4bch06(basic properties of circles 1)

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

∴


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


1. ( at centre twice at ☉

ce

)

2.

( at centre twice at ☉ce

)

3.


)

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

△


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)


2.


(adj. s on st. line)


4


3.

(arcs prop. to s at ☉ce)

4.


(arcs prop. to s at ☉

ce

)



ce

)

5.



ce

)

( sum of

△)

p. 38119

1.

(ext. , cyclic quad.)

( sum of

△)

2. (opp. s, cyclic quad.)



ACD = 90

Exercise

Exercise 6A (p.10 10)Level 11. ∵ ON AB (given)

∴

5

( in semi-circle)( sum of △)


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


∴

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


∴

8


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


∵ 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

∴


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


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


∵ 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 △)


(ext. of △)


(ext. of △)

5. Reflex

( at centre

twice at ☉

ce

)

12


(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

△


9.

Join OC.


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)


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 △)


)

(ext. of △)

(s at a pt.)


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 △)


∴ (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)


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



(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


∵ 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


ce

)

12. BAD = 90 ( in semi-circle)BAC = BAD – CAD

= 90 – 50= 40


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


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


Exercise 6D (p. 39)Level 11. (ext. , cyclic quad.)



∵ CD = CB (given)∴ BDC = x (base s, isos. △)

∴ ( sum

of △)

3. ACB = 90 ( in semi-circle)( sum of

△)

(opp. s, cyclic quad.)





( sum of △)

6. (a) (ext. of

△)

(b) (ext. , cyclic quad.)

(

sum of △)

20




(ext. , cyclic quad.)∵ AD = AB (given)


( 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 △)


∵ BC = CD (given)

∴ (equal chords, equal s)

( sum of △)

21


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 △)


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


15. ( sum of △)



∴

16.

Join BD.Let CBE = x.∵ CE = CB (given)


( sum of

△)


DBE = 90 ( in semi-circle) (adj. s

on st. line)

(ext. of △)

∴

17. (a) (ext. of

△)

(ext. of

△)

(b) (opp. s, cyclic quad.)

23


24


( sum of △)

25


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


∴

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



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







7.

∴

(

sum

of △)

8. (a) OC = OB (radii)OCB = OBC (base s,

isos. △)( sum

of △)

(b)


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



10.

Draw OM such that OM BC.∵ OM BC (constructed)∴ BM = MC (line from centre chord

bisects chord)

∴

Consider △OMC.(Pyth. theorem)


∴

11.

Join BD. ABD = 90 ( in semi-circle)


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)

∴


∴ The radius of the circle is 13 cm.

29

(line from centre chordbisects chord)


Level 2

16. (adj. s on st. line)

(alt. s, CD // BA)

(ext. of △)


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




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





(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


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)


∴ (ext. of △)

33



bisects


3. Answer: DBDC = 90 ( in semi-circle)

( sum of △)

AB = AD (given)ABD = ADB (base s, isos. △)

∴ ( sum of △)

4. Answer: D(ext. of △)


∴ ( 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)





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


(opp. s,cyclic quad.)

(opp. s,cyclic quad.)



∴ ( sum of △)

36


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)


ce

)

(ext. of △)

37




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)

4bch06(basic properties of circles 1)

Documents