4bch06(basic Properties Of Circles 1)

Certificate Mathematics in Action Full Solutions 4B

Basic Properties of Circles (I)

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

Activity

EM = OE

−OM

2

(Pyth. theorem)

= 10 2 −6 2 cm = 8 cm (line from centre ⊥ chord bisects chord) MF = EM =8 cm

Activity 6.1 (p. 14)

(c) ∠ AOB = 2∠ APB No matter where points B and P are, ∠ AOB = 2∠ APB. (or any other reasonable answers)

2. 3.

2

Activity 6.2 (p. 25) 1.

yes 2.

yes

3.

yes 4.

yes

p. 9 1. ∵ ∴

Activity 6.3 (p. 35) 3.

∠ A +∠ C =180 °

4.

The sum of the opposite angles of a cyclic quadrilateral is 180°.

∠ B +∠ D =180 °

ON = OM = 4 cm (given) CD = AB (chords equidistant from centre are equal) = 7 cm CN = ND (line from centre ⊥ chord bisects chord)

1  × 7  cm 2 

= 

= 3.5 cm

Follow-up Exercise p. 3



Element AB

•

•

Term minor arc

region BCE

•

•

major arc

AFB

AFB



region BECFA

AB

AB

•

•

= 8 cm

OQ = OP = 2 cm BC = AB = 8 cm

(given) (given) (chords equidistant from centre are equal)

BC = AB = 8 cm (chords equidistan centre are equal) QC = BQ

t from

(line from centre

⊥chord

∴ •

•

chord

•

•

major segment

•

•

minor segment

region OBEC

•

•

sector

AC

•

•

radius

1.

PB = AP (line from centre ⊥ chord bisects chord) = 4 cm (line from centre ⊥chord bisects chord) AB = ( 2 ×4) cm ∴ ∵ ∴

diameter

OB

p. 7

2.

MB = AM =8 cm

(line from centre ⊥ chord bisects chord) 1  =  × 8  cm 2  = 4 cm

bisects

chord)

(line from centre ⊥ chord bisects 3.

chord)

MB = AM MB = AM

(line from centre ⊥ chord bisects chord) (line from centre ⊥chord

bisects 2.

∠ OND = 90°

(line joining centre to mid-pt. of chord ⊥ chord)

Consider △OND. ∠OND +∠NOD +∠ODN =180 ° ∠ ( 90 °+∠NOD +35 ° =180 ° ∠NOD =55 ° sum of △)

∴

1

Consider △OEM.

AB = ( 2 ×5) cm =10 cm

CN = ND

(line from centre ⊥ chord bisects chord)

∴ ∵

3.

chord)

CD = ( 2 ×5) cm =10 cm AB = CD

Basic Properties of Circles (I) OM = ON = 2.5 cm

∴

(equal chords, equidistant from centre)

(∠s inthe

same segment)

(ext. ∠of

p. 18

1 ∠AOB 2 1 = ×60 ° 2 = 30 °

)

x= 1.

2.

3.

4.

(∠ at centre twice ∠ at ☉ )

4.

ce

x =2∠ APB =2 ×50 ° =100 °

(∠ at centre twice ∠ at ☉ )

x =2∠ APB =2 ×120 ° =240 °

(∠ at centre twice ∠ at ☉ )

ce

segment)

ce

p. 28 1. ∵

Reflex ∠ AOB = 360° – 140° (∠ s at a pt.) = 220°

∴ ∴

2.

☉ ) ce

p. 20

x =90 °

x =∠ BAC =40 °

2.

x =90 °

y =∠ ACB =20 °

3.

(∠ in semi-circle)

∠ APB = 90° (∠ in semi-circle) Consider △APB. ∠ APB +40 °+x =180 ° (∠ sum of △) 90 °+40 °+x =180 ° x =50 °

1.

(∠ s in the same

x =∠ ABD =50 °

1 reflex ∠AOB 2 (∠ at centre twice ∠ at 1 = ×220 ° 2 =110 °

6.

∠ABD = 50 °

sum of △)

x=

5.

Consider △ABD.

∠ABD +∠BAD +∠ADB =180 ° (∠ ∠ABD +( 40 ° + 20 °) +70 ° =180 °

3.

 

∠ DOC = ∠ AOB = 43°(given) (equal ∠ s, equal arcs)

=

DC AB x =4

∵

CD AB

(given)

∴ ∴

CD = AB

(equal arcs, equal chords)

∵ ∴ ∵ ∴ ∴

=

x =5

AB = DC

DC = AB

(given) (equal chords, equal ∠ s) (given)

=

(equal chords, equal arcs)

 x =65 °

DC AB y =6

(∠ s in the same segment)

(∠ in semi-circle) (∠ s in the same segment)

∠EAC = ∠CBE = 30 ° x = ∠ADC + ∠DAC

△

= 25 ° + 30 ° = 55 °

2

Certificate Mathematics in Action Full Solutions 4B (adj. ∠ s on st. line)

4.



ycm ∠B O C = AB ∠AO B Join OB. ∵ BC = ED

(given)

∴

∠BOC = ∠EOD = 55 °

∵

AB = BC

∴

∠AOB = ∠BOC = 55 °

∠ s)

∠ s)

100 °  y = 6   30 °  = 20 °

(equal chords, equal

(given) (equal chords, equal

x = ∠BOC +∠AOB =55 °+55 ° =110 °

p. 31 1.

x ∠A O

 

B C (arcs prop. to ∠s at centre)

= BAB

2 (80 °) 5 =32 °

x=



ycm ∠D O C = AB ∠AO B

(arcs prop. to ∠s at centre)

 48 °  y = 5   80 °  =3

2.



xcm ∠D O C = AB ∠AO B

(arcs prop. to ∠s at centre)

 50 °  x = 6   30 °  =10

∠BOC =180 °−∠AOB −∠DOC =180 °−30 °−50 ° =100 °

3

(arcs prop. to ∠s at centre)

 

x AB = ∠C F DC D

3.

Basic Properties of Circles (I) △)

(arcs prop. to ∠ s at ☉ce)

2.

(ext. ∠ , cyclic quad.)

y =∠CDE =113 °

6 (15 °) 2 = 45 °

3.



x cm ∠C E D = BC ∠BEC ce

y AB = ∠B E C B C

x + ∠CBA = 180 ° x = 180 ° − 127 ° = 53°

(opp. ∠ s, cyclic quad.)

(∠ in semi-circle) ∠ ACD = 90° (∠ sum of △) x + y + ∠ ACD = 180° y = 180° − 90° − 53° = 37°

(arcs prop. to ∠ s at ☉ ) ce

 30 °  x =10    40 °  = 7 .5

 

(arcs prop. to ∠ s at ☉ )

(opp. ∠ s, cyclic quad.)

= 87 °

x =

4.

x + ∠DAB = 180 ° x = 180 ° − 93 °

(arcs prop. to ∠ s at ☉ce)

6 (arcs prop. to ∠ s at ☉ ) y = 10 ( 40 °) ce = 24 °

5.

 

∠ A C B AB = ∠D B CC D

(arcs prop. to ∠ s at ☉ce)

Exercise 6A (p.10 10) Level 1

3 (arcs prop. to ∠ s at ☉ ) ∠ACB = ( 48 °) ce 4 = 36 ° x =180 °−∠ DBC −∠ ACB =180 °−48 °−36 ° =96 °

∵

ON ⊥ AB

NB = AN

∴

1  =  × 16  cm 2  = 8 cm

(line from centre bisects

(given)

⊥chord

chord)

(ext. ∠ , cyclic quad.)

x =180 °−∠ DEC −∠ ECD =180 °−81 °−65 ° =34 °

1.

(∠ sum of

△)

p. 38119 ∠DEC = ∠ABC 1. =81 °

Exercise

(∠ sum of Join OB. Consider △NOB.

4

Certificate Mathematics in Action Full Solutions 4B OB =

∴

NB

2

+ON

2

(Pyth. theorem)

= 8 2 + 6 2 cm =10 cm The radius of the circle is 10 cm.

∴

OM ⊥ AB

∴

∠ OMK = 90°

∠MKN

(line joining centre to mid-pt. of chord ⊥ chord)

=180 °−∠DKB =180 °−43 ° =137 °

line)

(adj. ∠ s on st.

