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2 Basic Properties of Circles
27
2 Basic Properties of Circles
Review Exercise 2 (p. 2.7)
1. (a)
°=
°=°+°+°+
75
3606513090
x
x (∠s at a pt.)
(b) )// s, (corr.60 RSPQCBP ∠°=∠
°=
∠∠=
60
s) opp. vert.( CBPy
2. (a)
°=∠
°=°+∠
55
180125
PQR
PQR (adj. ∠s on st. line)
°=
°=°+
∠=∠+
85
14055
x
x
PRBPQRx (ext. ∠ of △)
(b)
°=
°=°+°+
50
1809040
a
a (∠ sum of △)
°=
°+°=
°+=
84
3450
34ab (ext. ∠ of △)
3. Yes, EF is parallel to GH.
∠RSQ = 110° (corr. ∠s, AB // CD)
∵ °=°+°=∠+∠ 18011070RSQPQS
∴ EF // GH (int. ∠s supp.)
4. ∠BCD = 65° alt. ∠s, AB // CD
°=°+∠+∠ 18050BDCBCD ∠ sum of △
°=
°−°−°=∠
65
6550180BDC
∵ ∠BCD = ∠BDC = 65°
∴ BD = BC sides opp. equal ∠s
∴ △BCD is an isosceles
triangle.
5. ∵ △PQR is an equilateral triangle.
∴ ∠PRQ = 60°
∵ PR is the median of QS in △PQS.
∴ RS = QR = PR
∴ ∠RPS = ∠RSP (base ∠s, isos. △)
°=∠
°=∠
∠=∠+∠
30
602
RPS
RPS
PRQRSPRPS (ext. ∠ of △)
6. (a) In △ACE and △DCB,
2
5
2
32
=
+=
DC
AC
2
5
3
5.7
3
5.52
=
=
+=
CB
CE
∴ CB
CE
DC
AC=
∠ACE = ∠DCB common angle
∴ △ACE ~ △DCB ratio of 2 sides, inc. ∠
(b) ∵ △ACE ~ △DCB proved in (a)
∴
cm 25.6
cm 2
cm 5
cm 5.2
=
=
=
AE
AE
DC
AC
DB
AE
corr. sides, ~ △s
7. (a) In △ABC,
∵ ∠C = 90°
∴
cm 10
cm 5.75.12
theorem)(Pyth.
22
222
=
−=
=+
AC
ABBCAC
(b) In △CAD,
2
22222
cm 676
cm )2410(
=
+=+ ADAC
2
222
cm 676
cm 26
=
=CD
∵ AC2 + AD2 = CD2
∴ ∠CAD is a right angle. (converse of Pyth.
theorem)
(c) Area of quadrilateral ABCD
= area of △ABC + area of △CAD
2
2
2
cm 5.157
cm )1205.37(
cm 10242
1105.7
2
1
=
+=
××+××=
8. (a) In △ABC and △ACD,
°=∠=∠ 90CDABCA given
DACCAB ∠=∠ common angle
CABBCAABC ∠−∠−°=∠ 180 ∠ sum of △
DACCDA ∠−∠−°= 180 proved
ACD∠= ∠ sum of △
∴ △ABC ~ △ACD AAA
(b) (i) In △ABC,
∵ ∠ACB = 90°
∴
cm 17
cm 158 22
222
=
+=
+=
AB
BCACAB (Pyth. theorem)
NSS Mathematics in Action 5A Full Solutions
28
(ii) ∵ △ABC ~ △ACD
∴
cm 17
64
cm 8
cm 8
cm 17
=
=
=
AD
AD
AD
AC
AC
AB(corr. sides, ~ △s)
fig.) sig. 3 to(cor. cm 2.13
cm 17
6417
=
−=
−= ADABBD
9. (a) In △ADE and △CDE,
sidecommon
given
given 90
DEDE
CDAD
DECDEA
=
=
°=∠=∠
∴ △ADE ≅ △CDE RHS
(b) Yes, △ABD and △AED are congruent.
°=∠
°=°+°+∠
°=°+∠+∠
30
1809060
18090
ACB
ACB
BACACB (∠ sum of △)
∵ △ADE ≅ △CDE (proved in (a))
∴
°=
∠=∠
30
ECDEAD (corr. ∠s, ≅ △s)
∴
°=
°−°=∠
30
3060BAD
In △ABD and △AED,
sidecommon
proved 30
given 90
ADAD
EADBAD
AEDABD
=
°=∠=∠
°=∠=∠
∴ △ABD ≅ △AED AAS
Activity
Activity 2.1 (p. 2.22)
2. (b) 2=∠
∠
APB
AOB
(c) No matter where points B and P are,
∠AOB = 2∠APB.
3. (b) 2Reflex
=∠
∠
AQB
AOB
(c) No matter where points B and Q are,
reflex ∠AOB = 2∠AQB.
Activity 2.2 (p. 2.28)
1. (a) ∠AOB (i.e. c) is the angle at the centre subtended by
arc AB.
(b) ∠APB and ∠AQB are the angles at the circumference
in the same segment. They are both subtended by arc
AB.
2. (a) ∠APB = 2
c
( ∠ at centre twice ∠ at ⊙ce )
(b) ∠AQB =
2
c
( ∠ at centre twice ∠ at ⊙ce )
3. ∠APB = ∠AQB
Activity 2.3 (p. 2.34)
1. (a) Yes (b) Yes (c) Yes
2. Yes
Activity 2.4 (p. 2.46)
1. (b) ∠A + ∠C = 180° , ∠B + ∠D = 180°
2. ∠P + ∠R = 180° , ∠Q + ∠S = 180°
3. The sum of the opposite angles of a cyclic quadrilateral is
180°.
Activity 2.5 (p. 2. 53)
1. (a) Yes (b) Yes
2.
3. Yes
Classwork
Classwork (p. 2.11)
1. Element Term
AB • • minor arc
region BEC • • major arc
)
AFB • • chord
region OBFC • • major sector
region AFCEB • • major segment
)AB • • minor segment
region OBEC • • minor sector
2. (a) The two circles with the same centre are
concentric circles .
(b) The circle is the circumcircle of △ABC.
(c) The circle is the inscribed circle of △PQR.
2 Basic Properties of Circles
29
Classwork (p. 2.24)
(a)
°=
°×=
∠=
60
302
2 APBx (∠ at centre twice ∠ at ⊙ce)
(b)
°=
°×=
∠=
50
1002
1
2
1AOBx
(∠ at centre twice ∠ at ⊙ce
)
(c) Reflex
°=
°−°=∠
220
140360AOB (∠s at a pt.)
°=
°×=
∠=
110
2202
1
reflex 2
1AOBx
(∠ at centre twice ∠ at ⊙ce)
Classwork (p. 2.25)
1. (a) °= 90x (∠ in semi-circle)
(b) ∠APB = 90° (∠ in semi-circle)
°=
°=+°+°
°=+°+∠
50
1804090
18040
x
x
xAPB (∠ sum of △)
2. OA = OP (radii)
°=∠
∠=∠
55OPA
OPAOAP
(base ∠s, isos. △)
°=°+°=∠ 903555APB
∴ The line segment joining A and B is a diameter of the
circle. (converse of ∠ in semi-circle)
Classwork (p. 2.36)
(a) ∵ °=∠=∠ 43AOBDOC (given)
∴
4=
=
x
ABDC))
(equal ∠s, equal arcs)
(b) ∵ ))ABCD = (given)
∴ ABCD = (equal arcs, equal chords) ∴ 5=x
(c) ∵ AB = DC (given)
∴ °= 65x (equal chords, equal ∠s)
Classwork (p. 2.39)
(a)
32
)80(5
2
=
°=°
=∠
°
x
x
AB
BC
AOB
x)
) (arcs prop. to ∠s at centre)
3
80
485
cm
=
°
°=
∠
∠=
y
AOB
DOC
AB
y) (arcs prop. to ∠s at centre)
(b)
10
30
506
cm
=
°
°=
∠
∠=
x
AOB
DOC
AB
x) (arcs prop. to ∠s at centre)
°=
°−°−°=
∠−∠−°=∠
100
5030180
180 DOCAOBBOC (adj. ∠s on st. line)
20
30
1006
cm
=
°
°=
∠
∠=
y
AOB
BOC
AB
y) (arcs prop. to ∠s at centre)
(c)
5.7
40
3010
cm
=
°
°=
∠
∠=
x
BEC
CED
BC
x) (arcs prop. to ∠s at ⊙ce)
24
)40(10
6
=
°=°
=∠
°
y
y
BC
AB
BEC
y)
) (arcs prop. to ∠s at ⊙ce)
(d)
°=
°=∠
=∠
∠
36
)48(4
3ACB
CD
AB
DBC
ACB)
) (arcs prop. to ∠s at ⊙ce)
96
3648180
180
=
°−°−°=
∠−∠−°=°
x
ACBDBCx (∠ sum of △)
Classwork (p. 2.47)
(a)
°=
°=°+
°=∠+∠
120
18060
180
x
x
BCDDAB (opp. ∠s, cyclic quad.)
°=
°=°+
°=∠+∠
100
18080
180
y
y
CDAABC (opp. ∠s, cyclic quad.)
NSS Mathematics in Action 5A Full Solutions
30
(b)
°=
°=+°
°=∠+∠
87
18093
180
x
x
BCDDAB (opp. ∠s, cyclic quad.)
°=
∠=∠
113y
CDEABC (ext. ∠, cyclic quad.)
(c)
°=
°+°=
78
3246y (ext. ∠ of △)
°=
=
78
yx (ext. ∠, cyclic quad.)
Classwork (p. 2.55)
1. (a) ∵
°≠
°=
°+°=∠+∠
180
170
70100BCDBAD
∴ A, B, C and D are not concyclic.
(b)
°=∠
°=°+∠
°=∠+∠
80
180100
180
ABC
ABC
FBCABC (adj. ∠s on st. line)
∵ ADEABC ∠≠∠
∴ A, B, C and D are not concyclic.
(c)
°=∠
°=°+°+∠
20
18012040
ABD
ABD (∠ sum of △)
°=
°+°+°+°=∠+∠
180
)7040()5020(ADCABC
∴ A, B, C and D are concyclic. (opp. ∠s supp.)
2. (a) ∵ ∠PSQ =∠PRQ = 50° given
∴ PQRS is a cyclic
quadrilateral. converse of ∠s in the
same segment
°=
°=°+°+
°=∠+∠
60
180)7050(
180
x
x
RSPPQR (opp. ∠s, cyclic quad.)
(b)
°=
°+°=∠
90
6030QPS
∴ ∠QRK = ∠QPS = 90°
∴ PQRS is a cyclic
quadrilateral. ext. ∠ = int. opp. ∠
°=∠
°=°+°+∠
°=∠+∠+∠
50
1809040
180
PRQ
PRQ
QRKPRSPRQ
(adj. ∠s on st. line)
°=
∠=∠
50x
PRQPSQ
(∠s in the same segment)
Quick Practice
Quick Practice 2.1 (p. 2.13)
Join OQ.
OQ = 13 cm (radius)
In △OQN, ∵ ON 2 + NQ2 = OQ2 (Pyth. theorem) ∴
cm 12
cm 513 22
22
=
−=
−= ONOQNQ
∵ ON ⊥ PQ (given) ∴ PN = NQ (line from centre ⊥ chord bisects chord)
cm 24
cm )12(2
2
=
=
=
+=
NQ
NQPNPQ
Quick Practice 2.2 (p. 2.14) ∵ PN = QN (given) ∴ ON ⊥ PQ (line joining centre to mid-pt. of
chord ⊥ chord)
∠ONP = 90°
In △OPN, ∵ ON = PN ∴ ∠OPN = ∠PON (base ∠s, isos. △)
°=∠
°=°+∠
°=∠+∠+∠
45
180902
180
OPN
OPN
ONPPONOPN (∠ sum of △)
Quick Practice 2.3 (p. 2.14) ∵ PN = NR (given) ∴ ON ⊥ PR (line joining centre to mid-pt. of
chord ⊥ chord) ∴ △OPN is a right-angled triangle.
Let r cm be the radius of the circle. ∴ OP = r cm and ON = (r − 1) cm
In △OPN,
13
262
2512
5)1(
22
222
222
=
=
++−=
+−=
+=
r
r
rrr
rr
PNONOP (Pyth. theorem)
∴ The radius of the circle is 13 cm.
cm 12
cm )113(
=
−=ON
Quick Practice 2.4 (p. 2.17) ∵ M and N are the mid-points of PQ and RQ respectively. ∴ OM ⊥ PQ and ON ⊥ RQ (line joining centre to mid-pt.
of chord ⊥ chord)
cm 6
cm )33(
=
+=
+= MQPMPQ
∵ ON = OM (given) ∴ RQ = PQ (chords equidistant from centre are equal)
= 6 cm
2 Basic Properties of Circles
31
∴
cm 3
cm 62
1
2
1
=
×=
= RQRN
Quick Practice 2.5 (p. 2.17)
(a) ∵ AB = CD, OM ⊥ AB given
and ON ⊥ CD
∴ OM = ON equal chords, equidistant
from centre
∠MPN = ∠APC = 90° vert. opp. ∠s
∴ ∠MON = 90° ∠sum of polygon
∵ All four interior angles are equal to 90° and two
adjacent sides are equal.
∴ ONPM is a square.
