1 Basic Properties of Circles
1 Basic Properties of Circles
10. (a) In △ABE,
AB 2  AE 2  (6 2  8 2 ) cm 2  100 cm 2
BE 2  10 2 cm 2  100 cm 2
∵ AB 2  AE 2  BE 2
∴ △ABE is a right-angled triangle, where
(converse of Pyth. theorem)
CAD  90 .
Review Exercise 1 (p. 1.7)
1.
x  90  130  65  360
x  75
(s at a pt.)
2.
a  70
(vert. opp. ∠s)
90  70  b  180
b  20
(adj. s on st. line)
120  a  180
(int. ∠s, AB // CD)
(b) In △ACD,
3.
AC 2  AD 2  CD 2
81  64  16 x  x 2  225
x 2  16 x  80  0
( x  4)( x  20)  0
x  4 or
a  60
b 120
(alt. ∠s, AB // CD)
4.
DBP  60
y  60
(corr. ∠s, PQ // RS)
(vert. opp. ∠s)
5.
125  PQR  180
(adj. ∠s on st. line)
PQR  55
x  55  140
x  85
(ext. ∠ of △)
6.
In △XYZ,
a  90  40  180
12. Consider △ACE and △DCB.
AC (4  6) cm 10 5



DC
4 cm
4
2
EC (11  4) cm 15 5



BC
6 cm
6
2
AC EC

∵
and
DC BC
(common angle)
ACE  DCB
∴ △ACE ~ △DCB
(ratio of 2 sides, inc. ∠)
(∠ sum of △)
(ext. ∠ of △)
 84
13. (a) In △ABC,
AC 2  BC 2  AB 2
7.
8.
9.
x  20 (rejected)
11. Consider △PQD and △SRD.
PD = SD
(given)
∠DPQ = ∠DSR = 90°
(property of rectangle)
PQ = SR
(property of rectangle)
∴ △PQD  △SRD
(SAS)
a  50
In △PXY,
b  50  34
(Pyth.theorem)
(6  3) 2  (8  x) 2  15 2
∵ AB = AD and BC  CD
∴ AC⊥BD
∴ y  90
BC 
(prop. of isos. △)
BD = (1 + 1) cm
= 2 cm
∵ AB  BD  DA  2 cm
∴ x  60
(prop. of equil. △)
∵ DB = AB
∴ ADB  20
DBC  ADB  DAB
(base ∠s, isos. △)
(ext. ∠ of △)
 20  20
 40
∵ DB = DC
∴ x  40
(given)
(base ∠s, isos. △)
(Pyth.theorem)
AB 2  AC 2
 25 2  15 2 cm
 20 cm
(b) Consider △BDC and △BCA.
(given)
BDC  BCA = 90°
(common angle)
CBD  ABC
In △BDC,
DCB  180  90  CBD
( sum of △)
 90  ABC
In △BCA,
CAB  180  90  ABC
( sum of △)
 90  ABC
∴ DCB  CAB
∴ △BDC ~ △BCA
(AAA)
BD BC

∴
(corr. sides, ~△s)
BC BA
BD
20 cm

20 cm 25 cm
BD  16 cm
BCD  65
alt. s, AB // CD
ABC  CBD  BDC  180 int. ∠s, AB // CD
65  50  BDC  180
BDC  65
∵ BCD  BDC  65
∴ BC = BD
sides opp. equal s
∴ △BCD is an isosceles
triangle.
1
NSS Mathematics in Action (2nd Edition) 5A Full Solutions
14. (a) Consider △ABD and △AED.
∵ DE is an angle bisector
of ∠ADC.
∴ ADE  CDE
∵ AD = DC and
ADE  CDE
∴ DE ⊥ AC
∴ AED  90  ABD
BD = DE
AD = AD
∴ △ABD  △AED
Activity 1.4 (p. 1.61)
1. (a) Yes
(b) Yes
2.
prop. of isos. △
given
common side
RHS
3.
(b) ∵ △ABD  △AED
∴ ADB  ADE
(corr. ∠s,  △s)
CDE  ADE
ADB  ADE  CDE  180 (adj. s on st. line)
Yes
Classwork
Classwork (p. 1.10)
(a)
Element
3ADE  180
ADE  60
region ORST

 minor arc

QRST


 major arc

 minor sector

 major sector
region OQPT
Activity
PTSR
Activity 1.1 (p. 1.12)
1. (a)
(b)
(b) (i) Yes
(ii) Yes
2.
Yes
3.
ON  ON
OA = OB
∠ONA = ∠ONB = 90°
∴ △ONA  △ONB
∴ AN = NB
common side
radii
given
RHS
corr. sides,  △s
APB 
c
2
(∠ at centre twice ∠ at ⊙ce)
(b) AQB 
c
2
(∠ at centre twice ∠ at ⊙ce)
3.
APB  AQB
4.
The angles in the same segment of a circle are equal.
P  R  180 , Q  S  180
3.
The sum of the opposite angles of a cyclic quadrilateral is
180°.
AEB

 chord

 diameter
CEOD

 minor segment
region AEBC

 major segment
The green circle is the circumcircle of △ABC.
(ii) The green circle is the inscribed circle of △PQR.
Classwork (p. 1.28)
(a) x  2APB
(∠ at centre twice ∠ at ⊙ce)
 2  30
 60
Activity 1.3 (p. 1.52)
1. (b) A  C  180 , B  D  180
2.
Term
region AEBD
(b) (i)
(b) ∠APB and ∠AQB are the angles at the
circumference in the same segment. They are both
subtended by arc AB.
(a)
Element
Classwork (p. 1.11)
(a) (i) The purple circle and the orange circle are equal
circles.
(ii) The green circle and the orange circle are concentric
circles.
Activity 1.2 (p. 1.33)
1. (a) ∠AOB (i.e. c) is the angle at the centre subtended by
arc AB.
2.
Term
2
AOC
2
46

2
 23
(b)
x
(c)
x
(d)
x
AOB
2
100

2
 50
reflex BOA
2
220

2
 110
(∠ at centre twice ∠ at ⊙ce)
(∠ at centre twice ∠ at ⊙ce)
(∠ at centre twice ∠ at ⊙ce)
Classwork (p. 1.30)
1. (a) x  90
(∠ in semi-circle)
(b) APB  90
In △ABP,
BAP  180  APB  ABP
(d)
(∠ in semi-circle)
(∠ sum of △)
x  180  90  40
 50
(c)
(e)
(∠ in semi-circle)
BCA  90
In △ACB,
BCA  BAC  ABC  180 (∠ sum of △)
∵
(f)


AB

(arcs prop. to ∠s at ⊙ce)
ACB
CBD
(arcs prop. to ∠s at ⊙ce)


DC

DEC
CEB
(arcs prop. to ∠s at ⊙ce)
CB
x 30

10 40
x  7 .5
PSR  40  45
 85  90
∴ PR is not a diameter.
∵ SPQ  45  45
 90
∴ QS is a diameter. (converse of ∠ in semi-circle)
Classwork (p. 1.42)
(a) x  4


BCA BA

ABC
AC
x
12

50 10
x  60
CD
x 36

4 48
x3
90  x  2 x  180
3x  90
x  30
2.
1 Basic Properties of Circles


AEB AB

BEC
BC
y
6

40 10
y  24
(equal ∠s, equal arcs)
(arcs prop. to ∠s at ⊙ce)
(b)
x5
(equal arcs, equal chords)
(c)
x  65
(equal chords, equal ∠s)
(d)
x 135
(equal arcs, equal ∠s)
(e)
x6
(equal chords, equal arcs)
x  120
(f)
x4
(equal ∠s, equal chords)
ABC  ADC  180


Classwork (p. 1.54)
(a) BAD  BCD  180
(a)
DOC

AOB
AB
x 50

6 30
x  10
(b)
(c)




BOC BC

AOB
AB
x
2

80 5
x  32
DC

y  100
(b) ABC  FDC
(arcs prop. to ∠s at centre)
(ext. ∠, cyclic quad.)
x  75
DAB  180  EAB
 180  100
BOC BC

AOB
AB
x
4

90 6
x  60


(opp. ∠s, cyclic quad.)
y  80  180
Classwork (p. 1.45)
DC
(opp. ∠s, cyclic quad.)
x  60  180
DOC
AOB
 80
BCG  DAB
y  80
(arcs prop. to ∠s at centre)
(c)
(adj. ∠s on st. line)
(ext. ∠, cyclic quad.)
ADC  ABC  180
(opp. ∠s, cyclic quad.)
(46  y )  (32  35)  180
y  67
In △DBC,
DCE  DBC  BDC
(arcs prop. to ∠s at centre)
(ext. ∠ of △)
x  35  67
 102
Classwork (p. 1.64)
(a) BAD  60  30
 90
∵ BAD  BCK
∴ A, B, C and D are concyclic.
(arcs prop. to ∠s at centre)
AB
y 48

5 80
y3
3
(ext.   int. opp. )
NSS Mathematics in Action (2nd Edition) 5A Full Solutions
(b) ABC  FBC  180
(adj. s on st. line)
Quick Practice 1.3 (p. 1.16)
∵ PN = NR and PR  QS
(given)
∴ QS is the perpendicular bisector of the chord PR.
∴ QS passes through the centre.
(⊥bisector of chord
passes through centre)
i.e. QS is a diameter of the circle.
Let O and r cm be the centre and the radius of the circle
respectively.
ABC  100  180
ABC  80
∵ ABC  ADE
∴ A, B, C and D are not concyclic.
(c) In △ABD,
ABD  40  120  180
( sum of △)
ABD  20
ABC  ADC  (20  50)  (40  70)
∴
 180
A, B, C and D are concyclic.
(opp. s supp.)
(d) In △ABE,
BAE  70  90
(ext. ∠ of △)
BAE  20
∵ BAC  BDC
 20
∴ A, B, C and D are concyclic. (converse of s in the
same segment)
Join OP.
∴ OP  r cm and ON = (r  1) cm
In △OPN,
(Pyth. theorem)
OP2  ON 2  PN 2
r 2  ( r  1) 2  52
r 2  r 2  2r  1  25
2r  26
r  13
∴ The radius of the circle is 13 cm.
NS  QS  QN
Quick Practice
Quick Practice 1.1 (p. 1.14)
 (13  2  1) cm
 25 cm
Join OQ.
OQ  26 cm
In △ONQ,
∵
(radius)
ON 2  NQ 2  OQ 2
(Pyth. theorem)
NQ  OQ  ON
2
∴
Quick Practice 1.4 (p. 1.19)
∵ PQ = RS, OM⊥PQ and ON⊥RS
(given)
∴ OM = ON
(equal chords, equidistant from centre)
1
OM  MN
2
1
  6 cm
2
 3 cm
∵ OM⊥PQ
1
∴ MQ  PQ
(line from centre  chord bisects chord)
2
1
  8 cm
2
 4 cm
2
 26 2  10 2 cm
∵ ON  PQ
∴ PN = NQ
PQ  PN  NQ
 24 cm
(given)
(line from centre  chord bisects chord)
 2  24 cm
 48 cm
Quick Practice 1.2 (p. 1.15)
(a) ∵ PM = QM
∴ OM⊥PQ
(given)
(line joining centre to
mid-pt. of chord  chord)
Join OQ.
In △MOQ,
OQ 2  OM 2  MQ 2
i.e. ∠OMQ = 90°
∵ OR = OQ
(radii)
∴ ∠ORQ =∠OQR
(base ∠s, isos. △)
= 30°
In △MQR,
(∠ sum of △)
MRQ  RQM  RMQ  180
OQ  OM  MQ
2
(Pyth. theorem)
2
 3 2  4 2 cm
 5 cm
∴ The radius of the circle is 5 cm.
30  (30  OQP)  90  180
OQP  30
(b) ∵ ∠OQP =∠QPS = 30°
∴ OQ // PS
(alt. ∠s equal)
4
1 Basic Properties of Circles
Quick Practice 1.5 (p. 1.20)
(a) ∵ PQ = RS, OM⊥PQ and
ON⊥RS
∴ OM = ON
Quick Practice 1.8 (p. 1.31)
(∠ at centre twice ∠ at ⊙ce)
BOD  2BAD
Consider △OMK and △ONK.
∠OMK =∠ONK = 90°
OK = OK
OM = ON
∴ △OMK  △ONK
(b) ∵
∴
∵
∴
△OMK  △ONK
MK = NK
OM⊥PQ
PM = MQ
∵ ON⊥RS
∴ RN = NS
∵
∴
i.e.
QK
 2  56
given
equal chords,
equidistant from centre
 112
CBO  BOD  180
CBO  112  180
given
common side
proved
CBO  68
(∠ in semi-circle)
ACB  90
In △ACB,
CAB  ACB  CBA  180
( sum of △)
RHS
proved in (a)
corr. sides,  △s
CAB  90  68  180
CAB  22
line from centre  chord
bisects chord
Quick Practice 1.9 (p. 1.32)
Let ∠CBD = x.
∠COD = 2∠CBD
( at centre twice  at ☉ce)
= 2x
In △BCP,
x + 60° = ∠CPD
(ext.  of △)
In △ODP,
2x + 30° = ∠CPD
(ext.  of △)
∴ 2x + 30° = x + 60°
x = 30°
ABC  60  x
line from centre  chord
bisects chord
given
PQ = RS
2MQ = 2RN
MQ = RN
 MQ  MK
 RN  NK
 RK
 60  30
Quick Practice 1.6 (p. 1.29)
AOC
y
(∠ at centre twice ∠ at ⊙ce)
2
118

2
 59
 90
∴ AC is a diameter of the circle.
(converse of  in semi-circle)
Quick Practice 1.10 (p. 1.34)
(∠ in semi-circle)
BCD  90
In △BDC,
BDC  BCD  DBC  180
(∠ sum of △)
32  90  DBC  180
Reflex ∠AOC = 360° − 118°
(∠s at a pt.)
= 242°
reflex AOC
x
(∠ at centre twice ∠ at ⊙ce)
2
242

2
 121
DBC  58
ABD  ABC  DBC
 120  58
 62
ACD  ABD
 62
Quick Practice 1.7 (p. 1.31)
(∠ in semi-circle)
ACB  90
∵ CA  CD
(given)

CAB


CDA
∴
(base ∠s, isos. △)
x
In △ABC,
CBA  CAB  ACB  180 ( sum of △)
(∠s in the same segment)
Quick Practice 1.11 (p. 1.35)
Let CAD  x.
DEB  DAB
(s in the same segment)
x
AEB  90
( in semi-circle)
In △ACE,
ECA  CAE  AEC  180
( sum of △)
28  x  90  180
x  62
BCD  CBD  CDA
(int. ∠s, CB // DO)
(ext.  of △)
24  ( x  50)  (90  x)  180
2 x  164  180
2 x  16
x  8
∴ CAD  8
y  28  62
y  34
5
NSS Mathematics in Action (2nd Edition) 5A Full Solutions
(b) In △BAD,
BAD  BDA  DBA  180
Quick Practice 1.12 (p. 1.43)
(∠ sum of △)
BAD  90  60  180
BAD  30
DAC  BAC  BAD
 50  30
 20
Join OB and OD.
∵ AB = BC = CD = DE

