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3 Tangents to Circles
61
3 Tangents to Circles
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
Quick Practice 3.2 (p. 3.6)
(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
(tangent properties)
CY = (10 – x) cm
cm)10( x
CYCZ
(tangent properties)
AZ + CZ = 11 cm
4
11219
11)10()9(
x
x
xx
cm4BX
3 Tangents to Circles
63
Quick Practice 3.7 (p. 3.17)
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.
Quick Practice 3.8 (p. 3.24)
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.
NSS Mathematics in Action 5A Full Solutions
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 )
3 Tangents to Circles
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
NSS Mathematics in Action 5A Full Solutions
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.
3 Tangents to Circles
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)
NSS Mathematics in Action 5A Full Solutions
68
31
622
12
1BACOAB
6. Let the length of AR be x cm.
cm)8(
cm
x
APABBP
xARAP
(tangent properties)
cm)8( x
BPBQ
(tangent properties)
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
3 Tangents to Circles
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
(tangent properties)
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
NSS Mathematics in Action 5A Full Solutions
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
(tangent properties)
b
BCQC
(tangent properties)
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.)
3 Tangents to Circles
71
4. TA TB (tangent properties)TAB TBA (base s, isos. )
62
180562
180
TAB
TAB
ATBTBATAB ( sum of )
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, ~
NSS Mathematics in Action 5A Full Solutions
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 )
3 Tangents to Circles
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
ATBTBATAB ( sum of )
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 )
NSS Mathematics in Action 5A Full Solutions
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
ATBTBATAB ( sum of )
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 )
3 Tangents to Circles
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
BACACBABC ( sum of )
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.
NSS Mathematics in Action 5A Full Solutions
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.)
3 Tangents to Circles
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
(tangent properties)
cm)8( a
AXAZ
(tangent properties)
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
(tangent properties)
cm)5( r
CQCR
(tangent properties)
cm)217(
cm)]5()12[(
r
rr
CRARAC
NSS Mathematics in Action 5A Full Solutions
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
3 Tangents to Circles
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
BACACBABC ( sum of )
(corr. sides, ~ s)
NSS Mathematics in Action 5A Full Solutions
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
(tangent properties)
xb
ARACCR
xb
CRCQ
(tangent properties)
xba
xba
CQCBQB
)(
xba
QBPB
(tangent properties)
bacx
xbaxc
PBAPAB
2
)(
)(2
1bacx
3 Tangents to Circles
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
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