Consider quadrilateral OMKN. ∠ MON

2.

5.

∵ ∴

∴

Join OB.

OB = OC = ON + NC = (5 + 8) cm = 13 cm

(radii)

2

∵ ∴

3.

∵ ∴ ∵ ∴

2

2

− MB

2

= 5 2 −3 2 cm = 4 cm ON = MN −OM = (7 − 4) cm = 3 cm

(Pyth. theorem)

= 13 −5 cm =12 cm AN = NB (line from centre ⊥ chord bisects chord) AB =( 2 ×12 ) cm =24 cm 2

AM = MB

(given) (line joining from centre ⊥ chord bisects chord)

1  MB =  × 6  cm 2  = 3 cm

OM = OB

−ON

2

OM ⊥ AB AM = MB

Consider △OMB.

Consider △ONB. NB = OB

=360 °−∠ OMK −∠ ONK −∠ MKN =360 °−90 °−90 °−137 ° =43 °

(line from centre ⊥ chord bisects chord)

AB = (2 ×6) cm =12 cm

ON = OM CD = AB = 12 cm

(given) (chords equidistant from centre are equal) Join OD. Consider △OND.

OD = OB = 5 cm

ND = OD

∵ ∴ ∴

5

∵ ∴

CN = ND ON ⊥ CD

∴ ∵

∠ ONK = 90° AM = MB

(given) (line joining centre to mid-pt. of chord ⊥ chord) (given)

(radii) 2

−ON

= 5 −3 = 4 cm ON ⊥ CD CN = ND 2

4.

(Pyth. theorem)

2

2

(Pyth. theorem)

cm

(given) (line from centre ⊥ chord bisects chord) CD =( 2 ×4) cm =8 cm

Basic Properties of Circles (I)

6.

∵ ∴

CM = MD OM ⊥ CD

∴

∠ OMC = 90°

(given) (line joining centre to mid-pt. of chord ⊥ chord)

∠AOC = ∠OCM +∠OMC

△)

= 32 ° +90 ° =122 °

Consider △OAC. ∵ OC = OA (radii) ∴ ∠ OCA = ∠ OAC (base ∠ s, isos. △) ∠OCA +∠OAC +∠AOC =180 ° (∠ 2∠OCA +122 ° =180 ° ∠OCA = 29 ° sum of △)

1 AB 2 1 = ( AM + MB) 2 1 = (16 + 4) cm 2 = 10 cm

OB =

(ext. ∠ of 7.

∴

OM = OB − MB = (10 − 4) cm = 6 cm

Join OD. Consider △OMD.

OD = OB = 10 cm

(radii)

MD = OD

∵ ∴ ∴

8.

∵ ∴

2

−OM

2

(Pyth. theorem)

= 10 2 −6 2 cm = 8 cm OM ⊥ CD (given) CM = MD (line from centre ⊥ chord bisects chord) CD =( 2 ×8) cm =16 cm

BM = MC = 6 cm OM ⊥ BC

(given) (line joining centre to mid-pt. of chord ⊥ chord)

Consider △OMB. OM = OB

2

− BM

= 10 −6 = 8 cm Consider △OMD. 2

MD = OD

2

2

(Pyth. theorem)

cm

−OM

= 17 −8 =15 cm 2

2

2

2

(Pyth. theorem)

cm

6

Certificate Mathematics in Action Full Solutions 4B

∴

7

CD = MD −MC = (15 −6) cm = 9 cm

6 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.

Basic Properties of Circles (I)

∴

= 15° ∠MND = ∠OND +∠ONM =90 °+15 ° =105 °

∴

∠ OND = 90°

(line joining centre to mid-pt. of chord ⊥ chord)

Draw OM and ON such that OM ⊥ AB and ON ⊥ BC.

OB = OB

∠ABH = ∠CBH ∠OMB = ∠ONB = 90 ° common side given constructe ∴ ∴ ∴

SMEFSU08EX@F04 Construct the traingle ABC as shown.

AC 12 cm AC = 12 tan 60 ° cm

tan 60 ° =

d

△OBM ≅ △OBN OM = ON PQ = RS

AAS corr. sides, ≅ △s chords equidistant from centre are equal

= 12 3 cm ∴

Area of △ABC

1 × BC × AC 2 1  =  ×12 ×12 3  cm 2 2  =

= 72 3 cm 2 110.

Level 2 10. (a)

∵ ∴ ∴

AM = MB ∠ OMA = 90° ∠OMN

(given) (line joining centre to mid-pt. of chord ⊥ chord) =∠OMA −∠ AMN =90 °−75 ° =15 °

Let M be a point on AB such that OM ⊥ AB.

(b)

Join ON. ∵ CN = ND ∴ ∠ OND = 90° ∵ ∴ ∴

(given) (line joining centre to mid-pt. of chord ⊥ chord) CD = AB (given) ON = OM (equal chords, equidistant from centre) ∠ ONM = ∠ OMN (base ∠ s, isos. △)

8

Certificate Mathematics in Action Full Solutions 4B ∵

OM ⊥ AB

1  =  × 24  cm 2   = 12 cm

∴

chord

bisects

1  =  × 18  cm 2   = 9 cm

(line from centre

chord)

(given)

AN = NB

∴

⊥

(line from centre

ON ⊥ AB

(b) ∵

(constructed)

AM = MB

chord

bisects

⊥

chord)

Consider △OMA. OM = OA 2 − AM = 15 −12 = 9 cm Consider △OMC. 2

2

2

(Pyth. theorem)

cm

MC = MB + BC = (12 + 28 ) cm

Join OA. Consider △OAN. OA = r cm ON 2 + AN 2 = OA

= 40 cm OC = OM

2

+MC

= 9 +40 = 41 cm 2

12. (a)

2

2

(Pyth. theorem)

cm

(r −3) 2 + 9 2 = r 2

Consider △OAB and △OAC. common OA = OA

OB = OC AB = AC

∴ ∴ ∴ (b)

Consider △ABN and △ACN.

∠OAB = ∠OAC

(proved in (a)) (given) (common side)

AB = AC AN = AN ∴ ∴ ∴

(c)

side

radii given SSS corr. ∠ s, ≅ △s

△OAB ≅ △OAC ∠ OAB = ∠ OAC OA bisects ∠ BAC.

r

∴

△ABN ≅ △ACN (SAS) BN = CN (corr. sides, ≅ △s) ON ⊥ BC (line joining centre to mid-pt. of chord ⊥ chord)

14. ∵

2

(radius) 2

(Pyth. theorem)

−6r +90 = r r =15

2

OM ⊥ CD

(given)

CM = MD 1  =  × 12  cm 2   = 6 cm

∴

(line from centre chord

bisects

⊥

chord)

Let r cm be the radius of the circle. OM = AM −OA

= (18 − r ) cm

ON = AN − OA = (8 − 5) cm = 3 cm

Consider △ONC. NC = OC

2

−ON

= 5 −3 = 4 cm Consider △ANC. 2

AC =

AN

2

2

+NC

= 4 5 cm

9

(Pyth. theorem)

cm

= 8 2 +4 2 cm

13. (a)

2

ON = OY − NY = ( r −3) cm

Join OC. Consider △OCM. OC = r cm

OM

2

+ CM

2

(radius)

= OC 2

(18 − r ) 2 + 6 2 = r 2

2

(Pyth. theorem)

360 − 36 r + r 2 = r 2 r = 10

∴

MB = OB −OM

(Pyth. theorem)

6 =[ r −(18 −r )] cm

Basic Properties of Circles (I)

15.

=( 2r −18 ) cm =(2 ×10 −18) cm =2 cm

Let M be a point on AB such that OM ⊥ AB. ∵ OM ⊥ AB (constructed)

MB = AM

1  =  × 18  cm 2  = 9 cm

∴

(line from centre chord

bisects

⊥

chord)

Join OB. OB = 13 cm Consider △OMB. OM = OB

2

(radius) −MB

2

(Pyth. theorem)

= 13 2 −9 2 cm = 88 cm

Let N be a point on CD such that ON ⊥ CD. ∵ ON ⊥ CD (constructed)

NC = DN

1  =  × 24  cm 2  = 12 cm

∴

(line from centre chord

∵ ∴ ∴ ∴

bisects

⊥

chord)

∠ONK = ∠OMK = 90 ° ONKM is a rectangle. NK = OM KC = NC −NK = NC −OM

(property of rectangle)

=(12 − 88 ) cm =2.62 cm (cor. to 2 d.p.)

