4bch06(basic properties of circles 1)
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Certificate Mathematics in Action Full Solutions 4B
6 Basic Properties of Circles (I)• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
1
7 Basic Properties of Circles (I)
Activity
Activity 6.1 (p. 14)2. (c) AOB = 2APB 3. No matter where points B and P are, AOB = 2APB.
(or any other reasonable answers)
Activity 6.2 (p. 25)1. yes 2. yes 3. yes 4. yes
Activity 6.3 (p. 35)3.
4. The sum of the opposite angles of a cyclic quadrilateral is 180.
Follow-up Exercise
p. 3Element TermAB minor arc
region BCE major arc
AFB diameter
region BECFA chord
AB major segment
OB minor segment
region OBEC sector
AC radius
p. 7
1. (line from centre chord bisects
chord)
2. OND = 90 (line joining centre to mid-pt. of chord chord)
Consider △OND. ( sum
of △)
3. Consider △OEM.
(Pyth. theorem)
(line from centre chord bisects chord)
p. 91. ∵ ON = OM = 4 cm (given)
∴ CD = AB (chords equidistant from centre are equal)= 7 cm
CN = ND (line from centre chord bisects chord)
=
=
2. (line from centre chord bisects chord)
∴
∵ OQ = OP = 2 cm (given) (given) ∴ BC = AB (chords equidistant from centre are equal)
= 8 cm
∴
(line from centre chord bisects chord)
3. MB = AM (line from centre chord bisects chord)
∴
CN = ND (line from centre chord
bisects chord)
∴
∵ AB = CD
∴ (equal chords, equidistant from centre)
p. 18
2
Certificate Mathematics in Action Full Solutions 4B
1. ( at centre twice at ☉
ce
)
2.
( at centre twice at ☉ce
)
3.
( at centre twice at ☉ce
)
4. Reflex AOB = 360 – 140 (s at a pt.)= 220
( at centre twice at ☉
ce
)
5. ( in
semi-circle)
6. APB = 90 ( in
semi-circle)Consider △APB.
( sum of △)
p. 20
1. (s in the same segment)
2. ( in semi-circle)(s in the same segment)
3.
4. Consider △ABD.
( sum of
△)
(s in the same
segment)
p. 281. ∵ DOC = AOB = 43 (given)
∴ = (equal s, equal arcs)
∴
2. ∵ = (given)
∴ CD = AB (equal arcs, equal chords)∴
3. ∵ AB = DC (given)∴ (equal chords, equal s)∵ DC = AB (given)
∴ = (equal chords, equal arcs)
∴
3
△
7 Basic Properties of Circles (I)
4.
Join OB.∵ BC = ED (given)
∴ (equal chords, equal
s)
∵ AB = BC (given)
∴ (equal chords, equal
s)
p. 311.
(arcs prop. to ∠s at centre)
(arcs prop. to ∠s at centre)
2.
(arcs prop. to ∠s at centre)
(adj. s on st. line)
(arcs prop. to ∠s at centre)
4
Certificate Mathematics in Action Full Solutions 4B
3.
(arcs prop. to s at ☉ce)
4.
(arcs prop. to s at ☉ce)
(arcs prop. to s at ☉
ce
)
(arcs prop. to s at ☉ce)
(arcs prop. to s at ☉
ce
)
5.
(arcs prop. to s at ☉ce)
(arcs prop. to s at ☉
ce
)
( sum of
△)
p. 38119
1.
(ext. , cyclic quad.)
( sum of
△)
2. (opp. s, cyclic quad.)
(ext. , cyclic quad.)
3. (opp. s, cyclic quad.)
ACD = 90
Exercise
Exercise 6A (p.10 10)Level 11. ∵ ON AB (given)
∴
5
( in semi-circle)( sum of △)
7 Basic Properties of Circles (I)
Join OB.Consider △NOB.
(Pyth. theorem)
∴ The radius of the circle is 10 cm.
2.
Join OB.(radii)
Consider △ONB.(Pyth. theorem)
∵ AN = NB (line from centre chord bisects chord)
∴
3. ∵ AM = MB (line from centre chord bisects chord)
∴
∵ ON = OM (given)
∴
(chords equidistant from centre are equal)
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
(adj. s on st.
line)
Consider quadrilateral OMKN.
