4bch06(basic Properties Of Circles 1)

Certificate Mathematics in Action Full Solutions 4B

6 Basic Properties of Circles (I) • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

MF = EM = 8 cm

Activity

(line from centre ⊥ chord bisects chord)

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.

Activity 6.2 (p. 25) 1.

yes 2.

yes

3.

yes 4.

p. 9 1. ∵ ∴

yes

Activity 6.3 (p. 35)

∠A + ∠C = 180°

3.

= 3.5 cm

∠B + ∠D = 180°

4.

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 

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

•

•

Term minor arc

region BCE

•

•

major arc

AFB

AFB

•

•

diameter

region BECFA

•

•

chord

AB

•

•

major segment

PB = AP (line from centre ⊥ chord bisects chord) = 4 cm (line from centre ⊥ chord bisects chord) ∴ AB = (2 × 4) cm = 8 cm ∵ OQ = OP = 2 cm (given) (given) ∴ BC = AB (chords equidistant from centre are equal) = 8 cm (chords equidistant from ∴ BC = AB centre are equal) = 8 cm QC = BQ (line from centre ⊥ chord bisects chord) 1  =  × 8  cm 2  = 4 cm

OB

•

•

minor segment

(line from centre ⊥ chord bisects chord)

region OBEC

•

•

sector

AC

•

•

radius

2.

Follow-up Exercise p. 3 Element AB



AB

p. 7 1.

2.



MB = AM = 8 cm

(line from centre ⊥ chord bisects chord)

∠OND = 90°

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

3.

∴

= 10 2 − 6 2 cm = 8 cm

1

∴ ∵ ∴

Consider △OEM. EM = OE 2 − OM 2 (Pyth. theorem)

AB = (2 × 5) cm = 10 cm CN = ND

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

(line from centre ⊥ chord bisects chord)

Consider △OND. ∠OND + ∠NOD + ∠ODN = 180° ∠ ( sum of △) 90° + ∠NOD + 35° = 180° ∠NOD = 55° 3.

MB = AM MB = AM

p. 18

CD = ( 2 × 5) cm = 10 cm AB = CD OM = ON = 2.5 cm

(equal chords, equidistant from centre)

6 1 ∠AOB 2 1 = × 60° 2 = 30°

x= 1.

2.

3.

∴ (∠ at centre twice ∠ at ☉ )

∴

ce

2. (∠ at centre twice ∠ at ☉ )

x = 2∠APB = 2 × 120° = 240°

(∠ at centre twice ∠ at ☉ )

∵ ∴ ∵

AB = DC x = 65°

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

∴

DC = AB y=6

∴ 3.

ce

Reflex ∠AOB = 360° – 140° (∠s at a pt.) = 220° 1 x = reflex ∠AOB 2 (∠ at centre twice ∠ at ☉ ) ce 1 = × 220° 2 = 110°

5.

x = 90°

6.

∠APB = 90° Consider △APB. ∠APB + 40° + x = 180° 90° + 40° + x = 180° x = 50°

(equal arcs, equal chords)





DC = AB

(equal chords, equal arcs)

(∠ in semi-circle) (∠ in semi-circle) (∠ sum of △)

1.

x = ∠BAC = 40°

(∠s in the same segment)

2.

x = 90°

(∠ in semi-circle)

y = ∠ACB = 20°

(∠s in the same segment)

4.



ce

4.

3.



(equal ∠s, equal arcs)

(given)

∴

p. 20



CD = AB CD = AB x=5

∵ ∴

x = 2∠APB = 2 × 50° = 100°



DC = AB x=4

Basic Properties of Circles (I)

∠EAC = ∠CBE = 30° x = ∠ADC + ∠DAC = 25° + 30° = 55°

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

Consider △ABD. ∠ABD + ∠BAD + ∠ADB = 180° (∠ sum of △) ∠ABD + ( 40° + 20°) + 70° = 180° ∠ABD = 50° x = ∠ABD (∠s in the same segment) = 50°

p. 28 1. ∵

∠DOC = ∠AOB = 43° (given)

2

Certificate Mathematics in Action Full Solutions 4B

4.

Join OB. ∵ BC = ED (given) ∠ BOC = ∠ EOD ∴ (equal chords, equal ∠s) = 55° ∵ AB = BC (given) ∠ AOB = ∠ BOC ∴ (equal chords, equal ∠s) = 55° x = ∠BOC + ∠AOB = 55° + 55° = 110°

p. 31 1.

 

x BC = ∠AOB AB 2 x = (80°) 5 = 32°



y cm ∠DOC = AB ∠AOB

(arcs prop. to ∠s at centre)

(arcs prop. to ∠s at centre)

 48°  y = 5   80°  =3 2.



x cm ∠DOC = AB ∠AOB

(arcs prop. to ∠s at centre)

 50°  x = 6   30°  = 10 ∠BOC = 180° − ∠AOB − ∠DOC (adj. ∠s on st. line) = 180° − 30° − 50° = 100° y cm ∠BOC = (arcs prop. to ∠s at centre) AB ∠AOB



 100°  y = 6   30°  = 20°

3

7

Basic Properties of Circles (I)



(arcs prop. to ∠s at ☉ce)

∠CED ∠BEC

(arcs prop. to ∠s at ☉ce)

x AB = ∠CFD CD

3.

6 (15°) 2 = 45°

x=

4.



x cm BC

=

(arcs prop. to ∠s at ☉ ) ce



y AB = BC ∠BEC

 30°  x = 10   40°  = 7.5

(arcs prop. to ∠s at ☉ce)

6 (arcs prop. to ∠s at ☉ ) y = 10 (40°) ce = 24° 5.



∠ACB AB = ∠DBC CD

(arcs prop. to ∠s at ☉ce) 3 ( 48°) 4 = 36°

(arcs prop. to ∠s at ☉ ) ∠ACB = ce

x = 180° − ∠DBC − ∠ACB ∠ ( sum of △) = 180° − 48° − 36° = 96°

p. 38119 1.

2.

∠DEC = ∠ABC = 81° x = 180° − ∠DEC − ∠ECD = 180° − 81° − 65° = 34° x + ∠DAB = 180° x = 180° − 93° = 87° y = ∠CDE = 113°

3.

x + ∠CBA = 180° x = 180° − 127° = 53°

(ext. ∠, cyclic quad.) (∠ sum of △)

(opp. ∠s, cyclic quad.) (ext. ∠, cyclic quad.)

