4bch07(basic Properties Of Circles 2)

Certificate Mathematics in Action Full Solutions 4B

7 Basic Properties of Circles (II)

• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

Activity

Follow-up Exercise

p. 66

Activity 7.1 (p. 63) 1.

1.

∵

x = 25 °

(alt. ∠ s, TC // DE)

∠ OTB = 90°

(tangent ⊥ radius)

x + y = 90°

(tangent ⊥ radius)

y = 90° − 25° = 65°

2. 2.

3.

90°. IfF the angle between ST and PQA is not equal to 90°, a right-angled triangle with ST as the hypothenuse can be drawn. In siuch case, the opposite side perpendicular to PQ is shorter than ST. So, the line segment ST is shortest when ST ⊥ PQ.

OT = OC

=x

∠OTC + ∠OCT + ∠TOC = 180 ° (∠ sum of △)

180 ° − 70 ° 2 = 55 °

x =

∴

(a) OC (b) No

4.

∠OTB = 90 °

90°

(tangent ⊥ radius)

x + y = 90 ° y = 90 ° − x = 90 ° − 55 °

Activity 7.2 (p. 82) 1.

(radii) (base ∠ s, isos. △)

∠OTC = ∠OCT

(a)

(tangent ⊥ radius)

= 35 °

3.

∵

∠OTA = 90 ° AT

2

+ OT

2

= AO

(tangent 2

(Pyth. theorem)

12 2 + r 2 = ( 4 + r ) 2 144 + r 2 = 16 + 8r + r 2 (b) (i) Yes

r = 16 (ii) Yes

4.

2.

TG = TE

(given) (base ∠ s, isos. △)

∠TGE = ∠TEG = 60 ° ∠GTB = ∠TGE = 60 °

(alt. ∠ s, AB // CD)

∠OTB = 90 ° (tangent ⊥ radius) x + 60 ° = 90 ° x = 30 °

p. 69 1. 3.

Yes

x = 25 °

(tangent properties)

TQ = TP

(tangent properties)

∠OPT = 90 °

(tangent ⊥ radius)

y =6

2.

1

∠ PTO = ∠ QTO

⊥ radius)

x = 110 ° − 90 ° (ext. ∠ of △) = 20 °

7 ∠ QTO

(tangent properties))

=∠ PTO

Basic Properties of Circles (II)

of △)

y = 20 °

∠OPT = 90 °

3.

(tangent ⊥ radius)

x + 26 ° = 90 ° x = 64 °

TQ = TP

∴∵

(tangent ⊥ radius)

∠TQP = ∠TPQ = 64 °

y = 180 ° − 64 ° − 64 °

(tangent properties)

= 52 °

∠ QOP

4.

(base ∠ s, isos. △)

= 2∠ QRP x = 2 × 54 ° = 108 °

(∠ at centre twice ∠ at

⊙ ) ce

∠OPT = ∠OQT = 90 ° (tangent ⊥ radius) Consider quadrilateral OQTP. y = 360 ° − 90 ° − 90 ° − 108 ° = 72 °

∠ PQR

5.

=∠ APQ

AP = AQ ∠ APQ

=∠ AQP

alt. ∠s, AP // QR tangent properties base ∠s, isos. ∴ ∠ PQR = ∠ AQP ∴ PQ bisects ∠ RQA.

p. 75 1.

2.

∠ ATP

=∠ ABT

x = 70 °

(∠ in alt. segment)

∠ BTQ = ∠ BAT y = 45 °

(∠ in alt. segment)

∠ BAT

=∠ BTQ x = 50 °

(∠ in alt. segment)

∠ ABT

=∠ ATP

(∠ in alt. segment)

y = 74 °

3.

∠TBA = 180 ° − 35 ° − 42 ° = 103 °

∠ATQ = ∠ABT

(∠ sum of △)

(∠ in alt. segment)

x =103 °

4.

∠BAT = ∠BTP

=x

(∠ in alt. segment)

∠ATB + ∠TBA + ∠BAT = 180 ° ∠ ( sum x + 110 ° + x = 180 ° x = 35 °

2

Certificate Mathematics in Action Full Solutions 4B p. 77 1.

∠ CAT

=∠ CTQ a = 75 °

(∠ in alt. segment)

=∠ CTQ

(∠ in alt. segment)

∠ CBT

b = 75 ° ∠ BAT

2.

=∠ BTQ x = 35 °

∠ BOT

(∠ in alt. segment)

= 2∠ BAT

(∠ at centre twice ∠ at

y = 2x = 2 × 35 ° = 70 °

⊙ ) ce

3. ∠ OTQ

=90 °

(tangent

∠ ABT =∠ ATQ ∴ x =∠ ATO +∠ OTQ = 25 °+90 °

⊥radius)

(∠in alt. segment)

=115 °

∠STQ = ∠SRT

4.

=x

(∠ in alt. segment)

x + (62 ° + x ) + 40 ° = 180 ° ( ∠sum of

)

x = 39 °

p. 87 1.

∵ ∴

2.

∵ ∴

3.

∠BAD + ∠BCD = 100 ° + 70 ° = 170 ° ≠ 180 ° A, B, C and D are not concyclic.

∠ABC + 100 ° = 180 °

∴ 4.

∠ ADB = ∠ ACB = 50° (given) A, B, C and D are concyclic. (converse of ∠ s in the same segment)

∠ABC = 80 ° ∠ABC ≠ 100 °

(adj. ∠ s on st. line)

A, B, C and D are not concyclic.

∠ABD + 40 ° + 120 ° = 180 ° (∠ sum of △) ∠ABD = 20 °

∠ABC + ∠ADC = ( 20 ° + 50 °) + ( 40 ° + 70 °) ∴ 5.

∠BAD = 60 ° + 30 ° = 90 ° ∵ ∠BCE = ∠BAD = 90 ° ∴

3

= 180 °

A, B, C and D are not concyclic. (opp. ∠ s supp.)

A, B, C and D are concyclic.

(ext. ∠ = int. opp. ∠

7

Exercise

PA 2 + OA 2 = (82 + 62 ) cm 2

Exercise 7A (p. 70) Level 1 OC = OP 1. = 5 cm

= 100 cm 2 PO 2 = ( 4 + 6) 2 cm 2

(radii)

= 10 2 cm 2 = 100 cm 2

Consider △OPB.

OP

2

Basic Properties of Circles (II)

+ PB 2 = OB

2

(Pyth. theorem)

52 +12 2 = (5 + x ) 2 13 2 = (5 + x ) 2 x = 8 or x = −18 (rejected)

∴ ∴

PA + OA = PO2 OA ⊥ PA

∴

PA is the tangent to the circle at A.

2

2

converse of Pyth. theorem converse of tangent converse of tangent ⊥ radius ⊥ radius

2. converse of tangent ⊥ radius

OP = OC ∠OPC = ∠OCP 180 ° − 130 ° ∠OPC = 2 = 25 °

(radii) (base ∠s, isos. (∠sum of

)

)

∠ OPB = 90° (tangen(tangent ⊥ radius)t x + 25° = 90° x = 90° − 25° = 65° TB = TA ∠TBA = ∠TAB

3.

= 68 ° x + 68 ° + 68 ° = 180 ° x = 44 °

( tangent properties ) (base ∠s, isos. ) ( ∠sum of

4.

)

∠ATO = ∠BTO

(tangent properties)

∠ ATO + ∠ BTO + 130° = 180° 180° − 130° 2 = 25°

∠ ATO = ∠OAT

= 90 °

x = 180 ° − 90 ° − 25 ° = 65 °

( adj. ∠ s on st. line)

(tangent ⊥ radius)

(∠sum of ) 5.