(b) ∵ ONPM is a square. (proved in (a))
∴ PM = OM = 4 cm
∵ OM ⊥ AB (given) ∴ AM = MB (lines from centre ⊥ chord bisects
chord)
∴
cm 3
cm )411(cm 4
=
−=+
−=+
AP
AP
PMPBPMAP
Quick Practice 2.6 (p. 2.26)
°=
∠°−°=∠
242
pt.) aat s( 118360reflex AOC ∴ °=
°=
∠=
121
2422
reflex 2
x
x
AOCx (∠ at centre twice ∠ at ⊙ce)
∴ °=
°=
∠=
59
1182
2
y
y
AOCy (∠ at centre twice ∠ at ⊙ce)
Quick Practice 2.7 (p. 2.27)
∠CAD = 90° (∠ in semi-circle) ∵ AB = AD (given) ∴ ∠ABD = ∠ADB = y (base ∠s, isos. △)
In △ABD,
°=
°=
°=°+°++
°=∠+∠+∠
25
502
180)4090(
180
y
y
yy
BADADBABD
(∠ sum of △)
In △ACD,
°=
°=°++°
°=∠+∠+∠
65
1802590
180
x
x
ADCACDCAD
(∠ sum of △)
Quick Practice 2.8 (p. 2.27)
∠DOC = 2∠DBC ∠ at centre twice ∠ at ⊙ce
∠ODB = ∠DBC alt. ∠s, OD // BC
In △ODP,
°=∠
°=∠
°=∠+∠
°=∠+∠
30
903
902
90
DBC
DBC
DBCDBC
DOPODP
ext. ∠ of △
°=
°+°=
∠+°=∠
90
3060
60 DBCABC
∴ AC is a diameter of the circle. converse of ∠ in
semi-circle
Quick Practice 2.9 (p. 2.29)
°=
∠=∠
32
BDCBAC
(∠s in the same segment)
°=∠ 90BCD (∠ in semi-circle)
In △ABC,
°=
°−°−°=∠
28
32120180BCA (∠ sum of △) ∴
°=
°−°=∠
62
2890ACD
Quick Practice 2.10 (p. 2.30)
x
DABDEB
=
∠=∠ (∠s in the same segment)
∠AEB = 90° (∠ in semi-circle)
In △ACE,
°=
°=
°=°+
°=+°+°++°
°=∠+∠+∠
8
162
1801642
180)90()50(24
180
x
x
x
xx
AECCAEECA (∠ sum of △)
Quick Practice 2.11 (p. 2.37)
Join OB, OC, OE, OF and OG.
Let ∠AOB = x. ∵ AB = BC = CD = DE = EF = FG = GH (given) ∴ ∠AOB = ∠BOC = ∠COD = ∠DOE =
∠EOF = ∠FOG = ∠GOH = x (equal chords, equal ∠s)
°=
°=
°=∠+∠+∠
40
1203
120
x
x
CODBOCAOB
°=∠
°=∠+°
°=∠+
∠°=∠+∠+∠+
∠+∠+∠+∠+∠
80
360)40(7
3607
pt.) aat s( 360
AOH
AOH
AOHx
AOHGOHFOG
EOFDOECODBOCAOB
NSS Mathematics in Action 5A Full Solutions
32
Quick Practice 2.12 (p. 2.37) ∵ OP = OQ = OR, OP ⊥ AB,
OQ ⊥ BC and OR ⊥ AC given ∴ AB = BC = AC chords equidistant from
centre are equal ∴ )))ACBCAB == equal chords, equal arcs
Quick Practice 2.13 (p.2.40)
Let ∠COD = x. ∵ BC = CD (given) ∴ ∠BOC =∠COD = x (equal chords, equal ∠s)
)
)
BC
AD
BOC
AOD=
∠
∠ (arcs prop. to ∠s at centre)
xAOD7
10=∠
°=
°=
°=+++°
°=∠+∠+∠+∠
5.66
2287
24
3607
10132
360
x
x
xxx
DOACODBOCAOB (∠s at a pt.)
Quick Practice 2.14 (p.2.41)
xACB
CD
AB
DBC
ACB
2
3=∠
=∠
∠)
) (arcs prop. to ∠s at ⊙ce)
In △BCK,
°=
°=
°=+
∠=∠+∠
30
752
5
752
3
x
x
xx
AKBCBKBCK (ext. ∠ of △)
Quick Practice 2.15 (p.2.48) Join DC. ∵ AD = AB (given) ∴
x
ABDADB
=
∠=∠ (base ∠s, isos. △)
∠BDC = 90° (∠ in semi-circle)
°=
°=
°=°+
°=°++°+
°=∠+∠
34
682
1801122
180)90()22(
180
x
x
x
xx
ADCABC
(opp. ∠s, cyclic quad.)
Quick Practice 2.16 (p. 2.48)
∠FAD = x (ext. ∠, cyclic quad.)
In △FAD,
xADF
xADF
FADDFAADF
−°=∠
°=+°+∠
°=∠+∠+∠
137
18043
180 (∠ sum of △)
x
ADFEDC
−°=
∠=∠
137
(vert. opp. ∠s)
In △DCE,
°=
°=
=−°+°
∠=∠+∠
86
1722
)137(35
x
x
xx
BCDEDCCED (ext. ∠ of △)
Quick Practice 2.17 (p. 2.56)
(a) ∵
°≠
°=
°+°=∠+∠
180
210
100110FEBBAF
∴ A, B, E and F are not concyclic.
(b) ∵
°=
°+°=∠
110
2090CDF
∴ BEFCDF
BEF
∠≠∠
°=∠ 100
∴ 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 2.18 (p.2.57) ∵ ∠FAB + ∠BCD = 180° int. ∠s, AF // CD
and ∠BEF = ∠BCD ext. ∠, cyclic quad. ∴ ∠FAB + ∠BEF = 180° ∴ A, B, E and F are concyclic. opp. ∠s supp.
Quick Practice 2.19 (p.2.58)
(a) ∵ △ABC is an equilateral triangle.
∴ ∠ABC = 60° prop. of equil. △
∵ △CDF is an equilateral triangle.
∴ ∠DFC = 60° prop. of equil. △
∵ ∠ABC = ∠DFC
∵ B, C, E and F are concyclic. converse of ∠s in
the same segment
(b) ∵ B, C, E and F are concyclic. (proved in (a))
∴
°=
∠=∠
33
BFCBEC (∠s in the same segment)
∵ ∠BAC = 60° (prop. of equil. △)
∴
°=
°−°=
∠−∠=∠
27
3360
BECBACFCE (ext. ∠ of △)
2 Basic Properties of Circles
33
Further Practice
Further Practice (p. 2.15)
1. (a) ∵ CN = ND (given)
∴ ON ⊥ CD (line joining centre to mid-pt. of
chord ⊥ chord)
∴ ∠ONC = 90°
In △OND,
°=
°−°=
∠−∠=∠
55
3590
ODNONCNOD (ext. ∠ of △)
(b) In △OME,
cm 8
cm 61022
22
222
=
−=
−=
=+
OMOEEM
OEEMOM (Pyth. theorem)
∵ OM ⊥ EF (given)
∴
cm 8=
= EMMF (line from centre ⊥ chord bisects chord)
2. ∵ CD is the perpendicular bisector of the chord AB.
∴ CD is a diameter of the circle.
cm 20
cm )515(
=
+=
+= MDCMCD
Radius of the circle
cm 10
cm 202
1
2
1
=
×=
= CD
Let O be the centre of the circle.
cm 5
cm )510(
cm 10
=
−=
−=
=
MDODOM
OA
In △OAM,
cm) 310(or cm 752
2
cm 75
cm 510
theorem)(Pyth.
22
22
222
=
=
+=
=
−=
−=
+=
AM
MBAMAB
OMOAAM
AMOMOA
Further Practice (p. 2.18)
1. (a) ∵ ON = OM = 4 cm, OM ⊥ AB, ON ⊥ CD (given)
∴ CD = AB (chords equidistant from
= 7 cm centre are equal)
cm 5.3
cm 72
1
=
×=
= NDCN (given)
(b)
cm 4=
= APPB (given)
∴ OP ⊥ AB (line joining centre to mid-pt.
of chord ⊥ chord)
Also, OQ = OP
and OQ ⊥ BC (given)
∴ BC = AB (chords equidistant from
= 8 cm centre are equal)
cm 4
2
1
2
1
=
=
=
=
AB
BCQC
BQQC (line from centre ⊥ chord bisects chord)
(c) ∵ OM ⊥ AB, ON ⊥ CD (given)
∴ MB = AM (line from centre ⊥ chord
bisects chord)
∴
cm10
cm 52
AB
=
×=
and
cm10
cm 52
CD
=
×=
∵ AB = CD, OM ⊥ AB and ON ⊥ CD
∴ ONOM = (equal chords, equidistant
cm 5.2= from centre)
2.
Construct OM and ON such that OM ⊥ PQ and ON ⊥ RS.
Note that MON is a straight line, since PQ // RS.
cm 8
cm 162
1
2
1
=
×=
=
=
PQ
MQPM (line from centre ⊥ chord bisects chord)
∵ PQ = RS, OM ⊥ PQ and ON ⊥ RS
∴
cm 6
cm 122
1
=
×=
= ONOM
centre) from
t equidistan chords, (equal
In △PMO,
cm 10
cm 68 22
22
222
=
+=
+=
+=
OMPMOP
OMPMOP (Pyth. theorem)
∴ Radius of the circle cm 10=
NSS Mathematics in Action 5A Full Solutions
34
Further Practice (p. 2.30)
1. (a) Consider △ABD.
°=∠
°=°+°+°+∠
°=∠+∠+∠
50
18070)2040(
180
ABD
ABD
ADBBADABD (∠ sum of △)
°=
∠=
50
ABDx (∠s in the same segment)
(b)
°=
∠=∠
48
BDCBAC
(∠s in the same segment)
∠ABC = 90° (∠ in semi-circle)
In △ABC,
°=
°=+°+°
°=∠+∠+∠
42
1809048
180
x
x
BCAABCBAC (∠ sum of △)
2.
°=
∠=∠
40
BCDBOD (opp. ∠s of // gram)
°=
°×=
∠=∠
20
402
1
2
1BODBAD (∠ at centre twice ∠ at ⊙ce
)
∴ °=
∠=∠
20p
BADADO
(alt. ∠s, OD // AC)
3. ∠ADC = 90° (∠ in semi-circle)
In △APD,
°=∠
°=∠+°+°+°
°=∠+∠+∠
24
18090)3630(
180
APD
APD
APDPDAPAD (∠ sum of △)
∴ °=∠ 24BPC
Further Practice (p. 2.41)
1.
Join OB.
∵ AB = BC = DE (given)
∴ ∠AOB = ∠BOC = ∠DOE (equal chords, equal ∠s)
= 55°
∵ ))
CDAE = (given)
∴ ∠AOE = ∠COD (equal arcs, equal ∠s)
°=∠
°=∠
°=∠+°+∠+°+°
°=∠+∠
+∠+∠+∠
5.97
1952
360555555
360
COD
COD
CODCOD
EOADOE
CODBOCAOB (∠s at a pt.)
2.
Join PR.
°=∠ 90PRS (∠ in semi-circle)
1
3
21
=
+=
=∠
∠))
RS
PR
SPR
PSR (arcs prop. to ∠s at ⊙ce)
∴ SPRPSR ∠=∠
°=∠
°−°=∠
°=∠+∠+∠
45
901802
180
PSR
PSR
PRSSPRPSR
(∠ sum of △)
Further Practice (p. 2.49)
1. (a)
°=
∠=∠
81
ABCDEC (ext. ∠, cyclic quad.)
°=
°−°−°=
∠−∠−°=∠
34
6581180
180
x
ECDDECCDE (∠ sum of △)
(b)
Join OD.
∵ OD = OA (radii)
∴
°=
∠=∠
18
OADODA (base ∠s, isos. △)
∵ OD = OC (radii)
∴
°=
∠=∠
50
OCDODC (base ∠s, isos. △)
°=
°+°=∠
68
5018ADC
°=
°=°+
°=∠+∠
112
18068
180
x
x
ADCABC (opp. ∠s, cyclic quad.)
2. ∵ BC = CD (given)
∴ ))
CDBC = (equal chords, equal arcs)
)
)
CD
BC
CAD
BAC=
∠
∠ (arcs prop. to ∠s at ⊙ce)
128
=°
∠BAC
°=∠
°=∠
90
28
ACB
BAC
(∠ in semi-circle)
2 Basic Properties of Circles
35
°=
°−°−°=
∠−∠−°=∠
62
9028180
180 ACBBACABC
(∠ sum of △)
°=∠
°=°+∠
°=∠+∠
118
18062
180
ADC
ADC
ABCADC
(opp. ∠s, cyclic quad.)
Exercise
Exercise 2A (p. 2.18)
Level 1
1. ∵ ON ⊥ AB (given)
∴
cm 8
cm 162
1
chord) bisects chord centre from (line
=
×=
⊥= ANBN
Join OB.
Consider △NOB.
cm 10
cm 68 22
22
=
+=
+= ONBNOB
(Pyth. Theorem)
∴ The radius of the circle is 10 cm.
2.
Join OB.
cm 13
cm )85(
=
+=
+=
=
NCON
OCOB
(radii)
∵ AN = NB (given)
∴ ON ⊥ AB (line joining centre to mid-pt. of
chord ⊥ chord)
Consider △ONB.
cm 12
cm 513 22
22
=
−=
−= ONOBNB
(Pyth. theorem)
∵ AN = NB
∴
cm 24
cm 122
=
×=AB
3. ∵ AM = MB (line from centre ⊥ chord bisects chord)
∴
cm 12
cm 62
=
×=AB ∵ ON = OM, OM ⊥ AB and ON ⊥ CD (given)
∴
cm 12
equal) are centre fromt equidistan (chords
=
= ABCD
4. ∵ CN = ND (given)
∴ ON ⊥ CD (line joining centre to mid-pt. of chord
⊥ chord)
∴ ∠ONK = 90°
∵ AM = MB (given)
∴ OM ⊥ AB (line joining centre to mid-pt. of chord
⊥ chord)
∴ ∠OMK = 90°
°=
°−°=
∠−°=∠
137
43180
180 DKBMKN
(adj. ∠s on st. line)
Consider quadrilateral OMKN.
