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1 © Penerbitan Pelangi Sdn. Bhd.
Paper 1
1. P
Q
RO120°
7 cm
Arc PQR = 120°–––––360°
× 2 × 22–––7
× 7
= 14 2––3
cm
Perimeter of sector OPQR= RO + OP + arc PQR
= 7 + 7 + 14 2––3
cm
= 28 2––3
cm
2.
E F
O
14 cm
H
JG
Arc EJH = 1––4
× 2 × 22–––7
× 14
= 22 cm
Arc FGH = 1––2
× 2 × 22–––7
× 7
= 22 cm
Perimeter of the whole diagram= FE + arc EJH + arc FGH= 14 + 22 + 22= 58 cm
3.
240°
7 cm
OD
GEA
B
F
Arc ABD = 240°–––––360°
× 2 × 22–––7
× 7
= 29 1—3
cm
Arc EFG = 240°–––––360°
× 2 × 22–––7
× 14
= 58 2—3
cm
Perimeter of the shaded region= arc ABD + AE + DG + arc EFG
= 29 1––3
+ 7 + 7 + 58 2––3
= 102 cm
4.
60°
60°
Q
P OS
R
14 cm7 cm
7 cm
14 cm
Arc PQ = 60°–––––360°
× 2 × 22–––7
× 14
= 14 2––3
cm
Arc RS = 60°–––––360°
× 2 × 22–––7
× 7
= 7 1––3
cm
CHAPTER
7 Circles ICHAPTER
2
Mathematics SPM Chapter 7
© Penerbitan Pelangi Sdn. Bhd.
Perimeter of the whole diagram= arc PQ + arc RS + OP + OQ + OR + OS
= 14 2––3
+ 7 1––3
+ 14 + 14 + 7 + 7
= 64 cm
5.
60°
300°
7 cm
PR
Q
O
Area of sector OPQR
= 300°–––––360°
× 22–––7
× 72
= 128 1––3
cm2
6.
7 cmO
Q
T
S
RP
Area of semicircle OSR
= 1––2
× 22–––7
× 7––2
2
= 19 1––4
cm2
Area of semicircle PQR
= 1––2
× 22–––7
× 72
= 77 cm2
Area of the whole diagram
= 19 1––4
+ 77
= 96 1––4
cm2
7. 5 cm 5 cm
8 cm 6 cm
OL N
MArea of semicircle LMN
= 1––2
× 22–––7
× 52
= 39 2––7
cm2
LM 2 = LN2 − MN2
LM = 102 − 62
= 8 cm
Area of triangle LMN
= 1––2
× 6 × 8
= 24 cm2
Area of the shaded region
= 39 2––7
− 24
= 15 2––7
cm2
8. 7 cm7 cm
14 cm
M
T
NL K
H J
Area of square HJKL= 14 × 14= 196 cm2
Area of the two semicircles
= 22–––7
× 72
= 154 cm2
Area of the shaded region= 196 − 154= 42 cm2
9.
240°
7 cm
TO
S
P
R
Q
Area of sector OPQR
= 240°–––––360°
× 22–––7
× 72
= 102 2––3
cm2
Area of semicircle OSRT
= 1––2
× 22–––7
× 7––2
2
= 19 1––4
cm2
3
Mathematics SPM Chapter 7
© Penerbitan Pelangi Sdn. Bhd.
Area of the whole diagram
= 102 2––3
+ 19 1––4
= 121 11–––12
cm2
Paper 2
1.
P
Q
R7 cm
(a) Circumference of circle
= 2 × 22–––7
× 14
= 88 cm
Arc QR = 14
× 2 × 22–––7
× 7
= 11 cm
Perimeter of the shaded region = Circumference of circle + arc QR + PQ + PR = 88 + 11 + 7 + 7 = 113 cm
(b) Area of circle
= 22–––7
× 142
= 616 cm2
Area of quadrant PQR
= 1––4
× 22–––7
× 72
= 38 1—2
cm2
Area of the shaded region = area of circle – area of quadrant PQR = 616 − 38 1––
2 = 577 1––
2 cm2
2.
