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.

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. (a) OC 90°

Activity 7.2 (p. 82) 1.

(alt. ∠s, TC // DE)

∠OTB = 90° x + y = 90° y = 90° − 25° = 65°

(tangent ⊥ radius) (tangent ⊥ radius)

OT = OC (radii) ∠OTC = ∠OCT (base ∠s, isos. △) =x ∠OTC + ∠OCT + ∠TOC = 180° (∠ sum of △) 180° − 70° x = ∴ 2 = 55° ∠OTB = 90° x + y = 90° y = 90° − x = 90° − 55° = 35°

(b) No 4.

x = 25°

(a)

(tangent ⊥ radius)

(tangent ⊥ radius)

∠OTA = 90° AT + OT 2

3.

∵

2

= AO

(tangent ⊥ radius) 2

(Pyth. theorem)

12 2 + r 2 = ( 4 + r ) 2 144 + r 2 = 16 + 8r + r 2 r = 16

4. (b) (i) Yes (ii) Yes 2.

TG = TE ∠TGE = ∠TEG = 60° ∠GTB = ∠TGE = 60° ∠OTB = 90° x + 60° = 90° x = 30°

(given) (base ∠s, isos. △) (alt. ∠s, AB // CD) (tangent ⊥ radius)

p. 69 1.

2. 3.

1

Yes

∠PTO = ∠QTO x = 25°

(tangent properties)

TQ = TP y = 6

(tangent properties)

∠OPT = 90° x = 110° − 90° = 20°

(tangent ⊥ radius) (ext. ∠ of △)

7 ∠QTO = ∠PTO y = 20°

(tangent properties))

1.

∠OPT = 90° x + 26° = 90° x = 64°

3.

p. 77

(tangent ⊥ radius)

(tangent ⊥ radius) TQ = TP ∴∵ ∠TQP = ∠TPQ = 64° y = 180° − 64° − 64° (tangent properties) = 52° (base ∠s, isos. △) ∠QOP = 2∠QRP x = 2 × 54° = 108°

4.

2.

ce

3.

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

∠APQ = ∠AQP ∴ ∠PQR = ∠AQP ∴ PQ bisects ∠RQA.

p. 75 1.

2.

∠ATP = ∠ABT x = 70°

4.

(∠ in alt. segment)

∠BAT = ∠BTQ x = 35°

(∠ in alt. segment)

(∠ at centre twice ∠ at ⊙ ) ce

(tangent ⊥ radius) (∠ in alt. segment)

x = 39°

p. 87 1.

∵ ∴

2.

∵

(∠ in alt. segment)

∠BTQ = ∠BAT y = 45°

(∠ in alt. segment)

∠BAT = ∠BTQ x = 50°

(∠ in alt. segment)

∠ABT = ∠ATP y = 74°

(∠ in alt. segment)

∴ 3.

4. 3.

∠CBT = ∠CTQ b = 75°

∠STQ = ∠SRT (∠ in alt. segment) =x x + (62° + x ) + 40° = 180° ( ∠ sum of )

alt. ∠s, AP // QR tangent properties base ∠s, isos.

AP = AQ

(∠ in alt. segment)

∠OTQ = 90° ∠ABT = ∠ATQ ∴ x = ∠ATO + ∠OTQ = 25° + 90° = 115°

4.

∠PQR = ∠APQ 5.

∠CAT = ∠CTQ a = 75°

∠BOT = 2∠BAT y = 2x = 2 × 35° = 70°

(∠ at centre twice ∠ at ⊙ )

Basic Properties of Circles (II)

∠TBA = 180° − 35° − 42° = 103° ∠ATQ = ∠ABT x = 103°

(∠ sum of △) (∠ in alt. segment)

∠BAT = ∠BTP (∠ in alt. segment) =x ∠ATB + ∠TBA + ∠BAT = 180° (∠ sum of △) x + 110° + x = 180° x = 35°

5.

∠ADB = ∠ACB = 50° (given) A, B, C and D are concyclic. (converse of ∠s in the same segment) ∠BAD + ∠BCD = 100° + 70° = 170° ≠ 180° A, B, C and D are not concyclic.

∠ABC + 100° = 180° (adj. ∠s on st. line) ∠ABC = 80° ∠ABC ≠ 100° ∴ A, B, C and D are not concyclic. ∠ABD + 40° + 120° = 180° (∠ sum of △) ∠ABD = 20° ∠ABC + ∠ADC = ( 20° + 50°) + ( 40° + 70°) = 180° ∴ A, B, C and D are not concyclic. (opp. ∠s supp.) ∠BAD = 60° + 30° = 90° ∵ ∠BCE = ∠BAD = 90° ∴ A, B, C and D are concyclic.

(ext. ∠ = int. opp. ∠)

2

Certificate Mathematics in Action Full Solutions 4B

Exercise Exercise 7A (p. 70) Level 1 1.

converse of tangent ⊥ radius

OC = OP = 5 cm Consider △OPB. OP 2 + PB 2 = OB 2 5 + 12 = (5 + x ) 2

2

(radii)

(Pyth. theorem) 2

132 = (5 + x ) 2 x = 8 or x = −18 (rejected)

2.

3.

4.

OP = OC (radii) ∠OPC = ∠OCP (base ∠s, isos. ) 180° − 130° ∠OPC = (∠ sum of ) 2 = 25° ∠OPB = 90° (tangen(tangent ⊥ radius)t x + 25° = 90° x = 90° − 25° = 65° TB = TA ∠TBA = ∠TAB = 68° x + 68° + 68° = 180° x = 44°

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

∠ATO = ∠BTO

(tangent properties)

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

(adj. ∠s on st. line)

∠OAT = 90°

(tangent ⊥

x = 180° − 90° − 25° = 65° 5.

