isos.s,(base

(radii)

60

2180

)90()90(

a

aa

aaa

ACBABCOAC

60OCA

6.

63

63

2790

90

27

OTAOAT

OTOA

BTPOTPOTA

OTP

OBTBTP

)isos.s,(base

(radii)

radius)tangent(

)//s,alt.(

OBTP

36

6327

AOB

AOB

OATOBAAOB (ext. of )

7. ACO = 90 (tangent radius)

cm15

cm817 22

222

CA

CAOCOA (Pyth. theorem)

OC ABAC = CB (line from centre chord bisects chord)

cm30

cm152

AB

8. CAQ = ABC ( in alt. segment)BAC = 90 ( in semi-circle)Consider ABQ.

28

18034902

18034)(

180

ABC

ABC

CAQBACABC

AQCBAQABQ ( sum of )


75

9.

Join OB.

OABOBA

OAOB

ADBDAB

BDBA

)isos.s,(base

(radii)

)isos.s,(base

(given)

90OBD radius)(tangent Consider ABD.

30

180903

180)(

180

DAB

DAB

ADBOBDOBADAB

ADBABDDAB ( sum of )

10. Consider OBC.

46

180134

180

OCBOBC

OCBOBC

BOCOCBOBC ( sum of )

OBCABO (tangent properties)OCBACO (tangent properties)

88

462180

)(2180

22180

180

OCBOBC

OCBOBC

ACBABCBAC ( sum of )

11.

44

882

12

1BOCBAC ( at centre twice at ce)

Join BC.

ACBABC

ACAB

)isos.s,(base

(given)

68

180442

180

ABC

ABC


44

CABDBC ( in alt. segment)

In ABD,

24

180)4468(44

180

BDA

BDA

BDAABDBAC ( sum of )

12. 90TAO (tangent radius)

70

1802090

180

AOT

AOT

AOTOTATAO ( sum of )

AOTCBA 2

1( at centre twice

35

702

1 at ce)

35

CBACAT ( in alt. segment)

13. (a)

70

TCBTDA (corr. s, AD // BC)

44

7026

QTC

QTC

TDAQTCTQD (ext. of )

(b)

44

QTCTBC ( in alt. segment)

66

1807044

180

ATD

ATD

TCBTBCATD (ext. of )

14. (a)

Join AC. 90CAB ( in semi-circle)

36

BAQACB ( in alt. segment)

54

3690180

180 ACBCABABC ( sum of )

(b)

126

18054

180

ADC

ADC

ABCADC (opp. s, cyclic quad.)

15.


76

Join AC.

108

ABCACP ( in alt. segment)

45

902

12

1CODCAD ( at centre twice at ce)

27

18045108

180

APQ

APQ

APQCAPACP ( sum of )

16. (a) Consider PBC and PAB.PBC = PAB in alt. segmentBPC = APB common angle

PBC ~ PAB AAA

(b) PBC ~ PAB

PB

PA

PC

PB (corr. sides, ~ s)

PB

ACPC

PC

PB

Let PC = x cm.

(rejected)9or4

0)9)(4(

0365

536

6

56

2

2

xx

xx

xx

xx

x

x

The length of PC is 4 cm.

17.

circle)-semiin(902

3

)ats toprop.arcs(2

3 ce

ACB

CABABC

BC

AC

CAB

ABC

36

180902

3

180

CAB

CABCAB

ACBCABABC ( sum of )

54362

3ABC

36

1854BCT

BTCBCTABC (ext. of )

BCTCAB TC is the tangent to the circle at C.

(converse of in alt. segment)

18.

Join OP and OQ.OP AB tangent radiusOQ AC tangent radiusOP = OQ radiiAB = AC chords equidistant from centre are equal

19. Let BAC = x.xBACCDA given

90ACB in semi-circle

xCAD

ACBCDACAD

90

ext. of

xABC

ACBABCBAC

90

180 sum of

ABCCAD AD is the tangent to the circle at A. converse of in

alt. segment

20. CDTQTD alt. s, PQ // CD

CDTABC ext. , cyclic quad.ABCQTD

PQ is the tangent to the circle converse of in alt.at T. segment

21. BEABOF 2 at centre twice

60

302 at ce

60

BOFCOD vert. opp. s

30

FBDODC alt. s, CD // BE

90

1806030

180

OCD

OCD

CODODCOCD sum of

CD is the tangent to the converse of tangent circle at C. radius

Level 222. (a)

x

AQBDBC

(alt. s, BC // AQ)

34

BDCABD (alt. s, BA // CD)

34x

ABDDBCABC

(b)

34

ABDQAD ( in alt. segment)

x

AQDQADADB

34

(ext. of )

39

180)3434()34(

180

x

xx

ADCABC (opp. s, cyclic quad.)