60°
120°10 cm
14 cm
OP S
T
Q
R
(a) Arc PQR
= 360° – 120°–––––––––––360°
× 2 × 22–––7
× 14
= 58 23
cm
Perimeter of the whole diagram = OP + arc PQR + RO
= 14 + 58 23
+ 14
= 86 23
cm
(b) Area of sector OPQR
= 360° – 120°–––––––––––360°
× 22–––7
× 142
= 410 23
cm2
Area of sector SPT
= 60° 360°
× 227
× 102
= 52 821
cm2
Area of the shaded region
= 410 23
– 52 821
= 358 27
cm2
3.
O P
Q
R
S
T 60°
8 cm
8 cm 10 cm
4 cm
10 cm6 cm
(a) Area of sector ORS
= 60°–––––360°
× 22–––7
× 142
= 102 2––3
cm2
QT 2 = QO2 − OT2
QT = 102 − 62
= 8 cm
4
Mathematics SPM Chapter 7
© Penerbitan Pelangi Sdn. Bhd.
Area of triangle OTQ
= 1––2
× 6 × 8
= 24 cm2
Area of the shaded region
= 102 2––3
− 24
= 78 2––3
cm2
(b) Arc RS = 60°–––––360°
× 2 × 22–––7
× 14
= 14 2—3
cm
Arc PQ = 90°–––––360°
× 2 × 22–––7
× 10
= 15 5––7
cm
Perimeter of the whole diagram = arc RS + SO + OP + arc PQ + QR
= 14 2––3
+ 14 + 10 + 15 5––7
+ 4
= 58 8–––21
cm
4.
45°
60° 7 cm10 cm
A
B
C
D
O
(a) Arc AB = 45°–––––360°
× 2 × 22–––7
× 10
= 7.857 cm
Arc DO = 60°–––––360°
× 2 × 22–––7
× 7
= 7.333 cm
Perimeter of the whole diagram = OA + arc AB + BC + CD + arc DO = 10 + 7.857 + 3 + 7 + 7.333 = 35.19 cm
(b) Area of sector OAB
= 45°–––––360°
× 22–––7
× 102
= 39.286 cm2
Area of sector COD
= 60°–––––360°
× 22–––7
× 72
= 25.667 cm2
Area of the whole diagram = 39.286 + 25.667 = 64.95 cm2
5.
60°
7 cm
7 cm14 cm
120°
E O J
H
GF
K
(a) Arc EFG = 120°–––––360°
× 2 × 22–––7
× 14
= 29 1––3
cm
Arc HJ = 60°–––––360°
× 2 × 22–––7
× 7
= 7 1—3
cm
Perimeter of the whole diagram = arc EFG + GH + arc HJ + JO + OE
= 29 1––3
+ 7 + 7 1––3
+ 7 + 14
= 64 2––3
cm
(b) Area of the sector OEFG
= 120°–––––360°
× 22–––7
× 142
= 205 1––3
cm2
Area of semicircle OKG
= 1––2
× 22–––7
× 72
= 77 cm2
Area of sector OHJ
= 60°–––––360°
× 22–––7
× 72
= 25 2––3
cm2
Area of the shaded region
= 205 1––3
− 77 + 25 2––3
= 154 cm2
5
Mathematics SPM Chapter 7
© Penerbitan Pelangi Sdn. Bhd.
6.
120°7 cm
14 cm
45°R T O
PQ
S
(a) Area of sector OPQR
= 120°360°
× 227
× 142
= 205 13
cm2
Area of sector RST
= 45°360°
× 227
× 72
= 19 14
cm2
Area of the shaded region
= 205 13
− 19 14
= 231 112
cm2
(b) Arc PQR = 120°360°
× 2 × 227
× 14
= 29 13
cm
Arc ST = 45°360°
× 2 × 227
× 7
= 5 12
cm
Perimeter of the shaded region = arc PQR + RS + arc ST + TO + OP
= 29 13
+ 7 + 5 12
+ 7 + 14
= 62 56
cm
7. (a) Arc QR = 90°360°
× 2 × 227
× 7
= 11 cm
Arc QST = 12
× 2 × 227
× 14
= 44 cm
Perimeter of the shaded region = arc QR + RP + PT + arc QST = 11 + 7 + 21 + 44 = 83 cm
(b) Area of semicircle
= 12
× 227
× 142
= 308 cm2
Area of quadrant PQR
= 90°360°
× 227
× 72
= 38 12
cm2
Area of the shaded region
= 308 – 38 12
= 269 12
cm2
Paper 2
1.