°×−=
∠+∠+∠+∠
180)24(
MKNONKOMKMON
(∠ sum of polygon)
°=
°−°−°−°=∠
43
1379090360MON
5. ∵ AM = MB (given)
∴ OM ⊥ AB (line joining centre to mid-pt. of chord
⊥ chord)
cm 3
cm 62
1
=
×=MB
Consider △OMB.
cm 4
cm 3522
22
=
−=
−= MBOBOM
(Pyth. theorem)
cm 3
cm )47(
=
−=
−= OMMNON
Join OD.
Consider △OND.
cm 5=
= OBOD
(radii)
cm 4
cm 35 22
22
=
−=
−= ONODND
(Pyth. theorem)
∵ ON ⊥ CD (given)
∴ CN = ND (line from centre ⊥ chord bisects
chord)
∴
cm 8
cm 42
=
×=CD
NSS Mathematics in Action 5A Full Solutions
36
6. ∴ CM = MD (given)
∵ OM ⊥ CD (line joining centre to mid-pt. of chord
⊥ chord)
∴ ∠OMC = 90°
°=
°+°=
∠+∠=∠
122
9032
OMCOCMAOC
(ext. ∠ of △)
Consider △OAC.
∵ OC = OA (given)
∴ ∠OCA = ∠OAC (base ∠s, isos. △)
°=∠
°=°+∠
°=∠+∠+∠
29
1801222
180
OCA
OCA
AOCOACOCA
7. ∵ OM ⊥ AP (given)
∴ AM = MP (line from centre ⊥ chord bisects
chord)
∴
cm 12
cm 242
1
2
1
=
×=
= APMP
Join OP.
cm 15
cm 129 22
22
=
+=
+= MPOMOP
(Pyth. theorem)
∵ ON ⊥ BP (given)
∴ PN = NB (line from centre ⊥ chord bisects
chord)
∴
cm 9
cm 182
1
2
1
=
×=
= BPPN
Consider △ONP.
cm 12
cm 915 22
22
=
−=
−= PNOPON
(Pyth. theorem)
8.
cm 10
cm )416(2
1
)(2
1
2
1
=
+=
+=
=
MBAM
ABOB
∴
cm 6
cm )410(
=
−=
−= MBOBOM
Join OD.
Consider △OMD.
cm 10=
= OBOD
(radii)
cm 8
cm 610 22
22
=
−=
−= OMODMD
(Pyth. theorem)
∵ OM ⊥ CD (given)
∴ CM = MD (line from centre ⊥ chord bisects
chord)
∴
cm 16
cm 82
2
=
×=
= MDCD
9. (a) ∵ PQ is the perpendicular bisector of the chord RS.
∴ PQ is a diameter of the circle.
∵ PQ = 10 cm (given)
∴ radius of the circle
cm 5
cm 102
1
2
1
=
×=
= PQ
Join OR.
OR = 5 cm (radius)
cm 4
cm 35 22
22
=
−=
−= TROROT
(Pyth. theorem)
OP = 5 cm (radius)
∴
cm 9
cm )45(
=
+=
+= OTOPPT
(b)
cm 1
cm )910(
=
−=
−= PTPQTQ
2 Basic Properties of Circles
37
10. ∵ BM = MC = 6 cm (given)
∴ OM ⊥ BC (line joining centre to mid-pt. of chord
⊥ chord)
Consider △OMB.
cm 8
cm 610 22
22
=
−=
−= BMOBOM
(Pyth. theorem)
Consider △OMD.
cm 15
cm 817 22
22
=
−=
−= OMODMD
(Pyth. theorem)
∴
cm 9
cm )615(
=
−=
−= MCMDCD
11. Construct a circle with centre O lying on BH, such that the
circle cuts AB at two points P and Q, and cuts BC at two
points R and S as shown.
Draw OM and ON such that OM ⊥ AB and ON ⊥ BC.
∠ABH = ∠CBH given
∠OMB = ∠ONB = 90° by construction
OB = OB common side
∴ △OBM ≅ △OBN AAS
∴ OM = ON corr. sides, ≅ △s
∴ PQ = RS chords equidistant from centre
are equal
Level 2
12.
Construct OMN such that OM ⊥ CD and ON ⊥ AB.
OMN is a straight line.
OA = OC = 17 cm (radii) ∵ OM ⊥ CD (by construction) ∴
cm 15
cm 302
1
2
1
chord) bisects chord centre from (line
=
×=
=
⊥=
CD
MDCM
In △OCM,
cm 8
cm 1517 22
22
=
−=
−= CMOCOM
(Pyth. theorem)
∵ ON ⊥ AB (constructed) ∴
cm 8
cm 162
1
2
1
chord) bisects chord centre from (line
=
×=
=
⊥=
AB
NBAN
In △OAN,
cm 15
cm 817 22
22
=
−=
−= ANOAON
(Pyth. theorem)
∴ Distance between AB and CD
cm 7
cm )815(
=
−=
−= OMON
13.
Let M be a point on AB such that OM ⊥ AB.
∵ OM ⊥ AB (by construction)
∴
cm 12
cm 242
1
chord) bisects chord centre from (line
=
×=
⊥= MBAM
Consider △OMA.
cm 9
cm 121522
22
=
−=
−= AMOAOM
(Pyth. theorem)
Consider △OMC.
cm 40
cm )2812(
=
+=
+= BCMBMC
cm 41
cm 40922
22
=
+=
+= MCOMOC
(Pyth. theorem)
cm 26
cm )1541(
=
−=
−= ODOCCD
14. (a) Consider △ABP and △ACP.
AP = AP common side
BP = CP given
OP ⊥ BC line joining centre to
mid-pt.of chord ⊥ chord
∴ ∠APB = ∠APC
= 90°
∴ △ABP ≅ △ACP SAS
∵ AB = AC corr. sides, ≅ △s
∴ △ABC is an isosceles triangle.
NSS Mathematics in Action 5A Full Solutions
38
(b) ∵ OM ⊥ AB (given)
∴
cm 12
cm 62
chord) bisects2
chord centre from (line
=
×=
=
⊥=
AMAB
MBAM
cm 12=
= ABAC
(proved in (a))
cm 12
cm 62
2
=
×=
= BPBC (given)
∵ AC = BC, ON ⊥ AC and OP ⊥ BC
∴
cm 32
centre) fromt equidistan chords, (equal
=
= ONOP
15. (a) Consider △OAB and △OAC.
OA = OA common side
OB = OC radii
AB = AC given
∴ △OAB ≅ △OAC SSS
∴ ∠OAB = ∠OAC corr. ∠s, ≅ △s
∴ AO bisects ∠BAC.
(b) Consider △ABN and △CAN.
AB = AC given
∠OAB = ∠OAC proved in (a)
AN = AN common side
∴ △ABN ≅△CAN SAS
∴ BN = CN corr. sides, ≅ △s
∴ ON ⊥ BC line joining centre to
mid-pt. of chord ⊥ chord
(c)
cm 3
cm )58(
=
−=
−= OAANON
Consider △ONC.
cm 4
cm 3522
22
=
−=
−= ONOCNC
(Pyth. theorem)
Consider △ANC.
)cm 54(or cm 80
cm 48 22
22
=
+=
+= NCANAC (Pyth. theorem)
16. (a)
cm )3( −=
−=
r
NYOYON
(b) ∵ ON ⊥ AB (given)
∴ NBAN = (line from centre ⊥ chord
cm 9
cm 182
1
=
×=
bisects chord)
Consider △OAN.
OA = r cm (radius)
∴ 15
906
9)3(
22
222
222
=
=+−
=+−
=+
r
rrr
rr
OAANON (Pyth. theorem)
17. ∵ OM ⊥ CD (given)
∴ MDCM = (line from centre ⊥ chord
cm 6
cm 122
1
=
×=
bisects chord)
Let r cm be the radius of the circle.
cm )18( r
OAAMOM
−=
−=
Join OC.
Consider △OCM.
OC = r cm (radius)
10
36360
6)18(
22
222
222
=
=+−
=+−
=+
r
rrr
rr
OCCMOM (Pyth. theorem)
∴
cm 2
cm 18)10(2
cm 18)(2
cm )]18([
=
−×=
−=
−−=
−=
r
rr
OMOBMB
18.
Let M be a point on AB such that OM ⊥ AB. ∵ OM ⊥ AB (by construction)
∴ AMMB = (line from centre ⊥ chord
cm 9
cm 182
1
=
×=
bisects chord)
Join OB.
OB = 13 cm (radius)
Consider △OMB.
cm 88
cm 913 22
22
=
−=
−= MBOBOM
(Pyth. theorem)
Let N be a point on CD such that ON ⊥ CD.
∵ ON ⊥ CD (by construction)
∴
cm 12
cm 242
1
chord) bisects chord centre from (line
=
×=
⊥= DNNC
∵
°=
∠=∠
90
OMKONK
∴ ONKM is a rectangle.
2 Basic Properties of Circles
39
∴ NK = OM (property of rectangle)
∴
d.p.) 2 to(cor. cm 2.62
cm )8812(
=
−=
−=
−=
OMNC
NKNCKC
19. (a)
cm 6
cm )51(
=
+=
+= RQPRPQ
cm 3
cm 62
1
2
1
=
×=
= PQOP
cm 2
cm )13(
=
−=
−= PROPOR
In △ORM,
cm 3
cm 12 22
22
=
−=
−= OMORRM
(Pyth. theorem)
∵ OM ⊥ RS (given)
∴ RM = MS (line from centre ⊥ chord bisects
chord)
∴
cm 32
2
=
= RMRS
(b)
Join OD.
OD = OP = 3 cm (radii)
In △OND,
cm 8
cm 13 22
22
=
−=
−= ONODND
(Pyth. theorem)
∵ ON ⊥ CD (given) ∴ CN = ND (line from centre ⊥ chord bisects chord) ∴
cm) 24(or cm 82
2
=
= NDCD
20. (a) ∵ MCMN2
1= ∴ NCMN =
∵ MN = NC and BD ⊥ MC
∴ BD is the perpendicular bisector of chord MC.
∴ BD is a diameter of the circle.
(b) (i)
cm 8
cm 162
1
2
1
=
×=
= MCNC
In △BCN,
cm 6
cm 81022
22
=
−=
−= NCBCNB
(Pyth. theorem)
Join OC.
Let r cm be the radius of the circle.
cm r
OBOC
=
=
(radii)
cm )6( −=
−=
r
NBOBON
In △OCN,
d.p.) 2 to(cor. 33.8
12
100
10012
643612
8)6(
22
222
222
=
=
=
++−=
+−=
+=
r
r
r
rrr
rr
NC ON OC (Pyth. theorem)
∴ The radius of the circle is 8.33 cm.
(ii)
cm 3
50
cm 12
1002
2
=
×=
= OBBD
cm 13.33
cm 103
50 2
2
22
=
−
=
−= ADBDAB
(Pyth. theorem)
Exercise 2B (p. 2.30)
Level 1
1.
°=∠
°=°+∠
42
180138
ACB
ACB
(adj. ∠s on st. line)
°=
°×=
∠=
84
422
2 ACBx
(∠ at centre twice ∠ at ⊙ce)
2. ∠ACB = 90° (∠ in semi-circle)
∵ CA = CB (given)
∴ x = ∠CBA (base ∠s, isos. △)
°=
°=+°
°=+∠+∠
45
180290
180
x
x
xCBAACB (∠ sum of △)
NSS Mathematics in Action 5A Full Solutions
40
3.
°=
∠=∠
55
ABDACD
(∠s in the same segment)
°=
°=°+
°=∠+
70
12555
125
x
x
ACDx
(ext. ∠ of △)
4.
°=
∠=∠
25
DBEDAC
(∠s in the same segment)
°=
°+°=
∠+∠=
67
4225
ADCDACx
(ext. ∠ of △)
5.
°=
°×=
∠=∠
106
532
2 ACBAOB
(∠ at centre twice ∠ at ⊙ce)
°=
°=°+
°=∠+
12
118106
118
x
x
AOBx
(ext. ∠ of △)
6.
°=
°−°=
∠−∠=∠
58
3290
CADCDBACD
(ext. ∠ of △)
∠ACB = 90° (∠ in semi-circle)
∴
°=
°−°=
∠−∠=
32
5890
ACDACBx
7. °=∠ 90ADC (∠ in semi-circle)
°=
°−°−°=
∠−∠−°=∠
25
6590180
180 DACADCDCA
(∠ sum of △)
°=
∠=
25
DCAx
(∠s in the same segment)
8.
x
x
ACBAOB
4
)2(2
2
=
=
∠=∠ (∠ at centre twice ∠ at ⊙ce)
∵ OB = OA (radii)
∴ ∠OBA = x (base ∠s, isos. △)
°=
°=
°=++
°=∠+∠+
30
1806
1804
180
x
x
xxx
AOBOBAx
(∠ sum of △)
9.
Join OC.
°=
°×=
∠=∠
65
1302
1
2
1AOBACB
(∠ at centre twice ∠ at ⊙ce)
∵ OC = OA (radii) ∴
°=
∠=∠
20
OACOCA (base ∠s, isos. △) ∵ OB = OC (radii) ∴
°=
°−°=
∠−∠=
∠=
45
2065
OCAACB
OCBx
(base ∠s, isos. △)
10. ∠DAC = 90° (∠ in semi-circle)
°=
°−°−°=
∠−∠−°=∠
35
5590180
180 ACDDACADC
(∠ sum of △)
∵ AB = AD (given) ∴
°=
∠=∠
35
ADBABD
(base ∠s, isos. △)
°=
°−°=∠
°=∠+∠
20
3555
55
BAC
BACABD
(ext. ∠ of △)
11.