(opp. ∠s, cyclic quad.)

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

4

Certificate Mathematics in Action Full Solutions 4B OB = OC = ON + NC = (5 + 8) cm = 13 cm Consider △ONB.

3.

NB = OB 2 − ON 2

(Pyth. theorem)

= 13 2 − 5 2 cm = 12 cm ∵ AN = NB ∴ AB = (2 × 12) cm = 24 cm

(line from centre ⊥ chord bisects chord)

∵ ∴

Exercise

∵ ∴

Exercise 6A (p.10 10) Level 1 1.

∵

ON ⊥ AB NB = AN

∴

1  =  × 16  cm 2  = 8 cm

4.

(Pyth. theorem)

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

2

∵ ∴

CN = ND ON ⊥ CD

∴ ∵ ∴

∠ONK = 90° AM = MB OM ⊥ AB

∴ ∠OMK = 90° ∠MKN = 180° − ∠DKB = 180° − 43° = 137°

(line from centre ⊥ chord bisects chord)

(given) (chords equidistant from centre are equal)

(given) (line joining centre to midpt. of chord ⊥ chord) (given) (line joining centre to mid-pt. of chord ⊥ chord) (adj. ∠s on st. line)

Consider quadrilateral OMKN. ∠MON = 360° − ∠OMK − ∠ONK − ∠MKN = 360° − 90° − 90° − 137° = 43°

2.

5.

∵ ∴

∴ Join OB.

5

AM = MB AB = (2 × 6) cm = 12 cm ON = OM CD = AB = 12 cm

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

Join OB. Consider △NOB. OB = NB 2 + ON 2

(radii)

OM ⊥ AB AM = MB 1  MB =  × 6  cm 2  = 3 cm

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

7 Consider △OMB.

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

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

(Pyth. theorem)

Join OD. Consider △OND. OD = OB = 5 cm

(radii)

ND = OD 2 − ON 2

(Pyth. theorem)

= 5 − 3 cm = 4 cm ∵ ON ⊥ CD ∴ CN = ND ∴

1 AB 2 1 = ( AM + MB ) 2 1 = (16 + 4) cm 2 = 10 cm OM = OB − MB ∴ = (10 − 4) cm = 6 cm OB =

7.

2

Basic Properties of Circles (I)

2

CD = (2 × 4) cm = 8 cm

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

Join OD. Consider △OMD. OD = OB = 10 cm

(radii)

MD = OD 2 − OM 2

(Pyth. theorem)

= 10 2 − 6 2 cm = 8 cm ∵ OM ⊥ CD ∴ CM = MD

6.

∵ ∴

CM = MD OM ⊥ CD

∴ ∠OMC = 90° ∠AOC = ∠OCM + ∠OMC = 32° + 90° = 122°

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

8.

∴

CD = (2 × 8) cm = 16 cm

∵ ∴

BM = MC = 6 cm OM ⊥ BC

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

(given) (line joining centre to midpt. of chord ⊥ chord)

Consider △OMB. OM = OB 2 − BM 2

(Pyth. theorem)

= 10 2 − 6 2 cm = 8 cm Consider △OMD. MD = OD 2 − OM 2

(Pyth. theorem)

= 17 − 8 cm = 15 cm 2

2

6

Certificate Mathematics in Action Full Solutions 4B

∴

7

CD = MD − MC = (15 − 6) cm = 9 cm

7 9.

Construct a circle with centre O lying on BH, such that the circle cuts AB at two points P and Q, and cuts BC at two points R and S are shown.

Draw OM and ON such that OM ⊥ AB and ON ⊥ BC. OB = OB common side ∠ABH = ∠CBH given ∠OMB = ∠ONB = 90° constructed ∴ △OBM ≅ △OBN AAS ∴ OM = ON corr. sides, ≅ △s ∴ PQ = RS chords equidistant from centre are equal

Basic Properties of Circles (I)

∴

∠MND = ∠OND + ∠ONM = 90° + 15° = 105°

∴

∠OND = 90°

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

SMEFSU08EX@F04 Construct the traingle ABC as shown. AC tan 60° = 12 cm AC = 12 tan 60° cm = 12 3 cm 1 × BC × AC 2 1  =  × 12 × 12 3  cm 2 2   =

∴

Area of △ABC

= 72 3 cm 2

110.

Level 2 10. (a)

∵ ∴

∴

AM = MB ∠OMA = 90°

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

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

(b)

Join ON. ∵ CN = ND ∴ ∠OND = 90°

∵ ∴

CD = AB ON = OM

∴

∠ONM = ∠OMN = 15°

(given) (line joining centre to mid-pt. of chord ⊥ chord) (given) (equal chords, equidistant from centre) (base ∠s, isos. △)

8

Certificate Mathematics in Action Full Solutions 4B OM ⊥ AB AM = MB

(constructed) (line from centre ⊥ chord bisects chord) ∴ 1  =  × 24  cm 2  = 12 cm Consider △OMA. ∵

OM = OA 2 − AM 2

(Pyth. theorem)

= 15 − 12 cm = 9 cm Consider △OMC. MC = MB + BC = (12 + 28) cm = 40 cm 2

2

OC = OM 2 + MC 2

(Pyth. theorem)

= 9 2 + 40 2 cm = 41 cm 12. (a)

Consider △OAB and △OAC. common side OA = OA radii OB = OC given AB = AC ∴ ∴ ∴

(b)

Consider △ABN and △ACN. (proved in (a)) ∠OAB = ∠OAC (given) AB = AC (common side) AN = AN ∴ ∴ ∴

(c)

△OAB ≅ △OAC SSS ∠OAB = ∠OAC corr. ∠s, ≅ △s OA bisects ∠BAC.

△ABN ≅ △ACN BN = CN ON ⊥ BC

(SAS) (corr. sides, ≅ △s) (line joining centre to mid-pt. of chord ⊥ chord)

ON = AN − OA = (8 − 5) cm = 3 cm Consider △ONC. NC = OC 2 − ON 2

(Pyth. theorem)

= 5 − 3 cm = 4 cm Consider △ANC. 2

AC =

2

AN 2 + NC 2

= 8 2 + 4 2 cm = 4 5 cm 13. (a)

9

ON = OY − NY = ( r − 3) cm

(Pyth. theorem)

7 (b) ∵ ∴

ON ⊥ AB AN = NB 1  =  × 18  cm 2  = 9 cm

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

(radius)

( r − 3) 2 + 9 2 = r 2

(Pyth. theorem)

∴

15.