)

OA = OB = 6 cm PA2 + OA2 = (82 + 6 2 ) cm 2

radius) (∠ sum )

△ of radii

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

= 102 cm 2 ∴ ∴ ∴

3

= 100 cm 2 PA2 + OA2 = PO2 OA ⊥ PA PA is the tangent to the circle at A.

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

7 6.

∵ ∴ ∴ ∵ ∴ ∵ ∴

AP = AR AR = 3 cm RC = (8 − 3) cm = 5 cm RC = QC QC = 5 cm BP = BQ BQ = 2 cm BC = BQ + QC

(tangent properties)

Basic Properties of Circles (II)

1 ∠BOD 2 (∠ at centre twice ∠ at ⊙ ) ce 1 = × 58° 2 = 29° ∠ACB + ∠ABC + ∠BAC = 180° (∠ sum of △) ∠ACB = 180° − 90° − 29° = 61° ∠BAC =

(tangent properties) (tangent properties)

Level 2

= ( 2 + 5) cm = 7 cm

11.

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

(a)

ce

7.

= 130°

∠OAR = ∠OAP (tangent properties) = 27° ∠OAR + ∠AOB + ∠OBR = 180° ∠ ( sum of △) 27° + 116° + ∠OBR = 180°

(b)

∠OBR = 37°

∠OAT = ∠OBT = 90° (tangent ⊥ radius) Consider quadrilateral AOBT. ∵ Sum of the interior angles of quadrilateral AOBT = 360° ∴

∠OBQ = ∠OBR

= 50° (

(tangent properties) (tangent properties) (tangent properties) 12.

(tangent properties)

∴

8.

= 37° ∠QBR = 37° + 37° = 74°

OA = OB ∠OAB = ∠OBA = 34° ∠AOT = 34° + 34° = 68° ∠OAT = 90°

(radii) (base ∠s, isos. (ext. ∠ of

)

)

(tangent ⊥ radius)

∠AOT + ∠OAT + ∠ATO = 180° (∠ sum of

∴ 14.

∠PBT + 28° = 62°

(radii) ( base ∠s, isos.

)

)

(tangent ⊥ radius) ∠QBT = 180° (∠ sum of ) ∠QBT = 180° ∠QBT = 41°

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

( tangent properties)

(base ∠s, isos. ∠TAB = ∠TBA ∠ATB + ∠TAB + ∠TBA = 180° (∠ sum of △)

)

44° + 2∠TBA = 180° (tangent ⊥ radius)

∠TBA = 68° ∠BMQ + 25° = 68° (ext. ∠ of △)

(ext. ∠ of △)

∠PBT = 34° 10. ∠ABC = 90° ∠BOD = 180° − 90° − 32° = 58°

(∠ at centre twice ∠ atce

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

)

∠ATO = 22° OT = OP ∠OTP = ∠OPT = 62° ∠OTB = 90° ∴ ∠PTB = 90° − 62° = 28° ∠PBT + ∠PTB = ∠OPT

1 ∠TOQ 2 1 = × 68° 2 = 34° ∠OTB = 90° ∠TPB + ∠PTB + 34° + ( 90° + 15°) + ∵∴ ∠TPQ =

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

68° + 90° + ∠ATO = 180°

9.

∠ATB = 360° − 90° − 90° − 130°

∠BMQ = 43° ∠AMQ + ∠BMQ = 180° (adj. ∠s of ∠AMQ + 43° = 180° on st. line ) ∠AMQ = 137°

(tangent ⊥ radius) (∠ sum of △)

15. SMEFSU07EX@F01

4

Certificate Mathematics in Action Full Solutions 4B

Join OA, OB and OD. ∠OAE = ∠OBC = 90° (tangent ⊥ radius) ∵ ∠OAE + ∠OBC = 90° + 90° = 180° and AE // BF ∴ AOB is a straight line. ( tangent properties) ∠AOE = ∠DOE (tangent properties) ∠BOC = ∠DOC ∠s on ∠AOE + ∠DOE + ∠DOC + ∠BOC = 180(adj. ° 2∠DOE + 2∠DOC = 180°st. line) 2(∠DOE + ∠DOC ) = 180° 2∠EOC = 180° ∠EOC = 90° 16. (a)

(b)

Reflex ∠POT = 360° − 128° (∠s at a pt.) = 232° 1 ∠PQT = ( reflex ∠POT ) (∠ at centre twice 2 twice ∠ at ⊙ce) 1 = × 232° 2 = 116° ∠OTQ = 180° − ∠PQT = 180° − 116°

(int. ∠s, QP // TO)

= 64° OT ⊥ AB (tangent ⊥ radius) ∠QTA = 90° − ∠OTQ ∴ = 90° − 64° = 26° 17. (a)

∠PAT = ∠TCB (ext. ∠, cyclic quad.) = 100° ∠PTA = 180° − ∠TPA − ∠PAT (∠ sum of △) = 180° − 42° − 100° = 38°

(b) OT ⊥ PT (tangent ⊥ radius) ∠ATO = 90° − ∠PTA = 90° − 38° = 52° OA = OT (radii) ∠TAO = ∠ATO (base ∠s, isos. △) = 52° ∠TOA = 180° − 52° − 52° ∴ (∠ sum of △) = 76°

5

7 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 ∴ AC // TO

tangent properties radii base ∠s, isos. ∠ sum of

21. AP = AS, BP = BQ, CQ = CR and DR = DS (tangent properties) Perimeter of ABCD = AB + BC + CD + DA = 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 22.