16.

(a)

Join OD, OB and OA as shown. Let ∠ OAB = x, then ∠ OAD = 90° −x.

10

Certificate Mathematics in Action Full Solutions 4B ∵ OB = OA ∴ ∠ OBA = ∠ OAB

radii base ∠ s, isos. △

=x

∠AOB =180 ° −∠OAB −∠OBA =180 ° −2 x

∴

∠ sum of △

∵ OD = OA ∴ ∠ ODA = ∠ OAD = 90° −x

DL = EL FN = NG

∠AOD =180 ° −∠OAD −∠ODA =180 ° −2(90 ° − x )

∴

∠ sum of △

DE = 2( LM − EM ) = 2( MN − EM ) = 2( MN − MF ) = 2 FN = FG

=2x

∠AOB +∠AOD

∴

= (180 ° −2 x) +2 x =180 °

∴

(b) Draw OM ⊥ AB and ON ⊥ DA. ∵ OM ⊥ AB and ON ⊥ DA (constructed) ∴ AM = MB and DN = NA (line from centre ⊥ chord bisects chord)

1 AD 2 1  =  × 18  cm 2  = 9 cm

(adj. ∠ s on st. line) x =2∠ ACB =2 ×42 °

(∠ at centre twice ∠ at ☉ ) ce

=84 °

2.

AMON is a rectangle.

OM = NA = 9 cm

proved

Level 1 ∠ACB +138 ° =180 ° (adj. ∠ s on st. line) 1. ∠ACB = 42 °

NA =

∴

by (a)

Exercise 6B (p. 21)

BOD is a straight line.

∵

line from centre ⊥ chord bisects chord line from centre ⊥ chord bisects chord line from centre ⊥ chord bisects chord

(b) EM = MF

radii base ∠ s, isos. △

(property of rectangle)

∠ ACB = 90° (∠ in semi-circle) ∵ CA = CB (given) ∴ x = ∠ CBA (base ∠ s, isos. △) ∠ACB +∠CBA +x =180 ° (∠ sum of 2 x +90 ° =180 ° x = 45 ° △)

3.

2

= 15 2 −9 2 cm =12 cm AB = 2 AM = (2 ×12 ) cm = 24 cm

17. (a)

∵ ∴ ∵ ∴

11

(∠ s in the same segment)

∠DAC = ∠DBE = 25 °

(∠ s in the same segment)

x +∠ACD =125 ° (ext. ∠ of △) x +55 ° =125 ° x =70 °

Consider △OAM. AM = OA 2 −OM

∠ACD = ∠ABD = 55 °

(Pyth. theorem)

∠ ALM = ∠ BMN = ∠ CNG = 90°given LA // MB // NC corr. ∠ s equal LA // MB // NC and AB = BC given LM = MN intercept theorem

4.

x = ∠DAC +∠ADC = 25 °+42 °

(ext. ∠ of △)

= 67 °

5.

Reflex

∠AOC =2∠ABC

twice at ☉ ) ce

=2 ×140 ° =280 °

(∠ at centre

6

Basic Properties of Circles (I)

(∠at centre twice ∠at ) x =360 °−reflex ∠ AOC =360 °−280 °

(∠ s at a pt.)

=80 °

6.

∠ACD =180 °−∠CAD −∠CDA

sum of △)

=180 °−32 °−90 ° = 58 °

∠ACB = 90 °

(∠ in semi-circle)

x =∠ACB −∠ACD =90 °−58 ° =32 °

∴

7.

(∠

∠ ABC = 90°

(∠ in semi-circle)

∠BCA = ∠BDA =x

(∠ s in the same segment)

∠ BCA =180 °−∠ ABC −∠ BAC x =180 °−90 °−65 ° =25 °

(∠

sum of △)

8.

∠AOB = 2∠ACB

☉ )

(∠ at centre twice ∠ at

= 2 ×70 ° =140 °

△

ce

OB = OA ∠OBA = x (radii) (base ∠ s, isos.

)

x +∠OBA +∠AOB =180 ° 2 x +140 ° =180 ° x = 20 °

(∠ sum of

△)

12

Certificate Mathematics in Action Full Solutions 4B 9.

12.

Join OC.

65 ° + ∠BCA = 118 ° ∠BCA = 53 ° ∠AOB = 2∠BCA = 2 × 53 ° = 106 ° ∠AOB + ∠OAK = 118 ° ∠OAK = 118 ° −106 ° = 12 °

(ext. ∠of

1 ∠ACB = ∠AOB 2 1 = ×130 ° 2 = 65 °

(∠ at centre twice ∠ at

)

(∠at centre twice ∠at ) (ext. ∠of

☉ )

)

ce

OC = OA (radii) ∠ OCA = ∠ OAC (base ∠ s, isos. △) = 20° OB = OC (radii) x = ∠ OCB (base ∠ s, isos. △) = ∠ ACB − ∠ OCA = 65° −20° = 45 ° 10.

13. ∠ DAC, ∠ ACD, ∠ DAB, ∠ DBA, ∠ EFD and ∠ FED (any four of the above angles) Level 2 14. ∵ ∴ ∵

(∠ in semi-circle) ∠DAC = 90 ° ∠ACD =180 °−∠DAC −∠ADC (∠ =180 °−90 °−55 ° =35 °

∴

∠ABC = ∠ACB = 35 ° (given) (base ∠ s, isos.

△

)

= 20 °

(opp. ∠ s of // gram)

1 ∠BOD 2 (∠ at centre twice ∠ at ☉ ) ce 1 x = ×36 ° 2 = 18 °

∠BCD =

∠ ODC

=∠ BCD =18 °

(∠ in semi-circle) (∠

(∠ sum of △)

∠ABC + ∠BAD = 55 ° (ext. ∠ of △) ∠BAD = 55 ° − 35 °

∠ BOD = 36°

(base ∠ s, isos. △)

sum of △)

15. (a)

11.

∠BDC = ∠DCA =x ∠ADB = 90 °

(given) (base ∠ s, isos. △) (given)

∠DAC +∠DCA +∠ADC =180 ° x +x +(90 °+x ) =180 ° x =30 °

sum of △)

AB = AC

DC = DA ∠ DCA = x BD = BC

∠OAB + ∠AOC = 180 ° ∠AOC = 180 ° − 32° = 148 °

(int. ∠ s, BA // CO) ∴ Reflex ∠ AOC = 360° −∠ AOC = 360° −148° = 212 ° ∠ AOC =360 °−∠ AOC =360 °−148 ° =212 °

1 reflex ∠AOC 2 1 = ×212 ° 2 =106 °

∠ABC =

∠BKD =∠ODK +∠BOD (ext. ∠ of =18 °+36 ° =54 ° (ext. ∠ of △) △)

(b)

twice ∠ at ☉ ) ce

ce

(∠ at centre twice ∠ at ☉ )

(ext. ∠ of △)

(∠ s at a pt.)

(∠ s at a pt.)

(alt. ∠ s, DO // AC)

13

(int. ∠ s, BA // CO)

(∠ at centre

6

19. (a)

Basic Properties of Circles (I)

OABC is a parallelogram. OA = OC ∴ OABC is a rhombus.

(given) (radii)

(b) Reflex ∠ AOC = 360° −x ∠ s at a pt.

1 reflex ∠AOC 2 1 = (360 ° − x) 2 x =180 ° − 2

∠ABC =

16.

∠DCB + ∠CKB = 50°

(∠ ext. ∠ of △)

∠DCB = 50° − 28° = 22° ∠DAB = ∠DCB (∠ s in the same segment) = 22 ° (∠ in semi-circle) ∠ACB = 90 ° ∠CAB =180 °−∠ACB −∠CBA (∠

sum of △) ∴

twice ∠ at ☉

∴

=180 °−90 °−50 ° = 40 °

∠CAD =∠CAB −∠DAB =40 °−22 ° =18 °

20. (a)

∴ 21. (a)

(∠in semi - circle) ∠ ABD + ∠ BAD + ∠ ADB = 180°(∠ sum of △) ∠ ABD + (44° + ∠ BAC) + 90° = 180° ∠ ABD + ∠ ABD + 134° = 180° ∠ ABD

18.