Let M be a point on AB such that OM ⊥ AB. ∵ OM ⊥ AB (constructed) (line from centre ⊥ MB = AM chord bisects chord) ∴ 1  =  × 18  cm 2  = 9 cm Join OB. OB = 13 cm (radius) Consider △OMB. OM = OB 2 − MB 2

r − 6r + 90 = r r = 15 2

14. ∵

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

2

OM ⊥ CD CM = MD

(Pyth. theorem)

= 13 − 9 cm 2

2

= 88 cm Let N be a point on CD such that ON ⊥ CD. ∵ ON ⊥ CD (constructed) (line from centre ⊥ NC = DN chord bisects chord) ∴ 1  =  × 24  cm 2  = 12 cm ∵ ∠ONK = ∠OMK = 90° ∴ ONKM is a rectangle. ∴ NK = OM (property of rectangle) KC = NC − NK = NC − OM ∴

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

1  =  × 12  cm 2   = 6 cm Let r cm be the radius of the circle. OM = AM − OA = (18 − r ) cm ∴

Basic Properties of Circles (I)

= (12 − 88 ) cm = 2.62 cm (cor. to 2 d.p.) Join OC. Consider △OCM. OC = r cm OM 2 + CM 2 = OC 2 (18 − r ) 2 + 6 2 = r 2 360 − 36r + r 2 = r 2 r = 10 ∴ MB = OB − OM = [r − (18 − r )] cm = ( 2r − 18) cm = (2 × 10 − 18) cm = 2 cm

16. (radius) (Pyth. theorem)

(a)

Join OD, OB and OA as shown. Let ∠OAB = x, then ∠OAD = 90° − x. ∵ OB = OA ∴ ∠OBA = ∠OAB =x ∴ ∠AOB = 180° − ∠OAB − ∠OBA = 180° − 2 x ∵ OD = OA ∴ ∠ODA = ∠OAD

radii base ∠s, isos. △ ∠ sum of △ radii base ∠s,

10

Certificate Mathematics in Action Full Solutions 4B isos. △ = 90° − x ∠AOD = 180° − ∠OAD − ∠ODA ∴ ∠ sum of △ = 180° − 2(90° − x) = 2x ∠AOB + ∠AOD ∴ = (180° − 2 x) + 2 x = 180° ∴ BOD is a straight line. (b) Draw OM ⊥ AB and ON ⊥ DA. ∵ OM ⊥ AB and ON ⊥ DA (constructed) ∴ AM = MB and DN = NA (line from centre ⊥ chord bisects chord) 1 NA = AD 2 1  =  × 18  cm 2  = 9 cm ∵ AMON is a rectangle. ∴ OM = NA (property of rectangle) = 9 cm

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

Level 1 1.

2.

Consider △OAM. AM = OA 2 − OM 2

17. (a)

∵ ∴ ∵ ∴

(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.

5.

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

(adj. ∠s on st. line)

x = 2∠ACB = 2 × 42° = 84°

(∠ at centre twice ∠ at ☉ )

6.

DL = EL

11

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

ce

∠ACB = 90° ∵ CA = CB ∴ x = ∠CBA ∠ACB + ∠CBA + x = 180° 2 x + 90° = 180° x = 45°

(∠ in semi-circle) (given) (base ∠s, isos. △)

∠ACD = ∠ABD = 55° x + ∠ACD = 125° x + 55° = 125° x = 70°

(∠s in the same segment)

∠DAC = ∠DBE = 25° x = ∠DAC + ∠ADC = 25° + 42° = 67°

(∠s in the same segment)

∠AOC = 2∠ABC = 2 ×140° = 280° (∠ at centre twice ∠ at )

Reflex

x = 360° − reflex ∠AOC = 360° − 280° = 80°

(b) EM = MF

by (a) proved

Exercise 6B (p. 21)

3.

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

line from centre ⊥ chord bisects chord

(∠ sum of △)

(ext. ∠ of △)

(ext. ∠ of △)

(∠ at centre twice at ☉ ) ce

(∠s at a pt.)

∠ACD = 180° − ∠CAD − ∠CDA = 180° − 32° − 90° = 58° ∠ACB = 90°

(∠ sum of △) (∠ in semi-circle)

7

∴

7.

8.

Basic Properties of Circles (I)

x = ∠ACB − ∠ACD = 90° − 58° = 32°

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

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

(∠ sum of △)

∠AOB = 2∠ACB (∠ at centre twice ∠ at ☉ ) = 2 × 70° ce = 140° (radii) OB = OA (base ∠s, isos.△ ) ∠OBA = x x + ∠OBA + ∠AOB = 180° ∠ ( sum of △) 2 x + 140° = 180° x = 20°

12

Certificate Mathematics in Action Full Solutions 4B 9.

12.

Join OC. 1 ∠AOB 2 1 = ×130° 2 = 65° OC = OA ∠OCA = ∠OAC = 20° OB = OC x = ∠OCB = ∠ACB − ∠OCA = 65° − 20° = 45° ∠ACB =

10.

11.

(∠ at centre twice ∠ at ☉ ) ce

Level 2 14. ∵ ∴ ∵

(radii) (base ∠s, isos. △)

∴

(radii) (base ∠s, isos. △)

(∠ at centre twice ∠ at ) (ext. ∠ of

)

DC = DA ∠DCA = x BD = BC ∠BDC = ∠DCA =x ∠ADB = 90°

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

of △) (∠ sum of △) 15. (a)

x

(∠ at centre twice ∠ at ☉ )

∠OAB + ∠AOC = 180° (int. ∠s, BA // CO) ∠AOC = 180° − 32° = 148° (∠s at a pt.) (int. ∠s, BA // CO) ∴ Reflex ∠AOC = 360° − ∠AOC = 360° − 148° = 212°

∠AOC = 360° − ∠AOC = 360° − 148° = 212°

ce

(∠s at a pt.) 1 reflex ∠AOC 2 1 = × 212° 2 = 106°

∠ABC =

(alt. ∠s, DO // AC)

(b)

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

(∠ at centre twice ∠ at ☉ )

13

)

(∠ in semi-circle) ∠DAC + ∠DCA + ∠ADC = 180° (∠ sum x + x + (90° + x ) = 180° x = 30°

(opp. ∠s of // gram)

(ext. ∠ of △)

(ext. ∠ of

13. ∠DAC, ∠ACD, ∠DAB, ∠DBA, ∠EFD and ∠FED (any four of the above angles)

∠DAC = 90° (∠ in semi-circle) ∠ACD = 180° − ∠DAC − ∠ADC (∠ sum of △) = 180° − 90° − 55° = 35° (given) AB = AC (base ∠s, isos.△ ) ∠ABC = ∠ACB = 35° ∠ABC + ∠BAD = 55° (ext. ∠ of △) ∠BAD = 55° − 35° = 20° ∠BOD = 36° 1 ∠BCD = ∠BOD 2 1 = × 36° 2 = 18° ∠ODC = ∠BCD = 18°

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

(∠ at centre twice ∠ at ☉ ) ce

7

16.

