08[anal add math cd]

7/24/2019 08[Anal Add Math CD]

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Additional Mathematics SPM Chapter 8

Penerbitan Pelangi Sdn. Bhd.

1. (a) 180 = 180

180

= rad.

(b) 90 = 90

180

=1

2 rad.

(c) 120 = 120

180

=23

rad.

(d) 50 = 50

180

=5

18 rad.

(e) 3030= 30.5

180

=61

360 rad.

(f) 4525= 4525 180

=109

432 rad.

(g) 120.3 = 120.3

180

=401

600 rad.

2. (a) 2 rad. = 2 180

= 360

(b)1

4

rad. =1

4

180

= 45

(c) 0.35 rad. = 0.35 180

= 63

(d)5

6

rad. =5

6

180

= 150

3. s = r

= 10 1

4

= 7.854 cm

Arc lengthAB= 7.854 cm

4. s = r

= 8 35 180

= 4.887 cm

Arc length PQ= 4.887 cm

5. (a) s= r

22

7= r 1

3

r=22

7

3

= 22

7 3 7

22 = 3

Therefore, radius = 3 cm

(b) s= r

2.2 = 3

=2.2

3

= 0.7333 rad.

Therefore, BOC= 0.7333 rad.

6. (a) s = r

= 10 60 180

= 10.47 cm

Arc lengthAB= 10.47 cm

(b)

30

10 cm

C

O

B

A

CHAPTER

8 Circular Measure


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2



In BOC,

sin 30 =BC

10

BC= 10 sin 30

= 5 cm

AB= 2 5

= 10 cm

Perimeter of segmentAB= 10.47 + 10

= 20.47 cm

7.

12 cm

0.25 rad.

Q

O

PR

s= r

Arc length PQ = 12

0.5 = 6 cm

In POR,

0.25 rad. = 0.25 180

= 14.32

sin 14.32 =PR

12

PR= 12 sin 14.32

= 2.968 cm

PQ= 2 2.968

= 5.936 cm

Hence, the perimeter of segment PQ

= 6 + 5.936

= 11.94 cm

8. (a) OB=1

2

OD

=1

2

20

= 10 cm

Arc lengthAB= r

= 10 1

= 10 cm

(b) Arc length CD= r

= 20 1

= 20 cm

9. (a)

5cm

14.2cm

P

B

A

O

Q

OP: PB= 3 : 1

OB: OP= 4 : 3

OB

OP=

43

OB=4

3OP

=4

3

5

=20

3

In AOB, s= r

14.2 =203

= 14.2 3

20

= 2.13 rad.

Therefore, AOB= 2.13 rad.

(b) Arc length PQ =3

4

Arc lengthAB

= 34

14.2

= 10.65 cm

10. (a) AOC: BOC= 1 : 2

BOC=2

3

180

= 120

Arc lengthBC=120

180

(r)

=120

180

(12)

= 25.13 cm

(b) Arc lengthAC= r

= 1260 180 = 12.57 cm

The perimeter of sectorAOC= 2 12 + 12.57

= 36.57 cm

11. (a) Radius =1

2

16

= 8 cm

Area of minor sectorAOB=1

2r2

=1

2

821.2

= 38.4 cm2

(b) BOC= ( 1.2) rad.

Area of minor sectorBOC=1

2r2

=1

2

82 ( 1.2)

= 62.13 cm2


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12. Area of the sector =1

2r2

58 =1

2r2(0.57)

r2=2 58

0.57

r= 2 58

0.57

= 14.27 cm

13. Area of the sector =1

2r2

140 =1

2

202

=2 140

202

= 0.7 rad.

14. 30 = 30

180 =

6 rad.

The area of segment PQR

= Area of sector POR Area of POR

=1

2r2

12r2sin

=1

2r2( sin )

=1

2

(14)26 sin 30= 2.313 cm2

15. Given arc lengthAB: arc lengthBC= 1 : 2

AOB=1

2

45

= 22.5

= 22.5 180

rad.