OA = OB = 6 cm

radii

4

Certificate Mathematics in Action Full Solutions 4B 6.

∵ ∴ ∴

AP = AR AR = 3 cm RC = (8 − 3) cm

∵ ∴ ∵ ∴

= 5 cm RC = QC QC = 5 cm BP = BQ BQ = 2 cm

(tangent properties)

10. ∠ ABC = 90°

= 58 °

1 ∠BOD 2 1 = × 58 ° 2 = 29 °

∠BAC =

(tangent properties) (tangent properties)

BC = BQ + QC = ( 2 + 5) cm

(∠ at centre twice ∠ at ⊙ ) ce

∠ACB + ∠ABC + ∠BAC = 180° (∠ sum of △) ∠ACB = 180° − 90° − 29° = 61°

= 7 cm

7.

(tangent ⊥ radius) (∠ sum of △)

∠BOD = 180 ° − 90 ° − 32 °

∠OAR = ∠OAP

(tangent properties)

= 27 °

∠OAR + ∠AOB + ∠OBR 27 ° + 116 ° + ∠OBR ∠OBR

△)

Level 2

= 180 ° ∠ ( sum of 11. = 180 °

(

= 2∠ ACB

= 2 ×65 ° (∠ at centre twice ∠ at ⊙ ) ce

=130 °

= 37 °

(b)

∠OBQ = ∠OBR

∠ AOB

(a)

(tangent properties) (tangent properties)

∠OAT = ∠OBT = 90 ° (tangent ⊥ radius) Consider quadrilateral AOBT. ∵ Sum of the interior angles of quadrilateral ∴

AOBT

= 360 °

∠ ATB

= 360 ° − 90 ° − 90 ° −130 ° = 50 °

(tangent properties)

(tangent properties)

1 ∠TOQ 2 1 = × 68 ° 2 = 34 °

∠TPQ = 12.

=37 ° ∠ QBR

∴

8.

= 37 ° + 37 ° = 74 °

OA = OB

( ∠at centre

(radii)

∠OAB

= ∠OBA = 34 °

(base ∠s, isos.

∠AOT

= 34 ° + 34 ° = 68 ° = 90 °

(ext. ∠of

∠OAT ∠ AOT

+∠ OAT

+∠ ATO

68 ° + 90 ° + ∠ ATO ∠ ATO

9.

= 180 ° = 22 °

OT = OP

(radii)

∴

∠ PBT

+ 28 ° = 62 ° ∠ PBT

13. Let the radius of the circle be r cm. ∠ OTA = 90° (tangent ⊥ radius) OT 2 + AT 2 = AO 2 (Pyth. theorem) r 2 + 12 2 = (8 + r ) 2

14.

=∠ OPT = 34 °

r = 5 The radius of the circle is 5 cm.

TA = TB

∠TAB = ∠TBA ( tangent properties ) (base ∠s, isos.

(ext. ∠ of △)

(∠ sum of )

)

)

= 28 °

+∠ PTB

) OTB = 90 ° ∠

∴

∠PTB = 90 ° − 62 °

∠ PBT

)

r 2 + 144 = 64 + 16 r + r 2

⊥radius)

(tangent

)

(tangent ⊥ radius) ∠TPB + ∠PTB + ∠QBT = 180 ° 34 ° + (90 ° + 15 °) + ∠QBT = 180 ° ⊥radius) ∵∴ ∠QBT = 41 °

= 180 ° ( ∠sum of

∠OTP = ∠OPT

= 62 ° ∠OTB = 90 ° ( base ∠ s, isos.

5

(tangent

∠at

twice

ce

)

7 ∠ ATB

+∠ TAB

+∠ TBA (∠ = 1sum 80 ° of △)

44 ° + 2∠ TBA ∠ TBA ∠ BMQ ∠ AMQ

+∠ BMQ ∠ AMQ

−∠ PAT

= 180 ° − 42 ° −100 °

= 68 °

= 38 °

(ext. ∠ of △)

(∠

sum of △)

= 43 °

= 180 °

(adj. ∠ s of + 43 ° = 180 ° on st. line )

∠ AMQ

= 180 ° − ∠ TPA

= 180 °

+ 25 ° = 68 ° ∠ BMQ

∠ PTA

Basic Properties of Circles (II)

= 137 °

15. SMEFSU07EX@F01

(b) OT ⊥ PT

∠ATO = 90 ° − ∠PTA = 90 ° − 38 ° = 52 ° OA = OT ∠TAO = ∠ATO

(tangent ⊥ radius)

(radii) (base ∠ s, isos. △)

= 52 °

∴

∠ TOA

= 180 ° − 52 ° − 52 ° = 76 °

(∠ sum of △)

Join OA, OB and OD.

∠OAE = ∠OBC = 90 °

∵

(tangent ⊥ radius) ∠OAE + ∠OBC = 90 ° + 90 °

and AE // BF ∴ AOB is a straight line. ∠AOE = ∠DOE

= 180 °

∠BOC = ∠DOC ( tangent properties )

(adj. ∠ s on

st. line) (tangent properties ) ∠AOE + ∠DOE + ∠DOC + ∠BOC = 180 ° 2∠DOE + 2∠DOC = 180 ° 2(∠DOE + ∠DOC ) = 180 ° 2∠EOC = 180 ° ∠EOC = 90 ° 16. (a)

Reflex ∠POT

= 360 ° − 128 ° (∠ s at a = 232 °

pt.)

1 (∠ at centre twice ( reflex ∠POT ) twice ∠ at ⊙ce) 2 1 = × 232 ° 2 = 116 °

∠PQT =

(b)

∠OTQ

= 180 ° − ∠PQT = 180 ° − 116 °

OT ⊥ AB

= 64 °

∠ QTA

∴

(int. ∠ s, QP // TO

(tangent ⊥ radius)

= 90 ° − ∠ OTQ = 90 ° − 64 ° = 26 °

17. (a)

∠PAT

= ∠TCB = 100 °

(ext. ∠ , cyclic quad.)

6

Certificate Mathematics in Action Full Solutions 4B 18. SMEFSU07EX@F02

Join AB. Let ∠AOT

= θ.

∠BOT =θ OA = OB ∠OAB = ∠OBA 180 ° −∠AOB ∠OAB = 2 180 ° − 2θ = 2 = 90 ° −θ ∠CAB = 90 ° ∠CAO = ∠CAB −∠OAB = 90 ° −(90 ° −θ ) =θ ∠CAO = ∠AOT ∴

21. AP = AS, BP = BQ, CQ = CR and DR = DS (tangent properties) Perimeter of ABCD = AB + BC + CD + DA

AC // TO

tangent properties radii 22. base ∠s, isos.

∵ ∴ ∴ ∴

∠in semi - circle SMEFSU07EX@F03

alt. ∠ s equal

∠OBQ

∴

∠ABP

radii = ∠OPB = base θ ∠ s, isos. tangent ⊥ radius

= 90 ° = 90 ° − θ

= 180 ° − 90 ° − θ = ∠OBQ = 90 ° − θ

∠BPA

(tangent properties) (tangent properties)

= ∠ABP

AP = AB

∠sum of

PB = 2 cm AP =

= θ. = OP

= 90 ° − θ

PB + 9 cm = 7 cm + (6 cm − PB )

7

OQ ∠OQB ∠OPA ∠BPA

Consider △QOB.