∠DCB + ∠CKB = 50° ∠DCB = 50° − 28° = 22° ∠DAB = ∠DCB = 22° ∠ACB = 90° ∠CAB = 180° − ∠ACB − ∠CBA = 180° − 90° − 50° = 40° ∠CAD = ∠CAB − ∠DAB ∴ = 40° − 22° = 18° ∠BAC = ∠BDC

17.

(∠ ext. ∠ of △)

20. (a)

(∠s in the same segment)

(∠ in semi-circle) (∠ sum of △)

(∠s in the same segment) (alt. ∠s, DC // AB) (∠ in semi -circle)

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

(c)

∴

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

OK ⊥ EB BK = EK

(corr. sides, ~ △s)

given line from centre ⊥ chord bisects chord given common side SAS

(∠ in semicircle) ∠BDC = 180° – ∠CBD – ∠BCD (∠ sum of △) = 180° – 90° – 42° = 48° ∠KED = ∠KBD (corr. ∠s, ≅ △s) ∠KED + ∠KBD = ∠BDC (ext. ∠ of △) 2∠KED = 48 ∠KED = 24 ∴ ∠BAD = ∠BED (∠s in the same segment) = 24°

22.

(given) (radii)

(b) Reflex ∠AOC = 360° − x ∠s at a pt. 1 ∠ABC = reflex ∠AOC 2 ∠ at centre twice ∠ at ☉ 1 ce = (360° − x) 2 x = 180° − 2

∵ ∴

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

(b) ∠ABD = 90°

∠ABE =

19. (a)

∴

AK KD = BK KC 6 KD = 3 2 cm KD = 4 cm

∠BKD = ∠EKD = 90° KD = KD ∴ △BKD ≅ △EKD

∠BDC = ∠ABD ∠ADB = 90° ∠ABD + ∠BAD + ∠ADB = 180° (∠ sum of △) ∠ABD + (44° + ∠BAC) + 90° = 180° ∠ABD + ∠ABD + 134° = 180° ∠ABD = 23°

1 (∠ at centre twice ∠AOE 2 ∠ at ☉ce) 18. 1 = × 124° 2 = 62° ∠ACE + ∠BEC = ∠ABE (ext. ∠ of △) ∠BEC = 62° − 36° = 26° ∠BAD = ∠BEC (∠s in the same segment) = 26° ∠AKE = ∠BAD + ∠ABE ∴ (ext. ∠ of △) = 26° + 62° = 88°

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

(b)

21. (a)

Basic Properties of Circles (I)

Join AP. ∠APB = 90° ∠ACQ = ∠APQ = 90 ∴ QC ⊥ AB

∠ in semi-circle ∠s in the same segment

23.

(opp. ∠s of // gram)

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

14

Certificate Mathematics in Action Full Solutions 4B

Exercise 6C (p. 32) Level 1 1.

6.

Reflex ∠AOB = 360° – ∠AOB = 360° – 80° = 280° x cm ∠AOB = Reflex ∠AOB Major AB



(∠s at a pt.)



(arcs prop. to ∠s at centre)

4 ( 48°) 3 = 64°

∠AOB =

(arcs prop. to ∠s at centre)

(b)

 80°  x = 14   280°  =4

2.

(a)



∠AOB AB = ∠BOC BC

∠BAC = 180° – ∠ABC – ∠ACB (∠ sum of △) = 180° – 50° – 75° = 55° x cm ∠BAC (arcs prop. to = ∠ABC ∠s at ce ) AC ☉  55°  x = 10   50°  = 11

1 ∠AOB (∠ at centre twice ∠ at ☉ce) 2 1 = × 64° 2 = 32°

∠ACB =



3.

∠ACB = 90° ∠BAC = 180° – x – ∠ACB = 180° – x – 90° = 90° – x

7.

(∠ in semi-circle) (∠ sum of △)



(arcs prop. to ∠s at ☉ )

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

4.

ce

 

∠COD CD = ∠BOC BC

(arcs prop. to ∠s at centre)

4 (84°) 6 = 56° ∠BOD = ∠BOC + ∠COD = 84° + 56° = 140° ce 1 x = ∠BOD (∠ at centre twice ∠ at ☉ ) 2 ∴ 1 = ×140° 2 = 70° ∠COD =

5.

∴

15



∠ADC = ∠BAD AC AC = BD



BD

alt. ∠s, CD // AB equal ∠s, equal arcs

8.





AB : BC BC AB prop. to ∠s at = 27° : 54° = 1: 2

= ∠DEB : ∠BEC(arcs

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

(∠ in semi-circle)

∴



∠BAC BC = x AC

∠BEC = 180° – ∠EBC – ∠ECB (∠ sum of △) = 180° – 62° – 64° = 54° ∠DEB = ∠EBC − ∠EDB (ext. ∠ of △) = 62° – 35° = 27°

☉ce)

7 ∵ AB = AD (given) ∴ ∠ABD = ∠ADB (base ∠s, isos. △) ∠ABD + ∠ADB + ∠BAD = 180° (∠ sum of △) 2∠ABD + 120° = 180° ∠ABD = 30° ∠ACB = 180° − ∠BAC − ∠ABC (∠ sum of △) = 180° − 90° − 30° = 60° ∴

9.





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

∴

(equal ∠s, equal arcs) AC = CD AC = CD ∠BPD = ∠BPC + ∠CPD = 10° + 15° = 25° = ∠EPF

∵

∴

 

 

(equal ∠s, equal arcs) BD = EF BD = EF ∠APD = ∠APB + ∠BPC + ∠CPD = 5° + 10° + 15° = 30° = ∠FPG

 

AD = FG AD = FG

11.