16 cm
O
H
GE
F
45°10 cm
10 cm
(a) Arc EFG = 180°–––––360°
× 2 × 22–––7
× 10
= 31 3––7
cm
Arc HO = 45°–––––360°
× 2 × 22–––7
× 10
= 7 6––7
cm
Perimeter of the whole diagram = arc EFG + GH + arc HO + OE
= 31 3––7
+ 10 + 7 6––7
+ 10
= 59 2––7
cm
6
Mathematics SPM Chapter 7
© Penerbitan Pelangi Sdn. Bhd.
(b) Area of sector GOH
= 45°–––––360°
× 22–––7
× 102
= 39 2––7
cm2
Area of semicircle EFG
= 1––2
× 22–––7
× 102
= 157 1––7
cm2
FG 2 = EG2 − EF2
FG = 202 − 162 = 12 cm
Area of triangle EFG
= 1––2
× 16 × 12
= 96 cm2
Area of the shaded region
= 39 2––7
+ 157 1––7
− 96
= 100 3––7
cm2
2.
OE
F
D
180°60°
7 cm
7 cm
14 cmA B
C
(a) Arc BCD = 1––4
× 2 × 22–––7
× 14
= 22 cm
Arc AED = 1––2
× 2 × 22–––7
× 7
= 22 cm
Perimeter of the whole diagram = arc BCD + arc AED + AB = 22 + 22 + 14 = 58 cm
(b) Area of sector ABD
= 1––4
× 22–––7
× 142
= 154 cm2
Area of sector OAF
= 60°–––––360°
× 22–––7
× 72
= 25 2––3
cm2
Area of the shaded region
= 154 − 25 2––3
= 128 1––3
cm2
3.
O
J
EG F H60°120°120°
7 cm
7 cm 7 cm
7 cm
7 cm60°
M N
(a) Arc GJH = 1––2
× 2 × 22–––7
× 14
= 44 cm
Arc OM = arc NO
= 120°–––––360°
× 2 × 22–––7
× 7
= 14 2––3
cm
Perimeter of the shaded region = EG + arc GJH + HF + FN + arc NO + arc OM
+ ME
= 7 + 44 + 7 + 7 + 14 2––3
+ 14 2––3
+ 7
= 101 1––3
cm
(b) Area of semicircle OGJH
= 1––2
× 22–––7
× 142
= 308 cm2
Area of sector EOM = area of sector FON
= 120°–––––360°
× 22–––7
× 72
= 51 1––3
cm2
Area of the shaded region
= 308 − 2 51 1––3
= 205 1––3
cm2
7
Mathematics SPM Chapter 7
© Penerbitan Pelangi Sdn. Bhd.
4.
O
N
RM
Q
P
T
7 cm 7 cm
7 cm
7 cm
(a) Arc MNP = 270°–––––360°
× 2 × 22–––7
× 14
= 66 cm Arc MQ = arc PQ
= 90°–––––360°
× 2 × 22–––7
× 7
= 11 cm
Perimeter of the whole diagram = arc MNP + arc PQ + arc MQ = 66 + 2(11) = 88 cm
(b) Area of sector RMQ = area of sector TPQ
= 90°–––––360°
× 22–––7
× 72
= 38.5 cm2
Area of square QROT = 7 × 7 = 49 cm2
Area of the shaded region = (38.5 × 2) + 49 = 126 cm2
5.
O
S
T
R Q
P
NML
8 cm
8 cm
4 cm
4 cm
12 cm
(a) Arc LTR = 1––4
× 2 × 22–––7
× 8
= 12 4––7
cm
Arc RPM = 1––2
× 2 × 22–––7
× 4
= 12 4––7
cm
Perimeter of the shaded region = arc LTR + arc RPM + ML
= 12 4––7
+ 12 4––7
+ 8
= 33 1––7
cm
(b) Area of quadrant LTRM
= 1––4
× 22–––7
× 82
= 50 2––7
cm2
Area of semicircle RPM
= 1––2
× 22–––7
× 42
= 25 1––7
cm2
Area of the shaded region
= 50 2––7
+ 25 1––7
= 75 3––7
cm2
6.