1 ∠ABE = ∠AOE 2 1 = ×124 ° 2 = 62 °

(ext. ∠ of △)

∴

(ext. ∠ of △)

(corr. sides, ~ △s)

KD =4 cm

OK ⊥ EB BK = EK

given line from centre ⊥ chord bisects chord

side

△BKD ≅ △EKD

(b) ∠ ABD = 90° ∠ BDC ∠ CBD – ∠ BCD (∠ sum of △) = 180° – 90° – 42° = 48° ∠ KED = ∠ KBD ∠ KED + ∠ KBD = ∠ BDC 2∠ KED = 48 ∠ KED = 24 ∠ BAD =∠ BED ∴ =24 °

= 26 °

∠AKE =∠BAD +∠ABE =26 °+62 ° =88 °

vert. opp. ∠ s ∠ s in the same segment ∠ s in the same segment AAA

KD = KD

given common

(∠ at centre twice ∠ at ☉ce)

∠BEC = 62 ° − 36 ° = 26 ° ∠BAD = ∠BEC (∠ s in the same segment)

∵ ∴

(proved in (b))

∠BKD = ∠EKD = 90 °

= 23 °

∠ACE + ∠BEC = ∠ABE

∴

AK KD = BK KC 6 KD = 3 2 cm

(b)

∠BDC =∠ABD ∠ADB =90 ° (∠ s inthe same segment) (alt. ∠ s, DC // AB )

(opp. ∠ s of // gram)

x 180° − = x 2 3 x = 180 ° 2 x = 120 °

∠ AKB = ∠ DKC ∠ BAK = ∠ CDK ∠ ABK = ∠ DCK ∴ △AKB ~ △DKC

∠BAC =∠BDC 17.

ce

∠ ABC = x

(c)

∠ at centre

SAS (∠ in semicircle) = 180° –

(corr. ∠ s, ≅ △s) (ext. ∠ of △) (∠ s in the same segment)

22.

14

Certificate Mathematics in Action Full Solutions 4B

 

= 180° – x – 90° = 90° – x

∠ BAC BC = x AC

Join AP. ∠ APB = 90°

∠ACQ = ∠APQ = 90

∠s inthe ∴

∠ in semi-circle

same segment

QC ⊥ AB

4.

∠C O D C D = ∠BO C BC

4 (84 °) 6 = 56 °

∠BOD = ∠BOC + ∠COD = 84 ° + 56 °

Exercise 6C (p. 32) Level 1

Reflex ∠ AOB = 360° – ∠ AOB (∠ s at a pt.) = 360° – 80° = 280°

= 140 ° (∠ at centre twice ∠ at ☉ce) 1 x = ∠BOD 2 1 = ×140 ° 2 = 70 °

∴



x cm

(arcs prop. to ∠

 80°  x = 14   280 °  =4

∴

6. ∠ BAC = 180° – ∠ ABC – ∠ ACB = 180° – 50° – 75° = 55°



x cm AC

=

∠BAC ∠ABC

3.

15

∠ ACB = 90° ∠ BAC= 180° – x – ∠ ACB

(a)

(∠ sum of △)

AC

AC

= BD

alt. ∠ s, CD // AB

BD

∠ A O B A B (arcs prop. to ∠ s at centre) = ∠BO C BC 4 ( 48 °) 3 =64 °

(b)

∠ACB =

centre twice ∠ at ☉ce) (∠ in semi-circle) (∠ sum of △)

equal ∠ s, equal arcs

∠AOB =

(arcsce prop. to ⊙ ∠ s at )

 55°  x = 10   50°  = 11

   

∠ ADC = ∠ BAD

5.

centre)

2.

(arcs prop. to ∠ s at centre)

∠COD =

Join OA. ∠ ABQ = ∠ AOQ ∠ s in the same segment = 2∠ ABP ∠ at centre twice ∠ at ☉ce ∴ BP bisects ∠ ABQ.

∠AOB = Reflex ∠AOB Major AB

ce

 

23.

1.

(arcs prop. to ∠ s at ⊙ )

90° − x 4 = x 5 450° − 5 x = 4 x x = 50°

1 ∠AOB 2

(∠ at

6 ∵ ∴

1 ×64 ° 2 =32 ° =

Basic Properties of Circles (I)

AB = AD ∠ ABD = ∠ ADB

(given) (base ∠ s, isos. △)

∠ABD +∠ADB +∠BAD =180 ° 2∠ABD +120 ° =180 ° ∠ABD =30 °

sum of △)

∠ACB =180 °−∠BAC −∠ABC

7.

∠ BEC △)

= 180° – ∠ EBC – ∠ ECB

= 180° – 62° – 64° = 54° ∠ DEB = ∠ EBC − ∠ EDB = 62° – 35° = 27°

(ext. ∠ of △)

∴

= ∠ DEB : ∠ BEC

 

AB

AB : BC

BC

(arcs prop. to ∠ s at = 27° : 54° = 1: 2 8.

∠ BAC = 90° ∠ BAD = ∠ BAC + ∠ CAD = 90° + 30° = 120°

∴

AB

AB : BC

∵

⊙ce) ∴ ∵

∴ ∵

∴

10. ∵



(∠

= ∠ ACB : ∠ BAC

BC

(arcs prop. to ∠ s = 60° : 90° = 2:3 9.

(∠ in semi-circle)



sum of △)

(∠ sum of

=180 °−90 °−30 ° = 60 °

(∠

at ⊙ce)

∠ APC= ∠ APB + ∠ BPC = 5° + 10° = 125° = ∠ CPD

A C= C DAC = CD



(equal ∠ s, equal arcs)

∠ BPD = ∠ BPC + ∠ CPD = 10° + 15° = 25° = ∠ EPF

B D= E F

(equal ∠ s, equal arcs)



A D= F GAD = FG

(equal ∠ s, equal arcs)



(given)

BD = EF

∠ APD = ∠ APB + ∠ BPC + ∠ CPD = 5° + 10° + 15° = 30° = ∠ FPG

A B= C DAB = CD

∴ ∠ ADB = ∠ DAC and ∠ ACB = ∠ DBC (equal arcs, equal ∠ s) ∴ KD = KA and KC = KB (sides opp. equal ∠ s) ∴ △AKD and △BKC are isosceles triangles.

16



Certificate Mathematics in Action Full Solutions 4B 13. ∠ PRS = ∠ PQS

A B= B C= C D

∵

AB = BC = CD ∠ PRQ

(given) ∴ ∠ ACB = ∠ CAB and ∠ BDDBC = ∠ DBC (equal arcs, equal ∠ s) ∴ BC = BA and CD = CB (sides opp. equal ∠ s) ∴ △ABC and △BCD are isosceles triangles.



= 30°

= 75° −30° = 45°

(∠ s in the same segment) = ∠ QRS – ∠ PRS

Q R= P QQR = PQ (given)

∵

∠QSR = ∠PRQ = 45 ° ∴

∠ QSR = ∠ PRQ (equal arcs, equal ∠ s) = 45° ∠ RQS = 180° – ∠ QRS – ∠ QSR (∠ sum of △) = 180° −75° − 45° = 60 °

Level 2 11.