(given)
BD : DC  BAD : DAC (arcs prop. to ∠s at ⊙ce)
x
∴ AOB  BOC  COD  DOE 
2
(equal chords, equal ∠s)
AOB  BOC  COE  AOE  360 (∠s at a pt.)
 30 : 20
 3: 2
(a)
x  140
Quick Practice 1.13 (p. 1.44)
(a) Consider △ABC and △BAD.
ACB  BDA  90
 
∵ BC  AD
∴ BC = AD
AB = BA
∴ △ABC  △BAD
∠ in semi-circle
given
equal arcs, equal chords
common side
RHS
(b) ∵ △ABC  △BAD
∴ CAB  DBA
In △ABC,
CBA  90  CAB
CBD  CBA  DBA
proved in (a)
corr. ∠s,  △s
(arcs prop. to ∠s at ☉ce)
 
(b) ∵
CD  AB
(given)
∴
CD  AB
(equal chords, equal arcs)


CBD CD

ACB
AB
CBD  ACB
(arcs prop. to s at ☉ce)
 44
In △KBC,
AKB  CBK  KCB
∠ sum of △
(ext.  of △)
 44  44
 88
Quick Practice 1.17 (p. 1.54)
BAC  90
(∠ in semi-circle)
(alt. ∠s, AD // BC)
BCA  CAD
BAD  BCD  180 (opp. ∠s, cyclic quad.)
Quick Practice 1.14 (p. 1.46)
∵ BC = CD
(given)
∴ BOC  COD (equal chords, equal ∠s)
 66.5
(90  CAD )  (BCA  70)  180
2CAD  90  70  180
CAD  10


AOD AD

(arcs prop. to ∠s at centre)
BOC
BC
AOD 10

66.5
7
AOD  95
AOB  360  BOC  COD  AOD (∠s at a pt.)
 360  66.5  66.5  95
 132
Quick Practice 1.18 (p. 1.55)
FAD  ∠BCD = x
In △FAD,
ADF  DFA  FAD  180
ADF  43  x  180
ADF  137  x
EDC  ADF
 137  x
In △DCE,
CED  EDC  BCD
35  (137  x)  x
2 x  172
x  86
Quick Practice 1.15 (p. 1.47)
(a) ∠BDA = 90°
(∠ in semi-circle)


ACB AB

ABD
AD
ACB 2

66
3
ACB  44
∴
 (90  CAB )  CAB
 90
∴ CD is a diameter of the circle. converse of ∠ in
semi-circle
BAC BDC

BDA
AXB
BAC 5

90
9
BAC  50


Quick Practice 1.16 (p. 1.48)
x x
  x  80  360
2 2
2 x  280
(arcs prop. to s at ☉ce)
6
(ext. , cyclic quad.)
( sum of △)
(vert. opp. s)
(ext.  of △)
1 Basic Properties of Circles
Further Practice
Quick Practice 1.19 (p. 1.56)
Further Practice (p. 1.21)
1.
Join OB.
BAE  BDC
 62
∵ BO  AO
∴ ABO  BAO
 62
In △AOB,
BOE  ABO  BAO
(ext. , cyclic quad.)
Construct OM and ON such that OM  AB and ON  CD.
∵ AB // CD
∴ MON is a straight line.
∵ OM⊥AB
1
∴ AM  AB (line from centre  chord bisects chord)
2
1
  6 cm
2
 3 cm
OA = 5 cm
(radius)
In △OMA,
(radii)
(base. ∠s, isos. △)
(ext. ∠ of △)
 62  62
 124
BCD  BOE  180
(opp. s, cyclic quad.)
BCD  124  180
BCD  56
OA2  AM 2  OM 2
OM  OA  AM
2
Quick Practice 1.20 (p. 1.65)
(a) ∵ BAF  FEB  110  100
 5 2  3 2 cm
 210
 4 cm
∵ ON⊥CD
1
∴ ND  CD (line from centre  chord bisects chord)
2
1
  8 cm
2
 4 cm
OD = 5 cm (radii)
In △OND,
(Pyth. theorem)
OD2  ND 2  ON 2
 180
∴ A, B, E and F are not concyclic.
(b) ∵
CDF  90  20
 110
and ∠BEF = 100°
∴ CDF  BEF
∴ F, E, C and D are not concyclic.
(c) ∵ ADB  20 and ACB  20
∴ ADB  ACB
∴ A, B, C and D are concyclic. (converse of s in the
same segment)
Quick Practice 1.21 (p. 1.66)
∵ FAB  BCD  180
and BEF BCD
∴ FAB  BEF  180
∴ A, B, E and F are concyclic.
Quick Practice 1.22 (p. 1.67)
(a) ∵ AB = BC
∴ BAC = BCA
EFA = BAC
BDA = BCA
∴ EFA = BDA
i.e. EFA = EDA
∴ A, E, F and D are
concyclic.
(b) DAB  BEF
= 85°
∴
ON  OD2  ND 2
 52  42 cm
 3 cm
∴ The distance between AB and CD
 OM  ON
int. ∠s, AF // CD
ext. ∠, cyclic quad.
 (4  3) cm
opp. s supp.
 7 cm
2.
given
base s, isos. △
alt. ∠s, AC // EF
s in the same segment
(a) ∵ AM = BM
∴ OM⊥AB
∵ AB  CD, OM  AB
and ON  CD
∴ OM  ON
converse of s in the
same segment
MPN  APC  90
∴ MON  90
∵ All four interior angles
are equal to 90 and
two adjacent sides are
equal.
∴ ONPM is a square.
(s in the same segment)
DCB  DAB  180
(Pyth. theorem)
2
(opp. ∠s, cyclic quad.)
DCB  180  85
 95
ACD  DCB  ACB
 95  60
 35
7
given
line joining centre to
mid-pt. of chord 
chord
given
equal chords,
equidistant from centre
vert. opp. s
 sum of polygon
NSS Mathematics in Action (2nd Edition) 5A Full Solutions
(b) ∵
∴
∵
∴
∴
ONPM is a square.
PN  ON  4 cm
ON  CD
CN  ND
CP  PN  PD  PN
CP  3 cm
2.
2.
ACD  ABD
(alt. s, OD // AC)

BDE
(arcs prop. to ∠s at ⊙ce)
( sum of △)
 180  110  40
 30
(s in the same segment)
Further Practice (p. 1.56)
1.
Join AD.
∵ DO  AO
(radii)
∴ ADO  DAO
(base. ∠s, isos. △)
In △AOD,
DAO  ADO  AOD  180 (∠ sum of △)
2DAO  136  180


Further Practice (p. 1.48)
QOS
QRS

POS
PQRS
x
(3  4) cm

180 (2  3  4) cm
x  140
PS  QT
 
   
 
 
DAO  22
BAD  BAO  DAO
 70  22
(arcs prop. to ∠s at centre)
 48
BAD  C  180
(opp. ∠s, cyclic quad.)
C  180  48
 132
(diameters)
PQRS  QRST
∴
(equal chords, equal arcs)
major AD
EAB 2  2

110
92
EAB  40
In △ACE,
ACE  180  AEC  EAC
( at centre twice  at ⊙ce)
 35
∵
(arcs prop. to ∠s at ⊙ce)
(given)
BD  DE
EAB

AED
 55
( in semi-circle)
ADC  90
In △ACD,
∠CAD + ∠ADC + ∠ACD = 180°
( sum of △)
∠CAD + 90° + 55° = 180°
∠CAD = 35°
∵ AC is the angle bisector of ∠BAD.
∴ ∠BAC = ∠CAD
= 35°
∠BDC = ∠BAC
(s in the same segment)
1.
 

∵ BD = DE
∴
(opp. s of // gram)
p  20
∴
(∠ in semi-circle)
RQT
RST

PTS
PQRS
y
6 cm

90 (2  3  4) cm
y  60
(given)
(line from centre 
chord bisects chord)
CP  4 cm  (11  4) cm
Further Practice (p. 1.35)
1. BOD  BCD
 40
BOD
BAD 
2
40

2
 20
ADO  BAD


PTS  90
(proved in (a))
2.
PQ  QRS  QRS  ST
∵
∴
PQ  ST
 
 
(a) Consider △ABC and △ADC.
AB  AD
AB = AD
∵ CB  CD
∴ CB = CD
AC  AC
∴ △ABC  △ADC
∴ ABC  ADC
ABC  ADC  180
RST  RS  ST
 (4  2) cm
 6 cm
given
equal arcs, equal chords
given
equal arcs, equal chords
common side
SSS
corr. ∠s,  △s
opp. ∠s, cyclic quad.
2ADC  180
ADC  90
∴ AC is a diameter of
the circle.
Join PT.
8
converse of ∠ in
semi-circle
(b)
1 Basic Properties of Circles