(a)
°=
°−°=∠
°=∠+∠
148
32180
180
AOC
AOCOAB
(int. ∠s, BA // CO)
∴ Reflex
°=
°−°=
∠−°=∠
212
148360
360 AOCAOC
(∠s at a pt.)
(b)
°=
°×=
∠=∠
106
2122
1
reflex 2
1AOCABC
(∠ at centre twice ∠ at ⊙ce)
12. ∠BOD = 36° (opp. ∠s of // gram)
°=
°×=
∠=∠
18
362
1
2
1BODBCD
(∠ at centre twice ∠ at ⊙ce)
°=
∠=∠
18
BCDODC
(alt. ∠s, DO // AC)
°=
°+°=
∠+∠=∠
54
3618
BODODKBKD
(ext. ∠ of △)
13. (a) Reflex
°=
°×=
∠=∠
280
1402
2 ADCAOC
(∠ at centre twice ∠ at ⊙ce)
°=
°−°=∠
°=∠+∠
80
280360
360reflex
AOC
AOCAOC (∠s at a pt.)
2 Basic Properties of Circles
41
(b)
Join OB. ∵ OB = OA (radii) ∴
°=
∠=∠
28
OABOBA
(base ∠s, isos. △)
°=
°×=
∠=∠
40
802
1
2
1AOCABC
(∠ at centre twice ∠ at ⊙ce)
°=
°−°=
∠−∠=∠
12
2840
OBAABCOBC
∵ OC = OB (radii) ∴ °=
∠=∠
12
OBCBCO (base ∠s, isos. △)
14. ∠AEF, ∠AFE, ∠BDC, ∠DBC, ∠EAC and ∠CAF
(any four of the above angles)
15. Suppose ∠OBC = 2∠OAC.
°=
°×=
∠=∠
45
902
1
2
1AOBACB
(∠ at centre twice ∠ at ⊙ce)
OAC
OACAOKAKB
∠+°=
∠+∠=∠
90
(ext. ∠ of △)
OAC
OBCBCKBKA
∠+°=
∠+∠=∠
245
(ext. ∠ of △)
∴
°=∠
∠+°=∠+°
45
24590
OAC
OACOAC
If °=∠ 45OAC , then B and C will coincide, which
contradicts the assumption that B and C are distinct
points.
∴ ∠OBC ≠ 2∠OAC
(or any other reasonable answers)
Level 2
16.
°=
°−°=∠
°=∠+∠
22
2850
50
DCB
CKBDCB
(ext. ∠ of △)
°=
∠=∠
22
DCBDAB
(∠s in the same segment)
∠ACB = 90° (∠ in semi-circle)
°=∠
°=°+∠+°+°
°=∠+∠+∠
18
180)22(5090
180
CAD
CAD
CABCBAACB (∠ sum of △)
17. ∵ DC = DA (given)
∴ ∠DCA = x (base ∠s, isos. △)
∵ BD = BC (given)
∴
x
DCABDC
=
∠=∠
(base ∠s, isos. △)
∠ADB = 90° (∠ in semi-circle)
°=
°=+°++
°=∠+∠+∠
30
180)90(
180
x
xxx
ADCDCADAC
(∠ sum of △)
∴ °=∠ 30BDC
18.
°=
∠=∠
23
DBCDAC
(∠s in the same segment)
°=∠
°=∠+°+°
°=∠+∠+∠
90
1806723
180
ADC
ADC
ADCACDDAC
(∠ sum of △)
∴ AC is a diameter of the circle. (converse of ∠ in
semi-circle)
19. BCDBAD ∠=∠ (∠s in the same segment)
ABCBCD ∠=∠ (alt. ∠s, CD // AB) ∴ ABCBAD ∠=∠
∠ACB = 90° (∠ in semi-circle)
°=∠
°=°+∠+∠
°=°+∠+°+∠
°=∠+∠+∠
23
180134
18090)44(
180
ABC
ABCABC
BADABC
ACBBACABC
(∠ sum of △)
20.
°=
°×=
∠=∠
62
1242
1
2
1AOEABE
(∠ at centre twice ∠ at ⊙ce)
°=
°−°=∠
∠=∠+∠
26
3662BEC
ABEBECACE
(ext. ∠ of △)
°=
∠=∠
26
BECBAD
(∠s in the same segment)
∴
°=
°+°=
∠+∠=∠
88
6226
ABEBADAKE
(ext. ∠ of △)
21.
Join BD.
°=∠ 90ABD (∠ in semi-circle)
x
ABEABDEBD
−°=
∠−∠=∠
90
x
EBDECD
−°=
∠=∠
90
(∠s in the same segment)
NSS Mathematics in Action 5A Full Solutions
42
22. (a) OABC is a parallelogram. given
OA = OC radii
∴ OABC is a rhombus.
(b) Reflex xAOC −°=∠ 360 ∠s at a pt.
AOCABC ∠=∠ reflex 2
1 ∠ at centre twice
2180
)360(2
1
x
x
−°=
−°=
∠ at ⊙ce
(c) ∠ABC = x (opp. ∠s of // gram)
°=
°=
=−°
120
1802
3
2180
x
x
xx
(proved in (b))
23. (a) ∵ DK ⊥ BK
∴ BK = EK line from centre ⊥ chord
bisects chord
In △BKD and △EKD,
BK = EK proved
∠BKD = ∠EKD given
DK = DK common side
∴ △BKD ≅ △EKD SAS
(b) Let ∠BED = x.
x
KEDKBD
=
∠=∠ (corr. ∠s, ≅ △s)
x
BEDACEABE
+°=
∠+∠=∠
42
(ext. ∠ of △)
°=
°=++°
°=∠+∠
°=∠
24
9042
90
90
x
xx
KBDABE
ABD
(∠ in semi-circle)
∴
°=
∠=∠
24
BEDBAD
(∠s in the same segment)
24. (a) In △AKB and △DKC,
∠AKB = ∠DKC vert. opp. ∠s
∠BAK = ∠CDK ∠s in the same segment
∠ABK = ∠DCK ∠s in the same segment
∴ △AKB ~ △DKC AAA
CK
DK
BK
AK= corr. sides, ~ △s
∴ DKBKCKAK •=•
(b) DKBKCKAK •=• (proved in (a))
DK•=• cm) (3cm) (2cm) 6(
∴ cm 4=DK
25.
Join AP.
∠APB = 90° ∠ in semi-circle
°=
∠=∠
90
APQACQ
∠s in the same segment
i.e. QC ⊥ AB
26.
Join OA.
ABP
AOQABQ
∠=
∠=∠
2
ceat twicecentreat
segment same in the s ⊙∠∠
∠
∴ BP bisects ∠ABQ.
27. (a) ∵ EA = EB given ∴ ∠EAB = ∠EBA base ∠s, isos. △
∠BAD = ∠BED ∠s in the same segment
EAD
BADEAB
BEDEBABCE
∠=
∠−∠=
∠−∠=∠
ext. ∠ of △
∴ ∠BCE = ∠EAD
(b) ∠ABE = ∠ADE (∠s, in the same segment)
∴
AED
DAEAED
CDA
ADE
ABECBE
∠>
∠∠+∠=
∠∠=
∠−°=
∠∠−°=∠
) of (ext.
line) st.on s (adj.
180
line) st.on s (adj. 180 △
∴ It is not possible that △BCE ~ △EAD.
Exercise 2C (p. 2.42)
Level 1
1. Reflex
°=
°−°=
∠−°=∠
280
80360
360 AOBAOB
(∠s at a pt.)
4
280
8014
Reflex Major
cm
=
°
°=
∠
∠=
x
AOB
AOB
AB
x)
(arcs prop. to ∠s at centre)
2.
°=
°−°−°=
∠−∠−°=∠
55
7550180
180 ACBABCBAC
(∠ sum of △)
11
50
5510
cm
=
°
°=
∠
∠=
x
ABC
BAC
AC
x) (arcs prop. to ∠s at ⊙ce
)
2 Basic Properties of Circles
43
3. °=∠ 90ABC (∠ in semi-circle)
°−°=
°−°−°=
∠−∠−°=∠
x
x
ABCACBBAC
90
90180
180
(∠ sum of △)
50
4509
45450
5
490
=
°=°
°=°−°
=°
°−°
=∠
∠
x
x
xx
x
x
AB
CB
ACB
BAC))
(arcs prop. to ∠s at ⊙ce)
4.
°=
°+°=
∠+∠=∠
°=
°=∠
=∠
∠
140
5684
56
)84(6
4
CODBOCBOD
COD
BC
CD
BOC
COD))
(arcs prop. to ∠s at centre)
70
70
1402
1
2
1
=
°=
°×=°
∠=∠
x
x
BODBAD
(∠ at centre twice ∠ at ⊙ce)
5. BADADC ∠=∠ alt. ∠s, CD // AB
∴ ))
BDAC : equal ∠s, equal arcs
6.
(a)
°=
°=∠
=∠
∠
64
)48(3
4AOB
BC
AB
BOC
AOB))
(arcs prop. to ∠s at centre)
(b)
°=
°×=
∠=∠
32
642
1
2
1AOBACB
(∠ at centre twice ∠ at ⊙ce)
7.
°=
°−°−°=
∠−∠−°=∠
54
6462180
180 DCBDBCBDC
(∠ sum of △)
°=
°−°=
∠−∠=∠
27
3562
DEBDBCEDB (ext. ∠ of △) 2:1
54:27
)at s toprop. (arcs :: ce
=
°°=
∠∠∠= ⊙BDCEDBBCAB))
8. ∠BAC = 90° (∠ in semi-circle)
°=
°+°=
∠+∠=∠
120
3090
CADBACBAD
∵ AB = AD (given)
∴ ∠ABD = ∠ADB (base ∠s, isos. △)
°=∠
°=°+∠
°=∠+∠+∠
30
1801202
180
ABD
ABD
BADADBABD (∠ sum of △)
°=
°−°−°=
∠−∠−°=∠
60
3090180
180 ABCBACACB (∠ sum of △)
∴
3:2
90:60
)at s toprop. (arcs :: ce
=
°°=
∠∠∠= ⊙BACACBBCAB))
9. ∵ ))BAAC = (given)
∴
cm 6=
= BAAC
(equal arcs, equal chords)
°=∠ 90BAC (∠ in semi-circle)
In △ABC,
cm 72
cm 66 22
22
=
+=
+= ACABBC
(Pyth. theorem)
Perimeter of △ABC
fig.) sig. 3 to(cor. cm 20.5
cm )7266(
=
++=
10. ∵ CPD
BPCAPBAPC
∠=
°+°=
∠+∠=∠
105 ∴ 1=
∠
∠=
CPD
APC
CD
AC))
(arcs prop. to ∠s at ⊙ce)
))
CDAC =
∵
EPF
CPDBPCBPD
∠=
°=
°+°=
∠+∠=∠
25
1510
∴ 1=∠
∠=
EPF
BPD
EF
BD))
(arcs prop. to ∠s at ⊙ce)
))
EFBD =
∵
FPG
CPDBPCAPBAPD
∠=
°=
°+°+°=
∠+∠+∠=∠
30
15105
∴ 1=∠
∠=
EPG
APD
FG
AD))
(arcs prop. to ∠s at ⊙ce)
))
FGAD = (equal ∠s, equal arcs)
∴ ))
CDAC = , ))
EFBD = , ))
FGAD =
(any two of the above answers)
NSS Mathematics in Action 5A Full Solutions
44
11. ∵ 1==∠
∠))
CD
AB
DAC
ADB (arcs prop. to ∠s at ⊙ce)
∴ ∠ADB = ∠DAC
∴ KA = KD (sides opp. equal ∠s)
Similarly, KB = KC
∴ △AKD and △BKC are isosceles triangles.
∵ )))
CDBCAB == (given)
∴ AB = BC = CD (equal arcs, equal chords)
∴ △ABC and △BCD are isosceles triangles.
(any three of the above answers)
Level 2
12. ∠BAD = 90° (∠ in semi-circle)
°=
°−°=
∠−∠=∠
40
5090
CADBADBAC
12
50
4015
=
°
°=
∠
∠=
x
CAD
BAC
CD
BC))
(arcs prop. to ∠s at ⊙ce)
13.
Join BC.
°=
∠=∠
70
BDABCA
(∠s in the same segment)
°=
°−°−°=∠
°=∠+∠+∠
70
7040180
180
ABC
BCABACABC (∠ sum of △)
5.17
40
7010
=
°
°=
∠
∠=
x
BAC
ABC
BC
ADC))
(arcs prop. to ∠s at ⊙ce)
14.
Let AC and BD intersect at E.
°=∠ 90BAD (∠ in semi-circle)
∵ AB = AD (given)
∴ ∠ABD = ∠ADB (base ∠s, isos. △)
°=∠
°=°+∠
°=∠+∠+∠
45
180902
180
ADB
ADB
BADADBABD
(∠ sum of △)
°=∠
°=∠+°+°
°=∠+∠+∠
60
1807545
180
DAE
DAE
DAEAEDADE
(∠ sum of △)
4
45
60
3
=
°
°=
∠
∠=
x
x
ADB
DAC
AB
CD))
(arcs prop. to ∠s at ⊙ce)
15. (a) ∵ OD = OB (radii)
∴
°=
∠=∠
30
OBDODB
(base ∠s, isos. △)
°=
°+°=
∠+∠=∠
60
3030
ODBOBDBOC
(ext. ∠ of △)
(b)
°=
°×=∠
=∠
∠
40
603
2AOB
CB
BA
BOC
AOB))
(arcs prop. to ∠s at centre)
°=
°×=
∠=∠
20
402
1
2
1AOBADB
(∠ at centre twice ∠ at ⊙ce
)
16.