∠ in semi -circle SMEFSU07EX@F03 alt. ∠s equal

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 = ( x − 5) cm ∴ CR = (x − 5) cm AR = [10 − ( x − 5)] cm = (15 − x ) cm ∵ AP = AR ∴ x = 15 − x ∴ x = 7.5 ∴ AP = 7.5 cm 20. BP = BR CR = CQ BR + CR = BP + CQ = ∵ AP = BP + AB = BP + 9 cm = From (1), we have PB + 9 cm = ∴ PB = 11 ∴ AP = cm

Basic Properties of Circles (II)

Join OP. Let ∠OQB = θ . OQ = OP radii ∠OQB = ∠OPB = θ base ∠s, isos. tangent ⊥ radius ∠OPA = 90° ∠BPA = 90° − θ Consider △QOB. ∠OBQ = 180° − 90° − θ ∠ sum of = 90° − θ ∠ABP = ∠OBQ (tangent properties) (tangent properties)

6 cm 6 cm   (1) AQ (tangent properties) AC + CQ 7 cm + CQ 7 cm + (6 cm − PB ) 2 cm

vert. opp. ∠s

= 90° − θ ∠BPA = ∠ABP AP = AB 23. (a)

AB = BD BD = BC ∴ AB = BC

sides opp. equal ∠s tangent properties tangent properties

(b) ∠BAD = ∠BDA ∠BAD + ∠BDA = ∠DBC 2∠BDA = ∠DBC ∠BDA = ∠BDC = ∠BCD

base ∠s, isos. △ ext. ∠ of △

1 ∠DBC 2 base ∠s, isos. △

6

Certificate Mathematics in Action Full Solutions 4B ∠BDC + ∠BCD = ∠DBA

6.

2∠BDC = ∠DBA

ext. ∠ of △ 1 ∠BDC = ∠DBA 2 ∠DBC + ∠DBA = 180° adj. ∠s on st. line ∠ADC = ∠BDA + ∠BDC ∴ = 1 ∠DBC + 1 ∠DBA 2 2 1 = × 180° 2 = 90°

7.

Exercise 7B (p. 78) Level 1 1.

2.

3.

∠ABT = ∠ATP (∠ in alt. segment) x = 115° 　　 ∠BTQ = 180° − 132° (adj. ∠s on st. line) = 48° ∠BAT = ∠BTQ x = 48°

(b) ∵ ∴ ∴

TA = TB

(given) (base ∠s, isos. ) = 180° 8. = 180° ∠ △ ( sum of ) = 69° = ∠TAB (∠ in alt. segment) = 69°

∠ATB = 90°

(∠ in semi -circle) ∠TAB = 180° − ∠ATB − ∠TBA (∠ sum of ) = 25° ∠BTP = ∠TAB

(∠ in alt. segment)

= 25° ∠TBA = ∠BTP + ∠TPB 65° = 25° + x

(ext. ∠ of △)

9.

7

TP = TA (given) ∠TPA = ∠TAP ∠TPA + ∠TAP = 2∠TAP = ∠TAP = ∠BTP = = 37° + x + 74° = x =

∠ATQ 74° 37° ∠TAP 37° 180° 69°

(base ∠s, isos. △) (ext. ∠ of )

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

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

△BCD ~ △CAD BD CD = CD AD 9 12 cm = 12 9 cm + AB 9 cm + AB = 16 cm AB = 7 cm

Consider △BCT and △CAT. common angle ∠CTB = ∠ATC ∠ in alt. segment ∠TCB = ∠TAC ∠CBT = 180° − ∠CTB − ∠TCB ∠ sum of ∠ACT = 180° − ∠ATC − ∠TAC ∠ sum of ∴ ∠CBT = ∠ACT ∴ △BCT ~ △CAT AAA

∴

x = 40° 5.

(a)

(b) ∵

= 180° − 90° − 65°

(∠ in alt. segment)

(a) Consider △BCD and △CAD. ∠BDC = ∠CDA common angle ∠BCD = ∠CAD ∠ in alt. segment ∠DBC = 180° − ∠BDC − ∠BCD ∠ sum of ∠DCA = 180° − ∠CDA − ∠CAD ∠ sum of ∴ ∠DBC = ∠DCA ∴ △BCD ~ △CAD AAA

(∠ in alt. segment)

∠TAB = ∠TBA ∠TAB + ∠TBA + ∠ATB 2∠TAB + 42° ∠TAB ∠BTQ x

4.

∠TBC = ∠CTQ = 37° ∠BCT = ∠BTP = 75° ∠ACB = ∠TBC = 37° ∠ABT + ∠ACT = 180° (37° + x ) + (75° + 37°) = 180° x = 31°

△BCT ~ △CAT BT CT = CT AT 8 10 2 = x+8 10 2 x + 8 = 25 x = 17

TA = TB ∠TAB = ∠TBA ∠TAB + ∠TBA + ∠ATB 2∠TAB + 56° ∠TAB ∠ACB

( tangent properties) (base ∠s, isos. ) = 180° (∠ sum of ) = 180° = 62° = ∠TAB (∠ in alt. segment) = 62° ∠BAC + ∠ACB + ∠ABC = 180° (∠ sum of ) 32° + 62° + ∠ABC = 180° ∠ABC = 86°

7

10.

∠ADT = ∠ATQ ( ∠ in alt. segment) ∴ = 53° ∠DAT = ∠BCD (ext. ∠, cyclic quad.) = 60° ∠DAT + ∠ADT + ∠ATD = 180° (∠ sum of △) 14. 60° + 53° + ∠ATD = 180° ∠ATD = 67° ∴

11.

∴

15.

SMEFSU07EX@F04

Join BD. ∠ADB = = ∠BDC = =

∠PAB 66° ∠MCB 32°

∠ADC = ∠ADB + ∠BDC = 66° + 32° = 98° ∠ABC = 180° − ∠ADC (opp. ∠s, cyclic quad.) = 180° − 98° = 82° 12.