5.
BAC
BC

ADC
ABC
BAC
5

90
13  5
BAC  25
(arcs prop. to s at ☉ce)
 12 cm
∵ AB = CD, OP⊥AB and OQ⊥CD
∴ OP = OQ
(equal chords, equidistant
x  3
from centre)
(proved in (a))
(corr. s,  △s)
(ext. ∠, cyclic quad.)
 25  25
6.
∵ OP  AB and OQ  CD (given)
∴ PB  AP and CQ = QD (line from centre  chord
bisects chord)
∴ AB  2  5 cm
 10 cm
and CD  2  5 cm
 10 cm
∵ AB  CD, OP AB and OQ  CD
∴ OP = OQ
(equal chords, equidistant
x = 2.5
from centre)
7.
∵
 50
Exercise
Exercise 1A (p. 1.21)
Level 1
1. In △OEN,
EN  OE  ON
2
(Pyth.theorem)
2
 10 2  6 2 cm
∵
∴
∴
i.e.
2.
 8 cm
ON⊥EF (given)
NF = EN (line from centre ⊥ chord bisects chord)
= 8 cm
EF  2  8 cm  16 cm
x = 16
∴
ON  AB
(given)
1
BN  AB (line from centre  chord bisects chord)
2
1
  16 cm
2
 8 cm
HQ  PQ  PH
 (18  9) cm
 9 cm
∵ PH = HQ = 9 cm
∴ OH⊥PQ
∴
3.
(given)
(line joining centre to mid-pt.
of chord  chord)
CD  CQ  QD
 2  6 cm
∵ △ABC  △ADC
∴ DAC  BAC
 25
BCE  BAD
OE 2  ON 2  EN 2
∵ CQ = QD
∴ OQ⊥CD
Join OB.
In △NOB,
(line joining centre to mid-pt. of
chord  chord)
OB 2  BN 2  ON 2
x  90
OB  BN  ON
2
∵ CN = ND
∴ ON⊥CD
(Pyth. theorem)
2
 8 2  6 2 cm
 10 cm
∴ The radius of the circle is 10 cm.
(line joining centre to mid-pt. of chord
 chord)
∴ OND  90
In △OND,
OND  ODN  DON  180
8.
( sum of △)
90  35  x  180
x  55
4.
∵ OP  OQ, OP  AB and OQ  CD (given)
∴ CD  AB
(chords equidistant from
 7 cm
centre are equal)
∵ CQ = QD
(given)
1
∴ x  7
2
 3 .5
Join OB.
OB  OC
(radii)
 (5  8) cm
 13 cm
∵ AN = NB
∴ ON  AB
9
(given)
(line joining centre to mid-pt. of
chord  chord)
NSS Mathematics in Action (2nd Edition) 5A Full Solutions
In △ONB,
NB  OB  ON
2
2
(Pyth. theorem)
 132  52 cm
 12 cm
∵ AN = NB
∴ AB  2  12 cm
 24 cm
9.
(given)
OC 2  ON 2  CN 2
(Pyth. theorem)
 10 2  8 2 cm
 6 cm
(b) ∵ AB is a diameter.
(proved in (a))
∴ The radius of the circle
1
 AB
2
1
  12 cm
2
 6 cm
13. ∵ OQ⊥CD
(given)
∴ CQ = QD
(line from centre  chord bisects chord)
= 10 cm
∴ CD = 210 cm = 20 cm
∵ AB = CD, OP⊥AB and OQ⊥CD
∴ OQ = OP (equal chords, equidistant from centre)
1
= PQ
2
1
  10 cm
2
 5 cm
(given)
(line joining centre to mid-pt. of
chord  chord)
∴ ∠AMO = 90°
∵ AN = NC
∴ ON⊥AC
(radii)
ON  OC 2  CN 2
(a) ∵ AB is the perpendicular bisector of the chord CD.
∴ AB passes through the centre.
(⊥bisector of chord passes through centre)
i.e. AB is a diameter of the circle.
10. ∵ AM = MB
∴ OM⊥AB
OC  OA
1
 AB
2
1
  20 cm
2
 10 cm
In △CON,
(given)
(line joining centre to mid-pt. of
chord  chord)
∴ ∠ANO = 90°
AMO  MAN
ANO  MON  360 ( sum of polygon)
Join OD.
In △OQD,
90  45  90  MON  360
OD 2  OQ 2  QD 2
MON  135
OD  OQ  QD
2
11. ∵ CM = MD
∴ OM⊥CD
(given)
(line joining centre to mid-pt.
of chord  chord)
(Pyth. theorem)
2
 5 2  10 2 cm
 125 cm (or 5 5 cm)
∴ ∠OMC = 90°
∵ OB = OC
(radii)
∴ OBC  OCB
(base ∠s, isos. △)
 32  BCD
(ext. ∠ of △)
BCD  OBC  OMC
∴ The radius of the circle is 125 cm (or 5 5 cm).
14. ∵ AM = MB
∴ OM⊥AB
(given)
(line joining centre to mid-pt. of
chord  chord)
∴ ∠AMO = 90°
In △AOM,
BCD  (32  BCD )  90
2BCD  58
BCD  29
AO 2  AM 2  OM 2
OM 
12. ∵ ON⊥CD
(given)
1
∴ CN  CD (line from centre  chord bisects chord)
2
1
  16 cm
2
 8 cm
AO  AM
2
(Pyth.theorem)
2
 7.5 2  6 2 cm
 4.5 cm
∵ OM = ON, OM⊥AB and ON⊥CD
∴ CD = AB
(chords equidistant from centre
= (6 + 6) cm are equal)
 12 cm
Join OC.
10
1 Basic Properties of Circles
15. ∵ AM = MB
∴ OM  AB
17. Construct a circle with centre O lying on BH as shown,
such that the circle cuts AB at two points P and Q, and
cuts BC at two points R and S.
(given)
(line joining centre to mid-pt. of
chord  chord)
1
 6 cm
2
 3 cm
In △OMB,
MB 
OB 2  OM 2  MB 2
OM  OB  MB
2
(Pyth. theorem)
2
Draw OM and ON such that OM  AB and ON  BC.
Consider △OBM and △OBN.
ABH = CBH
given
OMB = ONB = 90
by construction
OB = OB
common side
∴ △OBM  △OBN
AAS
∴ OM = ON
corr. sides,  △s
∵ OM = ON, OM⊥AB
and ON⊥BC
∴ PQ = RS
chords equidistant from centre
are equal
 5 2  3 2 cm
 4 cm
ON  MN  OM
 (7  4) cm
 3 cm
Level 2
18.
Join OD.
OD  5 cm
In △OND,
(radii)
ND  OD2  ON 2
(Pyth. theorem)
 52  32 cm
 4 cm
∵ ON⊥CD
∴ CN = ND
= 4 cm
∴ CD  2  4 cm
(given)
(line from centre  chord bisects
chord)
Join OD.
1
OD = AB
2
1
=  10 cm
2
= 5 cm
∵ CM = MD
∴ OM⊥CD
 8 cm
16. (a) Consider △AOB and △DOC.
AO = DO
radii
BO = CO
radii
∵ OM = ON, OM⊥AB
and ON⊥CD
given
∴ AB = DC
chords equidistant
from centre are equal
∴ △AOB  △DOC
SSS
(radii)
(given)
(line joining centre to mid-pt. of
chord  chord)
∴ ∠OMD = 90°
In △OMD,
OD 2  OM 2  MD 2
OM  OD  MD
2
(Pyth. theorem)
2
 5 2  4 2 cm
(b) ∵ OB = OA
(radii)
∴ ∠OAB =∠OBA
(base ∠s, isos. △)
= 62°
In △OBA,
OBA  OAB  BOA  180 (∠ sum of △)
62  62  BOA  180
BOA  56
∵ △AOB  △DOC
(proved in (a))
∴ ∠COD = ∠BOA
(corr. sides,  △s)
= 56
 3 cm
AO = OD = 5 cm
In △AMD,
AD 2  AM 2  MD 2
AD 
(radii)
AM 2  MD 2
 (5  3) 2  4 2 cm
 80 cm (or 4 5 cm )
11
(Pyth. theorem)
NSS Mathematics in Action (2nd Edition) 5A Full Solutions
In △OMB,
19.
OM  OB 2  BM 2
(Pyth.theorem)
 10  6 cm
2
2
 8 cm
In △OMD,
Draw a line OMN such that OM  CD and ON  AB.
Join OA and OC.
OA = OC = 17 cm
(radii)
∵ OM  CD
(by construction)
1
∴ CM  CD (line from centre  chord bisects chord)
2
1
  30 cm
2
 15 cm
In △OCM,
OM  OC 2  CM 2
MD  OD 2  OM 2
(Pyth. theorem)
 17 2  8 2 cm
 15 cm
CD  MD  MC
 (15  6) cm
 9 cm
22.
(Pyth. theorem)
 17 2  152 cm
 8 cm
∵ ON  AB
(by construction)
1
∴ AN  AB (line from centre  chord bisects chord)
2
1
  16 cm
2
 8 cm
In △OAN,
Let M be a point on AB such that OM  AB.
∵ OM  AB (by construction)
∴ AM  MB (line from centre  chord bisects chord)
1
  24 cm
2
 12 cm
In △OMA,
ON  OA2  AN 2
OM  OA2  AM 2
(Pyth. theorem)
 17  8 cm
2
2
 15 2  12 2 cm
 15 cm
∴ Distance between AB and CD  ON  OM
 (15  8) cm
 7 cm
 9 cm
MC  MB  BC
 (12  28) cm
 40 cm
In △OMC,
20. Let r cm be the radius of the circle.
∵ AB⊥OC
(given)
1
∴ MB  AB (line from centre  chord bisects chord)
2
1
  30 cm
2
 15 cm
OM  OC  CM
 (r  9) cm
In △OMB,
OB 2  OM 2  MB 2
OC  OM 2  MC 2
(Pyth. theorem)
 9  40 cm
 41 cm
OD = OA = 15 cm
(radii)
∴ CD  OC  OD
 (41  15) cm
 26 cm
2
2
23. (a) ∵ CN = ND
∴ ON⊥CD
(Pyth.theorem)
r 2  (r  9) 2  15 2
∴ ∠ONE = 90°
∵ AM = MB
∴ OM⊥AB
r 2  r 2  18r  81  225
18r  306
r  17
∴ The radius of the circle is 17 cm.
21. ∵ BM = MC
∴ OM⊥BC
(Pyth. theorem)
∴
∴
∵
and
∴
given
line joining centre to mid-pt.
of chord  chord
given
line joining centre to mid-pt.
of chord  chord
∠OME = 90°
MON  90
 sum of polygon
AB  CD, OM  AB given
ON  CD
OM  ON
equal chords, equidistant
from centre
∵ All four interior angles are equal to 90
and two adjacent sides are equal.
∴ ONEM is a square.
(given)
(line joining centre to mid-pt. of
chord  chord)
12
1 Basic Properties of Circles
(b) In △OMA,
25.
AM  AO 2  OM 2 (Pyth.theorem)
 10 2  6 2 cm
 8 cm
ME = OM = 6 cm
MB = AM
= 8 cm
(property of square)
Draw a line MON such that OM  CD and ON  AB.
Join OA and OC.
Let r cm be the radius of the circle.
∵ OM  CD (by construction)
1
∴ CM  CD (line from centre  chord bisects chord)
2
1
  10 cm
2
 5 cm
∵ ON  AB (by construction)
1
∴ AN  AB (line from centre  chord bisects chord)
2
1
  24 cm
2
 12 cm
In △OCM,
BE  MB  ME
 (8  6) cm
 2 cm
24.
Let M be a point on AB such that OM  AB, and N be a
point on CD such that ON  CD.
∵ OM  AB
(by construction)
1
∴ MB  AB (line from centre  chord bisects chord)
2
1
  18 cm
2
 9 cm
Join OB.
OB = 13 cm (radius)
In △OMB,
OM  OB 2  MB 2
OC 2  OM 2  CM 2
r  OM  5
2
2
(Pyth. theorem)
2
r 2  OM 2  25  (1)
In △OAN,
OA 2  ON 2  AN 2
r  ON  12
2
2
(Pyth. theorem)
2
r 2  ON 2  144  (2)
∵ The distance between AB and CD is 17 cm.
∴ MN = 17 cm
Let ON = a cm, then OM = (17  a) cm.
By substituting (1) into (2), we have
OM 2  25  ON 2  144
(Pyth. theorem)
 132  9 2 cm
 88 cm
∵ ON  CD
(by construction)
1
∴ NC  CD (line from centre  chord bisects chord)
2
1
  24 cm
2
 12 cm
∠NOM = 90°
( sum of polygon)
∵ ONK  NKM  OMK  NOM  90
∴ ONKM is a rectangle.
∴ NK = OM
(property of rectangle)
∴ KC  NC  NK
OM 2  ON 2  119
(17  a) 2  a 2  119
289  34a  a 2  a 2  119
170  34a
a5
∴ The shortest distance from O on AB is 5 cm.
26. (a) (i)
 (12  88 ) cm
 2.62 cm (cor. to 2 d.p.)
∵
∴
AB⊥CD
(given)
ED = CE
(line from centre  chord
bisects chord)
1
  24 cm
2
 12 cm
In △AED,
AD 2  AE 2  ED 2
AE 
AD 2  ED 2
 20 2  12 2 cm
 16 cm
13
(Pyth.theorem)
NSS Mathematics in Action (2nd Edition) 5A Full Solutions
(ii)
(b) ∵ AB = AC and BP = PC
∴ AP  BC
(prop. of isos. △)
∵ OM  AB
(given)
∴ MB = AM
= 6 cm
(line from centre  chord
bisects chord)
AB  2  6 cm
Join OD.
Let r cm be the radius of the circle.
OD = OA = r cm
(radii)
OE = AE  OA
= (16  r) cm
In △OED,
OD2  OE 2  ED 2
r 2  (16  r ) 2  12 2
 12 cm
AC  AB
(proved in (a))
 12 cm
BC  2  6 cm
(given)
 12 cm
(Pyth. theorem)
∵ AC = BC, ON  AC and OP  BC
∴ OP = ON = 2 3 cm
r 2  256  32r  r 2  144
32r  400
(equal chords,
equidistant from centre)
r  12.5
∴ OA = 12.5 cm
Alternative Solution
(b) AB = 212.5 cm
= 25 cm
EB  AB  AE
 (25  16) cm
Join OB.
Let r cm be the radius of the circle and OP = a cm.
In △APB,
 9 cm
AB 2  AP 2  BP 2
AP 
(Pyth. theorem)
AB  BP
2
2
 12 2  6 2 cm
 108 cm
AO  AP  OP
Join BD.
In △EBD,
BD  EB 2  ED 2
r  108  a  (1)
In △OPB,
Pyth. theorem
OB 2  OP 2  BP 2
 9 2  12 2 cm
r  a 6
2
 15 cm
BD  AD 2  (15 2  20 2 ) cm 2
 625 cm
AB  25 cm 2
2
( 108  a ) 2  a 2  36
2
108  2 108 a  a 2  a 2  36
 625 cm 2
∵ BD 2  AD 2  AB 2
∴ △ABD is a right-angled
triangle, where AD⊥BD.
(Pyth. theorem)
2
r 2  a 2  36  (2)
By substituting (1) into (2), we have
2
2
2
72  2 108 a
converse of Pyth.
theorem
36  6 3a
6
a
3
27. (a) Consider △ABP and △ACP.
AP = AP
common side
BP = CP
given
AP  BC
line joining centre to
mid-pt. of chord chord
∴ APB = APC = 90
∴ △ABP  △ACP
SAS
∴ AB = AC
corr. sides,  △s
i.e. △ABC is an isosceles
triangle.
 12 (or 2 3 )
∴ OP = 12 cm (or 2 3 cm)
14
1 Basic Properties of Circles
28. (a)
PQ  PR  RQ
 (1  5) cm
∵ OB⊥CD
(by construction)
1
(line from centre ⊥ chord
∴ BD  CD
2
bisects chord)
1
  1 2m
2
6m
Let r m be the radius of the circle.
OA = OD = r m
(radii)
In △OBD,
(Pyth. theorem)
OD 2  OB 2  BD 2
 6 cm
1
OP  PQ
2
1
  6 cm
2
 3 cm
OR  OP  PR
 (3  1) cm
 2 cm
In △ORM,
RM  OR2  OM 2
r 2  (9  r ) 2  6 2
r 2  81  18r  r 2  36
18r  117
r  6 .5
∴ The radius of the circle is 6.5 m.
(Pyth. theorem)
 2 2  12 cm
 3 cm
∵ OM  RS
∴ MS = RM
 3 cm
∴ RS = 2RM
(b)
(given)
(line from centre ⊥ chord
bisects chord)
 2 3 cm
Draw horizontal line EF such that EF intersects AB at
G and AB⊥EF. Then, EF is the ceiling of the tunnel,
i.e. EF = 7.8 m.
Join OF.
∵ EF⊥OG
1
∴ GF = EF
(line from centre  chord
2
bisects chord)
1
=  7.8 m
2
= 3.9 m
OF = AO = 6.5 m
(radii)
In △GOF,
(b)
Join OD.
OD = OP = 3 cm
In △OND,
(radii)
ND  OD2  ON 2
(Pyth. theorem)
GO  OF 2  GF 2
 32  12 cm
 8 cm
∵ ON  CD
∴ CN = ND
∴
 8 cm
CD  2 ND
(Pyth. theorem)
 6.5 2  3.9 2 m
 5 .2 m
New height of the tunnel
 GB
(given)
(line from centre ⊥ chord
bisects chord)
 AB  AG
 [9  (6.5  5.2)] m
 7.7 m
 2 8 cm (or 4 2 cm)
29. (a) Denote the centre of the circle by O.
30. (a) ∵
∴
∵
∴
∴
Draw a line AB such that it passes through O and
AB⊥CD. Then, AB is the height of the tunnel,
i.e. AB = 9 m and CD = 12 m.
Join OD.
i.e.
15
1
MC
(given)
2
MN  NC
MN = NC and BD  MC
BD is the perpendicular bisector of the chord
CM.
BD passes through the centre of the circle.
(⊥bisector of chord passes through centre)
BD is a diameter of the circle.
MN 
NSS Mathematics in Action (2nd Edition) 5A Full Solutions
(b) (i)
(b) Let PK = a cm, then PH = (20  a) cm.
In △KCP,
1
MC
2
1
  16 cm
2
 8 cm
In △BCN,
NC 
KC 2  PK 2  CP 2
NB  BC 2  NC 2
CP 2  144  a 2
In △HCP,
(Pyth. theorem)
 (1)
CH 2  CP 2  PH 2
 10  8 cm
2
(Pyth. theorem)
12 2  a 2  CP 2
2
(Pyth. theorem)
20 2  CP 2  (20  a) 2
 6 cm
CP 2  400  (20  a) 2  (2)
By substituting (1) into (2), we have
144  a 2  400  ( 20  a ) 2
144  a 2  400  ( 400  40a  a 2 )
40a  144
Join OC.
Let r cm be the radius of the circle.
OC  OB
(radii)
 r cm
In △OCN,
OC 2  ON 2  NC 2
∴
Exercise 1B (p. 1.36)
Level 1
1. ACB  138  180
ACB  42
x  2ACB
 2  42
 84
(Pyth. theorem)
r 2  (r  6) 2  8 2
r 2  r 2  12r  36  64
12r  100
100
12
 8.33 (cor. to 2 d.p.)
The radius of the circle is 8.33 cm.
r
∴
(ii)
2.
BD  2OB
 2
100
cm
12
50
cm
3
AD = BC = 10 cm
(property of rectangle)
∠DAB = 90°
(property of rectangle)

3.
(∠ at centre twice ∠ at ⊙ce)
Reflex AOC  2ABC
(∠ at centre twice ∠ at ⊙ce)
 2  130
 260
x  360  reflex AOC
 360  260
 100
(∠s at a pt.)
AOB  2ACB
 2  44
(∠ at centre twice ∠ at ⊙ce)
 88
OB  OA
(radii)
OBA  OAB (base ∠s, isos. △)
x
In △AOB,
AOB  OBA  OAB  180 (∠ sum of △)
(Pyth. theorem)
2
 50 
    10 2 cm
 3 
 13.33 cm (cor. to 2 d.p.)
31. (a) ∵ CP = PD
∴ KP ⊥ CD
(adj. s on st. line)
∵
∴
In △DAB,
AB  BD 2  AD 2
a  3.6
PK  3.6 cm
88  x  x  180
2 x  92
x  46
given
line joining centre to
mid-pt. of chord  chord
∵ CP = PD and
CD  HK
∴ HK is the perpendicular
bisector of the chord CD.
∴ HK passes through the ⊥bisector of chord
centre of the semi-circle. passes through centre
∴ H is the centre of the
semi-circle.
4.
ACB  90
In △ABC,
ABD  BAC  ACB
4 x  x  90
3x  90
x  30
16
(∠ in semi-circle)
(ext. ∠ of △)
1 Basic Properties of Circles
5.
∵
∴
DA  DC
DAC  DCA
11. AOC  BAO
 44
AOC
ABC 
2
44