= (15 − x ) cm AP = AR x = 15 − x x = 7.5 AP = 7.5 cm

BR + CR = 6 cm BP + CQ = 6 cm   (1) AP = AQ (tangent properties ) BP + AB = AC + CQ BP + 9 cm = 7 cm + CQ From (1), we have

∴

Join OP. Let ∠OQB

= ( x − 5) cm CR = (x −5) cm AR = [10 − ( x − 5)] cm

20. BP = BR CR = CQ ∵

AB + ( BQ + CQ ) + CD + ( DS + AS ) AB + CD + ( BQ + AS ) + (CQ + DS ) AB + CD + ( BP + AP ) + (CR + DR ) AB + CD + AB + CD 2 AB + 2CD ( 2 × 12 + 2 × 8) cm 40 cm

∠sum of

19. Let AP = x cm. AP = AR (tangent properties) BP = BQ (tangent properties) CR = CQ (tangent properties) ∵ BP = (12 −x) cm ∴ BQ = (12 −x) cm CQ = [7 − (12 − x )] cm ∵ ∴

= = = = = = =

1 1 cm

vert. opp. ∠ s

sides 23. (a)

opp. equal

∠ s

AB = BD BD = BC ∴ AB = BC

(b) ∠ BAD = ∠ BDA

tangent properties tangent properties base ∠ s, isos. △

7 ∠BAD + ∠BDA = ∠DBC 2∠BDA = ∠DBC ∠BDA = ∠ BDC = ∠ BCD

∠ TBA

ext. ∠ of △

base ∠ s, isos. △

5.

∠TPA = ∠TAP

(base ∠ s, isos. △) ∠TPA + ∠TAP = ∠ATQ(ext. ∠ of ) 2∠TAP = 74 °

ext. ∠ of △

∠TAP = 37 ° ∠BTP = ∠TAP(∠ in alt. segment) = 37 ° 37 ° + x + 74 ° = 180 ° (adj. ∠ s on st. line) x = 69 °

∠ADC = ∠BDA + ∠BDC

1 1 ∠DBC + ∠DBA 2 2 1 = × 180 ° 2 = 90 °

∴ =

6.

(∠ in alt. segment) (alt. ∠ s, CA // TB)

　　

∠ABT + ∠ACT = 180 ° (37 ° + x ) + ( 75 ° + 37 °) = 180 ° x = 31 °

(∠ in alt. segment)

(opp. ∠ s, cyclic

quad.)

∠ BTQ

= 180 ° −132 ° (adj. ∠ s on st. line) = 48 ° ∠ BAT = ∠ BTQ ( ∠in alt. segment) x = 48 °

TA = TB

∠TAB = ∠TBA (given) (base ∠ s, isos. ) ∠TAB + ∠TBA + ∠ATB 2∠TAB + 42 ° ∠TAB ∠BTQ x

7.

(a)

Consider △BCD and △CAD.

∠BDC = ∠CDA ∠BCD = ∠CAD ∠DBC = 180 ° − ∠BDC − ∠BCD ∠DCA = 180 ° − ∠CDA − ∠CAD ∴ ∴ (b) ∵ ∴

= 180 ° = 180 ° (∠ sum of = 69 ° = ∠(∠ TABin alt. segment) = 69 ° segment)

∴

△)

4.

(∠ in alt. segment)

= 37 ° ∠BCT = ∠BTP

= 37 °

2.

3.

∠TBC = ∠CTQ

= 75 ° ∠ACB = ∠TBC

Exercise 7B (p. 78) Level 1

=∠ ATP x = 115 °

(ext. ∠ of △)

TP = TA (given)

1 ∠BDC = ∠DBA 2 ∠DBC + ∠DBA = 180 ° adj. ∠ s on st. line

∠ ABT

+∠ TPB

65 ° = 25 ° + x

∠BDC + ∠BCD = ∠DBA

1.

=∠ BTP

x = 40 °

1 ∠DBC 2

2∠BDC = ∠DBA

Basic Properties of Circles (II)

∠ATB

= 90 °

∠TAB

= 180 ° − ∠ATB − ∠TBA

8.

= 180 ° − 90 ° − 65 ° = 25 °

= 25 °

(∠ in alt. segment)

AAA

△BCD ~ △CAD

BD CD 9 12 9 cm + AB AB

CD AD 12 cm = 9 cm + AB = 16 cm = 7 cm =

Consider △BCT and △CAT.

∠CTB = ∠ATC

common angle

∠ in alt. segment ∠TCB = ∠TAC ∠CBT = 180 ° − ∠CTB − ∠TCB

∠ACT ∠sum of ∠sum of

( ∠in semi - circle) ( ∠sum of )

∠BTP = ∠TAB

(a)

∠ DBC = ∠ DCA △BCD ~ △CAD

common angle ∠in alt. segment ∠sum of ∠sum of

∴ ∴

(b) ∵

= 180 ° − ∠ATC − ∠TAC

∠ CBT = ∠ ACT △BCT ~ △CAT

AAA

△BCT ~ △CAT

8

Certificate Mathematics in Action Full Solutions 4B BT CT 8 10 2 x +8 x

∴

Join BD.

CT AT 10 2 = x +8 = 25 = 17 =

∠ ABC

)

12.

∠ACB = ∠TAB (∠ in alt. segment) = 62 ° line) ∠ BAC +∠ ACB +∠ ABC 32 °+62 °+∠ ABC ∠ ABC

=180 ° =180 °

∠DAT

+ ∠ADT

= 180 ° − ∠ ADC

(opp. ∠ s, cyclic quad.)

∠CTP = ∠TAC (∠ in alt. segment) = 37 ° ∠ATC = 180 ° − ∠ATQ − ∠CTP (adj. ∠ s, on st. = 180 ° − 52 ° − 37 ° = 91 ° ∠ )ATC + ∠ABC 91 ° + ∠ABC ∠ABC

= 180 ° = 180 ° = 89 °

∠ACB = ∠CAD AB = AC ( ∠in alt. segment) ∠ ABC = ∠ACB ∴ ∠ABC = ∠CAD (ext. ∠, cyclic∴ quad.) DE is the tangent to the

given base ∠ s, isos.

converse of ∠ in alt. segment

circle at A.

+ ∠ATD

= 180 ° ∠ ( sum of = 180 ° = 67 °

TC = TA

14. ∴

11. ∴

15.

given ∠TCA = ∠TAC base ∠ s, isos. △ given BT = BA base ∠ s, isos. △ ∠BTA = ∠BAT

∠BTA = ∠TCA PA is the tangent to the circle at T. converse of ∠ in alt. segment

BD = BC ∠BDC = ∠BCD

( given) (base ∠ s, isos.

Let ∠ BCD = t.

SMEFSU07EX@F04

(opp. ∠ s, cyclic quad.)

alt. ∠ s, CB // DE

13.

60 ° + 53 ° + ∠ATD ∠ATD

△)

(∠sum of

=86 °

10.