°=
∠=∠
30
PQSPRS
(∠s in the same segment)
°=
°−°=
∠−∠=∠
45
3075
PRSQRSPRQ 1==
∠
∠))
QP
QR
PRQ
QSR (arcs prop. to ∠s at ⊙ce)
°=
∠=∠
45
PRQQSR °=
°−°−°=
∠−∠−°=∠
60
4575180
180 QSRQRSRQS
(∠ sum of △)
17. (a)
2 Basic Properties of Circles
45
Join OB, OC and CD.
∵ )))CDBCAB == (given)
∴ ∠AOB = ∠BOC = ∠COD (equal arcs, equal ∠s)
°=∠
°=∠
°=∠+∠+∠
40
1203
120
BOC
BOC
CODBOCAOB
°=
°×=
∠=∠
20
402
1
2
1BOCBDC
(∠ at centre twice ∠ at ⊙ce)
(b) Reflex
°=
°−°=
∠∠−°=∠
240
120360
pt.) aat s(360 AODAOD
°=
°×=
∠∠∠×=∠
120
2402
1
)at twicecentreat ( reflex2
1 ce⊙AODACD
°=∠
°=°+°+∠
°=∠+∠+∠
40
18012020
180
CED
CED
ECDCDECED (∠ sum of △)
18. (a) (i)
Join OT, OP and OQ.
∵ RS = ST = TP = PQ = QR (given)
∠ROS = ∠SOT = ∠TOP =∠POQ = ∠QOR
(equal chords, equal ∠s)
∴
°=
°×=∠
72
3605
1ROS
(∠s at a pt.)
(ii)
°=
°×=
∠=∠
36
722
1
2
1ROSRPS
(∠ at centre twice ∠ at ⊙ce)
(b) Angle subtended at centre by any one side of the polygon
n
°=
360
Angle subtended at one of the vertices by any one side of
the polygon
n
n
°=
°×=
180
360
2
1
19. (a) OA = OC radii
AB = CB given
OB = OB common side
∴ △ABO ≅ △CBO SSS
(b) ∵ ∠AOB = ∠COB corr. ∠s, ≅ △s
∴
COB
AOBAOD
∠−°=
∠−°=∠
180
180
adj. ∠s on st. line
COD∠= adj. ∠s on st. line
∴ ))
DCAD = equal ∠s, equal arcs
20. ∵ 1==∠
∠))
CD
BC
DAC
CAB
arcs prop. to ∠s at ⊙ce
∴ ∠CAB = ∠DAC
∵ OC = OA radii
∴
DAC
CABACO
∠=
∠=∠
base ∠s, isos. △
∴ OC // AD alt. ∠s equal
21. (a) ∵ OE ⊥ BD given
∴ BE = ED line from centre ⊥ chord
bisects chord
AE = AE common side
∠AEB = ∠AED = 90° given
∴ △ABE ≅ △ADE SAS
(b) ∠BAC = ∠DAC corr. ∠s, ≅ △s
1=∠
∠=
DAC
BAC
CD
BC))
arcs prop. to ∠s at ⊙ce
∴ ))
CDBC =
22. (a) With the notations in the figure,
°=
°+°=
∠+∠=∠
50
3020
DBFBDFDFE
(ext. ∠ of △)
°=
°+°=
∠+∠=∠
90
5040
ACGCAGAGE
(ext. ∠ of △)
∴
°=
°−°−°=
∠−∠−°=
40
5090180
180 GFEFGEx
(∠ sum of △)
(b) )))))EADECDBCAB ::::
5:3:4:4:2
50:30:40:40:20
)at s toprop. (arcs
::::
ce
=
°°°°°=
∠
∠∠∠∠∠= ⊙ ACEDBECADBECADB
(c)
cm 9
cm 2
18circle theof nceCircumfere
53442
2
circle theof nceCircumfere
π
π
=
×=
++++=
)AB
NSS Mathematics in Action 5A Full Solutions
46
∴ Radius of the circle
cm 4.5
cm 2
9
=
=π
π
23. If △ABC is a right-angled triangle, then either one of ∠A,
∠B or ∠C must be equal to 90°.
∠A : ∠B : ∠C = a : b : c (arcs prop. to ∠s at ⊙ce)
and ∠A + ∠B + ∠C = 180° (∠ sum of △)
∴ °×++
=∠ 180cba
aA ,
°×++
=∠ 180cba
bB
and °×++
=∠ 180cba
cC
Since one of ∠A, ∠B or ∠C is equal to 90°,
one of the ratios cba
b
cba
a
++++, or
cba
c
++is equal
to 2
1.
∴ The student’s claim is correct.
(or any other reasonable answers)
Exercise 2D (p.2.49)
Level 1
1. °=∠ 95BCD (ext. ∠, cyclic quad.)
°=
°−°=
°=+∠
85
95180
180
x
xBCD (adj. ∠s on st. line)
2.
°=
°−°=∠
°=∠+∠
104
76180
180
BCD
BADBCD (opp. ∠s, cyclic quad.)
∵ CD = CB (given)
∴ ∠BDC = x (base ∠s, isos. △)
∴
°=
°=°+
°=∠++∠
38
1801042
180
x
x
BDCxBCD (∠ sum of △)
3. ∠ACB = 90° (∠ in semi-circle)
°=
°−°−°=∠
∠°=∠+∠+∠
50
4090180
) of sum (180
ABC
BACACBABC △
°=
°−°=
°=∠+
130
50180
180
x
ABCx (opp. ∠s, cyclic quad.)
4.
°=
∠=
46
ABDx (∠s in the same segment)
°=
°+°−°=
°=+°+
∠°=∠+
80
)4654(180
180)54(
quad.) cyclic s, (opp.180
y
xy
BCDy
5.
°=
°−°=∠
°=∠+∠
70
110180
180
EBC
CDEEBC (opp. ∠s, cyclic quad.)
°=
°−°=∠
°=∠+∠
60
120180
180
ECB
BAEECB (opp. ∠s, cyclic quad.)
°=
°−°−°=∠
∠°=∠+∠+∠
50
6070180
) of sum (180
BEC
ECBEBCBEC △
6. (a)
°=
°+°=
∠+∠=∠
96
6036
ACDCADADE (ext. ∠ of △)
(b)
°=
∠=∠
60
BCDDFE (ext. ∠, cyclic quad.)
°=
°−°−°=∠
∠°=∠+∠+∠
24
9660180
) of sum (180
DEF
DEFFDEDFE △
7. ∠BAP = ∠PQD ext. ∠, cyclic quad.
∠DCP + ∠PQD = 180° opp. ∠s, cyclic quad.
∴ ∠DCP + ∠BAP = 180°
∴ AB // CD int. ∠s supp.
8.
°=
°−°=∠
°=∠+∠
50
130180
180
FCD
DEFFCD
(opp. ∠s, cyclic quad.)
°=
∠=
50
FCDx (ext. ∠, cyclic quad.)
∠ABD = y (ext. ∠, cyclic quad.)
∵ ABAD = (given)
∴
y
ABDADB
=
∠=∠ (base ∠s, isos. △)
°=
°−°=
°=+
∠°=∠+∠+∠
65
501802
1802
) of sum (180
y
y
yx
ADBABDBAD △
9. (a)
°=
°×=
∠=∠
20
402
1
2
1AOBBPA (∠ at centre twice ∠ at ⊙ce
)
(b)
°=
°−°=∠
°=∠+∠
130
50180
180
BAP
BCPBAP (opp. ∠s, cyclic quad.)
°=
°−°−°=∠
∠°=∠+∠+∠
30
20130180
) of sum (180
ABP
APBBAPABP △
2 Basic Properties of Circles
47
10.
°=
°−°−°=∠
∠°=∠+∠+∠
35
30115180
) of sum (180
DAC
ACDADCDAC △
∠ABC + ∠ADC = 180° (opp. ∠s, cyclic quad.)
°=
°−°=∠
65
115180ABC
∵ CDBC = (given) ∴ ))
CDBC = (equal chords, equal arcs) 1==∠
∠))
CD
BC
DAC
BAC (arcs prop. to ∠s at ⊙ce)
°=
∠=∠
35
DACBAC
°=
°−°−°=∠
∠°=∠+∠+∠
80
3565180
) of sum (180
ACB
BACABCACB △
11.
Join AD.
∠ABC + ∠CDA = 180° opp. ∠s, cyclic quad.
∠ADE = 90° ∠ in semi-circle
∠ABC + ∠CDE
°=
°+°=
∠+∠+∠=
∠+∠+∠=
270
90180
)(
)(
ADECDAABC
ADECDAABC
12. (a) Consider △ABC and
△ADC.
AB = AD given
BC = DC given
AC = AC common side
∴ △ABC ≅ △ADC SSS
(b) ∵ △ABC ≅ △ADC (proved in (a))
∴ ∠ABC = ∠ADC (corr. ∠s, ≅ △s)
°=∠
°=∠
°=∠+∠
90
1802
180
ABC
ABC
ADCABC
(opp. ∠s, cyclic quad.)
∴ AC is a diameter of the circle.
(converse of ∠ in semi-circle)
(c) If △ABC and △ADC are right-angled isosceles
triangles, then
°=
°+°=
∠+∠=∠
90
4545
DACBACBAD
and
°=
°+°=
∠+∠=∠
90
4545
DCABCABCD
∴ In this case, BD is a diameter of the circle.
(converse of ∠ in semi-circle)
13. ∵ △ABD is an equilateral triangle.
∴ ∠BAD = 60°
∵ △BCD is a right-angled triangle with ∠C = 90°.
∴ ∠BCD = 90°
∵
°≠
°=
°+°=∠+∠
180
150
9060BCDBAD
∴ The polygon cannot be a cyclic quadrilateral.
(or any other reasonable answers)
Level 2
14. Reflex
°=
°×=
∠=∠
220
1102
2 ABCCOA (∠ at centre twice ∠ at ⊙ce)
°=
°−°=
∠−°=∠
140
220360
reflex 360 COACOA (∠s at a pt.)
°=
°−°=∠
°=∠+∠
40
140180
180
APB
COAAPB (opp. ∠s, cyclic quad.)
15.
°=
∠=∠
40
DAECEB (corr. ∠s, EC // AD)
∵ EC = EB (given)
∴ EBCECB ∠=∠ (base ∠s, isos. △)
°=∠
°=∠+°
°=∠+∠+∠
70
180240
180
EBC
EBC
EBCECBCEB (∠ sum of △)
°=
°−°=∠
°=∠+∠
110
70180
180
ADC
EBCADC (opp. ∠s, cyclic quad.)
16. (a) ∠KAD = ∠KCB ext. ∠, cyclic quad.
∠KDA = ∠KBC ext. ∠, cyclic quad.
∠AKD = ∠CKB common angle
∴ △KAD ~ △KCB AAA
(b)
cm 1
cm 3cm 4
cm 4)(2
cm 3
cm 3
cm 2
=
+=
+=
+
+=
+
=
DC
DC
DC
ABKA
KD
DCKD
KA
KB
KD
KC
KA (corr. sides, ~ △s)
17.
°=
∠=∠
65
DCEDAB (ext. ∠, cyclic quad.)
∠ABD = 90° (∠ in semi-circle)
°=
°−°−°=∠
∠°=∠+∠+∠
25
6590180
) of sum (180
ADB
DABABDADB △
°=
∠=∠
25
ADBCBD (alt. ∠s, BC // AD)
°=
°−°=∠
∠=∠+∠
40
2565BDC
DCECBDBDC (ext. ∠ of △)
NSS Mathematics in Action 5A Full Solutions
48
18.
ADC
ADCACD
CADADCACD
∠−°=
∠−°−°=∠
∠°=∠+∠+∠
140
40180
) of sum (180 △
ADCABC
ADCABC
∠−°=∠
°=∠+∠
180
180 (opp. ∠s, cyclic quad.)
°=∠+∠ 180ACDAED (opp. ∠s, cyclic quad.)
ADC
ADC
ACDAED
∠+°=
∠−°−°=
∠−°=∠
40
)140(180
180
∴
°=
∠+°+∠−°=∠+∠
220
)40()180( ADCADCAEDABC
19.
Join BD.
Let ∠CBE = x.
∵ CE = CB (given)
∴
x
CBECEB
=
∠=∠ (base ∠s, isos. △)
°−=∠
∠=∠+∠
27xBEA
CBEBAEBEA
(ext. ∠of △)
°=∠ 90DBE (∠ in semi-circle)
°=
°=
°=+°−++°
°=∠+∠
39
1173
180)27()90(
180
x
x
xxx
DECDBC
(opp. ∠s, cyclic quad.)
∴ °=∠ 39CBE
20. Let ∠CO′D = x.
∠ABC = ∠CO′D = x ext. ∠, cyclic quad.
∵ OC = OB radii
∴
x
OBCOCB
=
∠=∠
base ∠s, isos. △
∠OAD = ∠OCB = x ext. ∠, cyclic quad.
x
OBACDO
=
′∠=′∠
ext. ∠, cyclic quad.
∵ O′D = O′C radii
∴
x
CDODCO
=
′∠=′∠
base ∠s, isos. △
°=
°=
°=′∠+′∠+′∠
60
1803
180
x
x
DOCCDODCO
∠ sum of △
∴ ∠O′DC = ∠O′CD = ∠CO′D = 60°
∴ △CDO′ is an equilateral triangle.