2
 22
BCO  ABC
(given)
(base∠s. isos. △)
 35
CBA  90
In △CAB,
CAB  CBA  BCA  180
(∠ in semi-circle)
(∠ sum of △)
(35  x)  90  35  180
x  20
6.
(alt. ∠s, AB // CO)
(∠ at centre twice ∠ at ⊙ce)
(alt. ∠s, AB // CO)
 22
ABD  BDC
(alt. ∠s, BA // CD)
 40
ABC  90
(∠ in semi-circle)
In △ABE,
AEC  BAE  ABE
 44  22
 66
40  x  90
(ext. ∠ of △)
x  50
7.
BDC  BAC
(∠s in the same segment)
 44
ECD  EDC  AED
(ext. ∠ of △)
12. CDB  CAB
 32
ADC  ADB  CDB
 56  32
x  44  90
 88  90
∴ AC is not a diameter of the circle.
x  46
8.
CAE  CBE
13. AOC  2ABC
 2  25
(s in the same segment)
 25
ACB  CAD  ADC
∵
∴
 67
BDC  BAC
 40
ACD  BAC
 40
In △ECD,
BEC  EDC  ECD
(∠s in the same segment)
50  25  BCD  180
BCD  105
(∠ in semi-circle)
ACB  90
ACD  BCD  ACB
 105  90
 15
(alt. ∠s, BA // CD)
(ext.∠ of △)
x  40  40
 80
AOB
2
130

2
 65
OC = OA
OCA  OAC
 20
OB = OC
OBC  OCB
10. ACB 
∵
∴
∵
∴
(∠ at centre twice ∠ at ⊙ce)
 50
(given)
CD  CO
CDO  COD (base ∠s, isos. △)
 50
In △CDB,
CDB  CBD  BCD  180 (∠ sum of △)
(ext.  of △)
x  25  42
9.
(∠s in the same segment)
14. In △BDE,
DBE  BED  BDA
( at centre twice  at ☉ce)
(ext. ∠ of △)
DBE  32  62
DBE  30
(∠s in the same segment)
DAC  DBC
 30
CDA  90
(∠ in semi-circle)
In △ACD,
DAC  CDA  ACD  180 (∠ sum of △)
30  90  ACD  180
ACD  60
(radii)
(base s, isos. △)
(radii)
(base s, isos. △)
 ACB  OCA
 65  20
 45
15. DAB  90
(∠ in semi-circle)
∵ AB  AD
(given)
∴ ABD  ADB (base ∠s, isos. △)
In △ABD,
ABD  ADB  DAB  180 (∠ sum of △)
2ADB  90  180
ADB  45
17
NSS Mathematics in Action (2nd Edition) 5A Full Solutions
∴
ABD  ADB
19. In △ABE,
EAB  ABE  AED
 45
ACD  ABD
(∠s in the same segment)
 45
ACB  ADB
(∠s in the same segment)
∴
 73  30
 43
 45
ABD , ADB, ACD and ACB are 45°.
Level 2
16. BDC  BAC
(∠s in the same segment)
 52
∵ OC  OD
(radii)
∴ OCD  ODC (base ∠s, isos. △)
 52
Join BC.
CBD  DAC
 43
ABC  ABD  CBD
(∠ in semi-circle)
DCB  90
In △DBC,
OBC  DCB  BDC  180 (∠ sum of △)
 90
∴ AC is a diameter of the circle.
(converse of ∠ in semi-circle)
OBC  38
(int. ∠s, AC // OD)
20. (a) Let BAM  x .
∵ AM = BM
∴ ABM  x
In △MBA,
BMC  ABM  BAM
 2x
∵ BM = CM
∴ MBC  MCB
In △CBM,
BMC  MBC  MCB  180
CAB  130  180
CAB  50
(∠ in semi-circle)
ACB  90
In △ABC,
ACB  ABC  CAB  180 (∠ sum of △)
90  ABC  50  180
ABC  40
AOD
ABD 
(∠ at centre twice ∠ at ⊙ce)
2
130

2
 65
CBD  ABD  ABC
 65  40
 25
given
base ∠s, isos. △
ext. ∠ of △
given
base ∠s, isos. △
∠ sum of △
2 x  2MBC  180
MBC  90  x
ABC  ABM  MBC
 x  (90  x)
 90
(b) ∵
DAC  x .
DC = DA
(given)
DCA = x
(base s, isos. △)
BD = BC
(given)
BDC  DCA (base s, isos. △)
x
ADB = 90 ( in semi-circle)
In △ACD,
DAC  DCA  ADC  180 ( sum of △)
ABC  90
∴ AC is a diameter.
18. Let
∵
∴
∵
∴
∴
(∠s in the same segment)
 47  43
OBC  90  52  180
17. CAB  AOD  180
(ext. ∠of △)
EAB  47  77
EAB  30
DAC  DAB  EAB
proved in (a)
converse of ∠ in
semi-circle
∵ M is the mid-point of AC.
∴ M is the centre of the circle.
21. (a)
x  x  (90  x)  180
x  30
DAC  30
Join AB.
ABC  90
(∠ in semi-circle)
ABD  ACD
(∠s in the same segment)
x
CBD  ABC  ABD
 90  x
18
1 Basic Properties of Circles
(b) In △CBE,
(b) Reflex AOC  360  x
reflex AOC
ABC 
2
360  x

2
x
 180 
2
ABC = x
x
∴ 180   x
2
3
x  180
2
x  120
CBE  BCE  BEA (ext. ∠ of △)
(90  x)  BCE  75
BCE  x  15
BCD  ( x  15)  x
 2 x  15
(given)
BD  BC
(base ∠s, isos. △)
BDC  BCD
 2x  15
In △CBD,
BCD  BDC  CBD  180 ( sum of △)
∵
∴
2(2 x  15)  (90  x)  180
3x  60  180
x  40
( at centre twice  at ☉ce)
26. (a) ∵ DO  BE
∴ BK = EK
22. Let ABC  x .
(alt. ∠s, AB // CD)
BCD  x
(∠s in the same segment)
ADC  x
(∠ in semi-circle)
ACB  90
In △ACD,
CAD  ACD  ADC  180 (∠ sum of △)
Consider △BKD and △EKD.
BK = EK
BKD = EKD = 90°
DK = DK
∴ △BKD  △EKD
∴ KBD  KED
BAD  BED
∴ BAD  EBD
44  (90  x)  x  180
x  23
∴ ∠ABC = 23
AOE
23. ABE 
2
124

2
 62
In △BCE,
BCE  BEC  ABE
(s at a pt.)
(b) Let BAD = x.
In △ACD,
ADE  CAD  ACD
 x  42
BED  BAD  x
In △DKE,
( at centre twice  at ☉ce)
KED  KDE  BKD
(ext.  of △)
(opp. s of // gram)
line from centre  chord
bisects chord
proved
given
common side
SAS
corr. s,  △s
∠s in the same segment
(ext.  of △)
(s in the same segment)
(ext.  of △)
x  ( x  42)  90
BEC  62  36
2 x  48
 26
BAD  BED
(s in the same segment)
 26
In △ABK,
AKE  BAK  ABK
(ext. ∠ of △)
∴
x  24
BAD  24
27. With the notation in the figure,
 26  62
 88
24. (a) Consider △AKD and △BKC.
AKD = BKC
DAK = CBK
ADK = BCK
∴ △AKD ~ △BKC
(b) ∵ △AKD ~ △BKC
AK DK

∴
BK CK
6 cm
DK

3 cm 2 cm
DK  4 cm
vert. opp. s
s in the same segment
s in the same segment
AAA
(proved in (a))
(corr. sides, ~△s)
25. (a) ∵ OABC is a parallelogram.
and OA = OC
∴ OABC is a rhombus.
Join CD.
Let BAD  x .
∠BCD = ∠BAD
s in the same
=x
segment
In △DCQ,
QDC  QCD  CQD  180
∠ sum of △
QDC  x  90  180
QDC  90  x
ADE  90
CDE  ADE  ADC
 90  (90  x)
x
∠CAE = ∠CDE
=x
∴ BAD  CAE
given
radii
19
 in semi-circle
s in the same
segment
NSS Mathematics in Action (2nd Edition) 5A Full Solutions
28. (a) ∵ EA = EB
∴ EAB = EBA
BAD = BED
BCE  EBA  BED


given
base s, isos. △
s in the same segment
ext.  of △
AE
5.
(b) ABE = ADE
(s in the same segment)
∴ CBE  180  ABE
(adj. s on st. line)

BC
∵
∴
∴
6.
AOB
(arcs prop. to ∠s at centre)

reflex AOB
AB  BC  CD
(given)
∴
AB  BC  CD  2 cm
(equal chords, equal arcs)
∴
BCD  BC  CD
  
 
 55

BAC
ABC
∵
∴
∴
 
CB  AE
CB  AE  3 cm
y3


BDE BAE

CDB
CB
x
2

30 1
x  60
(given)
AB  BC  CD
AOB  BOC  COD (equal chords, equal ∠s)
AOC
DOC 
2
146
x
2
 73
∵
7.


BOD
BAD 
2
84  56
x
2
 70
4.


BAC BC

CAD
CD
x 18

8 9
x  16
(given)
(equal arcs, equal chords)
(arcs prop. to s at ☉ce)


y4
COD CD

BOC
BC
COD
4

84
6
COD  56
(arcs prop. to s at ☉ce)
BOC BC

(arcs prop. to s at centre)
AOB
AB
BOC 2

AOB 4
1
BOC  AOB
2
AOB  BOC  COD  180
(adj. s on st. line)
AOB 
y cm  (2  2) cm
3.
( sum of △)
AC
x 55

10 50
x  11
major AB
x
80

14 280
x4
2.
In △ABC,
BAC  180  ABC  ACB


(adj. s on st. line)
Exercise 1C (p. 1.49)
Level 1
1. Reflex AOB  360  80 (∠s at a pt.)
 280
AB
(arcs prop. to ∠s at ☉ce)
 180  50  75
 AED  DAE (ext.  of △)
 AED
∴ It is impossible that △BCE ~ △EAD.

ADE
CAD
CD
y 24

9
8
y  27
 EAB  BAD
 EAD
∴ BCE = EAD
 180  ADE
 CDA

1
AOB  51  180
2
3
AOB  129
2
AOB  86
(arcs prop. to s at centre)
8.
( at centre twice 
In △BCD,
BDC  180  DBC  DCB ( sum of △)
 180  62  64
 54
In △BDE,
EDB  DEB  DBC
at ☉ce)
(ext.  of △)
EDB  35  62

EDB  27
AB : BC  ADB : BDC
 27 : 54
 1: 2
(arcs prop. to ∠s at ⊙ce)
20
(arcs prop. to s at ☉ce)
9.
(a)


AOB AB

BOC
BC
AOB 4

48
3
AOB  64
(arcs prop. to s at centre)


AB

∴
BPD
1
EPF
(arcs prop. to s at ☉ce)
 30
 FPG


 
     
( at centre twice  at ☉ce)
AD
∴
APD
1
FPG

FG
(∠ in semi-circle)
BCA
ABC

BD  EF
APD  5  10  15
∴
∵
AOB
2
64

2
 32
10. BCA  90


 
BD
EF
(b) ACB 
1 Basic Properties of Circles
(arcs prop. to s at ☉ce)
(arcs prop. to s at ☉ce)
∴
AD  FG
∴
AC  CD , BD  EF , AD  FG
(any two of the above answers)
Level 2
13.
AC
x 90

9 36
x  22.5
In △ABC,
ABC  BCA  BAC  180 ( sum of △)
36  90  BAC  180


BC
BAC  54

BAC
ABC
Join BC.
(s in the same segment)
BCA  BDA
 70
In △ABC,
( sum of △)
ABC  BAC  BCA  180
(arcs prop. to s at ☉ce)
AC
y 54

9 36
y  13.5
11. (a) ∵
∴
(b)
 


BOD  AOC
(vert. opp. ∠s)
BD  AC
= 8 cm
(equal ∠s, equal arcs)


ADC
ABC
AC
(arcs prop. to s at ☉ce)

ABD
AD
ABC 8

ABD 4
ABC  2ABD
∵ ∠CBD = 90°
(∠ in semi-circle)
∴
ABD  ABC  90
14. ∵
∴


AC
CD
∴
∵

APC
1
CPD
(arcs prop. to s at ☉ce)
AB  AD
ABD  ADB
DCB
(given)
(base s, isos. △)
DAB  80

DAB
ABD
(arcs prop. to s at ☉ce)
AD
x  6 80

10
50
x  10
 CPD
∴
ABC
BAC
 50
In △ABD,
ABD  BDA  DAB  180 (∠ sum of △)
50  50  DAB  180
APC  5  10
 15


 

BC
x
70

10 40
x  17.5
ABD  2ABD  90
3ABD  90
ABD  30
12. ∵
ABC  180  40  70
 70
(arcs prop. to s at ☉ce)
15.
AC  CD
BPD  10  15
 25
 EPF
21


CAB
CB

CDA
CBA
x
4 cm

50 (4  4) cm
x  25
(arcs prop. to s at ☉ce)
NSS Mathematics in Action (2nd Edition) 5A Full Solutions


In △BCD,
CBD  BCD  BDC  180 ( sum of △)
DAC
DC
(arcs prop. to s at ☉ce)

CDA
CBA
DAC 12

50
8
DAC  75
In △CAD,
CAD  ADC  DCA  180 (∠ sum of △)


DA
30  BCD  20  180




BAD
ABC

BCD
ADC
(arcs prop. to s at ☉ce)
BAD 130

6 cm
60
75  50  DCA  180
DCA  55

BCD  130
BAD  13 cm
DCA
CDA
(arcs prop. to s
at ☉ce)
CBA
y 55

8 50
y  8 .8
(b)
16. PRS  PQS
(s in the same segment)
 30
PRQ  QRS  PRS
 75  30
19. ∵
 
 45
∵ PQ  QR
∴ PQ = QR
(equal arcs, equal chords)
∠RPQ = ∠PRQ
(base s, isos. △)
= 45°
In △PQR,
PQR  RPQ  PRQ  180
( sum of △)
∴
∵
∴
∴
(30  RQS )  45  45  180


ABD
AD

(arcs prop. to s at ☉ce)
ADC
ABC
ABD (13  4) cm

60
6 cm
ABD  90
∴ AD is a diameter of the circle.
(converse of  in semi-circle)


BAC BC

1
CAD
CD
BAC = CAD
OC = OA
OCA  OAC
 DAC
OC // AD
arcs prop. to s at ☉ce
radii
base s, isos. △
alt. s equal
20. (a)
RQS  60
  


17. ∵ AB = AD = CD
∴
AB  AD  CD
(given)


(equal chords, equal arcs)
  