∠ ABD = 180° – ∠ BAD − ∠ ADB (∠ sum of △) = 180° – (40° + 20°) – 70° = 50°

Join BC. ∠ CBD= ∠ CAD = 20° ∠ CBA = 20° + 50° = 70°



AD C ∠CBA = BC ∠BAC  7 0°  x = 1 0   4 0°  = 1 7.5

14. (a)

OD = OB (radii) ∠ ODB = ∠ OBD (base ∠ s, isos. △) = 30° ∴ ∠ BOA = ∠ OBD + ∠ ODB (ext. ∠ of △) = 30° + 30° = 60 °

 

(∠ s in the same segment) = ∠ CBD + ∠ ABD

(b)

(arcs prop. to ∠ s at ☉ ) ce



x cm CD

=

∠ BAC ∠ CAD

 40°  x = 15   50°  = 12

17

(arcs prop. to ∠ s at

☉ ) ce

15. (a) 12. ∠ BAD = 90° (∠ in semi-circle) ∠ BAC = ∠ BAD – ∠ CAD = 90° – 50° = 40°

∠ ADB AB = ∠ CDB BC 3 = 2 2 ∠ CDB = (30° ) 3 = 20°

OA = OC

radii

AB = CB OB = OB

given common

∴ (b) ∵ ∴

(arcs prop. to ∠ s at ☉ )

∴

ce

equal arcs



side

△ABO ≅ △CBO

SSS

∠ AOB = ∠ COB ∠ AOD = ∠ COD

corr. ∠ s, ≅ △s

A D= D CAD = DC

equal ∠ s,



6

16. ∵

B C= C D

∴ ∵ ∴

∠ CAB = ∠ DAC OC = OA ∠ ACO = ∠ CAB = ∠ DAC OC // AD

∴ 17. (a)

∵ ∴

equal arcs, equal ∠ s radii base ∠ s, isos. △

 π × 18  cm 2  

= 9π cm ∴

Radius of the circle =

9π cm 2π

=

4.5 cm

alt. ∠ s equal

OE ⊥ BD BE = ED



∠ BAC = ∠ DAC ∴

Circumference of the circle =

BC = CD given

AE = AE ∠ AEB = ∠ AED = 90° ∴ △ABE ≅ △ADE (b)

Basic Properties of Circles (I)

given line from centre ⊥ chord bisects chord common side given SAS corr. ∠ s, ≅ △s

B C= C DBC = 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° = 40 °



(b)

EA



AB BC CD D E AB :

BC :

CD :

DE :

EA = ∠ ADB : ∠ BEC : ∠ CAD : (arcs prop. to ∠ s at ☉ ∠ DBE : ∠ ACE (arcs prop. to ∠ s at ☉ce) = 20° : 40° : 40° : 30° : 50° = 2 : 4 : 4 :3:5

(cb)

to ∠ s at △ce)



AB 2 = C irc u m fne creeo f th ec irc le 2 + 4 + 4 + 3 + 5 (by (b))

18

Certificate Mathematics in Action Full Solutions 4B Exercise 6D (p. 39) Level 1 1. ∠BCD = 95 °

(ext. ∠ , cyclic quad.)

∠BCD + x = 180 ° x = 180 ° − 95 ° = 85 °

2.

(adj. ∠ s on st. line)

∵ ∴ ∴

∠BCD = 180 ° − 76 ° = 104 °

(opp. ∠ s, cyclic

CD = CB (given) ∠ BDC = x (base ∠ s, isos. △) ∠ BDC +x +∠ BCD =180 ° 2 x +104 ° =180 ° (∠ sum x =38 °

of △) 3.

∠ ACB = 90°

(∠ in semi-circle)

∠ABC + ∠ACB + ∠BAC =180 ° (∠ sum of

△)

∠ABC =180 °−90 °−40 ° =50 °

x + ∠ABC = 180 ° x = 180 ° − 50 °

(opp. ∠ s, cyclic quad.)

= 130 °

4.

x =∠ABD =46 °

(∠ s in the same segment)

y +∠BCD =180 ° (opp. ∠ s, cyclic quad.) y +(54 °+ x ) =180 ° y =180 °−(54 °+46 °) =80 °

5.

∠EBC + ∠CDE = 180 °

quad.)

∠EBC = 180 ° − 110° = 70°

∠ECB + ∠BAE = 180 °

quad.)

∠ECB = 180 ° − 120 ° = 60°

(opp. ∠ s, cyclic

(opp. ∠ s, cyclic

∠BEC + ∠EBC + ∠ECB =180 ° (∠ sum of △)

6.

(a) of △)

19

∠ BEC

∠DFE = ∠BCD (ext. ∠ , cyclic quad.) = 60 ° ∠DFE +∠FDE +∠DEF =180 ° (∠

sum of △) ∠ DEF

∠BCD + ∠BAD = 180 °

quad.)

(b)

=180 °−70 °−60 ° =50 °

∠ADE =∠CAD +∠ACD =36 °+60 ° =96 °

(ext. ∠

=180 °−60 °−96 ° =24 °

6

7.

∠FCD + ∠DEF = 180 ° ∠FCD = 180 ° − 130 ° = 50 °

quad.)

(opp. ∠ s, cyclic

(ext. ∠ , cyclic quad.)

x =∠FCD =50 °

∠ ABD = y ∵ AD = AB ∠ A DB =∠ABD ∴ =y

(ext. ∠ , cyclic quad.) (given) (base ∠ s, isos. △)

∠BAD +∠ABD +∠ADB =180 ° sum of △)

Basic Properties of Circles (I)

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

∵

∠ABC =180 °−115 ° =65 ° BC = CD

(given)

∠BAC =∠DAC ∴ (equal chords, equal ∠ s) =35 ° (∠ ∠ACB + ∠ABC + ∠BAC =180 ° sum of △)

∠ ACB

=180 °−65 °−35 ° =80 °

(∠

x + 2 y =180 ° 2 y =180 ° − 50 ° y = 65 °

8.

Join AD. ∠ ABC + ∠ CDA = 180° ∠ ADE = 90° ∠ ABC + ∠ CDE

opp. ∠ s, cyclic quad. ∠ in semi-circle

= ∠ABC +(∠CDA +∠ADE ) = (∠ABC +∠CDA ) +∠ADE =180 °+90 ° = 270 °

1 ∠AOB 2 1 = ×40 ° 2 = 20 °

∠APB = 9.

(a)

(∠ at centre twice ∠

at ☉ ) ce

(b) ∠BAP + ∠BCP =180 ° quad.)

(opp. ∠ s, cyclic

∠BAP =180 °−50 ° =130 ° ∠ABP +∠BAP +∠APB =180 ° (∠ sum of △)

10.

∠ ABP =180 °−130 °−20 ° =30 °

∠DAC + ∠ADC + ∠ACD =180 ° sum of △)

(∠

∠DAC =180 °−115 °−30 ° = 35 °

20

Certificate Mathematics in Action Full Solutions 4B Level 2 11.

Reflex ∠ AOC centre twice

= 2∠ ABC

(∠ at

= 2 × 110° ∠ at

☉ ) ce

= 220°

∠AOC =360 °−reflex ∠AOC =360 °−220 ° =140 °

pt.)

(∠ s at a

∴ ∠CPB +∠AOC =180 ° (opp. ∠ s, cyclic quad.) ∠ CPB =180 °−140 ° =40 ° 12.

∠COD = ∠BAD = 40 °

(corr. ∠ s, OC //

OD = OC ∠ ODC = ∠ OCD

(radii) (base ∠ s, isos. △)

AB)

∠COD +∠ODC +∠OCD =180 °

sum of △)

40 ° + 2∠ODC =180 ° ∠ODC = 70 °

(∠

∠ODC + ∠ABC =180 ° (opp. ∠ s, cyclic ∠ABC =180 ° − 70 ° =110 °

quad.) 13. (a)

∠ KAD = ∠ KCB ∠ KDA = ∠ KBC ∠ AKD = ∠ CKB ∴ △KAD ~ △KCB

ext. ∠ , cyclic quad. ext. ∠ , cyclic quad. common angle AAA

KA KD = KC KB KA KD = KD + DC KA + AB (b) 2 cm 3 cm = 3 cm + DC (2 + 4) cm 4 cm = 3 cm + DC DC =1 cm

(corr. sides,

~ △s)

14.

∠BAD = ∠BCE = 65 °

∠ ADB = 90°

(ext. ∠ , cyclic quad.) (∠ in semi-circle) (∠

∠ABD + ∠ADB +∠BAD =180 °

sum of △)

∠ABD =180 ° −90 ° −65 ° = 25 °

21

∠BDC = ∠ABD = 25 °

(alt. ∠ s, DC // AB)

∠DBC +∠BDC = ∠BCE ∠DBC = 65 ° −25 ° = 40 °

△)

(ext. ∠ of

6 15. ∠ACD + ∠ADC + ∠CAD =180 ° sum of △)

Basic Properties of Circles (I)

∠AED +∠ACD =180 ° (opp. ∠ s, cyclic quad.) ∠AED = 180 ° − ∠ACD = 180 − (140 ° − ∠ADC ) = 40 ° + ∠ADC

(∠

∠ACD =180 ° −40 ° −∠ADC =140 ° −∠ADC ∠ABC + ∠ADC = 180° (opp. ∠ s, cyclic quad.) ∠ABC = 180° − ∠ADC

∴

∠ ABC +∠ AED =(180 °−∠ ADC ) +( 40 °+∠ ADC ) =220 °

16.