23. In PBC,

80

180100

180

BCPBPC

BCPBPC

PBCBCPBPC ( sum of )

BPCPAB ( in alt. segment)BCPBAD (ext. , cyclic quad.)


77

80

BCPBPC

BADPABPAD

24.

BXYBYX

BXBY

)isos.s,(base

)properties(tangent

60

601802

180

BYX

BYX

XBYBXYBYX ( sum of )

Also, 60BXY

BYX is an equilateral triangle.Let XY = a cm,then BX = BY = a cm.

cm)8(andcm)5( aAXaCY

cm)5( a

CYCZ


cm)8( a

AXAZ


3

62

)5()8(7

a

a

aa

CZAZAC

cm3XY

25.

Construct straight line DOB.

90

90

90

OCQ

OAP

OBQOBP

radius)(tangent

radius)(tangent

radius)(tangent

1809090OBPOAPand 1809090OBQOCQ

APBO and CQBO are cyclic quadrilaterals.(opp. s supp.)

AOD = x (ext. , cyclic quad.)COD = y (ext. , cyclic quad.)

yx

CODAODz

26. (a) BR = BP tangent propertiesCR = CQ tangent properties

BC = BR + CRBC = BP + CQ

(b) AP = AQ (tangent properties)Perimeter of ABC

cm24

cm122

2

)()(

)(

AP

AQAP

CQACBPAB

CQBPACAB

BCACAB

(proved in (a))

27. (a) Consider OPA and QOB.PAO = OBQ = 90 tangent radius

APO

APO

APOPAOAOP

90

90180

180 sum of

APO

APO

AOPPOQBOQ

)90(90180

180 adj. s on st. line

OPA ~ QOB AAA(b) (i) OPA ~ QOB (proved in (a))

QB

OB

OA

PA

OA = OB (radii)

cm6

cm49

QBPAOA

OAOAQBPA

The radius of the circle is 6 cm.(ii) OA = OB = 6 cm

PC = PA = 9 cm (tangent properties)QC = QB = 4 cm (tangent properties)

The perimeter of the quadrilateral ABQP

cm38

cm)449966(

QBQCPCPAOBOA

28. (a) 90OQBOPB tangent radius

180)24(OPBQBP

OQBPOQ

sum of polygon

90

360909090

POQ

POQ

OPBQ is a parallelogram. opp. s equalOP = OQ radii

OPBQ is a square.(b) Let the radius of the circle be r cm.

PB = QB = r cm (property of square)AP = (12 – r) cm and CQ = (5 – r) cm

cm)12( r

APAR


cm)5( r

CQCR


cm)217(

cm)]5()12[(

r

rr

CRARAC


78

cm13

cm512 22

222

AC

BCABAC (Pyth. theorem)

2

13217

r

r

The radius of the circle cm2

29. (a) 90ONC tangent radius 90BAC given

BACONC ACBNCO common angle

CON ~ CBA AAA(b) OMA = ONA = 90 tangent radius

AM // NO int. s supp.AN // MO int. s supp.AM = AN tangent properties

AMON is a square.(c) Let the radius of the circle be r cm.

AM = AN = r cm (property of square)CON ~ CBA (proved in (a))

4.2

42466

6

4

r

rr

rrAC

NC

AB

NO

The radius of the circle cm4.2

30. (a) 90QAP (property of square)

90

180 QAPAPR (int. s, BA // RP)

AP is the tangent to the circle at P andAPR = 90.PR is a diameter of the circle.

(b)

90

90180AQS (int. s, BA // RP)

AQ is the tangent to the circle at Q andAQS = 90QS is a diameter of the circle.The intersection of PR and QS is the centre ofthe circle.