Join OB and OC.
ACB
AB
ACD AD
and


ACD
CBD
AD
CD
(arcs prop. to s at ☉ce)
∴ ∠ACB = ∠ACD = ∠CBD
In △BCD,
BCD  CBD  BDC  180 (∠ sum of △)
3CBD  30  180
CBD  50
18. (a)


BDC
BC

ADC
ABC
BDC
2 cm

60
( 4  2) cm
BDC  20


CBD
CD

ADC
ABC
CBD 3 cm

60
6 cm
CBD  30
∵ AB  BC  CD
(given)
∴ AOB = BOC = COD (equal arcs, equal s)
AOB  BOC  COD  120
3BOC  120
BOC  40
BOC
BDC 
2
40

2
 20
(arcs prop. to s at ☉ce)
( at centre twice  at ☉ce)
(b) Reflex ∠AOD  360  AOD
 360  120
(∠s at a pt.)
 240
reflex AOD
ACD 
( at centre twice  at ☉ce)
2
240

2
 120
In △CDE,
CED  CDE  ECD  180
( sum of △)
(arcs prop. to s at ☉ce)
CED  20  120  180
CED  40
22
1 Basic Properties of Circles
 
  
 
21. (a) ∵ AB = CD
∴
ACB  CBD
23. (a) With the notations in the figure,
equal chords, equal arcs
AC  CB  CB  BD
AC  BD
(b)
In △BDF,
DFE  BDF  DBF
(ext.  of △)
 20  30
 
 50
In △ACG,
AGE  CAG  ACG
Join AD.
∵
∴
∴
∴
AC  BD
(ext.  of △)
 40  50


ADC AC

1
DAB
BD
ADC  DAB
AE  DE
proved in (a)
∴
arcs prop. to s at ☉ce
sides opp. equal ∠s
(b)
 90
x  180  FGE  GFE
 180  90  50
 40
( sum of △)

AB : BC : CD : DE : EA
 ADB : BEC : CAD : DBE : ECA
22. (a)
(arcs prop.to s at ⊙ce )
 20 : 40 : 40 : 30 : 50
 2: 4: 4:3:5

(c)
Join AE and ED.
∠AED = 90°


AE


ABCD
( in semi-circle)
ABE
AED
(arcs prop. to s at ☉ce)
AE
50

(2  3  1) cm 90
5
AE   6 cm
9
10

cm
3

(b)
Exercise 1D (p. 1.56)
Level 1
1. A  C  180


x  125
B  D  180
y  90
2.
ABCD
BEC 3

90
6
BEC  45
BEC  OFE  BGO
C  A  180
C  76  180
(opp. ∠s, cyclic quad.)
C  104
CD  CB
CDB  CBD
x
In △BCD,
C  CDB  CBD  180
 110
BC
(opp. ∠s, cyclic quad.)
y  90  180
 50  60
BEC

AED
(opp. ∠s, cyclic quad.)
55  x  180
BAD
BCD

(arcs prop. to s at ☉ce)
AED
ABCD
BAD 4

90
6
BAD  60
With the notation in the figure,
BGO  GBA  BAG (ext.  of △)


AB
2

Circumfere nce of the circle 2  4  4  3  5
18 

Circumfere nce of the circle      cm
2

 9 cm
9
cm
∴ Radius of the circle 
2
 4.5 cm
∵
∴
(arcs prop. to s at ☉ce)
104  2 x  180
x  38
(ext.  of △)
45  OFE  110
OFE  65
23
(given)
(base ∠s, isos. △)
(∠ sum of △)
NSS Mathematics in Action (2nd Edition) 5A Full Solutions
3.
ABF  ADC
x  80
In △BCE,
CEB  ECB  EBC  180
(ext. ∠, cyclic quad.)
BCD  EAB
(ext. , cyclic quad.)
 95
BCD  DCG  180
(adj. s on st. line)
40  2EBC  180
EBC  70
ADC  EBC  180
ADC  70  180
ADC  110
95  y  180
y  85
4.
DBC  ADB
(alt. ∠s, AD // BC)
x
ABC  ADE
(ext. ∠, cyclic quad.)
9.
(a)
70  x  125
EBC  CDE  180
(opp. s, cyclic quad.)
EBC  180  110
 70
ECB  BAE  180
(opp. s, cyclic quad.)
(ext.  of △)
 96
(b) DFE  BCD (ext. , cyclic quad.)
 60
In △DEF,
DFE  FDE  DEF  180
( sum of △)
DEF  24
BOA
2
40

2
 20
( at centre twice  at ⊙ce)
(b) BAP  BCP  180
(opp. s, cyclic quad.)
BAP  50  180
BAP  130
In △ABP,
ABP  BAP  BPA  180
ABP  130  20  180
ABP  30
8.
(ext. ∠, cyclic quad.)
(corr. s, AD // EC)
 40
∵ EC = EB
∴ ECB  EBC
(given)
(base s, isos. △)
(ext. ∠, cyclic quad.)
 74
CDE  ABE
(ext. ∠, cyclic quad.)
∵
DAC  115  30  180
DAC  35
(given)
BC  CD
∴
BC  CD
 


corr. ∠s equal
int. ∠s supp.
( sum of △)
(equal chords, equal arcs)
Level 2
11. Let CAE  x .
∵ AC is an angle bisector of ∠EAB.
∴ BAC  CAE  x
ECD  EAB (ext. ∠, cyclic quad.)
 2x
∵ EC is an angle bisector of ∠ACD.
∴ ECA  ECD
 2x
( sum of △)
CEB  DAE
ABE  AFG
BAC BC
(arcs prop. to s at ☉ce)

1
DAC CD
BAC  DAC
 35
ABC + ADC = 180
(opp. s, cyclic quad.)
ABC  180  115
 65
In △ABC,
ACB  ABC  BAC  180
( sum of △)
ACB  65  35  180
ACB  80
60  96  DEF  180
BPA 
BED  100  180
BED  80
BAF  BED
10. In △CAD,
DAC  ADC  ACD  180
 36  60
(a)
(opp. ∠s, cyclic quad.)
(b) ∵ ∠CDE = ∠AFG
= 74°
∴ AF // CD
Alternative Solution
∵ ∠BAF + ∠BCD
= 80° + 100°
= 180°
∴ AF // CD
(a) In △ACD,
ADE  CAD  ACD
7.
BED  BCD  180
 74
ECB  180  120
 60
In △BCE,
BEC  EBC  ECB  180 ( sum of △)
BEC  70  60  180
BEC  50
6.
(opp. s, cyclic quad.)
 80
x  55
5.
( sum of △)
24
1 Basic Properties of Circles
In △ACD,
ACD  ADB  DAC  180 (∠ sum of △)
( 2 x  2 x)  60  x  180
5 x  120
x  24
In △ABD,
ABD  BDA  DAB  180 (∠ sum of △)
ABD  60  2 x  180
(s at a pt.)
 360  220
 140
APB  COA  180
APB  180  140
 40
(opp. s, cyclic quad.)
16. (a) Consider △KAD and △KCB.
KAD = KCB
ext. , cyclic quad.
KDA = KBC
ext. , cyclic quad.
AKD = CKB
common angle
∴ △KAD ~ △KCB
AAA
KA KD
(b)
(corr. sides, ~△s)

KC KB
2 cm
3 cm

3 cm  DC (2  4) cm
4 cm  3 cm  DC
ABD  60  2(24)  180
ABD  72
12. In △ACD,
ACD  ADC  CAD  180
(∠ sum of △)
ACD  ADC  40  180
ACD  140  ADC
(opp. s, cyclic quad.)
ABC  ADC  180
ABC  180  ADC
(opp. s, cyclic quad.)
AED  ACD  180
AED  180  ACD
∴
COA  360  reflex COA
DC  1 cm
17.
 180  (140  ADC )
 40  ADC
ABC  AED  (180  ADC )  (40  ADC )
 220
13.
Join DC.
reflex COE
2
280

2
 140
ADC  EDC  EDA
EDC 
Join AD.
ABC + CDA = 180
ADE = 90
ABC + CDE
 ABC  (CDA  ADE )
opp. s, cyclic quad.
 in semi-circle
( at centre twice  at ⊙ce)
 140  20
 120
ABC  ADC  180
(opp. ∠s, cyclic quad.)
ABC  120  180
 180  90
 270
ABC  60
14. FCD  DEF  180
FCD  180  130
(opp. s, cyclic quad.)
18.
 50
(ext. , cyclic quad.)
x  FCD
 50
ABD = ∠CFD = y
∵ AD  AB
∴ ADB  ABD
(ext. , cyclic quad.)
(given)
(base s, isos. △)
Join AD.
BAD  DCB  180
y
In △ABD,
BAD  ABD  ADB  180
( sum of △)
x  2 y  180
2 y  180  50
y  65
15. Reflex COA  2ABC
(opp. s, cyclic quad.)
BAD  180  80
 100
OAD  BAD  BAO
 100  35
 65
∵ OD = OA
∴ ∠ODA = ∠OAD
= 65°
ABC  ADC  180
( at centre twice  at ⊙ce)
ABC  (65  50)  180
 2 110
ABC  65
 220
25
(radii)
(base s, isos. △)
(opp. s, cyclic quad.)
NSS Mathematics in Action (2nd Edition) 5A Full Solutions
19.
Join AB.
FAB  DCB
 75
FEB  FAB
 75
AEB  FEA  FEB
Exercise 1E (p. 1.67)
Level 1
1. ∠ABC = ∠EAD
= 100°
∵ ∠ABC + ∠ADC
= 100° + 80°
= 180°
∴ A, B, C and D are concyclic.
Alternative Solution
BCD  ADC  180
(ext. , cyclic quad.)
(s in the same segment)
BCD  180  80
 100
∵ EAD  BCD
∴ A, B, C and D are concyclic.
 135  75
∴
 60
DFB  AEB
2.
(s in the same segment)
In △DCE,
DCE  CDE  CED  180
∵
∴
3.
 sum of △
In △BCD,
CBD  BDC  DCB  180
(ext.  of △)
4.
(a) ∵
 2x  y
ARB  RCB  RBC
 ( y  y)  x
APQ  ARQ  180
(2 x  y )  ( x  2 y )  180
3x  3 y  180
ADC  ABC
 90  90
 180
∴ ABCD is a cyclic quadrilateral.
(ext.  of △)
(b) ABD  ACD
 35
∴ x  90  35
 55
 x  2y
(b)
ABE  70
ABC  CDE
A, B, C and D are not concyclic.
( sum of △)
CBD  35  105  180
CBD  40
∵ CAD = CBD
∴ A, B, C and D are concyclic.
(converse of s in the same segment)
x  60
∵ ODC = OCD = COD = 60
∴ △CDO is an equilateral triangle.
APC  PBC  PCB
 ( x  x)  y
(ext.   int. opp. )
radii
base s, isos. △
3x  180
21. (a)
(int. ∠s, AD // BC)
ext. , cyclic quad.
ext. , cyclic quad.
ext. , cyclic quad.
radii
base s, isos. △
x
∵ OD = OC
∴ ODC  OCD
x
O DC  O CD  CO D  180
(opp. s supp.)
( sum of △)
DCE  60  50  180
DCE  70
∵ BAD  DCE
∴ A, B, C and D are not concyclic.
Alternative Solution
In △ABE,
ABE  BAE  AEB  180
( sum of △)
ABE  60  50  180
 60
20. Let COD = x.
ABC = COD = x
∵ OC = OB
∴ OCB  OBC
x
OAD = OCB = x
O CD  BA O 
(corr. ∠s, AD // BC)
opp. s, cyclic quad.
x  y  60
5.
In △ABC,
BAC  ABC  ACB  180  sum of △
BAC  2 x  2 y  180
BAC  2( x  y )  180
BAC  120  180
BAC  60
(s in the same segment)
(a) ∵ BAD + BCD
 (38  42)  (35  65)
 180
∴ ABCD is a cyclic quadrilateral. opp. s supp.
(b) CDB  CAB
x  38
6.
(s in the same segment)
(a) ∵ ADB  50 and ACB  50
∴ ABCD is a cyclic quadrilateral.
(b)
x  ADC  180
x  180  (50  70)
 60
26
opp. s supp.
converse of s in
the same segment
(opp. s, cyclic quad.)
1 Basic Properties of Circles
7.
(a) ∵ BAC = 90° and BDC = 90
∴ ABCD is a cyclic quadrilateral.
converse of s in
the same segment
(b) ADB  180  EDA  BDC (adj. s on st. line)
 180  55  90
 35
ACB = ADB
(s in the same segment)
x  35
8.
(a) In △BCD,
BCD  180  BDC  DBC
(s in the same segment)
 90
∵ FEC = FBC
∴ BCFE is a cyclic quadrilateral.
Level 2
15. (a) APB = 90
APM  APB  180
ext.  = int. opp. 
ext.  = int. opp. 
converse of s in the
same segment
 in semi-circle
adj. s on st. line
APM  90  180
APM  90
∵ APM = AOM
∴ A, O, P and M are concyclic. converse of s in
the same segment
(a) With the notation in the figure,
In △ABG,
ABG  180  BGA  GAB
(b) OAP  OMB
 28
∵ OA = OP
∴ OPA  OAP
 sum of △
 50
ABC  ABG  GBC
 50  35
 85
∵ ABC = ADF
∴ ABCD is a cyclic quadrilateral. ext.  = int. opp. 
(s in the same segment)
(ext. , cyclic quad.)
10. ABC = ADC
PQB = PDC
DPQ = PQB
∴ DPQ = ABQ
∴ A, B, Q and P are concyclic.
opp. s of // gram
ext. , cyclic quad.
alt. s, AD // BC
11. ∵ AB = BE
∴ BAE = BEA
BAE = BCD
∴ BEA = BCD
∴ B, C, D and E are concyclic.
given
base s, isos. △
opp. s of // gram
12. ∵ AB = AC
∴ ABC = ACB
∵ AM = MB and AN = NC
∴ MN // BC
AMN = ABC
∴ AMN = ACB
∴ B, C, N and M are concyclic.
given
base s, isos. △
given
mid-pt. theorem
corr. s, MN // BC
(s in the same segment)
(radii)
(base s, isos. △)
 28
 180  85  45
(b) CAD  CBD
 35
DCE  DAB
x  45  35
 80
given
base s, isos. △
radii
base s, isos. △
adj. s on st. line
FEC  180  90
 sum of △
 86
∵ BCD = EAD
∴ ABCD is a cyclic quadrilateral. ext.   int. opp. 
9.
AC = BC
CAB = CBA
OD = OA
ODA = OAD
CBA = ODA
OBCD is a cyclic quadrilateral.
14. ∵ AED = DBC
∴ ABDE is a cyclic quadrilateral.
 180  55  39
(b) ∠BAC = ∠BDC
x  55
13. ∵
∴
∵
∴
∴
∴
ext.  = int. opp. 
16. (a) Consider △BOM and △COM.
BM = CM
given
OM ⊥ BC
line joining centre to
mid-pt. of chord  chord
∴ ∠BMO = ∠CMO = 90°
OM = OM
common side
∴ △BOM  △COM
SAS
∴ ∠BOM = ∠COM
corr. s,  △s
(b) Let ∠BOM = ∠COM = a,
by (a)
then ∠BOC = 2a.
BOC
∠ at centre twice
BAC 
2
∠ at ⊙ce
2a