∴

∠ACB AB ∠ABC ∠ABC

= = = =

∠CAD AC ∠ACB ∠CAD

alt. ∠s, CB // DE given base ∠s, isos.

TC = TA ∠TCA = ∠TAC BT = BA ∠BTA = ∠BAT ∠BTA = ∠TCA PA is the tangent to the circle at T.

converse of ∠ in alt. segment given base ∠s, isos. △ given base ∠s, isos. △

converse of ∠ in alt. segment

BD = BC ∠BDC = ∠BCD Let ∠BCD = t. ∠ABD = ∠BCD + ∠BDC = 2t x = ∠ABD = 2t x + y + ∠BDC = 180° 2t + y + t = 180° y = 180° − 3t Take t = 20° , x = 2( 20°) = 40°

( given) (base ∠s, isos. (ext. ∠ of

)

)

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

(or any other reasonable answers)

Level 2 16.

∠CTP = ∠TAC (∠ in alt. segment) 17. = 37° ∠ATC = 180° − ∠ATQ − ∠CTP (adj. ∠s, on st. line) = 180° − 52° − 37° = 91° ∠ATC + ∠ABC = 180° (opp. ∠s, cyclic quad.) 91° + ∠ABC = 180° ∠ABC = 89°

13.

DE is the tangent to the circle at A.

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

(∠ in alt. segment) (∠ in alt. segment)

Basic Properties of Circles (II)

(∠ in alt. segment) ∠EDB = ∠ABE = 70° (∠ in alt. segment) ∠EBD = ∠EDC Consider △BCD. ∠EBD + ∠BCD + ∠BDC = 180° (∠ sum of △) ∠EDC + 50° + (70° + ∠EDC ) = 180° ∠EDC = 30° ∠CTP = 180° − ∠CTQ (adj. ∠s on st. line) = 180° − 145° = 35° ∠CAT = ∠CTP (∠ in alt. segment) = 35° ∠ACT = ∠TPC + ∠CTP (ext. ∠ of △) = 25° + 35° = 60° ∠CTA = 180° − ∠ACT − ∠CAT (∠ sum of △) = 180° − 60° − 35° = 85° ∠CTA + ∠ABC = 180° (opp. ∠s, cyclic quad.) 85° + ∠ABC = 180° ∠ABC = 95°

8

Certificate Mathematics in Action Full Solutions 4B

18. Let ∠ABC = θ. ∠CAP = ∠ABC =θ AC = CP ∠CAP = ∠CPA

(∠ in alt. segment) (given) (base ∠s, isos.

∠CPA = θ ∠BAC = 90° Consider △BPA. ∠ABP + ∠BPA + ∠PAB θ + θ + ( 90° + θ ) θ ∠ABC

)

∴

19. (a)

(b)

20. (a)

∠ABP = = ∠APB = = =

(∠ in semi-circle) = = = =

180° 180° 30° 30°

(∠ sum of △)

∠ACB (∠ in alt. segment) 39° 180° − ∠PAB − ∠ABP (∠ sum of △) 180° − 105° − 39° 36°

∠ADB = ∠ABP = 39° ∠DAE = ∠ACB = 39° ∠AEB = ∠ADB + ∠DAE = 39° + 39° = 78°

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

Consider △ABC and △BTC. ∠BAC = ∠TBC ∠ in alt. segment ∠ in semi-circle ∠ACB = 90° adj. ∠s on st. line ∠BCT = 180° − 90° = 90° ∠ACB = ∠BCT ∠ sum of △ ∠ABC = 180° − ∠BAC − ∠ACB ∠ sum of △ = 180° − ∠TBC − ∠BCT ∠ sum of △ = ∠BTC ∴ △ABC ~ △BTC AAA AB 2 = AC 2 + CB 2

(b) ∴ ∵

∴

AB =

5 + 3 cm 2

= 34 cm △ABC ~ △BTC AB AC = TB BC 34 cm 5 = TB 3 5TB = 3 34 cm TB =

21. SMEFSUO7EX@F05

9

2

3 34 cm 5

(Pyth. theorem)



Join AM, AN, AB, BM and BN. AMB

=

∠ANB 2 = ∠AMB 1

ANB ∠ANB = 2∠AMB ∠AMB + ∠ANB = 180° 3∠AMB = 180° ∠AMB = 60° PA = PB ∠PAB = ∠PBA ∠PAB = ∠AMB = 60° ∠APB + ∠PAB + ∠PBA ∠APB + 60° + 60° ∠APB

(arcs prop. to ∠s at ⊙ce)

(opp. ∠s, cyclic quad.)

(tangent properties) (base ∠s, isos. ) (∠ in alt. segment) = 180° = 180° = 60°

BA = BC ∠BAC = ∠BCA = x ∠ABT = ∠BAC + ∠BCA 22. = 2x ∠CTQ = ∠BCA = x ∠BAT = ∠CTQ = x Consider △ATB. ∠ATB + ∠ABT + ∠BAT = 78° + 2 x + x = x =

(given) (base ∠s, isos. △ ) (ext. ∠ of △ ) (alt. ∠s, AC // PQ) (∠ in alt. segment) 180° (∠ sum of △) 180° 34°

23. (a) Consider △PAT and △PTB. ∠APT = ∠TPB ∠PTA = ∠ABT ∠PAT = 180° − ∠PTA − ∠APT = 180° − ∠ABT − ∠TPB = ∠PTB ∴ △PAT ~ △PTB (b) Let PA = x cm. ∵ △PAT ~ △PTB

(∠ sum of △)

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

7 PA PT = PT PB x 6 = 6 5+ x 5 x + x 2 = 36

∴

x 2 + 5 x − 36 = 0 ( x − 4)( x + 9) = 0 x = 4 or x = −9 ( rejected) ∴ 24. (a)