AF
O
60°
7 cm 7 cm
5 cm5 cm
30° 30°E B
CD
(a) Arc CD = 60°–––––360°
× 2 × 22–––7
× 12
= 12 4––7
cm
Arc AB = arc EF
= 30°–––––360°
× 2 × 22–––7
× 5
= 2 13–––21
cm
Perimeter of the whole diagram = OA + arc AB + BC + arc CD + DE + arc EF
+ FO = 5 + 2 13–––
21 + 7 + 12 4––
7 + 7 + 2 13–––
21 + 5
= 41 17–––21
cm
(b) Area of sector OAB = area of sector OEF
= 30°–––––360°
× 22–––7
× 52
= 6 23–––42
cm2
8
Mathematics SPM Chapter 7
© Penerbitan Pelangi Sdn. Bhd.
Area of sector OCD
= 60°–––––360°
× 22–––7
× 122
= 75 3––7
cm2
Area of the whole diagram
= 75 3––7
+ 2 6 23–––42
= 88 11–––21
cm2
7.
H G 6 cm
3 cm
3 cm
3 cm
3 cm
6 cmD E C
A F B
(a) Arc DGA = 1––2
× 2 × 22–––7
× 3
= 9 3––7
cm
Arc FC = 90°–––––360°
× 2 × 22–––7
× 6
= 9 3––7
cm
Perimeter of the shaded region = CD + arc DGA + AF + arc FC
= 9 + 9 3––7
+ 3 + 9 3––7
= 30 6––7
cm
(b) Area of sector EFC
= 90°–––––360°
× 22–––7
× 62
= 28 2––7
cm2
Area of semicircle AGD
= 1––2
× 22–––7
× 32
= 14 1––7
cm2
Area of rectangle AFED = 3 × 6 = 18 cm2
Area of the shaded region
= 18 − 14 1––7
+ 28 2––7
= 32 1––7
cm2
8.
E
H
FO60°
7 cm 7 cm
7 cm
G
M N
(a) Arc OME = 1––2
× 2 × 22–––7
× 7––2
= 11 cm
Arc EH = 60°–––––360°
× 2 × 22–––7
× 7
= 7 1––3
cm
Perimeter of the combined region = arc OME + arc EH + HO
= 11 + 7 1––3
+ 7
= 25 1––3
cm
(b) Area of semicircle OME = area of semicircle ONF
= 1––2
× 22–––7
× 7––2
2
= 19 1––4
cm2
Area of sector OEH
= 60°–––––360°
× 22–––7
× 72
= 25 2––3
cm2
Area of the circle
= 22–––7
× 72
= 154 cm2
Area of the shaded region
= 154 − 25 2––3
− 2 × 19 1––4
= 89 5––6
cm2
9. D
E
O
A
B
C
45°9 cm
9 cm
9 cm
30°
9
Mathematics SPM Chapter 7
© Penerbitan Pelangi Sdn. Bhd.
(a) Arc AB = 30°–––––360°
× 2 × 22–––7
× 9
= 4 5––7
cm
Arc CD = 45°–––––360°
× 2 × 22–––7
× 18
= 14 1––7
cm
Perimeter of the whole diagram = OA + arc AB + BC + arc CD + DO
= 9 + 4 5—7
+ 9 + 14 1—7
+ 18
= 54 6—7
cm
(b) Area of sector OAB
= 30°–––––360°
× 22–––7
× 92
= 21 3–––14
cm2
Area of sector OCD
= 45°–––––360°
× 22–––7
× 182
= 127 2––7
cm2
Area of triangle OBE
= 1––2
× 9 × 9
= 40 1––2
cm2
Area of the shaded region
= 21 3–––14
+ 127 2––7
− 40 1––2
= 108 cm2
10.