∠DAT

= 66 ° + 32 ° = 98 °

= 180 ° − 98 ° = 82 °

2∠TAB + 56 ° = 180 ° ∠TAB = 62 °

∠ATQ 53 ° ∠BCD 60 °

(∠ in alt. segment)

∠ADC = ∠ADB + ∠BDC

∠TAB + ∠TBA + ∠ATB = 180 ° (∠ sum of

= = = =

(∠ in alt. segment)

= 66 ° ∠BDC = ∠MCB

= 32 °

TA = TB 9. ∠TAB = ∠TBA ( tangent properties ) (base ∠s, isos. )

∠ADT

∠ADB = ∠PAB

∠ABD = ∠BCD + ∠BDC = 2t x = ∠ABD = 2t x + y + ∠BDC

(ext. ∠of

= 180 °

2t + y + t = 180 ° y = 180 ° − 3t Take t = 20 ° , x = 2( 20 °) = 40 °

9

)

)

(∠in alt. segment) (adj. ∠s on st. line)

7 y = 180 ° − 3( 20 °) = 180 ° − 60 ° = 120 °

19. (a)

= 180 ° − ∠PAB

− ∠ABP

(∠

sum of △)

( ∠in alt. segment)

(b)

( ∠in alt. segment) Consider △BCD. sum ° of △) ∠EBD + ∠BCD + ∠BDC (∠ = 180 ∠EDC + 50 ° + ( 70 ° + ∠EDC ) = 180 ° ∠EDC = 30 °

∠CTP = 180 ° − ∠CTQ

20. (a)

= 180 ° − 60 ° − 35 ° = 85 ° = 180 °

85 ° + ∠ABC ∠ABC

= 180 ° = 95 °

(∠ in alt. segment) (alt. ∠ s, PD // BC)

= ∠ADB + ∠DAE = 39 ° + 39 °

(ext. ∠ of △)

Consider △ABC and △BTC. ∠BAC = ∠TBC ∠ in alt. segment ∠ in semi-circle ∠ACB =90 ° ∠BCT =180 °−90adj. ° ∠ s on st. line

=90 ° ∠ACB = ∠BCT

∠ sum of △ ∠ABC ∠ sum of △ =180 °−∠BAC −∠ CB of △ ∠A sum =180 °−∠TBC −∠BCT = ∠BTC ∴ △ABC ~ △BTC AAA

(∠

AB 2 = AC 2 + CB 2

(b)

∠CTA + ∠ABC

∠ABP 39 ° ∠ACB 39 °

= 78 °

(adj. ∠ s on st. line)

= 25 ° + 35 ° = 60 ° ∠CTA = 180 ° − ∠ACT − ∠CAT

∠ADB = = ∠DAE = =

∠AEB

= 180 ° − 145 ° = 35 ° ∠CAT = ∠CTP (∠ in alt. segment) = 35 ° ∠ACT = ∠TPC + ∠CTP (ext. ∠ of △)

sum of △)

(∠ in alt. segment)

= 39 °

= 180 ° − 105 ° − 39 ° = 36 °

Level 2 ∠EDB = ∠ABE 16. = 70 ° ∠EBD = ∠EDC

17.

∠ABP = ∠ACB ∠APB

(or any other reasonable answers)

Basic Properties of Circles (II)

(opp. ∠ s, cyclic

∴ ∵

quad.) 18. Let ∠ ABC = θ .

∴

∠CAP = ∠ABC =θ AC = CP ∠CAP = ∠CPA ( ∠in alt. segment)

AB =

(Pyth. theorem)

5 + 3 cm 2

2

= 34 cm △ABC ~ △BTC

AB AC = TB BC 34 cm 5 = TB 3 5TB = 3 34 cm TB =

3 34 cm 5

21. SMEFSUO7EX@F05

(given) (base ∠ s, isos. ∴

)

∠CPA = θ

∠BAC = 90 ° (∠ in semi-circle) Consider △BPA. ∠ABP + ∠BPA + ∠PAB = 180 ° θ + θ + (90 ° + θ) = 180 ° (∠ sum of θ = 30 ° ∠ABC = 30 °

Join AM, AN, AB, BM and BN.

△)

10



Certificate Mathematics in Action Full Solutions 4B

A M ∠B A N B2 = = ∠ A M 1 B ANB

PA PT = PT PB x 6 = 6 5+ x 5x + x 2 = 3 6

∴

(arcs prop. to ∠ s at

x 2 + 5x − 3 6 = 0 ( x − 4)(x + 9) = 0 x = 4 o r x = − 9 ( reje cte d )

⊙ ) ce

∴

PA

= 4 cm

∠ANB = 2∠AMB ∠AMB + ∠ANB = 180 ° (opp. ∠ s, cyclic quad.) PD = PC 24. (a) 3∠AMB = 180 ° ∠PDC = ∠PCD ( tangent properties ) ∠AMB = 60 ° (tangent properties ) PA = PB (base ∠s, isos. ) (base ∠s, isos. ) ∠PAB = ∠PBA 180 ° − ∠DPC ∠PCD = ( ∠in alt. segment) ∠PAB = ∠AMB 2 (∠ sum of △) = 60 ° 180 ° − 50 ° ∠APB + ∠PAB + ∠PBA = 180 ° ∠ ( sum of △ ∠APB + 60 ° + 60 ° = 180 ° ∠APB = 60 °

BA = BC ∠BAC = ∠BCA

(given) (base ∠ s, isos. △ )

= x (ext. ∠ of △ ) ∠ABT = ∠BAC + ∠BCA 22. = 2x (alt. ∠ s, AC // PQ) ∠CTQ = ∠BCA = x (∠ in alt. segment) ∠BAT = ∠CTQ = x Consider △ATB.

∠ATB + ∠ABT + ∠BAT = 180 ° ∠ ( sum of 78 ° + 2 x + x = 180 ° x = 34 °

23. (a)

Consider △PAT and △PTB.

∠APT = ∠TPB ∠PTA = ∠ABT ∠PAT = 180 ° − ∠PTA − ∠APT = 180 ° − ∠ABT − ∠TPB = ∠PTB common angle ∠in alt. segment ∠sum of

∠sum of ∴ △PAT ~ △PTB (b) Let PA = x cm. ∵ △PAT ~ △PTB

11

AAA

=

2

= 65 ° ∠CBD = ∠PCD = 65 ° ∠BDC = ∠BCN = 36 ° AB = AD ∠ABD = ∠ADB ( ∠in alt. segment) (given) (base ∠ s, isos.

(∠ in alt. segment)

)

∠ABC + ∠ADC = 180 ° (opp. ∠ s, cyclic quad.)

( ∠ABD + ∠DBC ) + ( ∠ADB + ∠BDC ) (∠ABD + ∠ADB ) + 65 ° + 36 ° 2∠ADB ∠ADB ∴

(b)

∠ADC

= ∠ADB

= 39 .5° + 36 ° = 75 .5°

∠BAC = ∠BCN = 36 °

∠AKD

+ ∠BDC

= 180 ° = 180 ° = 79 ° = 39 .5°

= ∠ABK + ∠BAK = 39 .5° + 36 ° = 75 .5°

25. SMEFSU07EX@F06

(∠ in alt. segment) (ext. ∠ of △)

7

Join AB.

∠ABC = ∠CAB = ( ∠in alt. ( ∠in alt.

Basic Properties of Circles (II)

Consider △APT. ∠APT + ∠PAT 42 ° + (∠BAC + ∠CAT ) 42 ° + (∠BAC + 25 °) ∠BAC

∠CAT ∠CBT segment) segment)

= ∠ATQ = 100 ° = 100 ° = 33 °

(ext. ∠ of △)

(b) Consider △ABT. ∠ABT + ∠ATB + ∠TAB = 180 ° ∠ ( 100 ° + ∠ATB + (33 ° + 25 °) = 180 ° ∠ATB = 22 °

Consider △ABT.

sum of △)

∠BAT + ∠ABT + ∠ATB = 180 ° (∠ sum of △

(∠CAT + ∠CAB ) + ( ∠ABC + ∠CBT ) + 70Exercise ° = 180 7C ° (p. 88) Level 1 ∠ABC + ∠CAB + ∠ABC + ∠CAB = 110 ° ∠BAD +∠BCD ∠ABC + ∠CAB1. =(a)55 ° = (38 ° + 42 °) +(35 ° +65 °) Consider △ABC. ∠ABC

+ ∠CAB + ∠ACB 55 ° + ∠ACB ∠ACB

= 180 ° ∠ ( sum of △) = 180 ° = 125 °

=180 °

∴ A, B, C and D are concyclic. opp. ∠ s supp. (b)

26. (a)

∠ABT

∠CBT

= ∠ATQ

= 100 ° = 180 ° − ∠PBC − ∠ABT = 180 ° − 55 ° − 100 ° = 25 °

line)

∠CAT = ∠CBT = 25 °

(∠ in alt. segment) 2. (adj.