PA = 4 cm

(∠CAT + ∠CAB ) + (∠ABC + ∠CBT ) + 70° = 180° ∠ABC + ∠CAB + ∠ABC + ∠CAB = 110° ∠ABC + ∠CAB = 55° Consider △ABC. ∠ABC + ∠CAB + ∠ACB = 180° (∠ sum of △) 55° + ∠ACB = 180° ∠ACB = 125° ∠ABT = ∠ATQ = 100° ∠CBT = 180° − ∠PBC − ∠ABT = 180° − 55° − 100° = 25° ∠CAT = ∠CBT = 25°

26. (a)

( tangent properties) PD = PC (base ∠s, isos. ) ∠PDC = ∠PCD 180° − ∠DPC ∠PCD = (∠ sum of △) 2 180° − 50° = 2 = 65° ∠CBD = ∠PCD (∠ in alt. segment) = 65° (∠ in alt. segment) ∠BDC = ∠BCN

Basic Properties of Circles (II)

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

(∠s in the same segment)

= 36° (given) AB = AD (base ∠s, isos. ) ∠ABD = ∠ADB ∠ABC + ∠ADC = 180° (opp. ∠s, cyclic quad.) (∠ABD + ∠DBC ) + (∠ADB + ∠BDC ) = 180° (∠ABD + ∠ADB ) + 65° + 36° = 180° 2∠ADB = 79° ∠ADB = 39.5° ∠ADC = ∠ADB + ∠BDC ∴ = 39.5° + 36° = 75.5° (b)

∠BAC = = ∠AKD = = =

∠BCN 36° ∠ABK + ∠BAK 39.5° + 36° 75.5°

(∠ in alt. segment) (ext. ∠ of △)

25. SMEFSU07EX@F06

Join AB. (∠ in alt. segment) ∠ABC = ∠CAT (∠ in alt. segment) ∠CAB = ∠CBT Consider △ABT. ∠BAT + ∠ABT + ∠ATB = 180° (∠ sum of △)

10

Certificate Mathematics in Action Full Solutions 4B Consider △APT. ∠APT + ∠PAT 42° + (∠BAC + ∠CAT ) 42° + ( ∠BAC + 25°) ∠BAC

= = = =

∠ATQ (ext. ∠ of △) 100° 100° 33°

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

Exercise 7C (p. 88) Level 1 1.

(a)

(b)

2.

(a)

∠BAD + ∠BCD = (38° + 42°) + (35° + 65°) = 180° ∴ A, B, C and D are concyclic. opp. ∠s supp.

∠CDB = ∠CAB ∠ ( s in the same segment) x = 38° ∠DBC

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

(b)

3.

(a)

∠ sum of △

converse of ∠s in the same segment

∠BAC = ∠BDC ∠ ( s in the same segment) x = 35° ∠CAD = ∠DBC = 90° ∴ A, B, C and D are concyclic.

converse of ∠s in the same segment

(b) ∠BAC = 180° − ∠EAB − ∠CAD (adj. ∠s on st. line) = 180° − 55° − 90° = 35° ∠BDC = ∠BAC (∠s in the same segment) x = 35° ∴ 4.

(a)

Mark the point F as shown in the figure.

11

7

Basic Properties of Circles (II)

Consider △ABF. ∠ABF = 180° − ∠BAF − ∠AFB ∠ sum of △ = 180° − 45° − 85° = 50° ∠ABC = ∠ABF + ∠FBC = 50° + 35° = 85° ∴ ∠ABC = ∠ADE ∴ A, B, C and D are concyclic. ext. ∠ = int. opp. ∠ (b) ∴ 5.

Consider △PBC. ∠PBC + ∠PCB = ∠APB 16° + ∠PCB = 62° ∠PCB = 46° ∠ACD = ∠BCD − ∠PCB = 82° − 46° = 36° ∴ ∠ABD = ∠ACD ∴ A, B, C and D are concyclic.

converse of ∠s in the same segment 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° ∠AQS = 180° − ∠SQC = 180° − 90° = 90° ∴ ∠AQS = ∠BPT = 90° ∴ A, P, R and Q are concyclic.

∠ in semi-circle ∠ in semi-circle adj. ∠s on st. line

∠ATQ = ∠ABT ∠ATQ = ∠CDT ∴ ∠ABT = ∠CDT ∴ A, B, C and D are concyclic. ∴ ABCD is a cyclic quadrilateral.

∠ in alt. segment alt. ∠s, PQ // CD

∴ ∴

7.

∴ ∴

9.

ext. ∠ of △

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

6.

8.

∠DAC = ∠DBC (∠s in the same segment) x = 35°

ext. ∠ = int. opp. ∠

ext. ∠ = int. opp. ∠

ext. ∠ = int. opp. ∠

ext. ∠ = int. opp. ∠

12

Certificate Mathematics in Action Full Solutions 4B 10. ∠BAD = 180° – (x + y) (∠ sum of △) ∠BAD + ∠BCD = 180° (opp. ∠s, cyclic quad.) 180° − ( x + y ) + z = 180° 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)

13

7

Basic Properties of Circles (II)

11.

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

Level 2 12. (a)

(b)

13. (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 segment) = 30° ∠APB = 90° ∠APM + ∠APB = 180° ∠APM + 90° = 180° ∠APM = 90° ∠AOM = 90° ∴ ∠APM = ∠AOM ∴ O, P, M and A are concyclic. OA = OP ∠OAP = ∠OPA ∠OAP = ∠OMP ∴ ∠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.

∠ in semi-circle adj. ∠s on st. line

given

converse of ∠s in the same segment radii base ∠s, isos. △ ∠s in the same segment ext. ∠, cyclic quad. adj. ∠s on st. line ext. ∠, cyclic quad.

opp. ∠s supp.