Area of sectorAOB

=1

2r2

=1

2

102

22.5

180= 19.63 cm2

16. (a) Given AOB: BOC= 1 : 2

BOC=2

3

90

= 60

=3 rad.

Given area of sectorBOC= 120 cm2

1

2r2= 120

12r23= 120

r2=6 120

r= 720 = 15.14 cm

(b) AOB= 30

=6 rad.

Area of segmentAB

= Area of sectorAOB Area of AOB

=1

2r2

12r2sin

=1

2r2( sin )

=12(15.14)26 sin 30

= 2.705 cm2

1. (a) POQ = 54

180

=3

10 rad

(b) Area =

1

2 r2

=1

282

3

103.142

= 30.1632 cm2

30.16 cm2

2. Major angleAOB= 2 0.412

Major arc lengthAB= 84.31 cm

r= 84.31

r(2 0.412) = 84.31

r=84.31

2 0.412 = 14.36 cm

3. (a) Let the radius of circle be r.

AO= r, CO=1

2r

In AOC,

cos AOC =CO

AO

=

12r

r

=1

2


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AOC= 60

= 60

180

=60 3.142

180

= 1.05 rad.

(b) Arc lengthAB= 1.05 r= 1.05

r(1.05) = 1.05

r= 1 cm

Area of the shaded region

= Area of sectorAOB Area of AOC

=1

2r2

12

base height

=12

(1)2(1.05) 1

2

1

2

3

2

= 0.308 cm2

4. (a) OR= OQ RQ = 13 1

= 12 cm

In OPR,

cos =OR

PO

=12

13

= 0.3948 rad.

(b) The area of the shaded region

= Area of sector POQ Area of POR

=

1

2r2

1

2 (RO) (PR)

=1

2

(13)2(0.3948) 1

2(12)(5)

= 3.361 cm2

5. (a)O

B

A C

ABC=

1

2

Minor angle ofAOC =12

+12 = 2

3

2 = 2

= 2 2

3

=43

rad.

(b) AOB=1

2AOC

=1

2

12

4

3

=1

3 rad.

Minor arc lengthAB= r

= 10 1

3

=10

3

= 10.47 cm

(c) Area of the shaded region

= Area of sectorBOC Area of BOC

=1

2r2

12r2sin

=1

2r2( sin )

=1

2

102

1

3

sin 60

= 50 13 sin 60 = 9.059 cm2

6. (a)OB

OQ

=3

1

OB

4

= 3

OB= 12 cm

AO= 12 cm

In POQ, cos 3 = OQOP

=4

OP

OP=4

cos

3

= 8 cm

AP =AO OP

= 12 8

= 4 cm

(b) In POQ,

PQ2= PO2 OQ2

= 82 42

= 48

PQ= 6.928 cm

QB= OB OQ

= 12 4

= 8 cm

Arc lengthAB = r

= 123 = 4

= 12.57 cm

PR2= PO2 RO2

= 132 122

PR= 5 cm


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The perimeter of the shaded region

= 6.928 + 8 + 4 + 12.57

= 31.5 cm

7. (a) PR=1

2PO

=1

2

(12)

= 6 cm

Arc lengthRS= 4

6 RPS= 4

RPS=2

3 radian

(b) Area of sector OPQ= 82 cm2

1

2r2 = 82

1

2

122 = 82

=164

144 =

4136

rad.

Arc length PQ= 12 4136

=41

3

cm


= Arc length PQ+ PS+ Arc lengthRS+RO+OQ

=41

3

+ 6 + 4 + 6 + 12

= 41.67 cm

8. (a) COR = ABD

=6 rad.

CO=1

2AO

=1

2

(10)

= 5 cm

RO=1

2PO

= 12

(10)

= 5 cm

Therefore, CORis a sector with Oas the centre

and radius 5 cm.