2
a
∵ ∠BAC = ∠BOM
∴ O, D, A and B are concyclic. ext.   int. opp. 
17. (a)
ext.  = int. opp. 
Join PB and let ARP = .
APQ = ARP = 
(given)
ABP = ARP = 
(s in the same segment)
APB = 90
( in semi-circle)
BPQ  APB  APQ
 90  
ext.  = int. opp. 
27
NSS Mathematics in Action (2nd Edition) 5A Full Solutions
In △PQB,
PQB  180  PBQ  BPQ ( sum of △)
 180    (90   )
(b)
 90
(b)
Join RB.
TQB = 90
TRB = 90
TQB + TRB = 90 + 90
= 180
∴ R, T, Q and B are concyclic.
18. Consider △ACB and △DBC.
AC = DB
ACB = DBC
BC = CB
∴ △ACB  △DBC
∴ BAC = CDB
∴ A, B, C and D are concyclic.
19. AQS = BRS
BRS = CPS
∴ AQS = CPS
∴ AQSP is a cyclic quadrilateral.
Join MP and NP.
Let ABD = x.
ACD  ABD
x
ACB  BCD  ACD
 90  x
∵ AM = MB and AP = PC
∴ MP // BC
∴ APM  ACB
 90  x
∵ NP  AC
∴ APN = 90
∴ NPM
proved in (a)
 in semi-circle
opp. s supp.
given
given
common side
SAS
corr. s,  △s
converse of s in
the same segment
s in the same segment
given
mid-pt. theorem
corr. s, MP // BC
proved in (a)
 APN  APM
 90  (90  x)
x
 ABD
22. (a)
ext. , cyclic quad.
ext. , cyclic quad.
ext.  = int. opp. 
20.
Join OC and let CAB = x.
∠COB = 2∠CAB
= 2x
∵ AD = BD
∴ DBA  DAB
x
∴ CDB  DAB  DBA
Join AE, BF, CG and DH.
EAC  CGE  180
opp. s, cyclic quad.
CGE  180  EAC
In AEFB,
EAB  BFE  180
opp. s, cyclic quad.
BFE  180  EAC
∴ CGE = BFE
BFE = BDH
ext. , cyclic quad.
∴ CDH = CGE
∴ C, G, H and D are concyclic. ext.  = int. opp. 
 xx
 2x
∵ ∠COB = ∠CDB
∴ B, C, D and O are concyclic.
 at centre twice 
at ☉ce
given
base s, isos. △
ext.  of △
converse of s in the
same segment
(b) With the notation in the figure,
21. (a) ∵ BAD + BCD = 90 + 90
= 180
∴ ABCD is a cyclic quadrilateral. opp. s supp.
∵ BAD = 90
given
∴ BD is a diameter of the circle. converse of  in
semi-circle
∵ N is the mid-point of BD.
∴ N is the centre of the circle.
∵ AP = PC
given
∴ NP  AC
line joining centre
to mid-pt. of chord
 chord
∵ BM = CM
∴ OM  BC
given
line joining centre to
mid-pt. of chord  chord
∴ OM is the
perpendicular bisector
of BC.
∴ OM passes through
the centre of the
circumcircle of BCDO. ⊥bisector of chord
passes through centre
28
1 Basic Properties of Circles
For III,
∵ ∠AOB = ∠COD
∴ AB = CD
∴ III is true.
∴ I, II and III are true.
Check Yourself (p. 1.75)
1. (a) 
(b) 
(c) 
(d) 
2.
∵ OM ⊥ AB
∴ BM  AM
(given)
(line from centre  chord bisects chord)
 2 cm
In △OMB,
7.
OB 2  OM 2  BM 2
OB  OM  BM
2
(Pyth. theorem)
2
x  12  ( 2 ) 2
 3
AB  2 AM
 2 2 cm
∵ AB = AC, OM ⊥ AB and ON ⊥ AC
∴ ON = OM
(equal chords, equidistant from centre)
= 1 cm
∴ y 1
3.
 
∵ BC = AB
∴
BC  AB
(equal chords, equal arcs)
x  22
BOC  2BAC
 114
∠ABC + ∠ADC = 180°
ABC  180  68
 112
(opp. s, cyclic quad.)
∠BCP = ∠ABC
(alt. ∠s, AB // DP)
∵ AB = AC, OM ⊥ AB and ON ⊥ AC
∴ ON = OM
(equal chords, equidistant from centre)
x3
x  112
5.
∠ADC + ∠ABC = 180°
(opp. s, cyclic quad.)
ADC  180  110
 70
In △ACD,
ACD  ADC  CAD  180 ( sum of △)
2.
ACD  70  50  180
ACD  60
∠AED + ∠ACD = 180°
(opp. s, cyclic quad.)
x  60  180
x  120
6.
For I,
∵ ∠ACB = ∠ADB
∴ ABCD is a cyclic quadrilateral.
(converse of s in the same segment)
For II,
In △BCD,
∠BCD + ∠BDC + ∠DBC = 180°
( sum of △)
∠BCD + 63° + 31° = 180°
∠BCD = 86°
∵ ∠BCD = ∠EAD
∴ ABCD is a cyclic quadrilateral.
(ext.  = int. opp. )
For III,
∵ ∠DAB + ∠DCB = 108° + 82°
= 190°
 180
∴ ABCD is not a cyclic quadrilateral.
∴ Only I and II are true.
Revision Exercise 1 (p. 1.76)
Level 1
1. ∵ AM = MB
∴ OM  AB
(line joining centre to mid-pt. of chord
 chord)
∵ AN = NC
∴ ON  AC
(line joining centre to mid-pt. of chord
 chord)
∴ y  90
( at centre twice  at ☉ce)
y  2  57
4.
AOB
2
54
x
2
 27
BCA 
( at centre twice  at ⊙ce)
ODC  OBC  BCA
(ext.  of △)
 42  27
 69
y  AOB  ODC
y  69  54
 15
(ext.  of △)


For I,
∠AOC = 2∠ABC
( at centre twice  at ☉ce)
∴ I is true.
For II,
∠BCD = ∠BAD
∵ OC = OD
∴ ∠OCD = ∠ODC
∴ ∠BAD = ∠ADC
∴ II is true.
(s in the same segment)
(radii)
(base s, isos. △)
(vert. opp. s)
(equal s, equal chords)
3.
AOB
AB

COD
CD
x
8

90 18
x  40
x
2
40

2
 20
y
29
(arcs prop. to s at centre)
( at centre twice  at ⊙ce)
NSS Mathematics in Action (2nd Edition) 5A Full Solutions
4.
∠CDB = ∠CAB
x  30
∵ AM = MB
∴ OM  AB
(s in the same segment)
8.
(line joining centre to mid-pt. of chord
 chord)
∠ABC = ∠ADE
x  86


BC

BAC
BCA
∴
( sum of △)
(arcs prop. to s at ☉ce)
Join BE.
∠AEB = 90°
BED  AED  AEB
x  110  180
x  70
 103  90
 13
BOD  2BED
 2  13
∠ACD = 90°
( in semi-circle)
In △ACD,
∠DAC + ∠ACD + ∠ADC = 180°
( sum of △)
y  90  70  180
7.
FG  OF 2  OG 2
10. In △AED,
∠EDF = ∠AED + ∠EAD
=e+a
(radii)
∠FCD = ∠BAD
( at centre twice  at ☉ce)
(ext.  of △)
(ext. ∠, cyclic quad.)
(property of rectangle)
11. DAE  60  x
∠CDB = ∠CAB = x
(s in the same segment)
EDA  80  x
In △AED,
∠EDA + ∠DAE = ∠AEB
(ext.  of △)
(80  x)  (60  x)  100
(Pyth. theorem)
2
 8 cm
∵ OG  FE
∴ GE = FG
= 8 cm
∴ FE  2  8 cm
( in semi-circle)
=a
In △FCD,
∠CFD + ∠CDF + ∠FCD = 180°
( sum of △)
∠CFD + (e + a) + a = 180°
CFD  180  2a  e
 10  6 cm
2
( sum of △)
 26
In △OCD,
∠ODE = ∠DOC + ∠DCO (ext.  of △)
= 26° + 24°
 50
y  20
Join OF.
OF  OB
1
 BC
2
1
  20 cm
2
 10 cm
OG  AB
 6 cm
In △OGF,
3
BAC
2
(arcs prop. to s at ⊙ce)
9.
(opp. s, cyclic quad.)
∠ADC + ∠ABC = 180°
ACB 
(opp. s, cyclic quad.)
3
65  BAC  BAC  180
2
5
BAC  115
2
BAC  46
AB
y 62

4 32
y  7.75
6.
 65
In △ABC,
ABC  ACB  BAC  180
(ext. ∠, cyclic quad.)
In △ABC,
∠BAC + ∠ABC + ∠ACB = 180°
∠BAC + 86° + 32° = 180°
∠BAC = 62°


ACB AB

BAC BC
3

2
∠AMD = 90°
(ext.  of △)
y  x  90
y  90  30
 60
5.
ABC  ADC  180
ABC  180  115
(given)
(line from centre  chord bisects
chord)
140  2 x  100
2 x  40
x  20
 16 cm
30
1 Basic Properties of Circles
15. AOB  BOD  180
DOC
( at centre twice  at ⊙ce)
2
90

2
 45
( in semi-circle)
ACB  90
In △BCE,
BCE  EBC  CEB  180 ( sum of △)
(adj. s on st. line)
AOB  180  140
 40
AOB
ACB 
( at centre twice  at ☉ce)
2
40

2
 20
CAD  ACB
(alt. s, DA // CB)
 20
In △AKO,
( sum of △)
AKO  KAO  AOK  180
12. DBC 
90  45  CEB  180
CEB  45
13. (a) ∠ABC + ∠OAB = 180°
∠ABC + 65° = 180°
∠ABC = 115°
∠BCD + ∠ABC = 180°
∠BCD + 115° = 180°
BCD  65
(int. ∠s, OA // CB)
AKO  20  40  180
AKO  120
(int. ∠s, AB // DC)
16. (a) ∵ CF = FD
∴ OF  CD
(b)
(given)
(line joining centre to
mid-pt. of chord  chord)
1
CD
2
1
  12 cm
2
 6 cm
In △OFD,
FD 
Join AD.
∠BAD + ∠BCD = 180° (opp. s, cyclic quad.)
(∠OAD + 65°) + 65° = 180°
∠OAD = 50°
∵ OD = OA
(radii)
∴ ∠ODA = ∠OAD
(base s, isos. △)
= 50°
∠ADC + ∠ABC = 180° (opp. s, cyclic quad.)
(∠ODC + 50°) + 115° = 180°
ODC  15
OF  OD 2  FD 2
 ( 48 ) 2  6 2 cm
 12 cm (or 2 3 cm)
(b)
14. (a) AB2 + BC2 = (242 + 72) cm2
= 625 cm2
AC2 = 252 cm2
= 625 cm2
∵ AB2 + BC2 = AC2
∴ ∠ABC = 90°
converse of Pyth. theorem
∴ AC is a diameter
converse of  in
of the circle.
semi-circle
Join AB.
∵ OF = OE, OE  AC
and OF  CD
∴ AC = CD
= 12 cm
∵ OE  AC
1
∴ AE  AC
2
1
  12 cm
2
 6 cm
OB  OD
(b) ∵ M is the mid-point of AC.
∴ M is the centre of the circle ABCD.
∴ MD  MC
(radii)
25
cm
2
 12.5 cm

DN  NC
15

cm
2
 7.5 cm
∴ MN  DC
∵
(proved in (a))
(proved in (a))
(chords equidistant
from centre are equal)
(given)
(line from centre ⊥
chord bisects chord)
(radii)
 48 cm
BE  OB  OE
(given)
 ( 48  12 ) cm
1
Area of △ ABE   BE  AE
2
1
  ( 48  12 )  6 cm 2
2
 10.39 cm 2 (cor. to 2 d.p.)
(line joining centre to mid-pt.
of chord  chord)
In △MDN,
MN  MD 2  DN 2
(Pyth. theorem)
(Pyth. theorem)
 12.5 2  7.5 2 cm
 10 cm
31
NSS Mathematics in Action (2nd Edition) 5A Full Solutions
∵ OM  AB
(by construction)
1
∴ AM  AB (line from centre ⊥ chord bisects chord)
2
1
  37 m
2
 18.5 m
OA = OP = 27 m
(radii)
In △OAM,
17.
Join BD and DC.
Consider △ABD and △ACD.
ABD = 90
ACD = 90
∴ ABD = ACD
AD = AD
AB = AC
∴ △ABD  △ACD
∴ BAD = CAD
∴ AD bisects BAC.
 in semi-circle
 in semi-circle
OM  OA2  AM 2
(Pyth. theorem)
 27 2  18.5 2 m
common side
given
RHS
corr. s,  △s
18. (a) In △ABC,
ABC  180  BCA  BAC
 386.75 m
∴ Height of the arch  OP  OM
 (27  386.75 ) m
 7.33 m (cor. to 3 sig. fig.)
22.
 sum of △
 180  28  36
 116
∵ ABC  ADE
∴ ABCD is a cyclic quadrilateral. ext.  = int. opp. 
Join OA and OC.
Draw a line OM such that OM  AD.
∵ OM  AD
(by construction)
∴ BM = MC
(line from centre  chord bisects
chord)
Let OM = a cm and BM = MC = b cm.
In △OAM,
(b) In △ACD,
CAD  ACD  ADE
(ext.  of △)
CAD  116  54
 62
(s in the same segment)
CBD  CAD
 62
19.
OM 2  AM 2  OA2


(Pyth. theorem)
a 2  (b  3) 2  7 2  (1)
In △OMC,
EBD DE
arcs prop. to s at ⊙ce

1
DAC
CD
∴ EBD  DAC
i.e. ∠PBQ = ∠PAQ
∴ P, A, B and Q are concyclic. converse of s in the
same segment
OM 2  MC 2  OC 2
a b 5
2
2
(Pyth. theorem)
2
a 2  25  b 2  (2)
By substituting (2) into (1), we have
(25  b 2 )  (b  3) 2  7 2
20. Draw a line segment PQ in rectangle ABCD, then draw
another line segment RS which is perpendicular to PQ as
shown in the following figure.
(or any other reasonable answers)
25  b 2  b 2  6b  9  49
6b  15
b  2.5
∴ BC = BM + MC
= (2.5 + 2.5) cm
 5 cm
23. (a) ∠ADC = 90°
( in semi-circle)
In △ACD,
∠DCA + ∠ADC + ∠CAD = 180° ( sum of △)
∠DCA + 90° + 54° = 180°
∠DCA = 36°
∠BAC + ∠APC = ∠DCA
(ext.  of △)
∠BAC + 20° = 36°
BAC  16
Level 2
21.
(b) ∠BDC = ∠BAC
(s in the same segment)
= 16°
In △CDK,
∠AKD = ∠CDK + ∠DCK
(ext.  of △)
= 16° + 36°
 52
Let P be the maximum point of the arch, and M be a point
on AB such that OP ⊥ AB.
Join OA and OP.
32
1 Basic Properties of Circles
24. ∵ OD = DE
(given)
∴ ∠DOE = ∠DEO
(base s, isos. △)
Let ∠DOE = ∠DEO = a.
DOA
( at centre twice  at ☉ce)
DCA 
2
a