Join BD. Let ∠ CBE = x. ∵ CE = CB

(given)

∠CEB = ∠CBE ∴ (base ∠ s, isos. △) =x ∠CBE + ∠BCE + ∠CEB = 180 ° (∠ sum

∠BCE = 180 ° − 2 x

of △)

(opp. ∠ s, cyclic quad.) ∠BDE +∠BCE =180 ° ∠BDE =180 ° −(180 °−2 x ) ∠ DBE = 90°

=2x

(∠ in semi-circle)

∠ABD + ∠DBE + ∠CBE =180 ° (adj. ∠ s

on st. line)

∠ABD =180 °−90 °− x =90 °−x ∠BAD +∠ABD = ∠BDE (ext. ∠ of △) 27 ° +(90 ° − x ) = 2 x 3 x =117 ° x = 39 ° ∴ 17. (a)

∠ CBE

=39 °

∠APC =∠ABC +∠PCB =( x + x ) + y =2 x + y

(ext. ∠

of △)

∠ARB =∠ACB +∠RBC =( y + y ) + x

(ext. ∠

= x +2 y of △) (b) ∠APC +∠ARB =180 ° (opp. ∠ s, cyclic quad.) ( 2 x + y ) +( x + 2 y ) =180 °

3 x +3 y =180 ° x + y = 60 °

22

Certificate Mathematics in Action Full Solutions 4B ∠BAC + ∠ABC + ∠ACB = 180 ° (∠ sum of △)

23

∠BAC +2 x +2 y =180 ° ∠BAC +2( x + y ) =180 ° ∠BAC +120 ° =180 ° ∠BAC = 60 °

6 Revision Exercise 6 (p. 47) Level 1

Basic Properties of Circles (I)

1 (∠ at centre twice reflex ∠ POR ∠ at ☉ce) 2 1 = ×222 ° 2 =111 °

∠PQR =

1.

(b)

∠ORQ +∠PQR =180 ° (int. ∠ s, OR // PQ)

∠ ORQ

=180 °−111 ° =69 °

Join OF. Draw ON such that ON ⊥ FE.

OF = OB

1 BC 2 1  =  × 20  cm 2  = 10 cm =

(radii)

ON = AB = 6 cm

(property of rectangle)

Consider △ONF. FN = OF

2

= 10 −6 =8 cm ON ⊥ FE FN = NE 2

∵ ∴ ∴

2.

2

2

(Pyth. theorem)

cm

(constructed) (line from centre ⊥ chord bisects chord) FE =( 2 ×8) cm

=16 cm

∠ AEC = 90°

(∠ in semi-circle)

∠EBC = ∠EAC (∠ s in the same segment) = 30 ° ∠AEB = ∠EDC +∠EBC (ext. ∠ of △) =35 ° +30 ° = 65 ° ∠ BEC

3.

−ON

(a)

=∠ AEC −∠ AEB =90 °−65 ° =25 °

∠POR +∠OPQ =180 ° (int. ∠ s, OR // PQ)

∠POR =180 °−42 ° =138 ° ∴ Reflex

∠ POR =360 °−∠ POR =360 °−138 ° =222 °

(∠ s at a pt.)

24

(opp. ∠ s, cyclic quad.)

Certificate Mathematics in Action Full Solutions 4B 4.

(arcs prop. to ∠ s at ☉ce)

∠ABC + ∠ADC =180 ° (62 ° + ∠CBD ) + (58 ° + ∠CDB ) =180 °

With the notations in the figure,

120 ° + ∴

1 ∠CDB + ∠CDB =180 ° 2 3 ∠CDB = 60 ° 2 ∠CDB = 40 ° ∠KDC = 40 °

1 ∠AOB 2 1 = ×54 ° 2 = 27 °

∠BCA =

(∠ at centre twice ∠ at ☉ce)

∠ONC = ∠OBC +∠BCA (ext. ∠ of △) = 42 ° + 27 ° = 69 °

∠OAC +∠AOB =∠ONC (ext. ∠ of △) ∠OAC =69 ° −54 ° =15 °

5.

∵ ∴



A D= D CAD = DC ∠ACD = ∠CAD =35 °

∠ BCA = 90°

(given)

(equal arcs, equal ∠ s) (∠ in semi-circle)

∠BCD = ∠BCA +∠ACD =90 ° +35 ° =125 °

∠BAD +∠BCD =180 ° (opp. ∠ s, (∠BAC +35 °) +125 ° =180 ° ∠BAC =20 ° cyclic quad.)

∠CDB BC = ∠CBD CD 2 = 1

6.

∴

25

 

∠CBD =

1 ∠CDB 2

(arcs prop. to ∠ s at ☉ce)

(opp. ∠ s, cyclic quad.)

6

7.

∠ABC + ∠ADC = 180 °

 

∠ABC = 180 ° − 115 ° = 65°

∠ACB AB = ∠BAC BC 3 = 2

3 ∠BAC 2 ∠ACB +∠ABC +∠BAC =180 ° 3 ∠BAC + 65 ° +∠BAC =180 ° ∠ ( sum 2 5 ∠BAC =115 ° 2 ∠BAC = 46 °

∴

∠ACB =

Basic Properties of Circles (I)

∠BOC = ∠ABO +∠BEC (ext. ∠ of △) = 28 ° +36 ° = 64 ° OC = OB ∠ OCB = ∠ OBC

(radii) (base ∠ s, isos. △)

∠BOC +∠OCB +∠OBC =180 ° (∠ sum 64 ° +2∠OCB =180 ° ∠OCB = 58 °

of △)

∠BAD +∠BCD =180 ° ∠BAD +(58 °+36 °) =180 ° ∠BAD =86 °

(opp. ∠ s, cyclic

quad.)

of △) 8.

(a)

OC = OB ∠ OCB =∠ OBC

(radii) (base ∠ s, isos. △) ∠OCB +∠OBC +∠BOC =180 ° 2∠OCB +80 ° =180 ° ∠OCB =50 °

(∠ sum of △)

1 (∠ at centre twice ∠BOC ∠ at ☉ce) 2 1 = ×80 ° 2 = 40 °

∠BAC = (b)

∠DAB + ∠BCD =180 ° (opp. ∠ s, cyclic quad.) (∠DAC +∠BAC ) +(∠OCB +∠OCD ) =180 ° 36 °+40 °+50 °+∠OCD =180 ° ∠OCD =54 ° 9.

Produce CO to cut AB at E. Join BC. ∠BEC = ∠OCD (alt. ∠ s, AB // DC)

=36 °

26

Certificate Mathematics in Action Full Solutions 4B 10.

∠ AOC +reflex ∠ AOC ∠ ABC +2∠ ABC

=360 ° =360 °

∠ ABC

=120 °

(∠ s at a pt.) 13.

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

1  MC =  ×6  cm 2  = 3 cm

Join BD and DC. ∠ ABD = 90° ∠ ACD = 90° ∴ ∠ ABD = ∠ ACD AD = AD AB = AC ∴ △ABD ≅ △ACD ∴ ∠ BAD = ∠ CAD ∴ AD bisects ∠ BAC.

Consider △OMC. OM = OC

2

− MC

2

(Pyth. theorem)

= 5 2 −32 cm = 4 cm Consider △OAM. AM = OA

2

−OM

2

(Pyth. theorem)

= 7 2 −4 2 cm = 33 cm AB = AM −BM

= 2 p +q

=( 33 −3) cm = 2.74 cm

common side given RHS corr. ∠ s, ≅ △s

14. ∠ ACD = p + q (ext. ∠ of △) ∠ BDC = p (∠ s in the same segment) r =∠ACD +∠BDC (ext. ∠ of △) = p +q + p

= AM −MC

∴

∠ in semi-circle ∠ in semi-circle

(cor. to 2 d.p.)