5. ∵ OM AB (given)∴ AM = MB (line joining from centre
chord bisects chord)
∴
Consider △OMB.
(Pyth. theorem)
Join OD.Consider △OND.
(radii)
(Pyth. theorem)
∵ ON CD (given)∴ CN = ND (line from centre chord
bisects chord)
6
Certificate Mathematics in Action Full Solutions 4B
∴
6. ∵ CM = MD (given)∴ OM CD (line joining centre
to mid-pt. of chord chord)
∴ OMC = 90(ext. of
△)
Consider △OAC.∵ OC = OA (radii)∴ OCA = OAC (base s, isos. △)
( sum
of △)
7.
∴
Join OD.Consider △OMD.
(radii)
(Pyth. theorem)
∵ OM CD (given)∴ CM = MD (line from centre chord
bisects chord)
∴
8. ∵ BM = MC = 6 cm (given)∴ OM BC (line joining centre to
mid-pt. of chord chord)Consider △OMB.
(Pyth. theorem)
Consider △OMD.(Pyth. theorem)
7
7 Basic Properties of Circles (I)
∴
8
Certificate Mathematics in Action Full Solutions 4B
9. 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 are shown.
Draw OM and ON such that OM AB and ON BC.
∴ △OBM △OBN AAS∴ OM = ON corr. sides, △s∴ PQ = RS chords equidistant from
centre are equal
Level 210. (a) ∵ AM = MB (given)
∴ OMA = 90 (line joining centre to mid-pt. of chord chord)
∴
(b)
Join ON.∵ CN = ND (given)∴ OND = 90 (line joining centre to
mid-pt. of chord chord)
∵ CD = AB (given)∴ ON = OM (equal chords,
equidistant from centre)
∴ ONM = OMN (base s, isos. △)= 15
∴
∴ OND = 90 (line joining centre to mid-pt. of chord chord)
SMEFSU08EX@F04
Construct the traingle ABC as shown.
∴ Area of △ABC 110.
Let M be a point on AB such that OM AB.
9
7 Basic Properties of Circles (I)
∵ OM AB (constructed)
∴
Consider △OMA.(Pyth. theorem)
Consider △OMC.
(Pyth. theorem)
12. (a) Consider △OAB and △OAC.
∴ △OAB △OAC SSS∴ OAB = OAC corr. s, △s∴ OA bisects BAC.
(b) Consider △ABN and △ACN.
∴ △ABN △ACN (SAS)∴ BN = CN (corr. sides, △s)∴ ON BC (line joining centre to
mid-pt. of chord chord)
(c)
Consider △ONC.(Pyth. theorem)
Consider △ANC.(Pyth. theorem)
13. (a)
(b) ∵ ON AB (given)
∴
Join OA.Consider △OAN.OA = r cm (radius)
(Pyth. theorem)
14. ∵ OM CD (given)
∴
Let r cm be the radius of the circle.
Join OC.Consider △OCM.OC = r cm (radius)
(Pyth. theorem)
∴ MB = OB OM
10
∴
Certificate Mathematics in Action Full Solutions 4B
15.
Let M be a point on AB such that OM AB.∵ OM AB (constructed)
∴
Join OB.OB = 13 cm (radius)Consider △OMB.
(Pyth. theorem)
Let N be a point on CD such that ON CD.∵ ON CD (constructed)
∴
∵
∴ ONKM is a rectangle.∴ NK = OM (property of rectangle)
∴
16.
(a) Join OD, OB and OA as shown.Let OAB = x,then OAD = 90 x.∵ OB = OA radii∴ OBA = OAB base s,
11
7 Basic Properties of Circles (I)
isos. △= x
∴
sum of △
∵ OD = OA radii∴ ODA = OAD base s,
isos. △= 90 x
∴
sum of △
∴
∴ BOD is a straight line.
(b) Draw OM AB and ON DA.∵ OM AB and ON DA (constructed)∴ AM = MB and DN = NA (line from centre
chord bisects chord)
∵ AMON is a rectangle.