(∠ s in the same

∠CDB =∠CAB x =38 °

(∠ s in the same segment)

(a)  ∠ DBC

=180 °−∠BDC −∠DCB =180 °−35 °−105 ° = 40 ° =∠DAC ∴ A, B, C and D are concyclic.

converse of ∠ s in the same segment

segment)

(b)

3.

∠ BAC =∠ BDC x =35 °

∠ sum of △

(∠ s in the same segment)

(a)  ∠ CAD = ∠ DBC = 90° ∴ A, B, C and D are concyclic. (b) ∠ BAC

converse of ∠ s in the same segment

=180 °−∠EAB −∠CAD =180 °−55 °−90 ° =35 °

(adj. ∠ s on

st. line)

∠ BDC

∴ segment) 4.

=∠ BAC x =35 °

(∠ s in the same

(a)

12

Certificate Mathematics in Action Full Solutions 4B

Mark the point F as shown in the figure.

13

7 Consider △ABF.

∠ABF =180 °−∠BAF −∠AFB =180 °−45 °−85 ° =50 °

∠ sum of △

= 50 °+35 ° =85 °

∴ ∠ ABC = ∠ ADE ∴ A, B, C and D are concyclic. ∠ DAC

∴ segment) 5.

x +y =z Take x = 30° and y = 50°, we have

z = 30 ° +50 ° =80 °

∴

x = 30°, y = 50°, z = 80° (or any other reasonable answers)

ext. ∠ = int. opp. ∠

=∠ DBC x =35 °

(∠ s in the same

Consider △PBC.

∠PBC +∠PCB = ∠APB 16 ° +∠PCB = 62 ° ∠PCB = 46 ° ∠ACD =∠BCD −∠PCB

∴ ∴

6. ∴ ∴ 7. ∴ ∴ 8.

10. ∠ BAD = 180° – (x + y)(∠ sum of △) ∠BAD +∠BCD =180 ° (opp. ∠ s, cyclic 180 ° −( x + y ) + z =180 ° quad.)

∠ABC = ∠ABF +∠FBC

(b)

Basic Properties of Circles (II)

=82 °−46 ° =36 ° ∠ABD = ∠ACD A, B, C and D are concyclic.

converse of ∠ s in the same segment

∠ ABC = ∠ ADC ∠ PQB = ∠ PDC ∠ DPQ = ∠ PQB ∠ DPQ = ∠ ABC A, B, Q and P are concyclic.

opp. ∠ s of // gram ext. ∠ , cyclic quad. alt. ∠ s, AD // BC

AD = AE ∠ ADE = ∠ AED ∠ ADE = ∠ ABC ∠ AED = ∠ ABC A, B, C and E are concyclic.

given base ∠ s, isos. △ opp. ∠ s of // gram

∠ BPT = 90° ∠ SQC = 90°

ext. ∠ = int. opp. ∠

ext. ∠ = int. opp. ∠ ∠ in semi-circle ∠ in semi-circle

∠AQS =180 °−∠SQC

∴ ∴ 9. ∴ ∴

ext. ∠ of △

=180 °−90 ° =90 °

∠ AQS = ∠ BPT = 90° A, P, R and Q are concyclic. ∠ ATQ = ∠ ABT ∠ ATQ = ∠ CDT ∴ ∠ ABT = ∠ CDT A, B, C and D are concyclic. ABCD is a cyclic quadrilateral.

adj. ∠ s on st. line

ext. ∠ = int. opp. ∠ ∠ in alt. segment alt. ∠ s, PQ // CD ext. ∠ = int. opp. ∠

14

Certificate Mathematics in Action Full Solutions 4B 11.

D is a point on the other side of AB such that AD ⊥ BD.

Level 2 12. (a)

(b)

∠ OAT = 90° ∠ OBT = 90° ∴ ∠ OAT + ∠ OBT = 180° ∴ O, B, T and A are concyclic.

tangent ⊥ radius tangent ⊥ radius opp. ∠ s supp.

 O, B, T and A are concyclic. (proved in (a)) ∴ ∠OTB =∠OAB (∠ s in the same =30 °

segment) 13. (a)

line

(b)

∠ APB = 90°

∠ in semi-circle

∠ AOM = 90° ∴ ∠ APM = ∠ AOM ∴ O, P, M and A are concyclic.

given

∠APM +∠APB =180 ° adj. ∠ s on st. ∠APM +90 ° =180 ° ∠APM = 90 °

OA = OP radii ∠ OAP = ∠ OPA base ∠ s, isos. △ ∠ OAP = ∠ OMP ∠ s in the same segment ∴ ∠ OPA = ∠ OMB

14. ∠ BRS = ∠ SQA ∠ SRC = 180° – ∠ BRS ∠ SRC = ∠ SPA ∴ 180° – ∠ SQA = ∠ SPA ∴ ∠ SQA + ∠ SPA = 180° ∴ A, Q, S and P are concyclic. ∴ AQSP is a cyclic quadrilateral.

15

converse of ∠ s in the same segment

ext. ∠ , cyclic quad. adj. ∠ s on st. line ext. ∠ , cyclic quad.

opp. ∠ s supp.

7

Basic Properties of Circles (II)

∠CAD =180 °−∠ACP −∠APB =90 °−∠APB

15. (a)

∠ sum of △

∴ ∠COD = 2(90 ° −∠APB )

=180 ° −2∠APB 18. (a) Join PB and let ∠ ARP = θ . ∠ PBA = ∠ ARP = θ

∠ sum of △

sum of △ ∠ BQC + ∠ BPC

In △AQP,

= ( x + y ) + (180 ° − x − y )

∠PQB = ∠PAQ +∠APQ = 90 ° −θ +θ = 90 °

∠ in alt. segment ∠ in alt. segment

 ∠ BQC = ∠ BQP + ∠ CQP ∴ ∠ BQC = x + y (ii) Consider △APD. ∠ APD =180 °−∠ PAD −∠ PDA =180 °−x −y

In △APB,

∠PAQ =180 °−∠APB −∠PBA =90 °−θ

∠ BQP = x ∠ CQP = y

∠ s in the same segment ∠ in semi-circle

∠ APB = 90°

(i)

=180 °

∴ B, P, C and Q are concyclic.

ext. ∠ of △

opp. ∠ s supp.

(b)

(b)

Join RB. ∠ TQB = 90° ∠ TRB = 90° ∠ TQB + ∠ TRB = 180° ∴ R, T, Q and B are concyclic. ∴ RTQB is a cyclic quadrilateral. 16. Consider △ACB and △DBC. AC = DB ∠ ACB = ∠ DBC BC = CB ∴ △ACB ≅ △DBC ∠ BAC = ∠ CDB ∴ A, B, C and D are concyclic. ∠ ACB = 90°

17. (a)

proved in (a) ∠ in semi-circle opp. ∠ s supp.

given given common side SAS corr. ∠ s, ≅ △s converse of ∠ s in the same segment

Join BC. ∠ CQP = y  B, P, C and Q are concyclic. ∴ ∠ CBP = ∠ CQP = y

ext. ∠ = int. opp. ∠

Revision Exercise 7 (p. 93) Level 1 1.

∠ in semi-circle adj. ∠ s on

st. line

(b) ∠ COD = 2∠ CAD

∠ s in the same segment

∴ ∠ CDA = ∠ CBP = y ∴ A, B, C and D are concyclic.