The area of the shaded region =1

2

526

= 6.545 cm2

(b) Arc length CR = 5 6

= 2.618 cm


= 2.618 + 2 5

= 12.62 cm

(c) AOP= 6

=5

6 rad.

Arc lengthAP= 1056 = 26.18 cm

1. (a) Radius = 19 42

=15

2

= 7.5 cm

(b) =s

r

=4

7.5

= 0.5333 rad.

2.O6.5 cm

BA

C

D

BOC= 60

=3 rad.

LetDbe the midpoint of chordAC, AOD=3 rad.

In AOD,

sin3

=AD

AO

=AD

6.5

AD= 6.5 sin3

= 5.629 cm

ChordAC= 2 5.629

= 11.26 cm

Arc lengthAC= 6.5 2

3

= 13.61 cm


= 11.26 + 13.61

= 24.87 cm


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3. Arc length PQ =28

11

r 27 =28

11

r =28

11

7

2

=98

11

Circumference of the circle = 2r

= 2 227 98

11

= 56 cm

4. Arc lengthAB: Arc lengthBC= 1 : 2

Arc lengthBC: Arc lengthAC=2

3

: 1

= 2 : 3

Hence, arc lengthAB: arc lengthBC: arc lengthAC

= 1 : 2 : 3

BOC=2

6

2

=2

3 rad.

The area of sectorBOC =12

r2

=1

2

722

3

= 51.31 cm2

5. (a) Arc lengthAB= 3

r3= 3 r= 3

3

= 9 cm

Therefore, radius = 9 cm

(b)

O 9 cmB

A

C

Let Cbe the midpoint of AB.

BOC=1

2

3 rad.

=6 rad.

In BOC, sin6

=BC

9

BC= 9 sin6

= 4.5 cm

ChordAB = 2 4.5

= 9 cm

6. Given OB= 10 cm

OA=1

2

(10)

= 5 cm

AOD=Arc lengthAD

OA

= 1 rad.

The area of the shaded region

= Area of sectorBOC Area of sector AOD

=1

2

1021 1

2

521

= 37.5 cm2

7. (a) Arc lengthRQ= 12 45 180 = 9.425 cm

The perimeter of sector PQR

= (2 12) + 9.425

= 33.43 cm

(b) The area of sector PQR

= 12 1224cm2

The volume of the slice of the cake

=1

2

1224

3

= 169.6 cm3

8. (a)Area of the minor sector

Area of the major sector

=

12r260

180

12r2

300 180 =

15

Area of minor sectorAOB: Area of major sector AOB

= 1 : 5

(b) Major arc lengthAB= 14 300

180

= 73.30 cm

The perimeter = 73.30 + (2 14)

= 101.3 cm

9. (a) AD=AB = 10 cm

GivenDC

AD

=1

5

DC=1

5AD

=1

5

10

= 2 cm


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(b) Arc lengthBD= 10 0.5

= 5 cm

BOC = 2 BAD

= 2 0.5

= 1 rad.

BC= r

= 122(1) = 6 cm


= 6 + 2 + 5

= 13 cm

10. (a) Given the perimeter of sectorBOC

= Arc lengthAB

2r+ r= ( )r

2 + =

2= 2

= 22

= 0.5708

(b) Arc lengthBC= r

= 6 0.5708

= 3.425 cm

Perimeter of sectorBOC

= 2 6 + 3.425

= 15.43 cm

11. (a) In sectorAOB,

= Arc lengthABOB

=8

12

= 0.6667 rad.