∴ (property of rectangle)
Consider △OAM.(Pyth. theorem)
17. (a) ∵ ALM = BMN = CNG = 90 given∴ LA // MB // NC corr. s
equal∵ LA // MB // NC and AB = BC given∴ LM = MN intercept
theorem
(b) EM = MF line from centre chord bisects chord
DL = EL line from centre chord bisects chord
FN = NG line from centre chord bisects chord
Exercise 6B (p. 21)
Level 1
1. (adj. s on st. line)
(
at centre twice at ☉ce
)
2. ACB = 90 ( in semi-circle)∵ CA = CB (given)∴ x = CBA (base s, isos. △)
( sum of △)
3. (s in the same segment)
(ext. of △)
4. (s in the same segment)
(ext. of △)
5. Reflex
( at centre
twice at ☉
ce
)
12
Certificate Mathematics in Action Full Solutions 4B
(s at a pt.)
6. ( sum
of △)
( in semi-circle)
∴
7. ABC = 90 ( in semi-circle)(s in the same segment)
( sum
of △)
8. ( at centre twice at ☉ce
)
( sum of △)
13
△
7 Basic Properties of Circles (I)
9.
Join OC.
( at centre twice at ☉
ce
)
OC = OA (radii)OCA = OAC (base s, isos. △)
= 20OB = OC (radii)
x = OCB (base s, isos. △)= ACB OCA= 65 20=
10. ( in semi-circle)(
sum of △)
(ext. of △)
11. BOD = 36 (opp. s of // gram)
( at centre twice at ☉
ce
)
(alt. s, DO // AC)
(ext. of △)
12.
13. DAC, ACD, DAB, DBA, EFD and FED(any four of the above angles)
Level 214. ∵ DC = DA (given)
∴ DCA = x (base s, isos. △)∵ BD = BC (given)
∴ (base s, isos. △)
( in semi-circle)
( sum
of △)
( sum of △)
15. (a)
(int. s, BA // CO)
(int. s, BA // CO) ∴ Reflex AOC = 360 AOC
= 360 148=
x
(s at a pt.)
(b)
( at centre
14
△
(ext. of △)
( at centre twice at ☉ce
)
(ext. of △)
(s at a pt.)
Certificate Mathematics in Action Full Solutions 4B
twice at ☉
ce
)
16.
( ext.
of △)
(s in the same segment)
( in semi-circle)( sum
of △)
∴
17.
ABD + BAD + ADB = 180 ( sum of △)ABD + (44 + BAC) + 90 = 180
ABD + ABD + 134 = 180
18.
(ext. of △)
(s in the same segment)
∴ (ext.
of △)
19. (a) OABC is a parallelogram. (given)OA = OC (radii)∴ OABC is a rhombus.
(b) Reflex AOC = 360 x s at a pt.
at centre
twice at ☉
ce
(c) ABC = x (opp. s of // gram)
∴ (proved in (b))
20. (a) AKB = DKC vert. opp. sBAK = CDK s in the same segmentABK = DCK s in the same segment∴ △AKB ~ △DKC AAA
(b) (corr. sides, ~ △s)
∴
21. (a) ∵ OK EB given∴ BK = EK line from centre
chord bisects chord
∴ △BKD △EKD SAS
(b) ABD = 90 ( in semi-circle)
BDC = 180 – CBD – BCD ( sum of △)= 180 – 90 – 42 = 48KED = KBD (corr. s,
△s)KED + KBD = BDC (ext. of △)
2KED = 48
15
( at centre twice at ☉ce)
7 Basic Properties of Circles (I)
KED = 24
∴ (s in the same segment)
22.
Join AP.APB = 90 in semi-circle
∴ QC AB
23.
Join OA.ABQ = AOQ s in the same segment
= 2ABP at centre twice at ☉ce
∴ BP bisects ABQ.
Exercise 6C (p. 32)Level 11. Reflex AOB = 360 – AOB (s at a pt.)