15.

11.

Let O be the centre of the circle and r cm be the radius. Join OA. OA = r cm (radius) OM =OC −MC Join BD. ∠ ABD = 90°

(∠ in semi-circle)

∠EBD =∠ABD −∠ABE =90 °−44 ° = 46 °

∠ ECD =∠EBD

∴ (∠ s in the same segment)

=46 °

12. (a)

(b)

∵ ∴ ∵ ∴

∠ AOC = ∠ ABC

= ( r −1) cm OM ⊥ AB

AM = MB

1  =  × 10  cm 2  = 5 cm

(given) (line from centre ⊥ chord bisects chord)

Consider △OAM.

AB // OC and OA // CB (given) OABC is a parallelogram. OA = OC (radii) OABC is a rhombus.

Reflex ∠ AOC = 2∠ ABC

27

∵

(property of rhombus) (∠ at centre twice ∠ at ☉ce)

OA 2 = AM 2 + OM

2

r 2 = 5 2 + ( r −1) 2

(Pyth. theorem)

r 2 = 25 + r 2 − 2r +1 2r = 26 r = 13 ∴

The radius of the circle is 13 cm.

6 Level 2 ∠AOD + ∠BOD = 180 ° 16. ∠AOD = 180 ° − 140 ° = 40°

(adj. ∠ s on st.

∠ ACB = 90° ∠ DBA = ∠ DCA

(∠ in semi-circle) (∠ s in the same segment) ∠CAB = ∠CEA + ∠DCA (ext. ∠

(alt. ∠ s, CD // BA)

∠ODC = ∠AOD = 40 °

1 ∠DCA = ∠AOD 2 1 = ×40 ° 2 = 20 °

(∠ at centre twice ∠ at ☉ce)

= 25 ° + ∠DCA

of △)

∠CAB +∠CBA +∠ACB =180 ° (∠ sum of △)

( 25 °+∠DCA ) +(37 °+∠DBA ) +90 ° =180 °

∠OKC = ∠ODC +∠DCA

∠AKO + ∠OKC = 180 ° ∠AKO = 180 ° − 60 °

2∠DCA = 28 °

(ext. ∠ of

= 40 ° + 20 ° = 60 °

△)

∠ AKD =180 °−74 °−54 ° =52 °

19. (a)

line)

Basic Properties of Circles (I)

(adj. ∠ s on st.

∠DCA =14 °

(b)

∠CAB =25 °+∠DCA =25 °+14 ° =39 °

= 120 °

line) 17. (a) ∠ABC +∠AFC =180 ° quad.)

(opp. ∠ s, cyclic

∠AFC =180 °−102 ° =78 °

∠ CDE

=∠ AFC =78 °

(ext. ∠ , cyclic

quad.)

(b)

∠COE = 2∠CDE = 2 ×78 ° =156 °

(∠ at centre twice ∠ at ☉ce)

∠BAF +∠ABO +∠COE +∠FEO =360 ° ∠BAF +102 °+156 °+38 ° =360 ° ∠BAF =64 °

18. (a)

∠ ADC = 90° (∠ ∠APD +∠ADC +∠BAD 20 °+90 °+(∠BAC +54 °) ∠BAC

in semi-circle) =180 ° ∠ ( =180 ° =16 °

sum of △) (b)

∠ BDC = ∠ BAC = 16°

(∠ s in the same segment)

∠ADB = ∠ADC −∠BDC = 90 ° −16 ° = 74 ° ∠AKD + ∠ADB + ∠CAD =180 ° (∠ sum of △)

28

 

Certificate Mathematics in Action Full Solutions 4B 20.

∠AOB AB = ∠BOC BC 1 = 2

22. (a)

∠BOC = 2∠AOB

Join MN. ∠ ABM = ∠ MNC (ext. ∠ , cyclic quad.) ∠ ADM = ∠ MNE (ext. ∠ , cyclic quad.) ∠ ABM + ∠ ADM = ∠ MNC + ∠ MNE = 180° (adj. ∠ s on st. line) ∴

∴

∠BAD +∠ABM +∠ADM +∠BMD =360 ° 65 °+180 °+∠BMD =360 ° ∠BMD =115 °

∴

∠AOC = ∠AOB + ∠BOC 90 ° = ∠AOB + 2∠AOB

1 ×90 ° 3 = 30 ° 1 (∠ B at centre twice ∠BCA = ∠AO 2 ∠ at ☉ce) 1 = ×30 ° 2 =15 °

∠AOB =

21.

(b)

Join BE.

 

∠BEC BC = ∠CAD CD 3 ∠ B E C= (2 8° ) 2 = 4 2°

 

2∠CAO =180 °−90 ° ∠CAO = 45 °

(arcs prop. to ∠ s at ☉ce)

∴

∠CEO =∠CAO +∠AOB =45 °+30 ° =75 °

(ext. ∠

of △) (arcs prop. to ∠ s at ☉ce)

∴ ∠BKE +∠KEB + ∠KBE =180 ° (∠ sum of △)

29

∵ OC = OA ∴ ∠ ACO = ∠ CAO

(radii) (base ∠ s, isos. △) ∠CAO + ∠ACO + ∠AOC =180 ° (∠ sum of △)

∠DBE DE = ∠CAD CD 4 ∠ D B E= (2 8° ) 2 = 5 6°

∠ BKE

(arcs prop. to ∠ s at centre)

=180 °−42 °−56 ° =82 °

23. (a)

∠ APD = ∠ CPB ∠ PAD = ∠ PCB ∠ PDA = ∠ PBC ∴ △PAD ~ △PCB

common angle ext. ∠ , cyclic quad. ext. ∠ , cyclic quad. AAA

(b)

∠ AKB = ∠ DKC ∠ BAK = ∠ CDK ∠ ABK = ∠ DCK ∴ △AKB ~ △DKC

vert. opp. ∠ s ∠ s in the same segment ∠ s in the same segment AAA

(c)

PA PD = PC PB PA PD = PD + DC PA + AB 6 cm 8 cm = 8 cm + DC (6 +10) cm 12 cm = 8 cm + DC

sides, ~ △s) ∴

DC =4 cm

(corr.

6

∠ ACE + ∠ BOE = 180°

AB BK = DC CK 10 BK = 4 3 cm 24. (a)

(b)

BK =7.5 cm

∵ ∴

AM = MB and CN = ND ∠ OMK = ∠ ONK = 90°

∵ ∴

AB = DC OM = ON

∴

OK = OK △OMK ≅ △ONK

∴ ∴ (c) ∴ ∴ isos. △ ∴ ∴ ∵ ∴ isos. △

25. (a)

(corr. sides, ~ △s)

∴

∴

given line joining centre to mid-pt. of chord ⊥ chord given equal chords, equidistant from centre common side RHS corr. sides, ≅ △s given KN – CN

KM = KN BM = CN KM – BM = KB = KC KB =KC AB =DC KA =KD ∠ KAD =

proved in (b) given ∠ KDA

base ∠ s,

∠ BCD + ∠ KDA = ∠ BCD +∠ KAD = 180° opp. ∠ s, cyclic quad. BC // AD int. ∠ s supp. AC = AB ∠ ACB =

cyclic quad. ∴ BC // ED (b) ∠ CED same segment = ∠ BDE ∴ FE ∠s

Basic Properties of Circles (I)

given ∠ ABC =

opp. ∠ s, cyclic quad.