∠ECP =180 °−∠ACB =90 °

∠ ADB = 90° ∴ ∠ ECP = ∠ ADB ∴ P, D, E and C are concyclic.

∠ in alt. segment

TB = TA (tangent properties) ∠ TBA = ∠ TAB = x (base ∠ s, isos. △) ∠TAB +∠TBA +∠ATB =180 ° (∠ sum of 2 x +58 ° =180 ° x =61 ° △)

∠ in semi-circle

∠ TBA = ∠ ACB ∠ ACB = ∠ CBF = y ∴ ∠ TBA = y But ∠ TBA = x = 61° y =61 ° ∴

ext. ∠ = int. opp. ∠ ∠ at centre twice ∠ at ⊙ce

2.

∠ OAB = 90° ∠ ODC = 90°

(∠ in alt. segment) (alt. ∠ s, AC // TF)

(tangent ⊥ radius) (tangent ⊥ radius)

16

∠

Certificate Mathematics in Action Full Solutions 4B Sum of interior angles of pentagon = 180° (5 – 2) = 540° ∴ ∠ AOD =540 °−90 °−90 °−100 °−128 ° =132 °

3.

(a)

∠ BAP = 90° (tangent ⊥ radius)

∠BAQ =180 °−∠ABQ −∠AQB =180 °−90 °−29 ° =61 °

Join OC.

∠COP =2∠CAP

(∠ sum of △)

∠ CAP =∠ BAP −∠ BAQ =90 °−61 ° =29 °

∠ at ⊙ )

=2 ×29 ° =58 °

(∠ at centre twice

ce

∠ OCP = 90° (tangent ⊥ radius) ∠ APC =180 °−∠ OCP −∠ COP =180 °−90 °−58 ° =32 °

(b)

(∠

sum of △)

4.

∠BAC = ∠BCQ =80 °

(∠ in alt. segment)

∠PAB +∠BAC +∠TAC =180 ° (adj. ∠ s on 36 °+80 °+x =180 ° x =64 °

st. line) ∠ ABC = ∠ TAC (∠ in alt. segment) ∴ y =64 ° TA = TC (tangent properties) ∠ TAC = ∠ TCA (base ∠ s, isos. △) z =180 °−∠ TAC −∠ TCA (∠ sum of △) =180 °−64 °−64 ° =52 ° Let ∠ OCA = a. OA = OC ∠ OAC = ∠ OCA = a ∠ OCB = 90°

5.

(radii) (base ∠ s, isos. △) (tangent ⊥ radius)

∠ACB = 90 ° −a (given) AB = AC (base ∠ s, isos. △) ∠ABC = ∠ACB =90 ° −a ∠OAC = ∠ABC + ∠ACB = (90 ° − a ) + (90 ° − a ) = 180 ° − 2a

∴ 180° – 2a = a ∴ a = 60° OCA =60 ° ∴ ∠ 6.

∠BTP = ∠OBT = 27 °

∠ OTP = 90°

17

(alt. ∠ s, TP // OB) (tangent ⊥ radius)

7

Basic Properties of Circles (II)

∠OTA = ∠OTP −∠BTP =90 °−27 ° = 63 °

OA = OT

∠OAT = ∠OTA = 63 °

(radii) (base ∠ s, isos. △)

∠AOB +∠OBA =∠OAT ∠AOB +27 ° =63 °

(ext. ∠ of △)

∠AOB =36 °

7.

∠ CAQ = ∠ ABC (∠ in alt. segment) ∠ BAC = 90° (∠ in semi-circle) Consider △ABQ. ∠ABQ +∠BAQ +∠AQC =180 °

∠ABC +(∠BAC +∠CAQ ) +34 ° =180 ° 2∠ABC +90 °+34 ° =180 ° ∠ABC =28 ° (∠ sum of △) 8.

Join OB.  BA = BD (given) ∴ ∠ DAB = ∠ ADB (base ∠ s, isos. △)  OB = OA (radii) ∴ ∠ OBA = ∠ OAB (base ∠ s, isos. △) ∠ OBD = 90° (tangent ⊥ radius) Consider △ABD. ∠ DAB + ∠ ABD + ∠ ADB = 180° (∠ sum of △) ∠DAB +(∠OBA +∠OBD ) +∠DAB =180 ° 3∠DAB +90 ° =180 ° ∠DAB =30 ° 9.

Consider △OBC.

∠OBC +∠OCB +∠BOC =180 ° ∠OBC +∠OCB +134 ° =180 ° ∠OBC +∠OCB = 46 °

(∠

sum of △)

∠ ABO = ∠ OBC (tangent properties) ∠ ACO = ∠ OCB (tangent properties) ∠BAC =180 °−∠ABC −∠ACB

=180 °−2∠OBC −2∠OCB =180 °−2(∠OBC +∠OCB ) =180 °−2 ×46 ° =88 °

(∠

sum of △)

18

Certificate Mathematics in Action Full Solutions 4B 10. (a) 13. (a)

∠ABC =180 ° −∠BCA −∠BAC

∠ sum of △

=180 ° −28 ° −36 ° =116 °

∠ ABC = ∠ ADE = 116° ∴ A, B, C and D are concyclic. Join AC. ∠ CAB = 90° (∠ in semicircle) ∠ ACB = ∠ BAQ (∠ in alt. segment) = 36° ∠ ABC =180 °−∠ CAB −∠ ACB △) =180 °−90 °−36 ° =54 ° ∠ ( sum of △)

(b)

∠ ADC +∠ ABC =180 ° ∠ ∠ ADC +54 ° =180 ° (opp. s, ∠ ADC

=126 °

cyclic quad.)

11.

(a)

∠TDA = ∠TCB = 70 °

(corr. ∠ s, AQ // BC)

∠TQD +∠QTC =∠TDA 26 °+∠QTC =70 ° ∠QTC = 44 ° (b)

∠TBC = ∠QTC = 44 °

(ext. ∠ of △

(∠ in alt. segment)

∠ATD +∠TBC +∠TCB =180 ° ∠ ( sum ∠ATD +44 °+70 ° =180 ° ∠ATD =66 °

of △) 12.

Join OP and OQ. OP ⊥ AB tangent ⊥ radius OQ ⊥ AC tangent ⊥ radius OP = OQ radii ∴ AB = AC chords equidistant from centre are equal

19

∴ ABCD is a cyclic quadrilateral. (b)

ext. ∠ = int. opp. ∠

∠CAD = ∠ADE −∠ACD =116 °−54 ° = 62 ° ∠CBD =∠CAD =62 °

segment)

(ext. ∠ of

(∠ s in the same

7 14. ∠ ACB = 90°

∴ ∠ CAD + ∠ CDA = ∠ BAC + ∠ ABC and ∠ CDA = ∠ BAC ∴ ∠ CAD = ∠ ABC ∴ AD is the tangent to the circle at A.

  



C D= D E

∴ ∠ EBD = ∠ DAC ∴ P, A, B and Q are concyclic. ∴ PABQ is a cyclic quadrilateral.

18. ∠ PAB = ∠ BPC ∠ BAD = ∠ BCP In △PBC,

(∠ in alt. segment) (ext. ∠ , cyclic quad.)

∠BPC +∠BCP +∠PBC =180 ° (∠ sum of △) ∠BPC +∠BCP +100 ° =180 ° ∠BPC +∠BCP =80 °

=90 °

∠PAD =∠PAB +∠BAD =∠BPC +∠BCP =80 °

given converse of ∠ in alt. segment 19.