= 360 – 80= 280
(arcs prop. to s at
centre)
2. BAC = 180 – ABC – ACB ( sum of △)= 180 – 50 – 75= 55
3. ACB = 90 ( in semi-circle)BAC = 180 – x – ACB ( sum of △)
= 180 – x – 90= 90 – x
(arcs prop. to s at ⊙
ce
)
4. (arcs prop. to s at centre)
∴
5. ADC = BAD alt. s, CD // AB
∴ AC = BD equal s, equal arcs
6. (a) (arcs prop. to s at centre)
16
⊙ce
( at centre twice at ☉ce)
Certificate Mathematics in Action Full Solutions 4B
(b) ( at centre
twice at ☉ce)
7. BEC = 180 – EBC – ECB ( sum of △)= 180 – 62 – 64= 54
DEB = EBC EDB (ext. of △)= 62 – 35= 27
∴ AB : BC = DEB : BEC
(arcs prop. to s at = 27 : 54 ⊙ce) =
8. BAC = 90 ( in semi-circle)BAD = BAC + CAD
= 90 + 30= 120
∵ AB = AD (given)∴ ABD = ADB (base s, isos. △)
( sum
of △)
( sum
of △)
∴ AB : BC = ACB : BAC
(arcs prop. to s= 60 : 90 at ⊙ce)=
9. ∵ APC = APB + BPC= 5 + 10= 125= CPD
∴ AC = CD(equal s, equal arcs)
∵ BPD = BPC + CPD= 10 + 15= 25= EPF
∴ BD = EF (equal s, equal arcs)
∵ APD = APB + BPC + CPD= 5 + 10 + 15= 30 = FPG
∴ AD = FG(equal s, equal arcs)
10. ∵ AB = CD(given)
∴ ADB = DACand ACB = DBC (equal arcs, equal s)∴ KD = KA and KC = KB (sides opp. equal s)∴ △AKD and △BKC are isosceles triangles.
17
7 Basic Properties of Circles (I)
∵ AB = BC = CD
(given)∴ ACB = CABand BDDBC = DBC (equal arcs, equal s)∴ BC = BA and CD = CB (sides opp. equal s)∴ △ABC and △BCD are isosceles triangles.
Level 211. ABD = 180 – BAD ADB ( sum of △)
= 180 – (40 + 20) – 70= 50
Join BC.CBD = CAD (s in the same segment)
= 20CBA = CBD + ABD
= 20 + 50= 70
(arcs prop. to s at ☉
ce
)
12. BAD = 90 ( in semi-circle)BAC = BAD – CAD
= 90 – 50= 40
(arcs prop. to s at ☉
ce
)
13. PRS = PQS (s in the samesegment)
= 30PRQ = QRS – PRS
= 75 30= 45
∵ QR = PQ (given)
∴ QSR = PRQ (equal arcs, equal s)
= 45RQS = 180 – QRS – QSR ( sum of △)
= 180 75 45=
14. (a) OD = OB (radii)ODB = OBD (base s, isos. △)
= 30∴ BOA = OBD + ODB (ext. of △)
= 30 + 30=
(b) (arcs prop. to
s at
☉
ce
)
15. (a)
∴ △ABO △CBO SSS
(b) ∵ AOB = COB corr. s, △s∴ AOD = COD
∴ AD = DC equal s,
equal arcs
18
Certificate Mathematics in Action Full Solutions 4B
16. ∵ BC = CD given
∴ CAB = DAC equal arcs, equal s∵ OC = OA radii∴ ACO = CAB base s, isos. △
= DAC∴ OC // AD alt. s equal
17. (a) ∵ OE BD given∴ BE = ED line from centre
chord bisects chord AE = AE common sideAEB = AED = 90 given∴ △ABE △ADE SAS
(b) BAC = DAC corr. s, △s
∴ BC = CD equal s,
equal arcs
18. (a) With the notations in the figure,
DFE = BDF + DBF (ext. of △)= 20 + 30= 50
AGE = CAG + ACG (ext. of △)= 40 + 50= 90
∴ x = 180 – FGE – GFE ( sum of △)= 180 – 90 – 50 =
(b) AB : BC : CD :
DE : EA
= ADB : BEC : CAD : (arcs prop. to s at ☉ DBE : ACE (arcs prop. to s at ☉ce) to s at △ce)
= 20 : 40 : 40 : 30 : 50=
(cb)
(by (b))
Circumference of the circle =
= 9 cm
∴ Radius of the circle =
=
19
7 Basic Properties of Circles (I)
Exercise 6D (p. 39)Level 11. (ext. , cyclic quad.)