= 3812

(b)Arc length CD

Arc lengthAB

=3

2

OD

OB

=3

2

OD

12

=3

2

OD= 18 cm

y= 18 12

= 6

(c) The area of the shaded region

= Area of sectorDOC Area of sector BOA

=1

2

1822

3

1

2

1222

3

= 60 cm2

12. (a)

RQ

8 cm8 cm

2 cm

18 cm

BA

SP

In PRS,

PS2= 182 22

PS= 17.89

Therefore,AB= 17.89 cm

(b) cos PRB=2

18

PRB= cos1 19 = 8337

(c) RPA= (90 8337) + 90

= 9623

Area of the shaded region= Area of trapezium ABSRQP

Area of sector APQ Area of sector BRQ

=1

2(8 + 10)(17.89)

12

829623 180

1

2

1028337 180

= 161.01 53.83 72.97

= 34.21 cm2

13. (a)

4 cm

5 cm

F

O

R

P

In FOR,

OR= 52 42

= 3 cm

PR = PO OR

= 5 3

= 2 cm

(b) In ROF, sin ROF=

45

ROF= 538

EOF= 2 538

= 10616

= 1.855 rad.

Arc lengthEPF= 5(1.855)

= 9.275 cm


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(c) Area of EOF =1

2

8 3

= 12 cm2

Area of sectorEOF=1

2

521.855

= 23.19 cm2

Area of segmentEPFR= 23.19 12

= 11.19 cm2

14. (a)

6 cm

6 cm6 cm

6 cm

4 cm4 cm

BA

C

P

RQ

In PBC,

PC= 102 62

= 8 cm

cos PBC=6

10

PBC= cos1 35 = 538

= 0.9273 rad.

PAC = 0.9273 rad.

ACB= 2 0.9273

= 1.287 rad.


= Area of ABC Area of sector QCR

Area of sector PBR Area of sector PAQ

=1

2

12 8 1

2

421.287 1

2

620.9273

1

2

620.9273

= 4.321 cm2

(b) Perimeter of the shaded region

= Arc length QR+ Arc length PQ+ Arc length PR

= 4(1.287) + 6(0.9273) + 6(0.9273) = 16.28 cm

15. (a)

7 cm2020

7 cm

BC

O

F

sin 20 =BF

BO

=BF

7

BF= 7 sin 20

= 2.394 cm

BC = 2 2.394

= 4.788 cm

Therefore, the radius of sectorADBCis 4.788 cm.

(b) Area of the shaded region

= Area of sector OACB Area of AOC

Area of BOC

=1

2

7280 180

12 72sin 40 1

2

72sin 40

= 2.712 cm2

(c) Minor angle ofACB= 2 70 = 140

Major angle ofACB= 360 140

= 220

= 220

180

= 3.84 rad.

Arc lengthADB= (4.788)(3.84)

= 18.39 cm

16. (a)

120

B

A

C

D

O

ABC=1

2

(120)

= 60

AOC= 2 60

= 120

The arc length of minor sectorADC

= r

= 9120 180 = 6 cm

(b) DOC =1

2AOC

=1

2

(120)

= 60

=3 rad.


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= Area of sectorDOC Area of DOC

=1

2

(9)231

2

(9)2sin 60

=27

2

81

23

2

= 272 8134 cm2

17. (a)

6 cm

8cm

B

A

C

O

In AOC,

tan AOC=8

6

AOC= tan18

6 BOC= 2 tan1 86 = 10616

= 1.855 rad.

The minor arc lengthBC= 6(1.855)

= 11.13 cm

(b) Area of AOC=1

2

8 6

= 24 cm2

Area of sectorBOC=1

2

621.855

= 33.39 cm2

Area of the shaded region = 2 24 33.39

= 14.61 cm2

18. (a)

C(3, 0)

B(3,)

A(0, 6)

O

415

x

y

E

x3 +

y6

= 1

Hence,A(0, 6) and C(3, 0).

AE=AO EO

= 6 15

4

=9

4 units

EB = 3 units

In ABE, tan ABE=AE

EB

=

94

3

=3

4

ABE= tan1 34 = 3652

= 0.6435 rad.

ABC=2

+ 0.6435

= 2.214 rad.