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

at ☉ce)

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

∴

Level 2

= ∠ACB : ∠BAC

∵

∵

Basic Properties of Circles (I)

(equal ∠s, equal arcs)

Join BC. ∠CBD = ∠CAD (∠s in the same segment) = 20° ∠CBA = ∠CBD + ∠ABD = 20° + 50° = 70°



ADC BC

=

∠CBA ∠BAC

12. ∠BAD = 90° ∠BAC = ∠BAD – ∠CAD = 90° – 50° = 40° x cm ∠BAC = CD ∠CAD  40°  x = 15   50°  = 12 13. ∠PRS = ∠PQS

 

AB = CD AB = CD

(given)

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

  

AB = BC = CD AB = BC = CD (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. ∵

ce

 70°  x = 10   40°  = 17.5



10. ∵

(arcs prop. to ∠s at ☉ )

(∠ in semi-circle)

(arcs prop. to ∠s at ☉ ) ce

(∠s in the same segment)

= 30° ∠PRQ = ∠QRS – ∠PRS = 75° − 30° = 45°

 

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

14. (a)

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

16

Certificate Mathematics in Action Full Solutions 4B = 30° + 30° = 60°



∴



(b)

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

(b) (arcs prop. to ∠s at ☉ ) ce

OA = OC AB = CB OB = OB ∴ △ABO ≅ △CBO ∠AOB = ∠COB ∠AOD = ∠COD

(b) ∵ ∴

 

arcs 16. ∵ ∴ ∵ ∴ ∴ 17. (a)

 

BC = CD BC = CD ∠CAB = ∠DAC OC = OA ∠ACO = ∠CAB = ∠DAC OC // AD ∵ ∴

OE ⊥ BD BE = ED

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

18. (a)

 

∠BAC = ∠DAC ∴

17

radii given common side SSS corr. ∠s, ≅ △s equal ∠s, equal

given equal arcs, equal ∠s radii base ∠s, isos. △ alt. ∠s equal given line from centre ⊥ chord bisects chord common side given SAS corr. ∠s, ≅ △s

BC = CD BC = CD equal ∠s, equal arcs

With the notations in the figure,

∠DFE = ∠BDF + ∠DBF = 20° + 30° = 50° ∠AGE = ∠CAG + ∠ACG = 40° + 50° = 90°







= 20° : 40° : 40° : 30° : 50° = 2 : 4 : 4 : 3:5

AD = DC AD = DC

∴



AB AB : BC BC : CD CD : DE DE : EA EA = ∠ADB : ∠BEC : ∠CAD : (arcs prop. to ∠s at ☉ce) ∠DBE : ∠ACE (arcs prop. to ∠s at ☉ce) to ∠s at △ce)

(cb)

15. (a)



x = 180° – ∠FGE – ∠GFE (∠ sum of △) = 180° – 90° – 50° = 40°

(ext. ∠ of △)

(ext. ∠ of △)



AB 2 = Circumference of the circle 2 + 4 + 4 + 3 + 5 (by (b)) 18   Circumference of the circle =  π ×  cm 2   = 9π cm 9π ∴ Radius of the circle = cm 2π = 4.5 cm

7 Exercise 6D (p. 39) Level 1 1.

2.

3.

4.

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

7. (ext. ∠, cyclic quad.) (adj. ∠s on st. line)

∠BCD + ∠BAD = 180° ∠BCD = 180° − 76° = 104° ∵ CD = CB ∴ ∠BDC = x ∠BDC + x + ∠BCD = 180° ∴ 2 x + 104° = 180° x = 38°

6.

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

(opp. ∠s, cyclic quad.) (ext. ∠, cyclic quad.)

∠ABD = y (ext. ∠, cyclic quad.) ∵ AD = AB (given) ∴ ∠ADB = ∠ABD (base ∠s, isos. △) =y ∠BAD + ∠ABD + ∠ADB = 180° (∠ sum of △) x + 2 y = 180° 2 y = 180° − 50° y = 65°

(opp. ∠s, cyclic quad.) (given) (base ∠s, isos. △) (∠ sum of △) 8.

∠ACB = 90° (∠ in semi-circle) ∠ABC + ∠ACB + ∠BAC = 180° (∠ sum of △) ∠ABC = 180° − 90° − 40° = 50° x + ∠ABC = 180° (opp. ∠s, cyclic quad.) x = 180° − 50° = 130° x = ∠ABD = 46°

Join AD. ∠ABC + ∠CDA = 180° ∠ADE = 90° ∠ABC + ∠CDE = ∠ABC + (∠CDA + ∠ADE ) = (∠ABC + ∠CDA) + ∠ADE = 180° + 90° = 270°

(∠s in the same segment)

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

5.

Basic Properties of Circles (I)

∠EBC + ∠CDE = 180° (opp. ∠s, cyclic quad.) ∠EBC = 180° − 110° = 70° ∠ECB + ∠BAE = 180° (opp. ∠s, cyclic quad.) ∠ECB = 180° − 120° = 60° ∠BEC + ∠EBC + ∠ECB = 180° (∠ sum of △) ∠BEC = 180° − 70° − 60° = 50°

(a)

(b)

∠ADE = ∠CAD + ∠ACD = 36° + 60° = 96°

∠APB = 9.

(a)

(b)

(ext. ∠ of △)

∠DFE = ∠BCD (ext. ∠, cyclic quad.) = 60° ∠DFE + ∠FDE + ∠DEF = 180° (∠ sum of △) ∠DEF = 180° − 60° − 96° = 24°

1 ∠AOB 2 1 = × 40° 2 = 20°

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

10.

(∠ at centre twice ∠ at ☉ ) ce

∠BAP + ∠BCP = 180° (opp. ∠s, cyclic quad.) ∠BAP = 180° − 50° = 130° ∠ABP + ∠BAP + ∠APB = 180° (∠ sum of △) ∠ABP = 180° − 130° − 20° = 30°

∠DAC + ∠ADC + ∠ACD = 180° (∠ sum of △) ∠DAC = 180° − 115° − 30° = 35° ∠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 △)

18

Certificate Mathematics in Action Full Solutions 4B ∠ACB = 180° − 65° − 35° = 80°

19

7 Level 2 11.

15.