31. (a)

Join OA, OB and OC.Consider ACO and BCO.

CA = CB givenOA = OB radiiCO = CO common side

ACO BCO SSS(b) ACBO is a cyclic

quadrilateral of thelarger circle.

180CBOCAO opp. s, cyclic quad.Also, CBOCAO corr. s, s

902

180

CBOCAO

AC and BC are tangentsto the smaller circle at Aand B respectively. converse of

tangent radius

32. (a)

AEDADT

AEAD

yx

ATEEATAED

isos.s,base

given

ofext.

yxADT (b)

y

ATDBTD

given

In CDT,

xDCT

yDCTyx

CTDDCTADT

ext. of

ACBBAT TA is the tangent to thecircle at A. converse of in

alt. segment

33. (a)

Join OC.

OCACOT

OACOCA

OAOC

CBTBAC

CATO //s,alt.

isos.s,base

radii

segmentalt.in

COTCBT C, O, B and T are concyclic. converse of s in

the same segment(b) BT is the tangent to the

circle at B.OBT = 90 tangent radiusOBTC is a cyclicquadrilateral.

90

18090

180

OCT

OCT

OBTOCT opp. s, cyclic quad.

CT is the tangent to thecircle at C. converse of

tangent radius

34. (a) ABC = 90 in semi-circleAM QS line joining centre

to mid-pt. ofchord chord

ABC = AMS = 90BRMA is a cyclic ext. = int. opp. squadrilateral.

(b)

PRBBAC

BACPBR

quad.cyclic,ext.

segmentalt.in


79

PBRPRB PRPB sides opp. equal s

35. (a) Consider OMP and PNQ.OAT = PBT = 90 tangent radiusOA // PB corr. s equalMOP = NPQ corr. s, OA // PB

OMP = PNQ = 90 givenOMP ~ PNQ AAA

(b) Let the radius of the middle circle be r cm.OMP ~ PNQ (proved in (a))

(rejected)4or4

16

322

616166

)8)(2()2)(8(2

8

2

8

2

2

22

rr

r

r

rrrr

rrrrr

r

r

r

PN

OM

PQ

OP

The radius of the middle circle is 4 cm.(c) Consider NPQ and CQT.

90QCT (tangent radius)

Also, 90PBT (proved in (a))BP // CQ (corr. s equal)

PNQQCT

cm2

cm)24(

NP

NP = CQNPQ = CQT (corr. s, BP // CQ)

NPQ CQT (ASA)

cm6

cm)24(

PQQT (corr. sides, s)

cm24

cm)62448(

QTYQPYXPOXOT

36. (a) Let Z be a point on OX such that SZ OX.OX PQ (tangent radius)SY PQ (tangent radius)SZ OX (by construction)

ZXYS is a rectangle.

ZSXY (property of rectangle)

yxOS

yx

SYOX

ZXOXOZ

(property of rectangle)

In OZS,

xyXY

xy

yxyxyxyxXY

XYyxyx

ZSOZOS

2

4

)2()2(

)()(

theorem)(Pyth.

22222

222

222

(b) From (a),

abBAacCAbcCB 2and2,2

abc

ab

abc

bc

abc

ac

abbcac

BACBCA

222

cab

111

Multiple Choice Questions (p. 3.41)1. Answer: D

47

ABDPAB (alt. s, PQ // BD)

47

PABADB ( in alt. segment)

86

472180

180 ADBABDBAD ( sum of )

94

18086

180

x

x

DCBBAD (opp. s, cyclic quad.)

2. Answer: C

102

GEDGBF (ext. , cyclic quad.)

x

FGBCBF

( in alt. segment)

42

18010236

180

x

x

CBFGBFABG (adj. s on st. line)

3. Answer: B

52

BAPACB ( in alt. segment)

90BAC ( in semi-circle)

38

1805290

180

ABC

ABC

ACBBACABC ( sum of )

38

ABCCAQ ( in alt. segment)

27

521338

13

x

x

ACBxCAQ (ext. of )

4. Answer: D

213

63

3

1

DBC

DA

CD

DCA

DBC

(arcs prop. to s at ce)

63ABD (s in the same segment) 37ACB ( in alt. segment)

59

18037)2163(

180

BAC

BAC


(corr. sides, ~ s)


80

5. Answer: BJoin AB. Refer to the figure.

37

CBEBAC ( in alt. segment)

59

3784180BAP (adj. s on st. line)

PAPB (tangent properties)

59

BAPABP (base s, isos. )

62

5959180x ( sum of )

6. Answer: C

Obviously, COB is the angle at centre subtended by .