D E ∠ E B Darcs prop. to ∠ s at ⊙ = C D ∠D A C

15.

∠ABC +∠ADC (opp. =18∠ 0 s,° cyclic quad.) ( x +34 °) +(34 °+x +34 °) =180 ° x =39 °

∠ in semi-circle

∠CAD +∠CDA = ∠ACB ext. ∠ of △ = 90 ° ∠ACD =180 °−∠ACB adj. ∠ s on st. line =180 °−90 ° =90 ° ∠BAC +∠ABC = ∠ACD ext. ∠ of △

Basic Properties of Circles (II)

ce

Join OB. ∠ OBP = ∠ OBQ = 90° ∠ OAP = ∠ OCQ = 90°

given

(tangent ⊥ radius) (tangent ⊥ radius)

∠AOB + ∠OBP + ∠APB + ∠OAP = 360 ° ∠AOB + 90 ° + x + 90 ° = 360 ° ∠AOB =180 ° − x

converse of ∠ s in the same segment

∠BOC + ∠OCQ +∠BQC + ∠OBQ = 360 ° ∠BOC +90 ° + y +90 ° = 360 ° ∠BOC =180 ° − y z =360 °−∠AOB −∠BOC (∠ s at a =360 °−(180 °− x ) −(180 °− y )

16. 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.

=x +y pt.) 20.

Level 2 17. (a)

(b)

△)

∠DBC = ∠AQB (alt. ∠ s, BC // AQ) =x ∠ABD = ∠BDC (alt. ∠ s, BA // CD) =34 ° ∴ ∠ABC = ∠DBC +∠ABD = x +34 ° ∠QAD = ∠ABD (∠ in alt. segment) =34 ° (ext. ∠ of ∠ADB = ∠QAD +∠AQD

BY =BX ∠ BYX =∠ BXY ∴

(tangent properties) (base ∠ s, isos. △)

180 ° − 60 ° (∠ sum of 2 = 60 °

∠ BYX = ∠BXY =

△)

∴ BYX is an equilateral triangle. Let XY = a cm, then BX = BY = a cm. CY = CZ (tangent properties) AZ = AX (tangent properties) CY =CZ = (5 −a ) cm ∴

= 34 ° + x and  ∴

AZ =CA −CZ =[7 −(5 −a )] cm = (2 +a ) cm AX = AB − BX = (8 − a ) cm AZ = AX 2+a= 8–a

20

Certificate Mathematics in Action Full Solutions 4B ∴ ∴

a= 3

XY =3 cm

21. (a)

Join OP and OQ. ∠ OPB = ∠ OQB = 90° tangent ⊥ radius ∠POQ +∠OQB +∠QBP + ∠OPB = 360 °

∠POQ +90 ° +90 ° +90 ° = 360 ° ∠POQ = 90 ° ∴ OPBQ is a parallelogram. opp. ∠ s equal OP = OQ radii ∴ OPBQ is a square. (b)

Let the radius of the circle be r cm. PB = QB = r cm (property of square)  AP = (12 – r) cm and CQ = (5 – r) cm AR = AP and CR = CQ (tangent properties) ∴

AC = AR + CR = [(12 − r ) + (5 − r )] cm = (17 − 2r ) cm

AC2 = AB2 + BC2

(Pyth. theorem)

AC = 12 +5 cm =13 cm 2

2

∴ 17 −2r =13

r =2

∴ The radius of the circle =2 cm 22. (a)

∠ ONC = 90° ∠ NCO = ∠ ACB ∠NOC

tangent ⊥ radius common angle

=180 °−∠ONC −∠NCO =90 °−∠NCO ∠ABC =180 °−∠BAC −∠ACB =90 °−∠ACB ∴ ∠ NOC = ∠ ABC ∴ △CON ~ △CBA AAA

21

(b)

∠ OMA = ∠ ONA = 90° AM // NO AN // MO AM = AN ∴ AMON is a square.

tangent ⊥ radius int. ∠ s supp. int. ∠ s supp. tangent properties

(c)

Let the radius of the circle be r cm. AM = AN = r cm (property of square)

7  △CON ~ △CBA

(proved in (a)) (b)

NO NC = AB AC r 6−r ∴ = 4 6 6r = 24 − 4r

Basic Properties of Circles (II)



△PAC ~ △PBA

∴

PC PA = PA PB

proved in (a) corr. sides, ~ △s

∴ PA2 = PC × PB (c)

r = 2.4

Let PC = x cm. Using the result of (b),

PA2 = PC × ( PC + BC)

∴ The radius of the circle =2.4 cm

6 2 = x( x + 5)

23. (a)

x 2 + 5 x − 36 = 0 ( x − 4)( x + 9) = 0 x = 4 or x = − 9 (rejected) ∴ PC =4 cm 26. (a) Join OA, OB and OC. Consider △ACO and △BCO. CA = CB given OA = OB radii CO = CO common side ∴ △ACO ≅ △BCO SSS

(b)

(ext. ∠ of

△) AE = AD ∠ AET = ∠ ADE ∴ ∠AET = x + y

(b)  ACBO is a cyclic quadrilateral of the larger circle. ∠ CAO + ∠ CBO = 180° opp. ∠ s, cyclic quad. ∠ CAO = ∠ CBO corr. ∠ s, ≅ △s ∴ ∠ CAO = ∠ CBO = 90° ∴ AC and BC are tangents to the smaller circle at A and B respectively. converse of tangent ⊥ radius 24. (a)

∠ADE =∠DAT +∠ATD =x +y

BR = BP CR = CQ  BC = BR + CR ∴ BC = BP + CQ

tangent properties tangent properties

AP = AQ Perimeter of △ABC = AB + AC +BC

(tangent properties)

(b)

(given) (base ∠ s, isos. △)

∠BTE =∠ATE given =y In △CET, ∠AET =∠ECT +∠CTE

ext. ∠ of

x + y =∠ECT + y △ ∠ ECT = x ∴ ∠ BAT = ∠ ACB ∴ TA is the tangent to the circle at A.

converse of ∠ in alt. segment

27.

= AB + AC +( BP +CQ ) =( AB +BP ) +( AC +CQ ) = AP + AQ

(proved in

= 2 AP

Join OC and CT. ∠ BAC = ∠ CBT OC = OA ∠ OCA = ∠ OAC ∠ COT = ∠ OCA ∴ ∠ CBT = ∠ COT ∴ C, O, B and T are concyclic.

=( 2 ×12 ) cm = 24 cm (a)) 25. (a)

Consider △PAC and △PBA. ∠ CAP = ∠ ABP ∠ CPA = ∠ APB

∠ in alt. segment common angle

∠ACP =180 °−∠CAP −∠CPA =180 °−∠ABP −∠APB =∠BAP

∠ sum of △

∴ △PAC ~ △PBA

28. ∴ ∴

∠ BDC = ∠ ABD ∠ DEF = ∠ ABD ∠ KAE = ∠ DEF ∠ BDC = ∠ KAE A, K, D and F are concyclic.

∠ in alt. segment radii base ∠ s, isos. △ alt. ∠ s, TO // CA converse of ∠ s in the same segment alt. ∠ s, CF // BA ext. ∠ , cyclic quad. corr. ∠ s, CA // DE ext. ∠ = int. opp. ∠

AAA

22

Certificate Mathematics in Action Full Solutions 4B 29. (a)

∠ ABC = ∠ ADC = 90° ∠ in semi-circle ∠FBE =180 °−∠ABC adj. ∠ s on st.

=90 °

line

line

∠EDF =180 ° −∠ADC adj. ∠ s on st. =90 ° ∴ ∠ FBE = ∠ EDF ∴ B, E, F and D are concyclic. ∴ BEFD is a cyclic quadrilateral.