(adj. s on st. line)
2. (opp. s, cyclic quad.)
∵ CD = CB (given)∴ BDC = x (base s, isos. △)
∴ ( sum
of △)
3. ACB = 90 ( in semi-circle)( sum of
△)
(opp. s, cyclic quad.)
4. (s in the same segment)
(opp. s, cyclic quad.)
5. (opp. s, cyclic quad.)
(opp. s, cyclic quad.)
( sum of △)
6. (a) (ext. of
△)
(b) (ext. , cyclic quad.)
(
sum of △)
20
Certificate Mathematics in Action Full Solutions 4B
7. (opp. s, cyclic quad.)
(ext. , cyclic quad.)
(ext. , cyclic quad.)∵ AD = AB (given)
∴ (base s, isos. △)
( sum of △)
8.
Join AD.ABC + CDA = 180 opp. s, cyclic quad.ADE = 90 in semi-circleABC + CDE
9. (a) ( at centre twice at
☉
ce
)
(b) (opp. s, cyclic quad.)
( sum of △)
10. ( sum of △)
(opp. s, cyclic quad.)
∵ BC = CD (given)
∴ (equal chords, equal s)
( sum of △)
21
7 Basic Properties of Circles (I)
Level 211. Reflex AOC 2ABC ( at centre twice
2 110 at ☉ce)
220(s at a
pt.)
∴ (opp. s, cyclic quad.)
12. (corr. s, OC // AB)
OD OC (radii)ODC = OCD (base s, isos. △)
( sum
of △)
(opp. s, cyclic quad.)
13. (a) KAD = KCB ext. , cyclic quad.KDA = KBC ext. , cyclic quad.AKD = CKB common angle∴ △KAD ~ △KCB AAA
(b) (corr. sides,
~ △s)
14. (ext. , cyclic quad.)
ADB = 90 ( in semi-circle)( sum
of △)
(alt. s, DC // AB)
(ext. of △)
22
Certificate Mathematics in Action Full Solutions 4B
15. ( sum of △)
(opp. s, cyclic quad.)
(opp. s, cyclic quad.)
∴
16.
Join BD.Let CBE = x.∵ CE = CB (given)
∴ (base s, isos. △)
( sum of
△)
(opp. s, cyclic quad.)
DBE = 90 ( in semi-circle) (adj. s
on st. line)
(ext. of △)
∴
17. (a) (ext. of
△)
(ext. of
△)
(b) (opp. s, cyclic quad.)
23
7 Basic Properties of Circles (I)
24
Certificate Mathematics in Action Full Solutions 4B
( sum of △)
25
7 Basic Properties of Circles (I)
Revision Exercise 6 (p. 47)Level 11.
Join OF.Draw ON such that ON FE.
(radii)
(property of rectangle)
Consider △ONF.(Pyth. theorem)
∵ ON FE (constructed)∴ FN = NE (line from centre
chord bisects chord)
∴
2. AEC = 90 ( in semi-circle)(s in the same segment)
(ext. of △)
3. (a) (int. s, OR // PQ)
∴ Reflex
(s at a pt.)
(b)
(int. s, OR // PQ)
26
( at centre twice at ☉ce)
Certificate Mathematics in Action Full Solutions 4B
4. With the notations in the figure,
(ext. of △)
(ext. of △)
5. ∵ AD = DC (given)
∴ (equal arcs, equal s)
BCA = 90 ( in semi-circle)
(opp. s,
cyclic quad.)
6.
∴
∴
27
( at centre twice at ☉ce)
(arcs prop. to s at ☉ce)
(opp. s, cyclic quad.)
(opp. s, cyclic quad.)
(arcs prop. to s at ☉ce)
7 Basic Properties of Circles (I)
7.
∴
(
sum
of △)
8. (a) OC = OB (radii)OCB = OBC (base s,
isos. △)( sum
of △)
(b)
(opp. s, cyclic quad.)
9.