CBI is not true.

For II, OC TC and OA TA (tangent radius) 90COA (given)

OC = OA (radii)COAT is a square.

45AOBCOB (property of square)CB = AB (equal s, equal chords)

Also, OT CA (property of square)CBO = ABO (prop. of isos. )II is true.

For III, CB = AB (proved)

BACBCA (base s, isos. )

BACTCB ( in alt. segment)

TCBBCA III is true.

The answer is C.

7. Answer: BJoin BD.

90ADB ( in semi-circle)

x

x

ADEADBBDC

90

90180

180 (adj. s on st. line)

x

ADEDBA

( in alt. segment)

902

)90(

xACE

xxACE

DBABDCACE (ext. of )

8. Answer: CFor I, consider CDA and CAB.ACD = BCA common angleDAC = ABC in alt. segment

CDA ~ CAB AAA

cm20cm25cm15

cm12

AB

ABBC

AB

AC

AD(corr. sides, ~ s)

For II, CDA ~ CAB

cm10

cm9cm25

cm15

cm15

CD

CDBC

CA

AC

CD(corr. sides, ~ s)

II is not true.For III, in ABC,

2

222

cm625

cm25

BC

2

2

22222

cm625

cm)2015(

BC

ABAC

ABC is a right-angled triangle with BAC = 90.(converse of Pyth. theorem)AB is a diameter of the circle.

Radius of the circle

cm10

cm202

12

1

AB

The radius of the circle is 10 cm.III is true.Only I and III are true.

9. Answer: A

x

APAR

xAP


xb

ARACCR

xb

CRCQ


xba

xba

CQCBQB

)(

xba

QBPB


bacx

xbaxc

PBAPAB

2

)(

)(2

1bacx


81

10. Answer: DFor I, QB is the tangent to the circle at B.

90OBQ (tangent radius)

OQOQ

OBOC

BOQCOQ

side)(common

(radii)

(given)

OCQ OBQ (SAS)

90

OBQOCQ (corr. s, s)

PQ is the tangent to the circle at C.(converse of tangent radius)

I is true.For II, DC // AB, but PCD may not be equal to CAD.

PQ may not be the tangent to the circle at C.II may not be true.

For III, PCD = CAD,PQ is the tangent to the circle at C.

(converse of in alt. segment)III must be true.Only I and III are true.

HKMO (p. 3.44)Refer to the figure.AB = BC = 20 cmX, Y and Z are the points of contact.

OX BC, OY BA and OZ CA (tangent radius)BOZ is a straight line perpendicular to CA.

3

4tan BAC

5

4sin BAC and

5

3cos BAC

5

4sin

AB

BZBAC

cm16

cm205

4

BZ

6

488

3)16(35165

3

cos

)90sin(

sin

r

r

rrr

rOZBZ

OYBAC

OZBZ

OYBAC

BO

OYOBY

Investigation Corner (p. 3.45)

(a) AB = CD = EFAP = FP (tangent properties)BQ = CQ (tangent properties)DR = ER (tangent properties)

PQ = QR = RPPQR is an equilateral triangle.

(b) OAP = OFP = 90 (tangent radius)APF = 60 (prop. of equil. )

302

60

FPOAPO (tangent properties)

60

3090180

180 APOOAPAOP ( sum of )

60

AOPFOP (tangent properties)

120

6060

FOPAOPAOF

(c)

m2

m)11(

EFCDAB

m3

2

m360

12012

DEBCAF

The length of the bold orange line

m)62(

m3

2

3

2

3

2222

DEBCAFEFCDAB

(d) Using similar steps as in (a) to (c), we can find that:

The shortest length

m)122(

m6)11(33

2

3 tangents to circles - … · 3 tangents to circles 61 3 tangents to circles activity activity 3.1...

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