∠ ACE + 2∠ CAE = 180° ∠ ACE + ∠ CAE = 180° – ∠ CAE (∠ ACE + ∠ CAE) +∠ CEA = 180° ∠ sum of △ (180° – ∠ CAE) + ∠ CEA = 180° ∠ CAE = ∠ CEA ∴ CA = CE sides opp. equal ∠ s 28. (a)

(b)

∵ CE = CD ∴ ∠ CED= ∠ CDE = ∠ ABC

given base ∠ s, isos. △ ext. ∠ , cyclic quad. ∴ ABE is an isosceles triangle. sides opp. equal ∠ s



Let ∠ ABD = x. ∵

C D= A DCD = AD (given)

∴ ∠ DBC= ∠ ABD =x ∠ ABE = ∠ ABD + ∠ DBC = 2x ∠ AEB = ∠ ABE = 2x ∠ EDC isos. △) = 2x ∠ DCB = 4x …… (1) ∠ BDC = 90°

(equal arcs, equal ∠ s)

(base ∠ s, isos. △) = ∠ AEB

(base ∠ s,

= ∠ AEB + ∠ EDC (∠ in semi-circle)

base ∠ s,

∠ ADE ext. ∠ ,

corr. ∠ s equal = ∠ CBD

∠ s in the

alt. ∠ s, BC // ED = FD sides opp. equal

26. ∠ NBP = ∠ MDP ∠ BNP = 180° – ∠ NBP – ∠ NPB = 180° – ∠ MDP – ∠ DPM = ∠ DMP = ∠ NMC ∴ QM = QN

ext. ∠ , cyclic quad. ∠ sum of △ given vert. opp. ∠ s sides opp. equal ∠ s

27.

Join BO and OE. ∠ BOE = 2 ∠ CAE

∠ at centre twice ∠ at ☉ce

30

Certificate Mathematics in Action Full Solutions 4B ∠ DBC + ∠ DCB + ∠ BDC =

△)

180°

∠ DCB = 180° – x – 90° = 90° – x …… (2) From (1) and (2), we have

4 x = 90 ° − x x =18 °

∠BAD + ∠DCB = 180 ° ∠BAD = 180 ° − 4 x = 180 ° − 4 × 18° = 108 °

(opp. ∠s, cyclic quad.) (by (1))

Multiple Choice Questions (p. 52) 1.

Answer: B ∵ OP ⊥ AB

AP = PB

∴

1  =  × 12  cm 2   = 6 cm

OP = OA 2 − AP 2

(given) (line from centre ⊥ chord bisects chord)

(Pyth. theorem)

= 10 2 −6 2 cm = 8 cm

Join OC. OC = OA = 10 cm ∵ OQ ⊥ CD

(radii)

CQ = QD

∴

1  =  ×16  cm 2  = 8 cm

OQ = OC

∴

2.

31

2

−CQ

2

(given) (line from centre ⊥ chord bisects chord) bisects

(Pyth. theorem)

= 10 2 −8 2 cm = 6 cm PQ =OP +OQ =(8 +6) cm =14 cm

Answer: A ∠ BCD = 90°

(∠ in semi-circle)

(∠ sum of

∠DCA =∠BCD −∠ACB =90 °−71 ° =19 ° ∠ABD = ∠DCA (∠ s in the same segment) =19 ° ∴

x =∠ABD +∠BAC =19 °+32 ° =51 °

(ext. ∠ of △)

6 3.

Answer: D ∠ BDC = 90°

(∠ in semi-circle) (∠ ∠BDC +∠DCB +∠DBC =180 ° sum of △)

Basic Properties of Circles (I)

∠BED + ∠EBD = ∠BDC ∠EBD = 43 ° − 24 ° =19 °

(ext. ∠ of △)

∠DCB =180 °−90 °−32 ° =58 ° (opp. ∠ s, ∠BAD + ∠DCB = 180 ° ∠BAD = 180 ° − 58 ° = 122 °

cyclic quad.)

AB = AD (given) ∠ ABD = ∠ ADB (base ∠ s, isos. △) ∴ ∠ABD + ∠ADB + ∠BAD =180 ° (∠ sum of △) 2∠ ABD +122 ° =180 ° ∠ ABD =29 ° 4.

Answer: D

∠ADE = ∠CAD +∠ACD (ext. ∠ of △) = 52 ° + x ∠CBF = ∠ADE (ext. ∠ , cyclic quad.) = 52 ° + x ∴ ∠CBE +∠BEC +∠BCE =180 ° (∠ sum of △)

5.

(52 °+x ) +30 °+ x =180 ° 2 x =98 ° x =49 °

Answer: C Reflex ∠ AOC = 360° – x

(∠ s at a pt.)

1 reflex ∠AOC (∠ at centre c e twice ∠ at ☉ ) 2 1 = (360 ° − x ) 2 x =180 ° − 2

∠ABC =

∴ ∠ABC + ∠BCA + ∠BAC =180 ° (∠ sum of △)

x  180 ° −  +35 ° + 28 ° =180 ° 2   x =126 ° 6.

Answer: C ∠ BDC = ∠ BAC = 46° ∠ ACB = ∠ ADB

(∠ s in the same segment) (∠ s in the same segment)

=x ∴ ∠PDC +∠PCD +∠CPD =180 ° (∠ sum of △) ( x +46 °) +( 48 °+x ) +34 ° =180 °

x =26 ° 7.

Answer: B

32

Certificate Mathematics in Action Full Solutions 4B Join AD. ∠ ADB = 90° (∠ in semi-circle) ∠ADC +∠ABC =180 ° (opp. ∠ s, (90 °+43 °) +(19 °+ x ) =180 ° x = 28 ° cyclic quad.) 8.

Answer: A

∠DBC + ∠DCB + ∠BDC =180 °

(∠

sum of △)

 

∠DBC =180 °−83 °−46 ° (arcs prop. to ∠ s =51 ° ce at ☉ )

∠ABD AD = ∠DBC DC 2 ∠ A B D= (5 1° ) 3 = 3 4°

(opp. ∠ s, cyclic quad.)

∠ABC +∠ADC =180 ° (34 °+51 °) +( 46 °+x ) =180 ° x =49 ° 9.

Answer: C

(opp. ∠ s, cyclic quad.)

∠ A D C+ ∠ A B C= 1 8 0°

 

∠ A D C= 1 8 0° − ∠ A B C(arcsceprop. to ∠ s at ☉ )

∠ A B C m in o rA C = ∠ A D C m ajo rA C ∠ABC 7+ 3 = 1 8 0° − ∠ A B C 6 + 8 5 ∠ A B C= (1 8 0° − ∠ A B C) 7 5 ∠ A B C= × 1 8 0° 12 = 7 5° 10. Answer: B

∠DCB + ∠BAD = 180 ° ∠DCB = 180 ° − 112 ° = 68°

∴

33

OD = OC ∠ ODC = ∠ OCD = 68° ∠ DOC = ∠ ABO =x

(opp. ∠ s, cyclic quad.)

(radii) (base ∠ s, isos. △) (corr. ∠ s, OD // BA)

6 ∴ ∠DOC + ∠ODC + ∠OCD =180 ° sum of △)

Basic Properties of Circles (I)

(∠

x +68 °+68 ° =180 ° x = 44 °

34

Certificate Mathematics in Action Full Solutions 4B 11. Answer: B

(opp. ∠ s, cyclic quad.)

∠ADC + ∠ABC =180 ° ∠ADC =180 ° − x

∠ACD + ∠AED =180 °

(opp. ∠ s, cyclic quad.)

∠ACD =180 ° − y ∠ACD +∠ADC +∠CAD =180 ° ∠ (180 °− y ) +(180 °−x ) +45 ° =180 ° ( x + y = 225 ° sum of △) 12. Answer: A With the notations in the figure, join FC.

∠BFC =∠BAC =a

(∠ s in the same

∠CFD = ∠CED =b

(∠ s in the same

segment)

segment)



For A, x = ∠ BFC + ∠ CFD =a+b For B, if x = y,

then

B C = DA F E

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°

35

(∠ in semi-circle)

1 ∠COD 2 1 = × 48 ° 2 = 24 °

∠CBD =

☉ )

(∠ at centre twice ∠ at

ce

x =∠ADB +∠CBD =90 °+24 ° =114 °

(ext. ∠ of △)

6

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.



∵

A C= C D= D BAC = CD = DB (given)

∴

∠COA = ∠COD = ∠DOB(equal arcs, equal ∠ s) 1 ×180 ° 3 = 60 °

= Join CD. OC = OD ∴ ∠ OCD = ∠ ODC

(radii) (base ∠ s, isos. △) ∠OCD +∠ODC +∠COD =180 ° (∠ sum of

∠OCD = 60 °

△)

∴ ∠ 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

60 ° ×area of circle 360 ° 60 ° 2= ×Q 360 ° Q =12 2=

∴

36