(b) Arc lengthAC= 154(2.214)

= 8.303 cm

(c) The area of the shaded region = Area of sectorABC Area of ABC

=1

2

154

2

2.214 12

154

2

sin (12652)

=1

2

154

2

(2.214 sin 12652)

= 9.942 cm2

19. (a) AOC=Arc lengthAC

AO

=

12r

r

=12

rad.

BOD= AOC

=12

rad.

(b) Given arc lengthBD= 2 Arc lengthAC

AO=1

2BO

=1

2

(8)

= 4 cm

Arc lengthAC= 4 12 = 2 cm

Arc lengthBD= 2 2

= 4 cm

The perimeter of the diagram

= Arc length AC+ Arc length BD+ AO+ CO

+DO+ BO

= 2 + 4 + 4 + 4 + 8 + 8

= 30 cm


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(c)

4rad.1

8 cm

O

B

ED

In BOE,

sin 14 180

= BE

8

BE = 8 sin 14 180

ChordBD= 2 8 sin 14 180

= 3.958 cm

20. (a) Area of sectorABC= 4 Area of sectorADE

12R2= 4 12r

2

R2r2

= 4

Rr2

= 4

Rr

= 2

R= 2r

Given arc lengthBC= 12 cm

R= 12.................

Area of sectorADE = 25

12r2= 25

12R2

2

= 25

1

2

R2

4 = 25

R2

8 = 25 ...............

From , =12

R

..............................

Substitute into ,

R2

812

R= 25 R=

25 8

12

= 52.36 cm

Therefore, the length ofAB= 52.36 cm

(b) Arc lengthBC= 12 cm

52.36 = 12

=12

52.36

= 0.2292 rad.

Therefore, BAC= 0.2292 rad.

R= Radius,ABr= Radius,AD


= Area of ABC Area of sector ADE

=12

52.362sin 0.2292 180 1

2

52.362

2

(0.2292)

= 232.9 cm2

21. (a)Arc lengthAB

OA

= AOB

AOB=OA

OA

= 1 radian

Given the area of sectorAOB= 32 cm2

12r2(1) = 32

r2= 64

r= 8

Therefore, OB= 8 cm

(b) Given CO=23BOand

COBO

=2

3

Arc length CD =2

3

Arc lengthAB

=2

3OA

=2

3

(8)

=16

3

cm

(c) Area of sector OCD

= 232Area of sectorAOB

=4

9

32

=128

9

cm2

1. (a) Area of AOB=1

2r2sin

=

12

4

2

sin 2

= 8 sin 2 180 = 7.274 cm2

(b) Area of sectorADE =1

2r2

=1

2

162BAO

=1

2

162 22

= 73.06 cm2


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Area of sectorBOC =1

2

42BOC

= 8 ( 2)

= 9.133 cm2


= Area of sectorADE Area of AOB Area of

sectorBOC

= 73.06 7.274 9.133

= 56.65 cm2

2. Perimeter of sectorAOC = Arc lengthBC

r+ r+ r( ) = r

2r+ r( ) = r

2 + =

2= 2 +

=2 +

2

= 2.571 rad.

= 14718

Area of sectorAOC=1

2

82( )

= 32 2 + 2

= 32 2

2

= 18.27 cm2

3. (a) Arc length CD = 3OC

Let OC = r

r = 3r

= 3

(b) Arc length CD = 2 Arc lengthAB

r= 2(10)

r = 20

OC= 20 cm


= Area of sector OCD Area of sector OAB

=12

2023 12

1023

=3

2

(202 102)

= 450 cm2

4. (a)

B

A

O P

A

B

PO

30 30

6060

60 60

9 cm9 cm

9 cm9 cm

Area of segmentAOB

= Area of sectorAPB Area of APB

=1

2

92120 180

12 92sin 120

=1

2

81 120 180

sin 120 = 49.76 cm2

Area of the shaded region = Area of circleAOB 2 Area of segmentAOB

= 92 2 49.76

= 154.98 cm2

(b) The perimeter of the shaded region

= The perimeter of a circle

= 2(9)

= 18 3.142

= 56.56 cm

5. (a)

O

8cm

30

F

ED

A CB

Since OA= AC= OC,

therefore AOCis an equilateral triangle. AOC = 60

=3 rad.