14

Certificate Mathematics in Action Full Solutions 4B ∠CAD = 180° − ∠ACP − ∠APB ∠ sum of △ = 90° − ∠APB ∠ COD = 2(90° − ∠APB ) ∴ = 180° − 2∠APB

15. (a)

18. (a)

(i)

∠BQP = x

∠ in alt. segment ∠ in alt. segment

∠CQP = y Join PB and let ∠ARP = θ. ∠PBA = ∠ARP = θ ∠APB = 90° In △APB, ∠PAQ = 180° − ∠APB − ∠PBA = 90° − θ In △AQP, ∠PQB = ∠PAQ + ∠APQ = 90° − θ + θ = 90°

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

∠s in the same segment ∠ in semi-circle

∠BQC + ∠BPC = ( x + y ) + (180° − x − y ) = 180° ∴ B, P, C and Q are concyclic.

∠ sum of △

ext. ∠ of △

∠ sum 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. 17. (a)

(b)

15

∠ACB = 90° ∠ECP = 180° − ∠ACB = 90° ∠ADB = 90° ∴ ∠ECP = ∠ADB ∴ P, D, E and C are concyclic. ∠COD = 2∠CAD

proved in (a) ∠ in semi-circle

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

opp. ∠s supp.

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

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

∠s in the same segment

ext. ∠ = int. opp. ∠

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

∠ in semi-circle adj. ∠s on st. line ∠ in semi-circle

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

∠ in alt. segment

2.

TB = TA ∠TBA = ∠TAB = x ∠TAB + ∠TBA + ∠ATB = 180° 2 x + 58° = 180° x = 61°

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

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

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

(∠ sum of △)

∠OAB = 90° (tangent ⊥ radius) ∠ODC = 90° (tangent ⊥ radius) Sum of interior angles of pentagon = 180° (5 – 2)

7

Basic Properties of Circles (II)

= 540° ∴ ∠AOD = 540° − 90° − 90° − 100° − 128° = 132° 3.

(a)

∠BAP = 90° (tangent ⊥ radius) ∠BAQ = 180° − ∠ABQ − ∠AQB (∠ sum of △) = 180° − 90° − 29° = 61°

∠CAP = ∠BAP − ∠BAQ = 90° − 61° = 29°

Join OC. ∠COP = 2∠CAP (∠ at centre twice ∠ at ⊙ ) ce = 2 × 29° = 58° ∠OCP = 90° (tangent ⊥ radius) ∠APC = 180° − ∠OCP − ∠COP ∠ ( sum of △) = 180° − 90° − 58° = 32°

(b)

4.

5.

6.

∠BAC = ∠BCQ = 80° ∠PAB + ∠BAC + ∠TAC = 180° 36° + 80° + x = 180° x = 64°

(∠ in alt. segment)

∠ABC = ∠TAC ∴ y = 64°

(∠ in alt. segment)

TA = TC ∠TAC = ∠TCA z = 180° − ∠TAC − ∠TCA = 180° − 64° − 64° = 52°

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

Let ∠OCA = a. OA = OC ∠OAC = ∠OCA = a ∠OCB = 90° ∠ACB = 90° − a AB = AC ∠ABC = ∠ACB = 90° − a ∠OAC = ∠ABC + ∠ACB = (90° − a ) + (90° − a) = 180° − 2a ∴ 180° – 2a = a ∴ a = 60° ∴ ∠OCA = 60° ∠BTP = ∠OBT = 27° ∠OTP = 90° ∠OTA = ∠OTP − ∠BTP = 90° − 27° = 63° OA = OT ∠OAT = ∠OTA = 63°

(adj. ∠s on st. line)

(∠ sum of △)

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

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

(radii) (base ∠s, isos. △)

16

Certificate Mathematics in Action Full Solutions 4B ∠AOB + ∠OBA = ∠OAT ∠AOB + 27° = 63° ∠AOB = 36° 7.

(ext. ∠ of △)

∠CAQ = ∠ABC (∠ in alt. segment) ∠BAC = 90° (∠ in semi-circle) Consider △ABQ. ∠ABQ + ∠BAQ + ∠AQC = 180° ∠ABC + (∠BAC + ∠CAQ ) + 34° = 180° (∠ sum of △) 2∠ABC + 90° + 34° = 180° ∠ABC = 28°

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.

17

Consider △OBC. ∠OBC + ∠OCB + ∠BOC = 180° (∠ sum of △) ∠OBC + ∠OCB + 134° = 180° ∠OBC + ∠OCB = 46° ∠ABO = ∠OBC (tangent properties) ∠ACO = ∠OCB (tangent properties) ∠BAC = 180° − ∠ABC − ∠ACB = 180° − 2∠OBC − 2∠OCB ∠ ( sum of △) = 180° − 2(∠OBC + ∠OCB ) = 180° − 2 × 46° = 88°

7 10. (a) (b)

Basic Properties of Circles (II)

∠CAD = ∠ADE − ∠ACD (ext. ∠ of △) = 116° − 54° = 62° ∠CBD = ∠CAD (∠s in the same segment) = 62°

Join AC. ∠CAB = 90° (∠ in semi-circle) ∠ACB = ∠BAQ (∠ in alt. segment) = 36° ∠ABC = 180° − ∠CAB − ∠ACB ∠ ( sum of △) = 180° − 90° − 36° = 54°

(b)

11.