Reflex ∠AOC = 2∠ABC

(∠ at centre twice = 2 × 110° ∠ at

☉ce) = 220° ∠AOC = 360° − reflex ∠AOC (∠s at a pt.) = 360° − 220° = 140° ∴ ∠CPB + ∠AOC = 180° (opp. ∠s, cyclic quad.) ∠CPB = 180° − 140° = 40° 12.

∠COD = ∠BAD = 40° OD = OC ∠ODC = ∠OCD ∠COD + ∠ODC + ∠OCD = 180° 40° + 2∠ODC = 180° ∠ODC = 70° ∠ODC + ∠ABC = 180° ∠ABC = 180° − 70° = 110°

13. (a)

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

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 14.

Basic Properties of Circles (I)

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

(corr. ∠s, OC // AB) (radii) (base ∠s, isos. △) (∠ sum of △)

(opp. ∠s, cyclic quad.)

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

(corr. sides, ~ △s)

∠BAD = ∠BCE (ext. ∠, cyclic quad.) = 65° ∠ADB = 90° (∠ in semi-circle) ∠ABD + ∠ADB + ∠BAD = 180° (∠ sum of △) ∠ABD = 180° − 90° − 65° = 25° ∠BDC = ∠ABD (alt. ∠s, DC // AB) = 25° ∠DBC + ∠BDC = ∠BCE (ext. ∠ of △) ∠DBC = 65° − 25° = 40°

20

Certificate Mathematics in Action Full Solutions 4B ∠AED + ∠ACD = 180° (opp. ∠s, cyclic quad.) ∠AED = 180° − ∠ACD = 180 − (140° − ∠ADC ) = 40° + ∠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 of △) ∠BCE = 180° − 2 x ∠BDE + ∠BCE = 180° (opp. ∠s, cyclic quad.) ∠BDE = 180° − (180° − 2 x) = 2x ∠DBE = 90° (∠ 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

∴

17. (a)

(b)

21

3 x = 117° x = 39° ∠CBE = 39° ∠APC = ∠ABC + ∠PCB = ( x + x) + y = 2x + y ∠ARB = ∠ACB + ∠RBC = ( y + y) + x = x + 2y ∠APC + ∠ARB = 180° ( 2 x + y ) + ( x + 2 y ) = 180° 3 x + 3 y = 180° x + y = 60°

(ext. ∠ of △) (ext. ∠ of △)

(opp. ∠s, cyclic quad.)

7

Basic Properties of Circles (I)

∠BAC + ∠ABC + ∠ACB = 180° (∠ sum of △) ∠BAC + 2 x + 2 y = 180° ∠BAC + 2( x + y ) = 180° ∠BAC + 120° = 180° ∠BAC = 60°

22

Certificate Mathematics in Action Full Solutions 4B Revision Exercise 6 (p. 47) Level 1

(b)

1.

Join OF. Draw ON such that ON ⊥ FE. OF = OB 1 = BC (radii) 2 1  =  × 20  cm 2  = 10 cm ON = AB (property of rectangle) = 6 cm Consider △ONF. FN = OF 2 − ON 2 = 10 − 6 cm = 8 cm ∵ ON ⊥ FE ∴ FN = NE 2

∴

2.

3.

∠AEC = 90° ∠EBC = ∠EAC = 30° ∠AEB = ∠EDC + ∠EBC = 35° + 30° = 65° ∠BEC = ∠AEC − ∠AEB = 90° − 65° = 25° (a)

(Pyth. theorem)

2

FE = ( 2 × 8) cm = 16 cm

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

(∠ in semi-circle) (∠s in the same segment) (ext. ∠ of △)

∠POR + ∠OPQ = 180°

(int. ∠s, OR // PQ)

∠POR = 180° − 42° = 138° ∠POR = 360° − ∠POR ∴ Reflex (∠s at a pt.) = 360° − 138° = 222° (∠ at centre twice ∠ at ☉ce)

23

1 reflex ∠POR 2 1 = × 222° 2 = 111° ∠ORQ + ∠PQR = 180° (int. ∠s, OR // PQ) ∠ORQ = 180° − 111° = 69°

∠PQR =

7 4.

With the notations in the figure,

Basic Properties of Circles (I)

(opp. ∠s, cyclic quad.)

(arcs prop. to ∠s at ☉ce)

1 ∠AOB 2 1 = × 54° 2 = 27° ∠ONC = ∠OBC + ∠BCA = 42° + 27° = 69° ∠OAC + ∠AOB = ∠ONC ∠OAC = 69° − 54° = 15° ∠BCA =

5.

 

AD = DC AD = DC ∠ACD = ∠CAD ∴ = 35° ∠BCA = 90° ∠BCD = ∠BCA + ∠ACD = 90° + 35° = 125° ∠BAD + ∠BCD = 180° (∠BAC + 35°) + 125° = 180° ∠BAC = 20° ∵

(∠ at centre twice ∠ at ☉ce)

(ext. ∠ of △)

(ext. ∠ of △)

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

(opp. ∠s, cyclic quad.)



6.



∠CDB BC = ∠CBD CD 2 = 1 1 ∴ ∠CBD = ∠CDB 2 ∠ABC + ∠ADC = 180° (62° + ∠CBD ) + (58° + ∠CDB) = 180° 1 120° + ∠CDB + ∠CDB = 180° 2 3 ∴ ∠CDB = 60° 2 ∠CDB = 40° ∠KDC = 40°

(arcs prop. to ∠s at ☉ce)

(opp. ∠s, cyclic quad.)

24

Certificate Mathematics in Action Full Solutions 4B

7.

∠ABC + ∠ADC = 180° ∠ABC = 180° − 115° = 65°

 

∠ACB AB = ∠BAC BC 3 = 2 3 ∠ACB = ∠BAC ∴ 2 ∠ACB + ∠ABC + ∠BAC = 180° 3 ∠BAC + 65° + ∠BAC = 180° 2 5 ∠BAC = 115° 2 ∠BAC = 46° 8.

(a)

OC = OB ∠OCB = ∠OBC

(∠ sum of △)

(radii) (base ∠s, isos. △)

∠OCB + ∠OBC + ∠BOC = 180° ∠ ( sum of △) 2∠OCB + 80° = 180° ∠OCB = 50° 1 ∠BOC (∠ at centre twice 2 ∠ at ☉ce) 1 = × 80° 2 = 40° ∠DAB + ∠BCD = 180° (opp. ∠s, cyclic quad.) (∠DAC + ∠BAC ) + (∠OCB + ∠OCD ) = 180° 36° + 40° + 50° + ∠OCD = 180° ∠OCD = 54° ∠BAC = (b)

9.