In AOB,

cos 30 =8

OA

OA =8

cos 30

= 9.238 cm


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(b) Arc lengthAC = r

= 9.238 60

180

= 9.674 cm


= Area of AOC Area of sector DOE

=12 9.238

2sin 60 1

2 82

3

= 3.443 cm2

6. (a)Arc lengthRS

Arc length PQ

=23

Arc lengthRS =23

8

=16

3

cm

(b) OQ=3

5RQ

=

3

5 10

= 6 cm

Arc length PQ= 8 cm

OQPOQ = 8

6 POQ = 8

POQ =8

6

=8

6

180

= 7624

(c) Area of sector POQ =1

2

r2

=1

2

(6)2 43 = 24 cm2

7. (a)A

B

O

Arc lengthAB = r

= OA

=Arc lengthAB

OA

Since arc lengthAB= OA,

therefore = 1 rad.

Given the area of the sector = 32 cm2

1

2

r21 = 32

r2 = 64

r= 8 r0

= 8 cm

Hence, OA= 8 cm

(b) AOB= 1 rad.

= 1 180

= 5718

(c)

PA B

O

8 cm

POA=1

2(5718)

= 2839

In APO,

sin 2839 =AP

8

AP = 8 sin 2839

= 3.836 cm

AB= 2 AP

= 2 3.836

= 7.672 cm

8. (a)

A B

P Q

rad.

6 cm10 cm

rad.

C

6

10 cm

6

R

A

P

6 cm5 cm

Q

B

S

In PAR,

AR = 10 sin6

= 5 cm


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In BSQ,

BS=AR

= 5 cm

sin =5

6

= sin156 = 5627

(b) (i) In PAR,

cos6

=PR

10

PR= 10 cos6

= 8.66 cm

In BSQ,

QS= 6 cos 5627

= 3.316 cm

AB= PQ PR SQ

= 16 8.66 3.316

= 4.024 cm

(ii) The area of trapeziumABQCP

=1

2

5 (4.024 + 16)

= 50.06 cm2

Area of sectorAPC

=1

2

1026

= 26.18 cm2

Area of sectorBQC

=1

2 62

5627

180 = 17.73 cm2


= 50.06 26.18 17.73

= 6.15 cm2

9. (a)

rad. 8 cm

3cm

xcm

B

A

O

D

C

Area of sector OBC

Area of sector OAD

=4

1

12(3 +x)2

1

2

32

= 4

(3 +x)2 = 36

3 + x = 6

x = 3 x

0

OB= 3 + 3

= 6 cm

(b) Arc lengthBC = 6

8 = 6

=8

6

= 43 rad.

(c) Arc lengthAD= 3 4

3

= 4 cm


= 4 + 8 + 3 + 3

= 18 cm

10. (a)x6 +

y8

= 1

A(0, 8) andB(6, 0).

D = 0 + 62 ,8 + 0

2 = (3, 4)

(b)

8 units

6 unitsO B

A

In AOB, AB= 10 units

DB= 5 units

5 units

3 units

5 units

O BP

D

In DOP,

sin =3

5

= 3652 ODB= 2 3652

= 7344

(c) Area of sector ODB

=1

2

527344 180

= 16.09 unit2

Area of ODB=1

2

DPOB

=1

2

4 6

= 12 unit2


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Area of the shaded region = 16.09 12

= 4.09 unit2

11. (a) x2+ y2= 100

Whenx= 0,y= 10

Wheny = 0,x= 10

A(10, 0),B(0, 10), C(10, 0),D(0, 10). Area of sector OADCB

=34

Area of circleABCD

=34

(10)2

= 75unit2

(b)

OC

D

y

A

B

P

x

x2+ y2= 100

In AOB,

AB2= 102+ 102

AB = 200 units

AP=200

2

units

Area of semicircleAPBO = 12r2

=1

2200

2

2

=502 unit2

Area of AOB=1

2

10 10

= 50 unit2

Area of segmentsBOand OA

= 502 50unit2


= Area of sector OADCB Area of segmentsBO

and OA

= 75 502 50 = 75

50

2 + 50

= 50+ 50

= 50(+ 1) unit2

12. (a)

O AD8 cm 8 cm

1.6 rad.