21. (a)
yx
yxx
PCBABCAPC
+=
++=
∠+∠=∠
2
)(
ext. ∠ of △
yx
xyy
RBCACBARB
2
)(
+=
++=
∠+∠=∠ ext. ∠ of △
(b) °=∠+∠ 180ARBAPC opp. ∠s, cyclic quad.
°=+
°=+
°=+++
60
18033
180)2()2(
yx
yx
yxyx
°=∠
°=°+∠
°=++∠
°=++∠
°=∠+∠+∠
60
180120
180)(2
18022
180
BAC
BAC
yxBAC
yxBAC
ACBABCBAC ∠ sum of △
Exercise 2E (p.2.58)
Level 1
1. (a) ∵ ∠BAD + ∠BCD
°=
°+°+°+°=
180
)6535()4238(
∴ A, B, C and D are concyclic. opp. ∠s supp.
(b)
°=
∠=∠
38x
CABCDB (∠s in the same segment)
2. (a) ∵
°=
°−°−°=
∠−∠−°=∠
40
10535180
180 DCBBDCDBC ∠ sum of △
∴ ∠DBC = ∠DAC
∴ A, B, C and D are concyclic. converse of
∠s in the
same segment
(b)
°=
∠=∠
35x
BDCBAC (∠s in the same segment)
3. (a) ∵ ∠BAC = ∠CDB = 90°
∴ A, B, C and D are concyclic. converse of ∠s
in the same
segment
(b)
°=
°−°−°=
∠−∠−°=∠
35
9055180
180 CDBEDAADB (adj. ∠s on st. line)
∠ACB = ∠ADB (∠s in the same segment)
°= 35x
4. (a) Refer to the figure.
Consider △ABG.
°=
°−°−°=
∠−∠−°=∠
50
4585180
180 GABBGAABG ∠ sum of △
°=
°+°=
∠+∠=∠
85
3550
GBCABGABC
∴ ∠ABC = ∠ADF
∴ A, B, C and D are concyclic. ext.∠ = int.
opp. ∠
2 Basic Properties of Circles
49
(b)
°=
∠=∠
35
CBDCAD (∠s in the same segment)
CADBAC
DABDCE
∠+∠=
∠=∠ (ext. ∠, cyclic quad.)
°=
°+°=
80
3545x
5. Consider △PBC.
°=∠
°=∠+°
∠=∠+∠
46
6216
PCB
PCB
APBPBCPCB ext. ∠ of △
°=
°−°=
∠−∠=∠
36
4682
PCBBCDACD
∴ ∠ABD = ∠ACD
∴ A, B, C and D are concyclic. converse of∠s in
the same segment
6. ∠ABC = ∠ADC opp. ∠s of // gram
∠PQB = ∠PDC ext. ∠, cyclic quad.
∠DPQ = ∠PQB alt. ∠s, AD // BC
∴ ∠DPQ = ∠ABC
∴ A, B, Q and P are concyclic. ext. ∠ = int. opp. ∠
7. ∵ AB = EB given
∴ ∠BAE = ∠BEA base ∠s, isos. △
∠BAE = ∠BCD opp. ∠s of // gram
∴ ∠BEA = ∠BCD
∴ B, C, D and E are concyclic. ext. ∠ = int. opp. ∠
8. ∠BPT = 90° ∠ in semi-circle
∠SQC = 90° ∠ in semi-circle
°=
°−°=
∠−°=∠
90
90180
180 SQCAQS adj. ∠s on st. line
∴ ∠AQR = ∠BPR = 90°
∴ A, P, R and Q are concyclic. ext. ∠ = int. opp. ∠
9. ∠BAD = 180° – (x + y) (∠ sum of △)
°=++−°
°=∠+∠
180)(180
180
zyx
BCDBAD (opp. ∠s, cyclic quad.)
zyx =+ (or any other equivalent
relationship)
10. ∵ AB = AC given
∴ ∠ABC = ∠ACB base ∠s, isos. △
∵ AM = MB and AN = NC given
∴ MN // BC mid-pt. theorem
∠AMN = ∠ABC corr. ∠s, MN // BC
∴ ∠AMN = ∠ACB
∴ B, C, N and M are concyclic. ext. ∠ = int. opp. ∠
11. D is any point on )AC , where AB is a diameter of the
circle and the mid-point of AB is the centre.
Level 2
12. (a) ∵ △ABC is an equilateral triangle.
∴ AB = BC = CA
Consider △ABD and △ACD.
AB = AC proved
BD = CD given
AD = AD common side
∴ △ABD ≅ △ACD SSS
Consider △ABE and △CBE.
AB = CB proved
AE = CE given
BE = BE common side
∴ △ABE ≅ △CBE SSS
(b) ∵ △ABD ≅ △ACD proved in (a)
∴ ∠ADB = ∠ADC corr. ∠s, ≅ △s
°=
∠=∠
°=∠+∠
90
180
ADCADB
ADCADB
adj. ∠s on st. line
Similarly, ∠BEA = ∠BEC = 90°
∴
°=
°+°=∠+∠
180
9090FECFDC
∴ D, C, E and F are concyclic. opp. ∠s supp.
∵ ∠ADB = ∠BEA = 90°
∴ A, B, D and E are concyclic. converse of ∠s of
the same segment
13. (a) ∠APB = 90° ∠ in semi-circle
°=∠
°=°+∠
°=∠+∠
90
18090
180
APM
APM
APBAPM adj. ∠s on st. line
°=∠ 90AOM given
∴ ∠APM = ∠AOM
∴ O, P, M and A are concyclic. converse of ∠s in
the same segment
(b) ∵ OPOA = radii ∴ ∠OAP = ∠OPA base ∠s, isos. △ ∠OAP = ∠OMP ∠s in the same segment
∴ ∠OPA = ∠OMB
14. ∠AQS = ∠BRS ext. ∠, cyclic quad.
∠BRS = ∠CPS ext. ∠, cyclic quad.
∴ ∠AQS = ∠CPS
∴ AQSP is a cyclic quadrilateral. ext. ∠ = int. opp. ∠
NSS Mathematics in Action 5A Full Solutions
50
15. (a)
Join PB and let ∠ARP = θ.
∠APQ = ∠ARP = θ given
∠PBA = ∠ARP = θ ∠s in the same
segment
∠APB = 90° ∠ in semi-circle
θ−°=
∠−∠=∠
90
APQAPBBPQ
°=
−°−−°=
∠−∠−°=∠
90
)90(180
180
θθ
BPQPBQPQB ∠ sum of △
(b)
Join RB.
∠TQB = 90° proved in (a)
∠TRB = 90° ∠ in semi-circle
∠TQB + ∠TRB = 90° + 90° = 180°
∴ R, T, Q and B are concyclic. opp. ∠s supp.
16. Consider △ACB and △DBC.
AC = DB given
∠ACB = ∠DBC given
BC = CB common side ∴ △ACB ≅ △DBC SAS
∴ ∠BAC = ∠CDB corr. ∠s, ≅ △s
∴ A, B, C and D are concyclic. converse of ∠s in
the same segment
17. (a) ∠ACB = 90° ∠ in semi-circle
°=
∠−°=∠
90
180 ACBECP adj. ∠s on st. line
°=∠ 90ADB ∠ in semi-circle
∴ ∠ECP = ∠ADB
∴ P, D, E and C are concyclic. ext. ∠ = int. opp. ∠
(b) ∠COD = 2∠CAD ∠ at centre twice
∠ at ⊙ce
APB
APBACPCAD
∠−°=
∠−∠−°=∠
90
180 ∠ sum of △
∴
APB
APBCOD
∠−°=
∠−°=∠
2180
)90(2
18. (a) ∵ ∠BAD + ∠BCD = 90° + 90°
= 180°
∴ BCDA is a cyclic quadrilateral. opp. ∠s supp.
∵ ∠BAD = 90° given
∴ BD is a diameter of the circle. converse of ∠s
in the same
segment
Since 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
(b)
Join MP and NP.
Let ∠ABD = x.
x
ABDACD
=
∠=∠ ∠s in the same segment
x
ACDBCDACB
−°=
∠−∠=∠
90
∵ AM = MB and AP = PC given
∴ MP // BC mid-pt. theorem
∴
x
ACBAPM
−°=
∠=∠
90
corr. ∠s, MP // BC
∴ NP ⊥ AC proved in (a)
∴ ∠APN = 90°
∴
ABD
x
x
APMAPN
NPM
∠=
=
−°−°=
∠−∠=
∠
)90(90
19. (a)
Join AE, BF, CG and DH.
EACCGE
CGEEAC
∠−°=∠
∠°=∠+∠
180
quad. cyclic s, opp.180
If A, E, F and B are concyclic,
then
EACBFE
BFEEAB
∠−°=∠
∠°=∠+∠
180
quad. cyclic s, opp.180
∴ ∠CGE = ∠BFE
∠BFE = ∠BDH ext. ∠, cyclic quad.
∴ ∠CDH = ∠CGE
∴ C, G, H and D are
concyclic. ext. ∠ = int. opp. ∠
(b)
Join PU, QT, RW and SV.
∠RSV = ∠PUT ext. ∠, cyclic quad.
∠VWR = ∠PQT ext. ∠, cyclic quad.
If R, S, W and V are
concyclic, then
∠RSV = ∠VWR ∠s in the same segment
∴ ∠PUT = ∠PQT
∴ P, Q, U and T are concyclic. converse of ∠s
in the same segment
2 Basic Properties of Circles
51
(opp. ∠s,
cyclic quad.)
Revision Exercise 2 (p. 2.66)
Level 1
1.
Join OF.
Draw ON such that ON ⊥ FE.
cm 10
cm 202
1
2
1
=
×=
=
=
BC
OBOF (radii)
cm 6=
= ABON (property of rectangle)
Consider △ONF.
cm 8
cm 610 22
22
=
−=
−= ONOFFN (Pyth. theorem)
∵ ON ⊥ FE (by construction)
∴ FN = NE (line from centre ⊥
chord bisects chord)
∴
cm 16
cm 82
=
×=FE
2. ∠AEC = 90° (∠ in semi-circle)
°=
∠=∠
30
EACEBC (∠s in the same segment)
°=
°+°=
∠+∠=∠
65
3035
EBCEDCAEB (ext. ∠ of △)
°=
°−°=
∠−∠=∠
25
6590
AEBAECBEC
3. ∠POR + ∠OPQ = 180° (int. ∠s, OR // PQ)
°=
°−°=∠
138
42180POR
°=
°−°=
∠−°=∠
222
138360
360Reflex PORPOR (∠s at a pt.)
°=
°×=
∠=∠
111
2222
1
reflex 2
1PORPQR (∠ at centre twice ∠ at ⊙ce
)
°=
°−°=∠
°=∠+∠
69
111180
180
ORQ
PQRORQ (int. ∠s, OR // PQ)
4. With the notations in the figure,
°=
°×=
∠=∠
27
542
1
2
1AOBBCA (∠ at centre twice ∠ at ⊙ce
)
°=
°+°=
∠+∠=∠
69
2742
BCAOBCONC (ext. ∠ of △)
°=
°−°=∠
∠=∠+∠
15
5469OAC
ONCAOBOAC (ext. ∠ of △)
5. 1==∠
∠))
BC
AB
CAB
ACB (arcs prop. to ∠s at ⊙ce)
°=
∠=∠
35
CABACB
°=∠ 90DCA (∠ in semi-circle)
°=
°+°=
∠+∠=∠
125
3590
ACBDCADCB
°=∠
°=°+°+∠
°=∠+∠
20
180125)35(
180
DAC
DAC
DCBDAB (opp. ∠s, cyclic quad.)
6. EABEDC ∠=∠ (∠s in the same segment)
EAB
EDCADCADE
EAB
EABDABDAC
∠−°=
∠−∠=∠
∠−°=
∠−∠=∠
80
60
EAB
EABEAB
ADEDACAEB
∠−°=°
∠−°+∠−°=°
∠∠+∠=∠
2140100
)80()60(100
) of (ext. △
∴ °=∠ 20EAB
7. 1
2==
∠
∠))
CD
BC
CBD
CDB (arcs prop. to ∠s at ⊙ce)
∴ CDBCBD ∠=∠2
1
∴ °=∠
°=∠
°=∠
°=∠+∠+°
°=∠+°+∠+°
°=∠+∠
40
40
602
3
1802
1120
180)58()62(
180
KDC
CDB
CDB
CDBCDB
CDBCBD
ADCABC
NSS Mathematics in Action 5A Full Solutions
52
8.
°=
°−°=∠
°=∠+∠
65
115180
180
ABC
ADCABC (opp. ∠s, cyclic quad.)
2
3=
=∠
∠))
BC
AB
BAC
ACB (arcs prop. to ∠s at ⊙ce)
∴ BACACB ∠=∠2
3
°=∠
°=∠
°=∠+°+∠
∠°=∠+∠+∠
46
1152
5
180652
3
) of sum (180
BAC
BAC
BACBAC
BACABCACB △
9. (a) ∵ OC = OB (radii) ∴ ∠OCB = ∠OBC (base∠s, isos. △)
°=∠
°=°+∠
°=∠+∠+∠
50
180802
180
OCB
OCB
BOCOBCOCB (∠ sum of △)
(b)
°=
°×=
∠=∠
40
802
1
2
1BOCBAC (∠ at centre twice ∠ at ⊙ce
)
quad.) cyclic s, (opp.180 ∠°=∠+∠ BCDDAB
°=∠
°=∠+°+°+°
°=∠+∠+∠+∠
54
180504036
180)()(
OCD
OCD
OCDOCBBACDAC
10.
Join BC.