Produce CO to cut AB at E. Join BC. ∠BEC = ∠OCD (alt. ∠s, AB // DC) = 36° ∠BOC = ∠ABO + ∠BEC (ext. ∠ of △) = 28° + 36° = 64° OC = OB (radii) ∠OCB = ∠OBC (base ∠s, isos. △) ∠BOC + ∠OCB + ∠OBC = 180° (∠ sum of △) 64° + 2∠OCB = 180° ∠OCB = 58° ∠BAD + ∠BCD = 180° (opp. ∠s, cyclic quad.) ∠BAD + (58° + 36°) = 180° ∠BAD = 86°

25

7

Basic Properties of Circles (I)

∠AOC + reflex ∠AOC = 360° ∠ ( s at a pt.) ∠ABC + 2∠ABC = 360° ∠ABC = 120°

10.

13.

Draw OM such that OM ⊥ BC. ∵ OM ⊥ BC (constructed) ∴ BM = MC (line from centre ⊥ chord bisects chord) 1  ∴ MC =  2 × 6  cm   = 3 cm Consider △OMC. OM = OC − MC 2

2

(Pyth. theorem)

= 5 − 3 cm = 4 cm Consider △OAM. 2

2

AM = OA 2 − OM 2

(Pyth. theorem)

= 7 2 − 4 2 cm

∴

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

= 33 cm AB = AM − BM = AM − MC

∠ in semi-circle ∠ in semi-circle 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 = 2p +q 15.

= ( 33 − 3) cm = 2.74 cm (cor. to 2 d.p.) 11.

Join BD. ∠ABD = 90° (∠ in semi-circle) ∠EBD = ∠ABD − ∠ABE = 90° − 44° = 46° ∠ECD = ∠EBD (∠s in the same segment) = 46° 12. (a)

(b)

∵ ∴ ∵ ∴

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

∠AOC = ∠ABC Reflex ∠AOC = 2∠ABC

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

Let O be the centre of the circle and r cm be the radius. Join OA. OA = r cm (radius) OM = OC − MC = ( r − 1) cm ∵ OM ⊥ AB (given) AM = MB (line from centre ⊥ chord bisects chord) ∴ 1  =  ×10  cm 2  = 5 cm Consider △OAM. 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.

26

Certificate Mathematics in Action Full Solutions 4B Level 2 16.

∠AOD + ∠BOD = 180° ∠AOD = 180° − 140° = 40° ∠ODC = ∠AOD = 40° 1 ∠DCA = ∠AOD 2 1 = × 40° 2 = 20° ∠OKC = ∠ODC + ∠DCA = 40° + 20° = 60° ∠AKO + ∠OKC = 180° ∠AKO = 180° − 60° = 120°

(adj. ∠s on st. line) (b) (alt. ∠s, CD // BA) (∠ at centre twice ∠ at ☉ce)

(ext. ∠ of △)

(adj. ∠s on st. line)

17. (a) ∠ABC + ∠AFC = 180° (opp. ∠s, cyclic quad.) ∠AFC = 180° − 102° = 78° ∠CDE = ∠AFC (ext. ∠, cyclic quad.) = 78°

(b)

18. (a)

(b)

∠COE = 2∠CDE (∠ at centre twice ∠ at ☉ce) = 2 × 78° = 156° ∠BAF + ∠ABO + ∠COE + ∠FEO = 360° ∠BAF + 102° + 156° + 38° = 360° ∠BAF = 64° ∠ADC = 90° (∠ in semi-circle) ∠APD + ∠ADC + ∠BAD = 180° ∠ ( sum of △) 20° + 90° + (∠BAC + 54°) = 180° ∠BAC = 16° ∠BDC = ∠BAC

(∠s in the same segment)

= 16° ∠ADB = ∠ADC − ∠BDC = 90° − 16° = 74° ∠AKD + ∠ADB + ∠CAD = 180° (∠ sum of △) ∠AKD = 180° − 74° − 54° = 52° 19. (a)

∠ACB = 90° ∠DBA = ∠DCA

(∠ in semi-circle) (∠s in the same segment) (ext. ∠ of △)

∠CAB = ∠CEA + ∠DCA = 25° + ∠DCA ∠CAB + ∠CBA + ∠ACB = 180° (∠ sum of △) ( 25° + ∠DCA) + (37° + ∠DBA) + 90° = 180° 2∠DCA = 28° ∠DCA = 14°

27

∠CAB = 25° + ∠DCA = 25° + 14° = 39°

7 20.

Basic Properties of Circles (I)

1 ∠AOB (∠ at centre twice 2 ∠ at ☉ce) 1 = × 30° 2 = 15°

∠BCA = ∴

(b)

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°

23. (a)

∵ OC = OA (radii) ∴ ∠ACO = ∠CAO (base ∠s, isos. △) ∠CAO + ∠ACO + ∠AOC = 180° (∠ sum of △) 2∠CAO = 180° − 90° ∠CAO = 45° ∠CEO = ∠CAO + ∠AOB ∴ (ext. ∠ of △) = 45° + 30° = 75° ∠APD = ∠CPB ∠PAD = ∠PCB ∠PDA = ∠PBC ∴ △PAD ~ △PCB

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

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

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

21. (b)

Join BE.





∠BEC BC = ∠CAD CD 3 ∠BEC = (28°) 2 = 42°

(c) (arcs prop. to ∠s at ☉ce)



AB BK = DC CK 10 BK = 4 3 cm ∴ BK = 7.5 cm



∠DBE DE = (arcs prop. to ∠CAD CD ∠s at ☉ce) 4 ∠DBE = (28°) 2 = 56° ∴ ∠BKE + ∠KEB + ∠KBE = 180° (∠ sum of △) ∠BKE = 180° − 42° − 56° = 82°

24. (a)

 

∠AOB AB = 22. (a) ∠BOC BC (arcs prop. to ∠s at centre) 1 = 2 ∠BOC = 2∠AOB ∠AOC = ∠AOB + ∠BOC 90° = ∠AOB + 2∠AOB 1 ∴ ∠AOB = 3 × 90° = 30°

PA PD = PC PB PA PD = (corr. sides, ~ △s) PD + DC PA + AB 6 cm 8 cm = 8 cm + DC (6 + 10) cm 12 cm = 8 cm + DC ∴ DC = 4 cm