2
ACB = 90°
( in semi-circle)
In △ECB,
BEC  ECB  EBC  180 ( sum of △)
∴
27.
Join MN.
ABM = MNC
(ext. , cyclic quad.)
AEM = MND
(ext. , cyclic quad.)
∵ MNC + MND = 180
(adj. s on st. line)
∴ ABM  AEM  180
In ABME,
BAE  ABM  AEM  BME  (4  2)  180
( sum of polygon)
65  180  BME  360
a

a    90   51  180
2


3a
 39
2
a  26
BEC  26
BME  115
28.
25.
Join BE.
Join OD.
∠OAD = ∠AOB
(alt. ∠s, AD // BO)
= 24°
∵ OD = OA
(radii)
∴ ∠ODA = ∠OAD
(base s, isos. △)
= 24°
In △ODA,
∠DOA + ∠OAD + ∠ODA = 180°
( sum of △)
∠DOA + 24° + 24° = 180°
∠DOA = 132°
BOD
BCD 
( at centre twice  at ☉ce)
2
132  24

2
 78
26. (a) ∠CFE = ∠ABC
= 102°
∠CDE + ∠CFE = 180°
∠CDE + 102° = 180°
CDE  78


BEC BC

CAD CD
BEC 3

28
2
BEC  42
(arcs prop. to s at ⊙ce)


DBE DE
(arcs prop. to s at ⊙ce)

CAD CD
DBE 4

28
2
DBE  56
In △BEK,
BKE  KEB  KBE  180 ( sum of △)
BKE  42  56  180
BKE  82
(ext. ∠, cyclic quad.)
(opp. s, cyclic quad.)
29. (a)
(b) ∠COE = 2∠CDE
( at centre twice  at ☉ce)
 2  78
= 156°
In ABOE,
∠BAF + ∠ABC + ∠COE + ∠OEF = (4  2)  180
( sum of polygon)
∠BAF + 102° + 156° + 38° = 360°
BAF  64


BDC
BC

CAE
CDE
x
4

48 3  1
x  48
(arcs prop. to s at ☉ce)
(b)
Join AB and AD.
∠BAC = ∠BDC
(s in the same segment)
= 48°
In ABCE,
∠BCE + ∠BAE = 180° (opp. s, cyclic quad.)
∠BCE + (48° + 48°) = 180°
∠BCE = 84°
33
NSS Mathematics in Action (2nd Edition) 5A Full Solutions


(b) Let ABD = x.
CAD
CD

(arcs prop. to s at ☉ce)
CAE
CDE
CAD 3

48
4
CAD  36
∠BAD = ∠BAC + ∠CAD
= 48° + 36°
= 84°




BAE

BCE
BAD

84
84
BCD
BAE
BCD
∴
(arcs prop. to s at ⊙ce)
 
BAE : BCD  1 : 1
30. (a) Consider △PAD and △PCB.
APD = CPB
PAD = PCB
PDA = PBC
∴ △PAD ~ △PCB
(b) Consider △AKB and △DKC.
AKB = DKC
BAK = CDK
ABK = DCK
∴ △AKB ~ △DKC
common angle
ext. , cyclic quad.
ext. , cyclic quad.
AAA
vert. opp. s
s in the same segment
s in the same segment
AAA
33. (a)
(c) ∵ △PAD ~ △PCB
(proved in (a))
PA PD

∴
(corr. sides, ~ △s)
PC PB
6 cm
8 cm

8 cm  DC (6  10) cm
12 cm  8 cm  DC
Join OC.
DC  4 cm
∵ △AKB ~ △DKC
AB BK

∴
DC CK
10 cm
BK

4 cm
3 cm
BK  7.5 cm
31. NBP  MDP
BNP  180  NBP  NPB
 180  MDP  DPM
 DMP
 NMC
∴ QM  QN
32. (a) ∵ CD  CE
∴ ∠CDE = ∠CED
∠CDE = ∠ABC
∴ ∠ABE = ∠AEB
∴ AB  AE
i.e. △ABE is an isosceles
triangle.


DBC DC
(arcs prop. to s at ⊙ce)

1
ABD AD
∴ DBC  ABD
x
ABE  ABD  DBC
 2x
∵ AE = AB
(proved in (a))
∴ AEB  ABE
(base s, isos. △)
 2x
∵ CD = CE
(given)
∴ CDE  CED
(base s, isos. △)
 2x
( in semi-circle)
BDC  90
In △BDE,
DBE  BDE  BED  180 ( sum of △)
x  (90  2 x)  2 x  180
5 x  90
x  18
In △ABD,
BAD  ABD  BDE
(ext.  of △)
BAD  18  90  2(18)
BAD  108
∵


AOC
ABC
(arcs prop. to s at centre)

AOD
ABCD
AOC
11

90
2 11
1
AOC   90
2
 45
∵ OA = OC
(radii)
∴ OAC  OCA (base s, isos. △)
In △OAC,
OAC  OCA  AOC  180
( sum of △)
2OAC  45  180
OAC  67.5
(proved in (b))
(corr. sides, ~ △s)
ext. , cyclic quad.
 sum of △
 sum of △
vert. opp.s
sides opp. equal s
(b)
given
base s, isos. △
ext. ∠, cyclic quad.
sides opp. equal s
 
Join OB and AD.
∵ AB  BC
∴ AOB = BOC
34
(equal arcs, equal s)
1 Basic Properties of Circles
AOC
2
45

2
 22.5
AOB
ADB 
2
22.5

2
 11.25
∴
37. (a)
AOB 
( at centre twice  at ⊙ce)
Join BE.
In BCDE,
∠DEB + ∠BCD = 180°
(opp. s, cyclic quad.)
∠DEB + 80° = 180°
∠DEB = 100°
AOB
AEB 
( at centre twice  at ☉ce)
2
70

2
 35
∠AED = ∠AEB + ∠DEB
= 35° + 100°
 135


CAD CD
(arcs prop. to s at ⊙ce)

ADB
AB
CAD 2

11.25 1
CAD  22.5
In △ADE,
AED  EAD  ADE  180
( sum of △)
AED  22.5  11.25  180
AED  146.25
34. BDC = ABD
DEF = ABD
KAE = DEF
∴ KDC = KAF
∴ A, K, D and F are concyclic.
35. (a)
ABC  ADC  90
EBF  180  ABC
 90
EDF  180  ADC
 90
∵ EBF  EDF
∴ BEFD is a cyclic
quadrilateral.
(b)
alt. s, BA // CF
ext. , cyclic quad.
corr. s, CA // DE
With the notation in the figure,
∠EMO = ∠AEB + ∠OAE (ext.  of △)
= 35° + a
∠DEB + ∠EBC = 180°
(int. ∠s, ED // BC)
100° + ∠EBC = 180°
∠EBC = 80°
OBM  EBC  OBC
 80  b
EMO  AOB  OBM (ext.  of △)
35  a  70  (80  b)
35  a  150  b
∴ a  b  115
ext.  = int. opp. 
 in semi-circle
adj. s on st. line
adj. s on st. line
converse of s
in the same segment
(b) BDC  BAC  35
(s in the same segment)
DBF  DEF  27
(s in the same segment)
In △BDF,
( sum of △)
DBF  BDF  BFD  180
38. (a)
 50
In △ACE,
EAC  ECA  AEC  180  sum of △
37  13  AEC  180
27  (35  90)  BFD  180
BFD  28
∴ CFD  28
36. (a) ∵
∴
∵
∴
∴
 
 AD  DC
 AD  DC
 AD  DE
 DE  DC
DEC  DCE
opp. s of // gram
ADC  ABC
∵
given
equal arcs, equal chords
given
AEC  130
ADC  AEC  50  130
 180
∴ A, E, C and D are concyclic. opp. s supp.
(b)
base s, isos. △
(b) ∵ AD = ED
given
∴ DAE  DEA
base s, isos. △
DEC  DCE
proved in (a)
BCD  DAE  180 opp. s, cyclic quad.
(BCE  DCE)  DAE  180
BCE  180  DAE  DCE
 180  DEA  DEC
 CEB
∴ BC = BE
adj. s on st. line
sides opp. equal s
35
Join DE.
CDE  EAC
 37
∵ CE = CD
∴ CED  CDE
 37
(s in the same segment)
(given)
(base s, isos. △)
NSS Mathematics in Action (2nd Edition) 5A Full Solutions
40. (a) ∠BAC = 90°
 in semi-circle
∠ACD = 90°
 in semi-circle
∵ ∠BAC + ∠ACD = 90° + 90°
= 180°
∴ AB // CD
int. ∠s supp.
In △CDE,
ECD  CDE  CED  180 ( sum of △)
ECD  37  37  180
ECD  106
(int. s, BA // CD)
BCD  ABC  180
(b) (i)
(ECB  ECD )  50  180
ECB  106  50  180
ECB  24
39. (a) ∠BAC = ∠ABC = ∠ACB = 60° prop. of equil. △
∠CDE = ∠ABC = 60°
ext. ∠, cyclic quad.
∠CED = ∠BAC = 60°
ext. ∠, cyclic quad.
∵ ∠CDE = ∠CED = ∠DCE
∴ △CDE is an equilateral
triangle.
(b)


DOE DE

(arcs prop. to s at centre)
COE
CE
DOE 2

2a
1
DOE  4a
In △COD,
∠CDO + ∠DCO + ∠COD = 180° ( sum of △)
a + a + (2a + 4a) = 180°
a = 22.5°
∴ ABC  22.5
Join CM such that CM ⊥ AB, and CM intersects
DE at G.
Draw a line from O to AC such that OF  AC.
In △ACM,
CM
sin MAC 
AC
CM
sin 60 
AC
3
CM

2
10 cm
(ii) ∠ACF = 90°
∠DOF = 4a
 4  22.5
= 90°
∵ ∠ACF = ∠DOF
∴ ACFO is a cyclic
quadrilateral.
 in semi-circle
ext.   int. opp. 
(c) ∵ ∠ACF = 90°
∴ AF is a diameter of the circle
converse of 
passing through A, C, F and O. in semi-circle
(∵ AC = AB = 10 cm)
CM  5 3 cm
In △ACM,
∠MCA + ∠CAM + ∠AMC = 180°
( sum of △)
∠MCA + 60° + 90° = 180°
∠MCA = 30°
∵ OF  AC
(by construction)
1
∴ AF  AD
(line from centre ⊥ chord
2
bisects chord)
 4 cm
FC  AC  AF
41. (a) Let ∠DCE = a.
∠DAB = ∠DCE = a
ext. ∠, cyclic
quad.
 at centre twice
 at ☉ce
opp. s, cyclic
quad.
∠DOB = 2∠DAB
= 2a
∠BED + ∠DOB = 180°
∠BED + 2a = 180°
∠BED = 180  2a
In △CDE,
∠CDE + ∠DEC + ∠DCE = 180°  sum of △
∠CDE + (180°  2a) + a = 180°
∠CDE = a
∵ ∠CDE = ∠DCE
∴ △CDE is an isosceles
sides opp. equal s
triangle.
 (10  4) cm
 6 cm
In △COF,
FC
OC
6 cm
cos 30 
OC
cos OCF 
3 6 cm

2
OC
12
OC 
cm
3
 4 3 cm
OM  CM  OC
 (5 3  4 3 ) cm
 3 cm
∴ The shortest distance from O on AB is
Let ∠ABC = a.
∠BCD = ∠ABC = a
(alt. ∠s, AB // CD)
∠EDC = ∠BCD = a (alt. ∠s, BC // DE)
∠ADC = ∠ABC = a
(s in the same segment)
∠COE = 2∠CDE ( at centre twice  at ☉ce)
= 2a
3 cm.
36
1 Basic Properties of Circles
(b) (i)
∠ODC = ∠EDC
Consider △ALM and
△CNM.
1
AB
AL
 2
1
CN
CD
2
AB

CD
AM

CM
LAM = NCM
∴ △ALM ~ △CNM
(angle bisector)
=a
∠OBE + ∠EDO = 180° (opp. s, cyclic quad.)
∠OBE + (a + a) = 180°
∠OBE = 180  2a
∠ABO = ∠OBE
(angle bisector)
= 180°  2a
∵ ∠ABC = ∠EDC
(ext. , cyclic quad.)
∴
∠ABO + ∠OBE = ∠EDC
(180°  2a) + (180°  2a) = a
5a = 360°
a = 72°
∴ EAB  a
from (a)(i)
s in the same segment
ratio of 2 sides, inc. 
(b)
 72
(ii)
Join OM, OR and OW.
∵ PM = MQ
∴ OM  PQ
Join OA.
∠ODA + ∠ODE = 180° (adj. s on st. line)
∠ODA + 2a = 180°
∠ODA = 180°  2(72°)
= 36°
∵ OA = OD
(radii)
∴ ∠OAD = ∠ODA
(base s, isos. △)
= 36°
∠OAB = ∠EAB  ∠OAD
= 72°  36°
= 36°
∵ ∠OAB = ∠OAD
∴ OA is the angle bisector of ∠EAB.
42. (a) (i)
Consider △ABM and
△CDM.
ABM = CDM
BAM = DCM
AMB = CMD
∴ △ABM ~ △CDM
AB
AM

∴
CD CM
AB
CD

i.e.
AM
CM
OL  AB
AL = LB
1
 AB
2
∵ ON  CD
∴ CN = ND
1
 CD
2
Join LM and NM.
(ii) ∵
∴
given
line joining centre to
mid-pt. of chord  chord
RLO  RMO
∵
 90  90
 180
∴ RLOM is a cyclic
quadrilateral.
ROM = RLM
∵ WNO  WMO
 90  90
opp. s supp.
s in the same segment
 180
∴ MONW is a cyclic
quadrilateral.
WOM = WNM
∵ △ALM ~ △CNM
∴ ALM = CNM
∴ ROM = WOM
s in the same segment
s in the same segment
vert. opp. s
AAA
opp. s supp.
s in the same segment
proved in (a)(ii)
corr. s, ~ △s
(c) Consider △ROM and
△WOM.
ROM  WOM
OM  OM
OMR  OMW  90
∴ △ROM  △WOM
∴ RM = WM
∴ M is also the mid-point
of RW.
corr. sides, ~ △s
given
line from centre 
chord bisects chord
proved in (b)
common side
proved in (b)
ASA
corr. sides,  △s
Multiple Choice Questions (p. 1.83)
given
line from centre 
chord bisects chord
1.
Answer: B
∵ OC  AB
∴ CB = AC = 2 cm
(given)
(line from centre  chord
bisects chord)
In △OAC,
OC  OA2  AC 2
 4  2 cm
2
 12 cm
37
2
(Pyth. theorem)
NSS Mathematics in Action (2nd Edition) 5A Full Solutions
For III,
ADB  ACB  90
∵ AC = CB
∴ CBA = CAB
In △ABC,
CBA  CAB  ACB  180
1
 AB  OC
2
1
  (2  2)  12 cm 2
2
Area of △AOB 
 2 12 cm 2
 4 3 cm 2
2.
Answer: B
For I,
∵ CQ = QD
∴ PQ  CD
5.
∴ I is true.
For II,
∵ AB = CD, OP  AB and OQ  CD
∴ OP = OQ
(equal chords, equidistant from centre)
1
∴ OQ  PQ
2
∴ II is true.
For III,
∵ OP  AB
(given)
1
∴ AP  AB (line from centre  chord bisects chord)
2
1
  12 cm
2
 6 cm
In △AOP,
4.