°=
°×=
∠=∠
45
902
1
2
1DOCDBC (∠ at centre twice ∠ at ⊙ce
)
°=∠ 90ACB (∠ in semi-circle)
In △BCE,
°=∠
°=∠+°+°
∠°=∠+∠+∠
45
1804590
) of sum (180
CEB
CEB
CEBEBCBCE △
11.
Join DC.
∵ OC = OD (radii)
∴ ∠OCD = ∠ODC (base ∠s, isos. △)
°=∠+∠ 180BCDADC (int. ∠s, AD // BC)
°=∠
°=∠+°
°=∠+°+∠+°
58
180264
180)36()28(
OCD
OCD
OCDODC
∴
°=
°+°=∠
94
5836BCD
°=∠
°=°+∠
°=∠+∠
86
18094
180
BAD
BAD
BCDBAD (opp. ∠s, cyclic quad.)
12.
Draw OM such that OM ⊥ BC.
∵ OM ⊥ BC (by construction)
∴ BM = MC (line from centre ⊥ chord bisects chord)
∴
cm 3
cm 62
1
=
×=MC
Consider △OMC.
cm 4
cm 35 22
22
=
−=
−= MCOCOM (Pyth. theorem)
Consider △OAM.
cm 33
cm 47 22
22
=
−=
−= OMOAAM (Pyth. theorem)
∴
d.p.) 2 to(cor.cm 74.2
cm )333(
=
−=
−=
−=
MCAM
BMAMAB
13. (a) ∵ CF = FD (given)
∴ OF ⊥ CD (line joining centre to
mid-pt. of chord ⊥ chord)
cm 2
cm 42
1
2
1
=
×=
= CDFD
In △OFD,
cm 1
cm 2)5( 22
22
=
−=
−= FDODOF (Pyth. theorem)
2 Basic Properties of Circles
53
(line from centre ⊥
chord bisects chord)
(line from centre ⊥
chord bisects chord)
(b)
Join AB.
∵ OE = OF = 1 cm (proved in (a))
and OE ⊥ AC and OF ⊥ CD (proved in (a))
∴
cm 4=
= CDAC
equal) are centre from
t equidistan (chords
∵ OE ⊥ AC (given)
∴
cm 2
cm 42
1
2
1
=
×=
=
=
AC
ECAE
cm 5=
= ODOB (radii)
cm )15( −=
−= OEOBBE
d.p.) 2 to(cor.cm 24.1
cm 2)15(2
1
2
1 of Area
2
2
=
×−×=
××= AEBEABE△
14. (a)
2
222
cm 169
cm 13
=
=AE
2
22222
cm 169
cm ]5)66[(
=
++=+ DEAD
∵ 222 DEADAE +=
∴ °=∠ 90ADE (converse of Pyth. theorem)
∴ AE is a diameter of the circle.
(converse of ∠ in semi-circle)
∴ The centre of the circumcircle lies on AE.
(b) ∵ CB ⊥ AD and AB = BD (given)
∴ CB is the perpendicular bisector of AD.
The centre lies on CB.
The centre also lies on AE.
∴ The intersection of CB and AE, i.e. C, is the
centre of the circumcircle.
15.
Join BD and DC.
∠ABD = 90° ∠ in semi-circle
∠ACD = 90° ∠ in semi-circle
∴ ∠ABD = ∠ACD
ACAB
ADAD
=
=
given
sidecommon
∴ △ABD ≅ △ACD RHS
∴ ∠BAD = ∠CAD corr. ∠s, ≅ △s
∴ AD bisects ∠BAC.
16. ∠ADC = p + q (ext. ∠ of △)
∠BCD = p (∠s in the same segment)
qp
pqp
BCDADCr
+=
++=
∠+∠=
2
(ext. ∠ of △)
17.
Let O be the centre of the circle and r cm be the radius.
Join OA.
cm rOA =
cm )1( −=
−=
r
MCOCOM
∵ ABOM ⊥ (given)
∴
cm 5
cm 102
1
=
×=
= MBAM
Consider △OAM.
13
262
1225
)1(5
22
222
222
=
=
+−+=
−+=
+=
r
r
rrr
rr
OMAMOA (Pyth. theorem)
∴ The radius of the circle is 13 cm.
18. (a)
°=
°−°−°=
∠−∠−°=∠
116
3628180
180 BACBCAABC ∠ sum of △
°=∠=∠ 116ADEABC
∴ A, B, C and D are concyclic. ext. ∠ = int. opp. ∠
∴ ABCD is a cyclic quadrilateral.
(b)
°=
°−°=
∠−∠=∠
62
54116
ACDADECAD (ext. ∠ of △)
°=
∠=∠
62
CADCBD (∠s in the same segment)
19. 1==∠
∠))
CD
DE
DAC
EBD arcs prop. to ∠s at ⊙ce
DACEBD ∠=∠
∴ P, A, B and Q are concyclic. converse of ∠s in the
same segment
NSS Mathematics in Action 5A Full Solutions
54
(∠ at centre twice
∠ at ⊙ce)
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)
Level 2
21.
°=
°−°=∠
°=∠+∠
40
140180
180
AOB
DOBAOB (adj. ∠s on st. line)
°=
∠=∠
40
AOBOBC (alt. ∠s, DA // CB)
°=
°×=
∠=∠
20
402
1
2
1AOBBCA
°=
°+°=
∠+∠=∠
60
2040
BCAOBCOKC (ext. ∠ of △)
°=
°−°=∠
°=∠+∠
120
60180
180
AKO
OKCAKO (adj. ∠s on st. line)
22. (a)
°=
°−°=∠
°=∠+∠
78
102180
180
AFC
AFCABC (opp. ∠s, cyclic quad.)
°=
∠=∠
78
AFCCDE (ext. ∠, cyclic quad.)
(b)
°=
°×=
∠=∠
156
782
2 CDECOE
(∠ at centre twice ∠ at ⊙ce)
ABOBAF ∠+∠
°=∠
°=°+°+°+∠
∠°×−=∠+∠+
64
36038156102
polygon) of sum (180)24(
BAF
BAF
FEOCOE
23. (a) °=∠ 90ADC (∠ in semi-circle)
°=∠
°=°+∠+°+°
°=∠+∠+∠
16
180)54(9020
180
BAC
BAC
BADADCAPD (∠ sum of △)
(b)
°=
∠=∠
16
BACBDC (∠s in the same segment)
°=
°−°=
∠−∠=∠
74
1690
BDCADCADB (ext. ∠ of △)
°=
°−°−°=∠
∠°=∠+∠+∠
52
5474180
) of sum (180
AKD
CADADBAKD △
24. (a) °=∠ 90ACB (∠ in semi-circle)
DCADBA ∠=∠ (∠s in the same segment)
°=∠+∠+∠ 180ECBCBECEB (∠ sum of △)
°=∠
°=∠
°=°+∠+°+∠+°
14
282
180)90()37(25
DCA
DCA
DCADBA
(b)
°=∠
°=°+°+∠
∠°=∠+∠+∠
53
1809037
) of sum (180
CFB
CFB
BCACBDCFB △
25.
Join MN.
∠ABM = ∠MNC (ext. ∠, cyclic quad.)
∠AEM = ∠MND (ext. ∠, cyclic quad.)
∠MNC + ∠MND = 180° (adj. ∠s on st. line)
°=
∠+∠=∠+∠
180
MNDMNCAEMABM
∴ ABMBAE ∠+∠
°=∠
°=∠+°+°
°×−=∠+∠+
115
36018065
180)24(
BME
BME
BMEAEM (∠ sum of polygon)
26.
Join BE.
))
CD
BC
CAD
BEC=
∠
∠ (arcs prop. to ∠s at ⊙ce)
°=
°=∠
42
)28(2
3BEC
))
CD
DE
CAD
DBE=
∠
∠ (arcs prop. to ∠s at ⊙ce)
°=
°=∠
56
)28(2
4DBE
∴
°=
°−°−°=∠
∠°=∠+∠+∠
82
5642180
) of sum (180
BKE
KBEKEBBKE △
2 Basic Properties of Circles
55
(∠ at centre twice
∠ at ⊙ce)
27. (a)
21
1
+=
=∠
∠))
ABC
AB
AOC
AOB (arcs prop. to ∠s at centre)
°=
°×=
∠=∠
30
903
1
3
1AOCAOB
∴
°=
°×=
∠=∠
15
302
1
2
1AOBBCA
(b) ∵ OC = OA (radii)
∴ ∠ACO = ∠CAO (base ∠s, isos. △)
°=∠
°−°=∠
∠°=∠+∠+∠
45
901802
) of sum (180
CAO
CAO
AOCACOCAO △
∴
°=
°+°=
∠∠+∠=∠
75
3045
) of (ext. △AOBCAOCEO
28. (a) ∠APD = ∠CPB common angle
∠PAD = ∠PCB ext. ∠, cyclic quad.
∠PDA = ∠PBC ext. ∠, cyclic quad.
∴ △PAD ~ △PCB AAA
(b) ∠AKB = ∠DKC vert. opp. ∠s
∠BAK = ∠CDK ∠s in the same segment
∠ABK = ∠DCK ∠s in the same segment
∴ △AKB ~ △DKC AAA
(c) ∵ △PAD ~ △PCB (proved in (a))
∴DC
DC
ABPA
PD
DCPD
PA
PB
PD
PC
PA
+=
+=
+
+=
+
=
cm 8cm 12
cm 10)(6
cm 8
cm 8
cm 6
(corr. sides, ~ △s)
∴ cm 4=DC
∵ △AKB ~ △DKC (proved in (b)) ∴
cm 3cm 4
cm 10 BK
CK
BK
DC
AB
=
= (corr. sides, ~ △s)
∴ cm 5.7=BK
29. (a) ∵ AM = MB and CN = ND given
∴ ∠OMK = ∠ONK = 90° line joining
centre to mid-pt
of chord ⊥ chord
∵ AB = DC given
∴ OM = ON equal chords,
equidistant from
centre
OK = OK common side
∴ △OMK ≅ △ONK RHS
(b) ∴ KM = KN corr. sides, ≅△s
Also, BM = CN given
∴ KM – BM = KN – CN
∴ KB = KC
(c) KB = KC proved in (b)
Also, AB = DC given
∴ KA = KD
∴ ∠KAD = ∠KDA base ∠s, isos. △
∠BCD + ∠KAD = 180° opp. ∠s, cyclic quad.
∴ ∠BCD + ∠KDA = 180°
∴ BC // AD int. ∠s supp.
30. MDPNBP ∠=∠ ext. ∠, cyclic quad.
NMC
DMP
DPMMDP
NPBNBPBNP
∠=
∠=
∠−∠−°=
∠−∠−°=∠
180
180
s opp. vert.
of sum
of sum
∠
∠
∠ △△
∴ QNQM = sides opp. equal ∠s
31. (a)
Join OC.
112
11
++
+=
=∠
∠))
ABCD
ABC
AOD
AOC (arcs prop. to ∠s at centre)
∴
°=
°×=∠
45
904
2AOC
∵ OA = OC (radii)
∴ OCAOAC ∠=∠ (base ∠s, isos. △)
°=∠
°=°+∠
∠°=∠+∠+∠
5.67
180452
) of sum (180
OAC
OAC
AOCOCAOAC △
(b)
Join OB and AD.
∵ ))BCAB =
∴ ∠AOB = ∠BOC (equal arcs, equal ∠s)
∴
°=
°×=
∠=∠
5.22
452
1
2
1AOCAOB
NSS Mathematics in Action 5A Full Solutions
56
°=
°×=
∠=∠
25.11
5.222
1
2
1AOBADB (∠ at centre twice ∠ at ⊙ce)
°=
°×=∠
=∠
∠
5.22
25.111
2CAD
AB
CD
ADB
CAD))
(arcs prop. to ∠s at ⊙ce)
°=∠
°=°+°+∠
∠°=∠+∠+∠
25.146
18025.115.22
) of sum (180
AED
AED
EDAEADAED △
32.
Join BO and OE.
Let ∠CAE = x.
∠BOE = 2∠CAE ∠ at centre twice
= 2x ∠ at ⊙ce
∠ACE + ∠BOE = 180° opp. ∠s, cyclic quad.
∠ACE = x2180 −°
xCEA
CEAxx
CEACAEACE
=∠
°=∠++−°
°=∠+∠+∠
180)2180(
180 ∠ sum of △
∴ CEACAE ∠=∠
∴ CECA = sides opp. equal ∠s
33. (a) ∵ CDCE = given
∴
ABC
CDECED
∠=
∠=∠
quad. cyclic , ext.
isos. s, base
∠
∠ △
∴ ABAE = sides opp. equal ∠s
∴ △ABE is an isosceles
triangle.
(b) Let ∠ABD = x.
∵ 1==∠
∠))
AD
DC
ABD
DBC (arcs prop. to ∠s at ⊙ce)
∴
x
ABDDBC
=
∠=∠
x
DBCABDABE
2=
∠+∠=∠
x
ABEAEB
2=
∠=∠ (base ∠s, isos. △)
x
AEBEDC
2=
∠=∠ (base ∠s, isos. △)
)1(4 KKx
EDCAEBDCB
=
∠+∠=∠
(ext. ∠ of △)
°=∠ 90BDC (∠ in semi-circle)
)2(90
90180
) of sum (180
KKx
xDCB
BDCDCBDBC
−°=
°−−°=∠
∠°=∠+∠+∠ △
From (1) and (2), we have
°=°=∠
°=
−°=
72)18(4
18
904
DCB
x
xx
°=
°−°=∠
°=∠+∠
108
72180
180
BAD
DCBBAD (opp. ∠s, cyclic quad.)
34. ∠BDC = ∠ABD alt. ∠s, CF // BA
∠DEF = ∠ABD ext. ∠, cyclic quad.