∠ADC + ∠ABC = 180° ∠ADC + 54° = 180° (opp. ∠s, cyclic quad.) ∠ADC = 126°

(a)

∠TDA = ∠TCB (corr. ∠s, AQ // BC) = 70° ∠TQD + ∠QTC = ∠TDA (ext. ∠ of △) 26° + ∠QTC = 70° ∠QTC = 44°

(b)

∠TBC = ∠QTC (∠ in alt. segment) = 44° ∠ATD + ∠TBC + ∠TCB = 180° ∠ ( sum of △) ∠ATD + 44° + 70° = 180° ∠ATD = 66°

12.

Join OP and OQ. OP ⊥ AB tangent ⊥ radius OQ ⊥ AC tangent ⊥ radius OP = OQ radii ∴ AB = AC chords equidistant from centre are equal 13. (a)

∠ABC = 180° − ∠BCA − ∠BAC = 180° − 28° − 36° = 116° ∠ABC = ∠ADE = 116° ∴ A, B, C and D are concyclic.

∠ sum of △

ext. ∠ = int. opp. ∠

∴ ABCD is a cyclic quadrilateral.

18

Certificate Mathematics in Action Full Solutions 4B 14. ∠ACB = 90° ∠CAD + ∠CDA = ∠ACB = 90° ∠ACD = 180° − ∠ACB = 180° − 90° = 90° ∠BAC + ∠ABC = ∠ACD = 90° ∴ ∠CAD + ∠CDA = ∠BAC + ∠ABC and ∠CDA = ∠BAC ∴ ∠CAD = ∠ABC ∴ AD is the tangent to the circle at A.

 

DE

15.

=

∠EBD ∠DAC

 

CD

∠BPC + ∠BCP + ∠PBC = 180° ∠BPC + ∠BCP + 100° = 180° ∠BPC + ∠BCP = 80° ∠PAD = ∠PAB + ∠BAD = ∠BPC + ∠BCP = 80°

∠ in semi-circle ext. ∠ of △ adj. ∠s on st. line

ext. ∠ of △ 19.

given

converse of ∠ in alt. segment

arcs prop. to ∠s at ⊙ce

given CD = DE ∴ ∠EBD = ∠DAC ∴ P, A, B and Q are concyclic. converse of ∠s in the same segment ∴ PABQ is a cyclic quadrilateral. 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. 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° ∠ADB = ∠QAD + ∠AQD (ext. ∠ of △) = 34° + x ∠ABC + ∠ADC = 180° (opp. ∠s, ( x + 34°) + (34° + x + 34°) = 180° cyclic quad.) x = 39°

18. ∠PAB = ∠BPC ∠BAD = ∠BCP In △PBC,

19

(∠ sum of △)

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

Join OB. ∠OBP = ∠OBQ = 90° (tangent ⊥ radius) ∠OAP = ∠OCQ = 90° (tangent ⊥ radius) ∠AOB + ∠OBP + ∠APB + ∠OAP = 360° ∠AOB + 90° + x + 90° = 360° ∠AOB = 180° − x ∠BOC + ∠OCQ + ∠BQC + ∠OBQ = 360° ∠BOC + 90° + y + 90° = 360° ∠BOC = 180° − y z = 360° − ∠AOB − ∠BOC (∠s at a pt.) = 360° − (180° − x) − (180° − y ) = x+ y BY =BX ∠BYX =∠BXY

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

180° − 60° (∠ sum of △) 2 = 60° ∴ 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 AZ = CA − CZ ∴ = [7 − (5 − a )] cm = (2 + a ) cm ∴

∠BYX = ∠BXY =

and

AX = AB − BX = (8 − a) cm

∴ ∴ ∴

AZ = AX 2+a= 8–a a= 3 XY = 3 cm

7

Basic Properties of Circles (II)

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. OP = OQ ∴ OPBQ is a square. (b)

opp. ∠s equal radii

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 2 AC = AB2 + BC2 (Pyth. theorem) AC = 12 2 + 5 2 cm = 13 cm ∴ 17 − 2r = 13 r=2 ∴ The radius of the circle = 2 cm ∠ONC = 90° ∠NCO = ∠ACB ∠NOC = 180° − ∠ONC − ∠NCO = 90° − ∠NCO ∠ABC = 180° − ∠BAC − ∠ACB = 90° − ∠ACB ∴ ∠NOC = ∠ABC ∴ △CON ~ △CBA

tangent ⊥ radius common angle

(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)

22. (a)

AAA

20

Certificate Mathematics in Action Full Solutions 4B △CON ~ △CBA NO NC = AB AC r 6−r ∴ = 4 6 6r = 24 − 4r r = 2.4

(proved in (a))

PC PA = PA PB ∴ PA2 = PC × PB ∴

(c)

corr. sides, ~ △s

Let PC = x cm. Using the result of (b), PA 2 = PC × ( PC + BC ) 6 2 = x( x + 5)

∴ The radius of the circle = 2.4 cm

x + 5 x − 36 = 0 ( x − 4)( x + 9) = 0 x = 4 or 2

23. (a)

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)

24. (a)

(b)

25. (a)

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 27. tangent ⊥ radius BR = BP CR = CQ BC = BR + CR ∴ BC = BP + CQ

tangent properties tangent properties

AP = AQ Perimeter of △ABC = AB + AC + BC = AB + AC + ( BP + CQ ) = ( AB + BP ) + ( AC + CQ ) = AP + AQ = 2 AP = ( 2 × 12) cm = 24 cm

(tangent properties)

Consider △PAC and △PBA. ∠CAP = ∠ABP

21

△PAC ~ △PBA

proved in (a)

(ext. ∠ of △) (given) (base ∠s, isos. △)

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

given

ext. ∠ of △

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

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

(proved in (a))

∠ in alt. segment ∠CPA = ∠APB common angle ∠ACP = 180° − ∠CAP − ∠CPA ∠ sum of △ = 180° − ∠ABP − ∠APB = ∠BAP ∴ △PAC ~ △PBA AAA (b)

(b)

∠ADE = ∠DAT + ∠ATD = x+ y AE = AD ∠AET = ∠ADE ∴ ∠AET = x + y

28.