Produce CO to cut AB at E. Join BC.(alt. s, AB // DC)
(ext. of △)
OC = OB (radii)OCB = OBC (base s, isos. △)
( sum
of △)
(opp. s, cyclic
quad.)
28
( at centre twice at ☉ce)
Certificate Mathematics in Action Full Solutions 4B
10.
Draw OM such that OM BC.∵ OM BC (constructed)∴ BM = MC (line from centre chord
bisects chord)
∴
Consider △OMC.(Pyth. theorem)
Consider △OAM.(Pyth. theorem)
∴
11.
Join BD. ABD = 90 ( in semi-circle)
(s in the same segment)
12. (a) ∵ AB // OC and OA // CB (given)∴ OABC is a parallelogram.∵ OA = OC (radii)∴ OABC is a rhombus.
(b) AOC = ABC (property of rhombus)
Reflex AOC = 2ABC ( at centre twice at ☉ce)
(s
at a pt.)
13.
Join BD and DC.ABD = 90 in semi-circleACD = 90 in semi-circle∴ ABD = ACD
AD = AD common sideAB = AC given
∴ △ABD △ACD RHS∴ BAD = CAD corr. s, △s∴ AD bisects BAC.
14. ACD = p + q (ext. of △)BDC = p (s in the same segment)
(ext. of △)
15.
Let O be the centre of the circle and r cm be the radius.Join OA.OA = r cm (radius)
∵ OM AB (given)
∴
Consider △OAM.(Pyth. theorem)
∴ The radius of the circle is 13 cm.
29
(line from centre chordbisects chord)
7 Basic Properties of Circles (I)
Level 2
16. (adj. s on st. line)
(alt. s, CD // BA)
(ext. of △)
(adj. s on st. line)
17. (a) (opp. s, cyclic quad.)
(ext. , cyclic
quad.)
(b)
18. (a) ADC = 90 ( in semi-circle)( sum
of △)
(b) BDC = BAC (s in the same segment)
= 16
( sum of △)
19. (a) ACB = 90 ( in semi-circle)
DBA = DCA (s in the samesegment)
(ext. of
△)
( sum of △)
(b)
30
( at centre twice at ☉ce)
( at centre twice at ☉ce)
Certificate Mathematics in Action Full Solutions 4B
20.
Join MN.ABM = MNC (ext. , cyclic quad.)ADM = MNE (ext. , cyclic quad.)ABM + ADM = MNC + MNE
= 180 (adj. s on st. line)∴
21.
Join BE.
∴ ( sum of △)
22. (a) (arcs prop. to s at centre)
∴
∴
(b) ∵ OC = OA (radii)∴ ACO = CAO (base s, isos. △)
( sum of △)
∴ (ext. of △)
23. (a) APD = CPB common angle PAD = PCB ext. , cyclic quad.PDA = PBC ext. , cyclic quad.
∴ △PAD ~ △PCB AAA
(b) AKB = DKC vert. opp. sBAK = CDK s in the same segmentABK = DCK s in the same segment∴ △AKB ~ △DKC AAA
(c) (corr. sides,
~ △s)
∴
31
( at centre twice at ☉ce)
(arcs prop. to s at ☉ce)
(arcs prop. to s at ☉ce)
7 Basic Properties of Circles (I)
(corr. sides, ~
△s)
∴
24. (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 BM = CN given∴ KM – BM = KN – CN∴ KB = KC
(c) KB =KC proved in (b)AB =DC given
KA =KD KAD = KDA base s,
isos. △∴ BCD + KDA
= BCD +KAD= 180 opp. s, cyclic quad.
∴ BC // AD int. s supp.
25. (a) ∵ AC = AB given∴ ACB =ABC base s, isos. △
= ADE ext. , cyclic quad.
∴ BC // ED corr. s equal
(b) CED = CBD s in the same segment= BDE alt. s, BC // ED
∴ FE = FD sides opp. equal s
26. NBP = MDP ext. , cyclic quad.BNP = 180 – NBP – NPB sum of △
= 180 – MDP – DPM given= DMP= NMC vert. opp. s
∴ QM = QN sides opp. equal s
27.