∠KAE = ∠DEF corr. ∠s, CA // DE
∴ ∠BDC = ∠KAE
∴ A, K, D and F are concyclic. ext. ∠ = int. opp. ∠
35. (a) °=∠=∠ 90ADCABC ∠ in semi-circle
°=
∠−°=∠
90
180 ABCFBE adj. ∠s on st. line
°=
∠−°=∠
90
180 ADCEDF adj. ∠s on st. line
∴ EDFFBE ∠=∠
∴ BEFD is a cyclic
quadrilateral. converse of ∠s
in the same segment
(b) °=∠=∠ 35BACBDC (∠s in the same segment)
°=∠=∠ 35BDCBFE (∠s in the same segment)
°=
°+°=
∠+∠=∠
62
3527
CFECEFDCF (ext. ∠ of △)
°=∠
°=∠+°
∠=∠+∠
28
9062
CFD
CFD
ADCCFDDCF (ext. ∠ of △)
36. (a) ∵ ))DCAD = given
∴ DCAD = equal arcs, equal chords
∵ DADE = given
∴ DCDE =
∴ DCEDEC ∠=∠ base ∠s, isos. △
(b) ∵ AD = ED given
∴ DEADAE ∠=∠ base ∠s, isos. △
DCEDEC ∠=∠ proved in (a)
°=∠+∠+∠ 180CEBDECDEA adj. ∠s on st. line
DCEDAE
DECDEACEB
∠−∠−°=
∠−∠−°=∠
180
180
°=∠+∠ 180DAEBCD opp. ∠s, cyclic quad.
DCEDAEBCE
DAEDCEBCE
∠−∠−°=∠
°=∠+∠+∠
180
180)(
∴ BCECEB ∠=∠
∴ BC = BE sides opp. equal ∠s
37. (a)
°=
∠=∠
50
ABCADC opp. ∠s of // gram
°=∠+∠+∠ 180AECACECAE ∠ sum of △
°=
°−°−°=∠
130
1337180AEC
2 Basic Properties of Circles
57
∵
°=
°+°=∠+∠
180
13050AECADC
∴ A, E, C and D are concyclic. opp. ∠s supp.
(b)
Join DE.
°=
∠=∠
37
CAECDE (∠s in the same segment)
∵ DC = AB (opp. sides of // gram)
and AB = EC (given)
∴ DC = EC
∴
°=
∠=∠
37
CDEDEC (base ∠s, isos. △)
°=∠
°=°+°+∠
∠°=∠+∠+∠
106
1803737
) of sum (180
ECD
ECD
CEDCDEECD △
°=∠
°=°+°+∠
°=°+∠+∠
°=∠+∠
24
18050106
18050)(
180
ECB
ECB
ECDECB
ABCBCD (int. ∠s, CD // BA)
38. (a) (i) Consider △ABM and △CDM.
∠ABM = ∠CDM ∠s in the same segment
∠BAM = ∠DCM ∠s in the same segment
∠AMB = ∠CMD vert. opp. ∠s
∴ △ABM ~ △CDM AAA
∴ CM
AM
CD
AB= corr. sides, ~ △s
(ii) ∵ OL ⊥ AB given
∴ AL = LB
ABAL2
1=
∵ ON ⊥ CD given
∴ CN = ND
CDCN2
1=
Join LM and NM.
Consider △ALM and △CNM.
CM
AM
CD
AB
CD
AB
CN
AL
=
=
=
2
1
2
1
from (a) (i)
∠LAM = ∠NCM ∠s in the same segment
∴ △ALM ~ △CNM ratio of 2 sides, inc. ∠
(b)
Join OM, OR and OW.
∵ PM = MQ given
∴ OM ⊥ PQ line joining centre to
mid-pt. of chord ⊥ chord
∵
°=
°+°=
∠+∠
180
9090
RMORLO
∴ RLOM is a cyclic opp. ∠s supp.
quadrilateral.
∠ROM = ∠RLM ∠s in the same segment
∵
°=
°+°=
∠+∠
180
9090
WMOWNO
∴ WNOM is a cyclic
quadrilateral. opp. ∠s supp.
∠WOM = ∠WNM ∠s in the same segment
∵ △ALM ~ △CNM proved in (a)(ii)
∴ ∠ALM = ∠CNM corr. ∠s, ~ △s
∴ ∠ROM = ∠WOM
(c) Consider △ROM and △WOM.
WOMROM ∠=∠ proved in (b)
OMOM = common side
°=∠=∠ 90OMWOMR proved in (b)
∴ △ROM ≅ △WOM ASA
∴ RM = WM corr. sides, ≅ △s
∴ M is also the mid-point of RW.
Multiple Choice Questions (p. 2.73)
1. Answer: B
∵ OC ⊥ AB (given)
∴ AC = CB = 2 cm (line from centre ⊥ chord bisects
chord)
cm 12
cm 2422
22
=
−=
−= ACOAOC
(Pyth. theorem)
Area of △AOB
2
2
2
cm 34
cm 122
cm 12)22(2
1
2
1
=
=
×+×=
××= OCAB
2. Answer: B
For I, PQ ⊥ CD (given)
∴ I is true.
For II, ∵ AB = CD = 12 cm, OP ⊥ AB and OQ ⊥ CD
∴ OP = OQ (equal chords, equidistant from
centre)
∴ PQOQ2
1=
∴ II is true.
line from centre ⊥
chord bisects chord
line from centre ⊥
chord bisects chord
NSS Mathematics in Action 5A Full Solutions
58
For III, ∵ OP ⊥ AB (given)
∴ AP = PB (line from centre ⊥ chord bisects
chord)
∴
cm 6
cm 122
1
2
1
=
×=
= ABAP
Consider △AOP.
cm 8
cm 610 22
22
=
−=
−= APOAOP
(Pyth. theorem)
cm 18
cm 16
cm 82
2
≠
=
×=
= OPPQ
∴ III is not true.
∴ Only I and II are true.
3. Answer: A
∠BCD = 90° (∠ in semi-circle)
°=
°−°=
∠−∠=∠
19
7190
ACBBCDDCA
°=
∠=∠
19
DCAABD
(∠s in the same segment)
∴
°=
°+°=
∠+∠=
51
3219
BACABDx
(ext. ∠ of △)
4. Answer : D
For I, ))
CBDCCBDC :: ≠ in general
∴ I may not be true.
For II,
chords) equal arcs, (equal
2
1
2
1
CBAC
CB
CBCB
DCADAC
=
=
+=
+=
)
)))))
∴ AC : CB = 1 : 1
∴ II must be true.
For III, °=∠=∠ 90ACBADB (∠ in semi-circle)
∵ AC = CB (proved in II)
∴ ∠ABC = ∠CAB (proved in II)
∴ ∠ABC + ∠CAB
°=∠
°=°+∠
°=+∠
45
180902
180
CAB
CAB
ACB (∠ sum of △)
∴ 1:245:90: =°°=∠∠ CABADB
∴ III must be true. ∴ Only II and III must be true.
5. Answer: B
°=
°−°=∠
∠=∠+∠
19
2443EBD
BDCEBDBED
(ext. ∠ of △)
Join AD.
°=∠ 90ADB (∠ in semi-circle)
°=
°=+°+°+°
°=∠+∠
28
180)19()4390(
180
x
x
ABCADC
(opp. ∠s, cyclic quad.)
6. Answer: C For option C,
x
ADBACB
=
∠=∠
(∠s in the same segment)
θ=
∠=∠ BACBDC (∠s in the same segment)
In △CDP,
zyx
xzxy
DCPPDCCPD
−−−°=
°=++++
∠°=∠+∠+∠
2180
180)()(
) of sum ( 180
θ
θ
△
∴ C must be true.
7. Answer: C
∠ADC + ∠ABC = 180° (opp. ∠s, cyclic quad.)
°=
°×=∠
∠−°=∠
+
+=
∠−°
∠
=∠
∠
∠−°=∠
75
18012
5
)180(7
5
86
37
180
180
ABC
ABCABC
ABC
ABC
ABC
ADC
ADC
ABC
ABCADC
))
(arcs prop. to ∠s at ⊙ce)
8. Answer: A
Join BD.
°=∠ 90ADB (∠ in semi-circle)
°=
°×=
∠=∠
24
482
1
2
1CODCBD
(∠ at centre twice ∠ at ⊙ce)
°=
°+°=
∠+∠=
114
2490
CBDADBx
(ext. ∠ of △)
2 Basic Properties of Circles
59
9. Answer: B
yDCB
BADDCB
−°=∠
°=∠+∠
180
180
(opp. ∠s, cyclic quad.)
∵ OD = OC (radii)
∴
y
OCDODC
−°=
∠=∠
180
(base ∠s, isos. △)
x
ABODOC
=
∠=∠
(corr. ∠s, OD // BA)
∴
°−=
°=−°+−°+
°=∠+∠+∠
1802
180)180()180(
180
yx
yyx
OCDODCDOC (∠ sum of △) ∴ B must be true.
10. Answer: B
°=∠+∠ 180ABCADC (opp. ∠s, cyclic quad.)
yACD
AEDACD
xADC
−°=∠
°=∠+∠
−°=∠
180
180
180
(opp. ∠s, cyclic quad.)
°=+
°=°+−°+−°
°=∠+∠+∠
225
18045)180()180(
180
yx
xy
CADADCACD
11. Answer: B
°=∠=∠ 52CABCDB ∴ ABCD is a cyclic quadrilateral.
(converse of ∠s in the same segment)
°=∠
°=°+∠+°
°=∠+∠
56
18072)52(
180
ADB
ADB
CBAADC
(opp. ∠s, cyclic quad.)
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.)
For II, ∠ABE = ∠DEB (given)
∠DEB = ∠EFC (corr. ∠s, BE // CF)
∴ ∠ABE = ∠EFC
∴ BCFE is a cyclic quadrilateral.
(ext. ∠ = int. opp. ∠)
For III, ∠DFC = ∠ABE (ext. ∠, cyclic quad.)
∠ABE + ∠BAD = 180° (int. ∠s, AD // BE)
∴ ∠DFC + ∠BAD = 180°
∴ ACFD is a cyclic quadrilateral. (opp. ∠s supp.)
HKMO (p. 2.76)
1.
5
34 22
22
=
+=
+= BCABAC
(Pyth. theorem)
Consider △AMN and △ABC.
∠AMN = ∠NBC (ext. ∠, cyclic quad.)
∠ANM = ∠ACB (ext. ∠, cyclic quad.)
∠MAN = ∠BAC (common angle)
∴ △AMN ~ △ABC (AAA)
∴ AC
AN
BC
MN
AB
AM== (corr. sides, ~ △s)
Area of △AMN MNAM ××=2
1
Area of △ABC BCAB ××=2
1
∴ 2
4
1
4
1
2
12
1
of Area
of Area
=
×=
××
××=
AC
AN
BC
MN
AB
AM
BCAB
MNAM
ABC
AMN△△
∴
2
5
2
1
5
2
1
=
=
=
AN
AN
AC
AN
2
3
2
54
=
−=
−= ANABNB
2
53
2
45
94
9
32
3 2
2
22
=
=
+=
+
=
+= BCNBNC (Pyth. theorem)
∵ ∠NBC = 90° (given)
∴ NC is a diameter of the circle. (converse of
∠ in semi-circle)
∴ Radius of circle BNMC
4
53
2
53
2
1
2
1
=
×=
= NC
NSS Mathematics in Action 5A Full Solutions
60
2. Let O be the centre of the circle.
Join OM, OA and OF. Let MN intersect OA at P.
Let BC = 2a.
Then AB = AC = BC = 2a,
and ∠B = ∠C = ∠BAC = 60° (prop. of equil. △)
∵ AM = MB = a and AN = NC = a (given)
∴ MN // BC and MN = BC2
1 = a (mid-pt. theorem) ∴
°=
∠=∠
60
BAMP
(corr. ∠s, MN // BC)
Since △ABC is an equilateral triangle, O is not only the
circumcentre of △ABC, but also its incentre, i.e. AO
bisects ∠BAC.
∴ °=°
=∠ 302
60MAP and AP ⊥ MN
In △AMP,
2
60cosa
AMMP =°=
and 2
360sin
aAMAP =°=
In △AMO,
3
32
3
2
30cos
aaAMAO ==
°= ∴ 6
3
2
3
3
32
a
aa
APAOPO
=
−=
−=
In △POF,
2
5
6
3
3
3222
22
a
aa
OPOFPF
=
−
=
−=
(Pyth. theorem)
∴ a
aa
PFMPMF
+=
+=
+=
2
51
2
5
2
2
512
51
+=
+
==a
a
MN
MFd
3. Let O be the centre of the circle.
Radius of circle
cm 5
cm 102
1
=
×=
Join OA, OB and OC.
Construct OD ⊥ BC and OE ⊥ AB.
OC = OB (radii)
CD = BD (line from centre ⊥ chord
bisects chord)
OD = OD (common side)
∴ △OCD ≅ △OBD (SSS)
∴ ∠COD = ∠BOD (corr. ∠s, ≅ △s)
OA = OB (radii)
AE = BE (line from centre ⊥ chord
bisects chord)
OE = OE (common side)
∴ △OAE ≅ △OBE (SSS)
∴ ∠AOE = ∠BOE (corr. ∠s, ≅△s)
Let ∠COD = α and ∠AOE = β.
Then, CD = 5 sin α and AE = 5 sin β
)sin(sin10
sin10sin10
22
βα
βα
+=
+=
+=+ CDAEBCAB
AB + BC attains its maximum when α = β = 45°.
∴
cm 210
cm 2
2
2
210
)45sin45(sin10
=
+=
°+°≤x
∴ The greatest possible value of x is 210 cm.
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