6.
(Pyth. theorem)

Answer: B
∠BCD + ∠BAD = 180°
(opp. s, cyclic quad.)
∠BCD + 100° = 180°
∠BCD = 80°
In △BCD,
∠CBD + ∠BCD + ∠BDC = 180°
( sum of △)
∠CBD + 80° + 70° = 180°
∠CBD = 30°
∵ AD = CD
∴
 


AD  CD
ABD
AD

CBD
CD
ABD
1
30
ABD  30
7.

  
 

ADC  AD  DC
1
1
CB  CB
2
2
 CB
∴ AC  CB
∴ AC : CB = 1 : 1
∴ II must be true.
(arcs prop. to s at ☉ce)
DA  6 cm
DC : CB  DC : CB in general.
∴ I may not be true.
For II,

ABC
ADC
DA 4 cm  10 cm
Answer: B
In △ABE,
∠EAB + ∠ABE = ∠AED (ext.  of △)
∠EAB + 32° = 80°
∠EAB = 48°
∠CDB = ∠CAB
(s in the same segment)
= 48°
In △CDB,
∠DCB + ∠CDB + ∠DBC = 180°
( sum of △)
∠DCB + 48° + 55° = 180°
DCB  77
∵

(opp. s, cyclic quad.)
DA 4 cm 120

5 cm
60
2
Answer : D
For I,


M inor AC
 8 cm
∴ III is not true.
∴ The answer is B.
3.
Answer: B
∠ADC + ∠ABC = 180°
∠ADC + 120° = 180°
∠ADC = 60°
M ajor AC
 10  6 cm
2
( sum of △)
2CAB  90  180
CAB  45
∴ ADB : CAB  90 : 45  2 : 1
∴ III must be true.
∴ The answer is D.
(given)
(line joining centre to mid-pt. of
chord  chord)
OP  OA2  AP 2
( in semi-circle)
(proved in II)
(base s, isos. △)
(equal arcs, equal chords)
38
(equal chords, equal arcs)
(arcs prop. to s at ☉ce)
Answer: B
∠ADC + ∠ABC = 180°
(opp. s, cyclic quad.)
∠ADC = 180  x
∠ACD + ∠AED = 180°
(opp. s, cyclic quad.)
∠ACD = 180  y
In △ACD,
∠ACD + ∠ADC + ∠CAD = 180°
( sum of △)
(180°  y) + (180°  x) + 45° = 180°
x  y  2 2 5
1 Basic Properties of Circles
8.
Answer: A
11. Answer: B
DCB  BAD  180
DCB  180  y
∵ OD = OC
∴ ODC  OCD
Join BD.
ADB  90
COD
CBD 
2
48

2
 24
x  ADB  CBD
 90  24
 114
9.
(opp. s, cyclic quad.)
(radii)
(base s, isos. △)
 180  y
DOC  ABO
(corr. s, BA // OD)
x
In △OCD,
∠DOC + ∠ODC + ∠OCD = 180°
( sum of △)
x  (180  y)  (180  y)  180
( in semi-circle)
( at centre twice  at ⊙ce)
x  2 y  180
∴ B must be true.
(ext.  of △)
12. Answer: D
For I,
ADE + DEB = 180
(int. s, AD // BE)
∵ ABE = DEB
(given)
∴ ADE + ABE = 180
∴ ABED is a cyclic quadrilateral.
(opp. s supp.)
∴ I is true.
For II,
DEB = EFC
(corr. s, BE // CF)
∵ ABE = DEB
(given)
∴ ABE = EFC
∴ BCFE is a cyclic quadrilateral. (ext.  = int. opp. )
∴ II is true.
For III,
ABE + BAD = 180
(int. s, AD // BE)
∵ DFC = ABE
(ext. , cyclic quad.)
∴ DFC + BAD = 180
∴ ACFD is a cyclic quadrilateral.
(opp. s supp.)
∴ III is true.
∴ The answer is D.
Answer: D
Join AQ.
∠AQB = 90°
∠AQP = ∠AQB  ∠PQB
= 90°  44°
= 46°
∠QAB = ∠QPB
= 72°
∴ y = ∠QAB + ∠AQP
= 72° + 46°
 118
( in semi-circle)
(s in the same segment)
(ext.  of △)
13. Answer: A
∵ O is the circumcentre of △ABC.
∴ There is a circle with centre O passing through A, B
and C.
AOB
∴ ACB 
( at centre twice  at ☉ce)
2
110

2
 55
∵ ∠ACB = ∠ADB
∴ ABCD is a cyclic quadrilateral.
(converse of s in the same segment)
∠DAB = ∠DCB
(s in the same segment)
= 38°
∵ OA = OB
(radii)
∴ ∠OAB = ∠OBA
(base s, isos. △)
In △AOB,
∠OAB + ∠OBA + ∠AOB = 180°
( sum of △)
2∠OAB + 110° = 180°
∠OAB = 35°
∴ ∠OAD  DAB  OAB
10. Answer: C
ACB  ADB (s in the same segment)
x
BDC  BAC (s in the same segment)

In △CDP,
CPD  PDC  DCP  180
( sum of △)
y  ( x   )  ( z  x)  180
  180  2 x  y  z
∴ C must be true.
 38  35
 3
39
NSS Mathematics in Action (2nd Edition) 5A Full Solutions
HKMO (p. 1.85)
1. ∠AOB = 2∠ACB
( at centre twice  at ☉ce)
= 2 
∠OBC = ∠AOB
(alt. ∠s, AO // BC)
= 2 
In △DBC,
∠BCD + ∠CDB + ∠DBC = 180°
( sum of △)
   111  2   180
4.
Join BE.
∠BEC = 90°
∠BEA + ∠BEC = 180°
∠BEA + 90° = 180°
∠BEA = 90°
3   69
  23
( in semi-circle)
(adj. s on st. line)
BE 2  AE 2  AB 2
2.
BE 
(Pyth. theorem)
AB 2  AE 2
 82  4 2
Join OE.
∵ OE = AE
∴ ∠EOA = ∠EAO
= 20°
∠OED = ∠EOA + ∠EAO
= 20° + 20°
= 40°
∵ OD = OE
∴ ∠ODE = ∠OED
= 40°
In △AOD,
∠DOC = ∠DAO + ∠ODA
x° = 20° + 40°
x  60
3.
 48
Let ∠BEF = b.
In △EBF,
∠EBF + ∠BEF + ∠BFE = 180°
∠EBF + b + 90° = 180°
∠EBF = 90  b
∠CEF  BEC  BEF
 90  b
∠ECF  BFE  CEF
(base s, isos. △)
(ext.  of △)
(radii)
(base s, isos. △)
(ext.  of △)
 90  (90  b)
b
Consider △EBF and △CEF.
∠BFE = ∠EFC = 90°
∠EBF = ∠CEF = 90  b
∠BEF = ∠ECF = b
∴ △EBF ~ △CEF
(AAA)
Let BF = 3x units and FC = x units.
EF BF

(corr. sides, ~△s)
CF EF
2
2
EF  3x
(ext.  of △)
Let ∠CAD = b.
In △CAD,
∠ACD + ∠CAD + ∠ADC = 180°
( sum of △)
∠ACD + b + 90° = 180°
∠ACD = 90  b
∠BAC = 90°
( in semi-circle)


BAC


CAD
∠BAD
 90  b
(ext.  of △)
∠ABD  ADC  BAD
EF  3 x
EC CF

BE EF
EC
x

48
3x
 90  (90  b)
b
Consider △ABD and △CAD.
∠BDA = ∠ADC = 90°
∠BAD = ∠ACD = 90  b
∠ABD = ∠CAD = b
∴ △ABD ~ △CAD
AD CD
∴ BD  AD
a 4

1 a
a2  4
a  2 or  2( rejected )
( sum of △)
EC 
(corr. sides, ~△s)
48
3
4
Exam Focus
(AAA)
Exam-type Questions (p. 1.87)
1. (a) (i) ∵ H is the orthocentre of △ABC.
∴ ∠APB = 90°
∴ AB is a diameter of the circle passing
through A, P and B.
(converse of  in semi-circle)
∵ E is the mid-point of AB.
∴ E is the centre of the circle.
∴ E is the circumcentre of △APB.
(corr. sides, ~△s)
40
1 Basic Properties of Circles
(ii) ∵
∴
∴
E is the circumcentre proved in (a)(i)
of △APB.
EB = EP
∠EBP = ∠EPB
base s, isos. △
BP
BA

BC BQ
BP 2 BC

BC
BQ
(b)
2 BC 2  BP  BQ
1
BC 2   BP  BQ
2
Join EF and DF.
∵ G is the centroid of △ABC.
∴ AF = FC and CD = DB
∵ AF = FC and AE = EB
∴ FE // CB
∵ CF = FA and CD = DB
∴ FD // AB
∴ FDBE is a parallelogram.
∠EFD = ∠EBP
∠EBP = ∠EPB
∴ ∠EFD = ∠EPB
∴ D, P, E and F are concyclic.
2.
(b) From (a),
1
BC 2   BP  BQ
2
1
 BP  BQ
BC 2
 2
BC
BC
1 BP
BC  
 BQ
2 BC
1 2
   BQ
2 3
1
 BQ
3
∴ BQ = 3BC
Let BC = a, then
2
BP  a and BQ = 3a.
3
mid-pt. theorem
mid-pt. theorem
opp. s of // gram
proved in (a)(ii)
ext.   int. opp. 
(a) Consider △ADE and △CBE.
∠ADE = ∠CBE
s in the same segment
DE = BE
given
∠DEA = ∠BEC
vert. opp. s
∴ △ADE  △CBE ASA
(b)
With the notation in the figure,
1
 BP  CE
Area of △ BPC
 2
Area of △ PQC 1
 PQ  CE
2
BP

PQ
2
a
 3
2
3a  a
3
2

7
∴ Area of △BPC : area of △PQC  2 : 7
Let F be a point on AC such that EF ⊥ AC.
∵ AE = CE
(corr. sides, △s)
and EF ⊥ AC
∴ AF = FC
(prop. of isos. △)
∵ EF is the perpendicular bisector of AC.
∴ The straight line passing through E and F passes
through the centre of the circle.
(⊥bisector of chord passes through centre)
∴ The centroid of △AEC lies on the straight line
passing through the point E and the centre of the
circle.
3.
corr. sides, ~△s
4.
(a) Consider △ABP and △QBC.
∠BAP = ∠BQC
s in the same segment
∠ABP = ∠QBC
common angle
∠APB = 90°
 in semi-circle
∠APQ = 180°  ∠APB adj. s on st. line
= 90°
∠ACQ = ∠APQ
s in the same segment
= 90°
∠QCB = 180°  ∠ACQ adj. s on st. line
= 90°
∴ ∠APB = ∠QCB
∴ △ABP ~ △QBC
AAA
(a) ∵ I is the incentre of △ABC.
∴ ∠ACD = ∠DCB
∠FIE = ∠DCB
alt. ∠s, EI // BC
∠ABD = ∠ACD
s in the same
segment
∴ ∠FIE = ∠FBE
∴ B, I, F and E are concyclic. converse of s in
the same segment
(b) ∵ I is the incentre of △ABC.
∴ ∠FBI = ∠IBC
∠EIB = ∠IBC
alt. ∠s, EI // BC
∠EFB = ∠EIB
s in the same
segment
∴ ∠EFB = ∠FBI
∴ EF // BI
alt. ∠s equal
41
NSS Mathematics in Action (2nd Edition) 5A Full Solutions
5.
Answer: A
7.
Join EC.
In △ABC,
∠BAC + ∠ACB + ∠ABC = 180°
∠BAC + 47° + 70° = 180°
∠BAC = 63°
(given)
(line joining centre to mid-pt. of chord
 chord)
(radii)
OA = OC
= (8 + 9) cm
= 17 cm
In △AOD,
AD 2  OA2  OD 2
( sum of △)
(Pyth. theorem)
AD  17  8 cm
2
2
 15 cm


OD
OA
8

17
AOD  61.9275 
Area of the shaded region ADC
= Area of sector AOC  area of △AOD
AOD 1


  (17 2 ) 
  15  8 cm 2
360
2


cos AOD 
BCE
BE
(arcs prop. to s at ☉ce)

BAC
BEC
BCE
2

63
2 1
BCE  42
ACE  ADE  180 (opp.s, cyclic quad.)
(ACB  BCE )  ADE  180
(47  42)  ADE  180
ADE  91
6.
Answer: A
∵ AD = BD
∴ OD  AB
 96 cm 2 (cor. to the nearest cm 2 )
8.
Answer: B
Join AD.
AOD
( at centre twice  at ☉ce)
2
100

2
 50
∵ BD is the angle bisector of ∠ABC.
∴ ∠CBD = ∠ABD
= 50°
∵ OA = OD
(radii)
∴ ∠OAD = ∠ODA
(base s, isos. △)
In △AOD,
∠ODA + ∠OAD + ∠AOD = 180°
( sum of △)
2∠ODA + 100° = 180°
∠ODA = 40°
∠ABC + ∠ADC = 180°
(opp. s, cyclic quad.)
(50° + 50°) + (14° + 40° + ∠BDO) = 180°
BDO  26
ABD 
Answer: A
For I,
∵ BD is the angle bisector of ∠OBC.
∴ ∠CBE = ∠OBD
∵ OD = OB
(radii)
∴ ∠ODB = ∠OBD (base s, isos. △)
∴ ∠ODE = ∠CBE
∠OED = ∠CEB
(vert. opp. s)
In △DOE,
∠DOE  180  OED  ODE ( sum of △)
 180  CEB  CBE
( sum of △)
 BCE
∴ △OED ~ △CEB
(AAA)
∴ I is true.
For II,
∵ There is not sufficient data to show
△OBD ~ △CBD.
∴ II may not be true.
For III,
∵ There is not sufficient data to show
△BOE ~ △DCE.
∴ III may not be true.
∴ The answer is A.
42