Join BO and OE.BOE = 2 CAE at centre
twice at ☉ce
ACE + BOE = 180 opp. s,cyclic quad.
ACE + 2CAE = 180 ACE + CAE = 180 – CAE(ACE + CAE) +CEA = 180 sum of △
(180 – CAE) + CEA = 180CAE = CEA
∴ CA = CE sides opp.equal s
28. (a) ∵ CE = CD given∴ CED = CDE base s, isos. △
= ABC ext. , cyclic quad.
∴ ABE is an isosceles triangle. sides opp. equal s
(b) Let ABD = x.
∵ CD = AD (given)
∴ DBC = ABD (equal arcs, equal s)= x
ABE = ABD + DBC= 2x
AEB = ABE (base s, isos. △)= 2x
EDC = AEB (base s, isos. △)= 2x
DCB = AEB + EDC= 4x (1)
BDC = 90 ( in semi-circle)
32
Certificate Mathematics in Action Full Solutions 4B
DBC + DCB + BDC = 180 ( sum of △)DCB = 180 – x – 90
= 90 – x (2)From (1) and (2), we have
Multiple Choice Questions (p. 52)1. Answer: B
∵ OP AB (given)
∴
(Pyth. theorem)
Join OC.OC = OA (radii)
= 10 cm∵ OQ CD (given)
∴
(Pyth. theorem)
∴
2. Answer: ABCD = 90 ( in semi-circle)
(s in the same segment)
∴ (ext. of △)
33
(line from centre chord bisects chord)
(line from centre chord bisects chord)
bisects
7 Basic Properties of Circles (I)
3. Answer: DBDC = 90 ( in semi-circle)
( sum of △)
AB = AD (given)ABD = ADB (base s, isos. △)
∴ ( sum of △)
4. Answer: D(ext. of △)
(ext. , cyclic quad.)
∴ ( sum of △)
5. Answer: CReflex AOC = 360 – x (s at a pt.)
∴ ( sum of △)
6. Answer: CBDC = BAC
= 46ACB = ADB
= x∴ ( sum of △)
7. Answer: B
(ext. of △)
34
( at centre twice at ☉c e)
(opp. s, cyclic quad.)
(s in the same segment)
(s in the same segment)
Certificate Mathematics in Action Full Solutions 4B
Join AD.ADB = 90 ( in semi-circle)
(opp. s,
cyclic quad.)
8. Answer: A( sum
of △)
9. Answer: C
10. Answer: B(opp. s, cyclic quad.)
OD = OC (radii)∴ ODC = OCD (base s, isos. △)
= 68DOC = ABO (corr. s, OD // BA)
= x
35
(arcs prop. to s at ☉ce)
(opp. s,cyclic quad.)
(opp. s,cyclic quad.)
(arcs prop. to s at ☉ce)
7 Basic Properties of Circles (I)
∴ ( sum of △)
36
Certificate Mathematics in Action Full Solutions 4B
11. Answer: B
(
sum of △)
12. Answer: AWith the notations in the figure, join FC.
(s in the same
segment)
(s in the same
segment)
For A, x = BFC + CFD = a + b
For B, if x = y,
then BCD = AFE (equal
s, equal arcs)which is not always true.
For C,∵ x = a + b and x = y is not always true.∴ y = a + b is not always true.
For D, join BC and CD.∵ x + BCD = 180 (opp. s, cyclic quad.)∴ x + y = 180 is false.
13. Answer: A
Join BD.ADB = 90 ( in semi-circle)
( at centre twice at ☉
ce
)
(ext. of △)
37
(opp. s, cyclic quad.)
(opp. s, cyclic quad.)
7 Basic Properties of Circles (I)
HKMO (p. 54)Let O be the centre of the circle.With the notations in the figure, join OC and OD.
∵ AC = CD = DB (given)
∴
Join CD.OC = OD (radii)
∴ OCD = ODC (base s, isos. △)( sum of
△)
∴ OCD = COA∴ CD // AB (alt. s equal)Consider △CAD and △COD.∵ They have the same base and the same height.∴ Area of △CAD = area of △COD Shaded area = area of sector OCD
38
(equal arcs,equal s)
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