08[anal add math cd]
TRANSCRIPT
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Additional Mathematics SPM Chapter 8
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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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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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Additional Mathematics SPM Chapter 8
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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
The perimeter of the shaded region
= 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
The perimeter of the shaded region
= 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
The perimeter of the shaded region
= 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
The perimeter of the shaded region
= 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 the shaded region
= 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 the shaded region
= 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
(c) Area of the shaded region
= 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 the shaded region
= 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 the shaded region
= 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
(c) Area of the shaded region
= 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
Area of the shaded region
= 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
The perimeter of the shaded region
= 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 the shaded region
= 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.
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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
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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 the shaded region
= Area of sector OABC 2.34
=12
4.4462 220 180
2.34 = 35.61 cm2