(b)

converse of ∠ s in the same segment

∠ BDC = ∠ BAC = 35° (∠ s in the same segment) ∠ BFE = ∠ BDC = 35° (∠ s in the same segment)

∠DCF = ∠CEF +∠CFE = 27 ° +35 ° = 62 °

△)

∠DCF +∠CFD =∠ADC 62 °+∠CFD =90 °

(ext. ∠ of

(ext. ∠ of

∠CFD =28 °

△) 30. (a)

∠ ABC = 90° CM ⊥ QS

∠ in semi-circle line joining centre to mid-pt. of chord ⊥ chord

∴ ∠ ABC = ∠ CMS = 90° ∴ B, R, M and C are concyclic. ext. ∠ = int. opp. ∠ ∴ BRMC is a cyclic quadrilateral. (b)

∠ PBR = ∠ BCA ∠ BCA = ∠ BRP

∴ ∠PBR = ∠BRP

PB = PR

∠ in alt. segment ext. ∠ , cyclic quad. sides opp. equal ∠ s

Multiple Choice Questions (p. 98) 1. Answer: C ∠ ABO = ∠ CBO = 34° ∠ OAC = ∠ OAB

(tangent properties) (tangent properties) =x (tangent properties)

∠ OCB = ∠ OCA = 25° ∠ABC +∠BAC +∠ACB =180 ° 2 ×34 °+2 x +2 ×25 ° =180 ° x =31 °

(∠

sum of △) 2.

Answer: C ∠ GBF cyclic quad.) = 102° ∠ CBF

23

= ∠ GED

(ext. ∠ ,

= ∠ FGB

(∠ in alt.

segment) =x ∠ABG +∠GBF +∠CBF =180 ° (adj. ∠ s on 36 °+102 °+ x =180 ° x = 42 °

st. line)

7 3. Answer: B ∠ CDB = ∠ CAB = 52° ∴ A, B, C and D are concyclic.

∠ADB = ∠PAB (∠ in alt. segment) = 47 ° ∠BAD =180 °−∠ABD −∠ADB (∠ =180 °−2 ×47 ° =86 °

(converse of ∠ s in the same segment)

∴ ABCD is a cyclic quadrilateral. ∠ADC +∠CBA =180 ° (52 °+∠ADB ) +72 ° =180 ° ∠ADB =56 °

Basic Properties of Circles (II)

sum of △)

(opp. ∠ s,

∠ BAD +∠ DCB =180 ° 86 °+x =180 ° x =94 °

cyclic quad.)

(opp. ∠ s, cyclic

quad.) 4.

Answer: C 7.

Answer: B ∠ ACB segment) = 52° ∠ BAC = 90°

= ∠ BAP

(∠ in alt.

(∠ in semi-circle)

∠ABC +∠BAC +∠ACB =180 ° (∠ sum of ∠ABC +90 ° +52 ° =180 ° ∠ABC =38 °

Join QS. BP = ∠ BPQ = △)

△) BQ (tangent properties) ∠ BQP (base ∠ s, isos.

180 ° −90 ° 2 = 45 °

∠BQP =

△)

CQ = ∠ CQR =

∠CAQ + x =13 °+∠ACB 38 °+ x =13 °+52 ° x =27 °

(∠ sum o f △) CR (tangent properties) ∠ CRQ (base ∠ s, isos.

8.

180 ° −124 ° (∠ sum of △) 2 = 28 °

∴

180 °(5 − 2) 5 =108 °

(∠ in alt. segment)

PC = QC ∠ CPQ = ∠ CQP

(∠ in alt. segment)

∠CQP =

∠PSR =∠PSQ +∠QSR =45 °+28 °

∠ PRQ =∠ CQP =36 °

Answer: A TC = TB ∠ TCB = ∠ TBC ∠ TCB = ∠ ABC ∠ TBC = ∠ BAC ∠ ACB =180 °−∠ BAC

(tangent properties) (base ∠ s, isos. △) (alt. ∠ s, CT // AB) (∠ in alt. segment) −∠ ABC (∠ =180 °−∠ TBC −∠ TCB

(tangent properties) (base ∠ s, isos. △)

180 ° − ∠BCD 2 180 ° −108 ° = 2 = 36 °

=73 °

5.

Answer: D

(ext. ∠ of △)

∠BCD =

∠CQR =

∠PSQ = ∠BQP = 45 ° ∠QSR = ∠CQR = 28 °

(∠ in alt. segment)

∠CAQ = ∠ABC = 38 °

9.

(∠ sum of △)

(∠ in alt. segment)

Answer: B With the notations in the figure, join AB.

=∠ BTC =32 ° sum of △) 6.

Answer: D

∠PAB = ∠ABP = 47 °

(alt. ∠ s, PQ // BD)

segment)

∠ BAC

= ∠ CBE

(∠ in alt.

= 37°

24

Certificate Mathematics in Action Full Solutions 4B ∠BAP =180 ° −84 ° −37 ° =59 °

line)

PB = PA

∠ABP = ∠BAP = 59 °

 

(adj. ∠ s on st. 10. Answer: D

(tangent properties) (base ∠ s, isos. △)

x =180 °−59 °−59 ° (∠ sum of △) =62 °

C D ∠ D B C1 = = D A ∠ D C A3 63 ° 3 = 21 °

∠DBC =

(arcs prop. to ∠ s at ⊙ce)

(arcs prop. to ∠ s at ⊙ce)

∠ ABD = 63° (∠ s in the same segment) ∠ ACB = 37° (∠ in alt. segment) ∠ABC +∠ACB +∠BAC =180 ° (∠ (63 °+21 °) +37 °+∠BAC =180 ° ∠BAC =59 ° sum of △) 11. Answer: D ∠ ACE = 90° (∠ in semi-circle) ∠ ECD = x (∠ in alt. segment) ∠BAC +∠ACD =180 ° (int. ∠ s, AB // (52 °+ x) +(90 °+ x) =180 ° x =19 ° PD) 12. Answer: B

Join BD. ∠ BDA = 90° ∠ ABD = 53°

(∠ in semi-circle) (∠ in alt. segment) ∠BDC =180 °−90 °−53 ° (adj. ∠ s on st.

line)

=37 °

x +37 ° =53 ° x =16 °

(ext. ∠ of △)

13. Answer: C Let PB = a cm.

AC = 6 2 +8 2 cm =10 cm PB = QB QC = RC AP = AR  QC = (6 – a) cm and AP = (8 – a) cm

25

(Pyth. theorem) (tangent properties) (tangent properties) (tangent properties)

7



Basic Properties of Circles (II)

AC = AR + RC 10 = (8 − a) + (6 − a) a =2

∴

AR =(8 −2) cm =6 cm

26

Certificate Mathematics in Action Full Solutions 4B HKMO (p. 99)

Join OA, OB and OP. Draw a line OM which is perpendicular to CB. Let OM = l cm and radius of the circle = r cm. CM = MB (line from centre ⊥ chord bisects chord)

9 cm 2 = 4.5 cm

MB =

Consider △OMB. OM2 + MB2 = OB2 (Pyth. theorem) l2 + 4.52 = r2 2 2 2 l = r – 4.5 …… (1) Consider △OMP and △OPA. (4.5 + d)2 + l2 = r2 + 62 4.52 + 9d + d2 + l2 = r2 + 36…… (2) By substituting (1) into (2), we have 4.5 2 + 9d + d 2 + (r 2 − 4.5 2 ) = r 2 + 36

d 2 + 9d − 36 = 0 (d − 3)(d + 12) = 0 d = 3 or d = − 12 (rejected)

27