(b)

(corr. sides, ~ △s)

∵ ∴

AM = MB and CN = ND ∠OMK = ∠ONK = 90°

∵ ∴

AB = DC OM = ON

∴

OK = OK △OMK ≅ △ONK

∴ ∴

KM = KN BM = CN KM – BM = KB = KC

∴

KB =KC AB =DC KA =KD

∴

(c)

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

proved in (b) given

28

Certificate Mathematics in Action Full Solutions 4B ∴ isos. △ ∴

25. (a)

∠KAD =

∠KDA ∠KDA

∴

∠BCD + = ∠BCD +∠KAD = 180° BC // AD

∵ ∴

AC = AB ∠ACB =∠ABC

given base ∠s, isos. △ = ∠ADE ext. ∠,

cyclic quad. ∴ BC // ED (b)

∠CED = ∠CBD = ∠BDE ∴ FE

base ∠s,

opp. ∠s, cyclic quad. int. ∠s supp.

corr. ∠s equal ∠s in the same segment alt. ∠s, BC // ED = FD sides opp. equal

∠s 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 ∠ACE + ∠BOE = 180°

∠ at centre twice ∠ at ☉ce 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. ∵ CD = AD CD = AD ∴ ∠DBC = ∠ABD =x

29

(given) (equal arcs, equal ∠s)

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

(base ∠s, isos. △) (base ∠s, isos. △)

(∠ in semi-circle)

7 ∠DBC + ∠DCB + ∠BDC =

180°

(∠ sum of

△) ∠DCB = 180° – x – 90° = 90° – x …… (2) From (1) and (2), we have 4 x = 90° − x x = 18° ∠BAD + ∠DCB = 180° (opp. ∠s, cyclic quad.) ∠BAD = 180° − 4 x (by (1)) = 180° − 4 × 18° = 108°

∴

Basic Properties of Circles (I)

x = ∠ABD + ∠BAC = 19° + 32° = 51°

(ext. ∠ of △)

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 CQ = QD ∴

(radii)

1  =  × 16  cm 2   = 8 cm

OQ = OC 2 − CQ 2

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

(Pyth. theorem)

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

2.

2

Answer: A ∠BCD = 90° ∠DCA = ∠BCD − ∠ACB = 90° − 71° = 19° ∠ABD = ∠DCA = 19°

(∠ in semi-circle)

(∠s in the same segment)

30

Certificate Mathematics in Action Full Solutions 4B 3.

Answer: D ∠BDC = 90° (∠ in semi-circle) ∠BDC + ∠DCB + ∠DBC = 180° (∠ sum of △) ∠DCB = 180° − 90° − 32° = 58° ∠BAD + ∠DCB = 180° (opp. ∠s, cyclic quad.) ∠BAD = 180° − 58° = 122° 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 △) (52° + x) + 30° + x = 180° 2 x = 98° x = 49°

5.

Answer: C Reflex ∠AOC = 360° – x 1 ∠ABC = reflex ∠AOC 2 1 = (360° − x) 2 x = 180° − 2 ∴ ∠ABC + ∠BCA + ∠BAC = 180° x  180° −  + 35° + 28° = 180° 2  x = 126°

6.

Answer: C ∠BDC = ∠BAC

7.

31

(∠ sum of △)

(∠s in the same segment)

=x ∠PDC + ∠PCD + ∠CPD = 180° ( x + 46°) + (48° + x ) + 34° = 180° x = 26°

Answer: B ∠BED + ∠EBD = ∠BDC ∠EBD = 43° − 24° = 19°

(∠ at centre twice ∠ at ☉c e)

(∠s in the same segment)

= 46° ∠ACB = ∠ADB

∴

(∠s at a pt.)

(∠ sum of △)

(ext. ∠ of △)

7 Join AD. ∠ADB = 90° ∠ADC + ∠ABC = 180° (90° + 43°) + (19° + x) = 180° x = 28° 8.

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

Answer: A ∠DBC + ∠DCB + ∠BDC = 180° (∠ sum of △) ∠DBC = 180° − 83° − 46° = 51°

 

∠ABD AD = ∠DBC DC 2 ∠ABD = (51°) 3 = 34° ∠ABC + ∠ADC = 180° (34° + 51°) + (46° + x) = 180° x = 49° 9.

Basic Properties of Circles (I)

Answer: C ∠ADC + ∠ABC = 180°

(arcs prop. to ∠s at ☉ce)

(opp. ∠s, cyclic quad.)

(opp. ∠s, cyclic quad.)



∠ADC = 180° − ∠ABC



(arcs prop. to ∠s ∠ABC minor AC = at ☉ce) ∠ADC major AC ∠ABC 7+3 = 180° − ∠ABC 6 + 8 5 ∠ABC = (180° − ∠ABC ) 7 5 ∠ABC = × 180° 12 = 75° 10. Answer: B ∠DCB + ∠BAD = 180° (opp. ∠s, cyclic quad.) ∠DCB = 180° − 112° = 68° OD = OC (radii) ∴ ∠ODC = ∠OCD (base ∠s, isos. △) = 68° ∠DOC = ∠ABO (corr. ∠s, OD // BA) =x ∴ ∠DOC + ∠ODC + ∠OCD = 180° (∠ sum of △) x + 68° + 68° = 180° x = 44°

32

Certificate Mathematics in Action Full Solutions 4B 11. Answer: B ∠ADC + ∠ABC = 180°

(opp. ∠s, cyclic quad.)

∠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 (∠s in the same segment) =a ∠CFD = ∠CED (∠s in the same segment) =b For A, x = ∠BFC + ∠CFD =a+b For B, if x = y,

 

then BCD = AFE 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° 1 ∠CBD = ∠COD 2 1 = × 48° 2 = 24° x = ∠ADB + ∠CBD = 90° + 24° = 114°

33

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

(ext. ∠ of △)

7

Basic Properties of Circles (I)

HKMO (p. 54) Let O be the centre of the circle. With the notations in the figure, join OC and OD.

  

∵ AC = CD = DB AC = CD = DB (given) ∴ ∠COA = ∠COD = ∠DOB (equal arcs, equal ∠s) 1 = ×180° 3 = 60° Join CD. OC = OD (radii) ∴ ∠OCD = ∠ODC (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° 2= × area of circle 360° 60° 2= ×Q ∴ 360° Q = 12

34