∴ ∴

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

converse of ∠ in alt. segment

∠ 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. ∠

7 29. (a)

∠ABC = ∠ADC = 90° ∠FBE = 180° − ∠ABC = 90° ∠EDF = 180° − ∠ADC = 90° ∴ ∠FBE = ∠EDF ∴ B, E, F and D are concyclic.

Basic Properties of Circles (II)

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

∠ in semi-circle adj. ∠s on st. line adj. ∠s on st. line 3.

converse of ∠s in the same segment

∴ BEFD is a cyclic quadrilateral. (b)

30. (a)

∠BDC = ∠BAC = 35° (∠s in the same segment) ∠BFE = ∠BDC = 35° (∠s in the same segment) ∠DCF = ∠CEF + ∠CFE (ext. ∠ of △) = 27° + 35° = 62° ∠DCF + ∠CFD = ∠ADC (ext. ∠ of △) 62° + ∠CFD = 90° ∠CFD = 28° ∠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° ∠ ( sum of △) 2 × 34° + 2 x + 2 × 25° = 180° x = 31° 2.

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

= ∠GED

(ext. ∠,

= ∠FGB

(∠ in alt.

22

Certificate Mathematics in Action Full Solutions 4B 4.

Answer: B ∠CDB = ∠CAB = 52° ∴ A, B, C and D are concyclic. ∴ ABCD is a cyclic quadrilateral. ∠ADC + ∠CBA = 180° (52° + ∠ADB ) + 72° = 180° ∠ADB = 56°

5.

(converse of ∠s in the same segment)

∠BAD + ∠DCB = 180° 86° + x = 180° x = 94° 7.

(opp. ∠s, cyclic quad.)

Answer: C

8. Join QS. BP = ∠BPQ = 180° − 90° ∠BQP = 2 = 45° CQ = ∠CQR = 180° − 124° ∠CQR = 2 = 28° ∠PSQ = ∠BQP = 45° ∠QSR = ∠CQR = 28° ∠PSR = ∠PSQ + ∠QSR ∴ = 45° + 28° = 73° 5.

6.

23

BQ (tangent properties) ∠BQP (base ∠s, isos. △) (∠ sum o f △) CR (tangent properties) ∠CRQ(base ∠s, isos. △) (∠ sum of △) (∠ in alt. segment) (∠ in alt. segment)

Answer: A TC = TB ∠TCB = ∠TBC ∠TCB = ∠ABC ∠TBC = ∠BAC ∠ACB = 180° − ∠BAC − ∠ABC = 180° − ∠TBC − ∠TCB = ∠BTC = 32°

9.

Answer: B ∠ACB segment) = 52° ∠BAC = 90° ∠ABC + ∠BAC + ∠ACB = 180° ∠ABC + 90° + 52° = 180° ∠ABC = 38° ∠CAQ = ∠ABC = 38° ∠CAQ + x = 13° + ∠ACB 38° + x = 13° + 52° x = 27° Answer: D 180°(5 − 2) ∠BCD = 5 = 108° PC = QC ∠CPQ = ∠CQP 180° − ∠BCD ∠CQP = 2 180° − 108° = 2 = 36° ∠PRQ = ∠CQP = 36°

Answer: D ∠PAB = ∠ABP (alt. ∠s, PQ // BD) = 47° ∠ADB = ∠PAB (∠ in alt. segment) = 47° ∠BAD = 180° − ∠ABD − ∠ADB (∠ sum of △) = 180° − 2 × 47° = 86°

= ∠BAP

(∠ in alt.

(∠ in semi-circle) (∠ sum of △)

(∠ in alt. segment) (ext. ∠ of △)

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

(∠ in alt. segment)

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

(tangent properties) (base ∠s, isos. △) (alt. ∠s, CT // AB) (∠ in alt. segment) (∠ sum of △)

(opp. ∠s, cyclic quad.)

∠BAC

= ∠CBE

(∠ in alt.

segment) = 37° ∠BAP = 180° − 84° − 37° = 59° PB = PA ∠ABP = ∠BAP = 59° x = 180° − 59° − 59° = 62°

(adj. ∠s on st. line) (tangent properties) (base ∠s, isos. △) (∠ sum of △)

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

Basic Properties of Circles (II)

10. Answer: D



CD DA

=

∠DBC 1 = ∠DCA 3

(arcs prop. to ∠s at ⊙ce)

63° (arcs prop. to ∠s at ⊙ce) 3 = 21° ∠ABD = 63° (∠s in the same segment) ∠ACB = 37° (∠ in alt. segment) ∠ABC + ∠ACB + ∠BAC = 180° ∠ ( sum of △) (63° + 21°) + 37° + ∠BAC = 180° ∠BAC = 59° ∠DBC =

11. Answer: D ∠ACE = 90° ∠ECD = x ∠BAC + ∠ACD = 180° (52° + x) + (90° + x) = 180° x = 19°

(∠ in semi-circle) (∠ in alt. segment) (int. ∠s, AB // PD)

12. Answer: B

Join BD. ∠BDA = 90° ∠ABD = 53° ∠BDC = 180° − 90° − 53° = 37° x + 37° = 53° x = 16°

(∠ in semi-circle) (∠ in alt. segment) (adj. ∠s on st. line) (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 AC = AR + RC 10 = (8 − a) + (6 − a ) a=2 AR = (8 − 2) cm ∴ = 6 cm

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

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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 MB = cm 2 = 4.5 cm Consider △OMB. OM2 + MB2 = OB2 (Pyth. theorem) 2 2 l + 4.5 = 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)

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