8 cm

C

B

AOB =1.62

= 0.8 rad.

(b)

O D

R

C

8 cm

8 cm

0.8 rad.

In ROD,

cos 0.8 =RO8

RO= 8 cos 0.8

= 8 cos 0.8 180 = 5.574 cm

OC= 2 5.574

= 11.15 cm

13. (a)

O

rcm

0.2

rad.

10 cm

RS

T

QP

The length ofRS

The length of ST

=1

2

SOT= 2 0.2

= 0.4 rad.


15/16

15



Area of sector ORST = Area of sector OPQ

1

2r2(0.6) =

12

(10)2(0.4)

r2 =100 0.4

0.6

r= 8.165 cm

OT= 8.165 cm

(b) Arc length ST= 8.165 0.4

= 3.266 cm

(c) Area of sector OST =1

2

8.1652 0.4

= 13.333 cm2

Area of shaded region PQTS

= Area of sector OPQ Area of sector OST

= 20 13.333

= 6.667 cm2

14. (a)

O

120

10 cm

P

BA

BPO= APO= 120

POB= 30, POA= 30

AOB= 60

6010 cm10 cm

60

O BC

P

In POC,

sin 60 =OC10

OC= 10 sin 60

= 8.66 cm

OB= 2 8.66

= 17.32 cm

Area of sector OAB=1

2(17.32)23

= 157.1 cm2

(b) Arc lengthAB= 17.323 = 18.14 cm

Perimeter of sector PAB = 10 + 10 + 18.14

= 38.14 cm

(c) Area of OAB=1

2

17.322 sin 60

= 129.9 cm2

Area of segmentAB

= Area of sector OAB Area of OAB

= 157.1 129.9

= 27.2 cm2

15. (a)

O

B

C

DH

G

F

0.48 rad.E

A

2xcm

xcm

Perimeter of sectorEOF

=x+x+ 0.48x

= 2.48x Perimeter of sectorAOB

= 2x+ 2x+ (2x)(0.48)

= 4.96x

2.48x+ 4.96x= 74.4

7.44x= 74.4

x =74.4

7.44

= 10

(b) Area of the shaded region

= Area of sectorEOF+ Area of sector AOB

= 12

(10)2(0.48) + 12

(20)2(0.48)

= 120 cm2

(c) Arc length CD= 40(0.48)

= 19.2 cm

16. (a)

P

O

B

A C

40

6.5 cm6.5 cm

40


16/16

16



P

ROA

7070

40

6.5 cm

In APR,

cos 70 =AR6.5

AR= 6.5 cos 70

= 2.223 cm

AO= 2 2.223

= 4.446 cm

Therefore, the radius of sector OABC is

4.446 cm.

(b) Arc lengthABC= 4.446 220 180

= 17.07 cm

Arc lengthAOC= 6.5 80 180

= 9.076 cm

Perimeter of the shaded region

= 17.07 + 9.076

= 26.15 cm

(c) Area of sector PAOC

=1

2

6.5280 180

= 29.5 cm2

(d) Area of PAO+ Area of POC

= 2 12 6.52sin 40

= 27.16 cm2

Area of segmentsAOand OC

= 29.5 27.16

= 2.34 cm2


= Area of sector OABC 2.34

=12

4.4462 220 180

2.34 = 35.61 cm2

08[anal add math cd]

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