OR P45°
212 cm–
212 cm–
Q
(a) Arc PRQ = 360° – 45°––––––––––360°
× 2 × 22–––7
× 21–––2
= 57 3—4
cm
Perimeter of the major sector OPQ = perimeter of sector OPRQ = arc PRQ + QO + OP
= 57 3––4
+ 21–––2
+ 21–––2
= 78 3––4
cm
(b) Area of the circle
= 22–––7
× 21–––2
2
= 346 1––2
cm2
Area of the semicircle
= 1––2
× 22–––7
× 21–––4
2
= 43 5–––16
cm2
Area of sector OPQ
= 45°–––––360°
× 22–––7
× 21–––2
2
= 43 5–––16
cm2
Area of the shaded region
= 346 1––2
− 3 43 5–––16
− 43 5–––16
= 173 1––4
cm2
11.
OR
T 140°12 cm
3 cmQP
S
(a) Arc PST = 140°–––––360°
× 2 × 22–––7
× 15
= 36 2––3
cm
Arc RT = arc PQ
= 1––2
× 2 × 22–––7
× 6
= 18 6––7
cm
Perimeter of the shaded region = arc PST + arc TR + RO + OQ + arc QP
= 36 2––3
+ 18 6––7
+ 3 + 3 + 18 6––7
= 80 8–––21
cm
(b) Area of sector OPST
= 140°–––––360°
× 22–––7
× 152
= 275 cm2
10
Mathematics SPM Chapter 7
© Penerbitan Pelangi Sdn. Bhd.
Area of the semicircle
= 1––2
× 22–––7
× 62
= 56 4––7
cm2
Area of the shaded region
= 275 − 2 56 4––7
= 161 6––7
cm2
12.
O
R
QS
30°
7 cm 14 cm
7 cm
(a) Arc QRS = 360° – 30°––––––––––360°
× 2 × 22–––7
× 14
= 80 2––3
cm
Arc OQ = 1––2
× 2 × 22–––7
× 7
= 22 cm
Perimeter of the shaded region = arc QRS + SO + arc OQ
= 80 2––3
+ 14 + 22
= 116 2––3
cm
(b) Area of sector OQRS
= 330°–––––360°
× 22–––7
× 142
= 564 2––3
cm2
Area of the semicircle
= 1––2
× 22–––7
× 72
= 77 cm2
Area of the shaded region
= 564 2––3
− 77
= 487 2––3
cm2
13.
E F O A
B
45°9 cm
9 cm
9 cm9 cm
D
(a) Area of sector AED
= 45°–––––360°
× 22–––7
× 272
= 286 11–––28
cm2
Area of quadrant OBF
= 1––4
× 22–––7
× 92
= 63 9–––14
cm2
Area of triangle AOB
= 1––2
× 9 × 9
= 40 1––2
cm2
Area of the shaded region
= 286 11–––28
− 63 9–––14
− 40 1––2
= 182 1––4
cm2
(b) AB2 = OA2 + OB2
AB = 92 + 92
= 12.73 cm BD = 27 − 12.73 = 14.27 cm
Arc ED = 45°–––––360°
× 2 × 22–––7
× 27
= 21 3–––14
cm
Arc BF = 90°–––––360°
× 2 × 22–––7
× 9
= 14 1––7
cm
Perimeter of the shaded region = BD + arc DE + EF + arc BF
= 14.27 + 21 3–––14
+ 9 + 14 1––7
= 58.63 cm
11
Mathematics SPM Chapter 7
© Penerbitan Pelangi Sdn. Bhd.
14.
60°30° 7 cm
14 cm
T
O
P
Q S
R
(a) Arc TS = 1––4
× 2 × 22–––7
× 14
= 22 cm
Arc QP = 30°–––––360°
× 2 × 22–––7
× 7
= 3 2––3
cm
Perimeter of the whole diagram = OT + arc TS + SQ + arc QP + PO
= 14 + 22 + 7 + 3 2––3
+ 7
= 53 2––3
cm
(b) Area of quadrant OST
= 1––4
× 22–––7
× 142
= 154 cm2
Area of sector OQR
= 60°–––––360°
× 22–––7
× 72
= 25 2––3
cm2
Area of the shaded region
= 154 − 25 2––3
= 128 1––3
cm2
15. A O
B E
C D
F
60°60°60°
(a) Arc AB = arc EF
= 60°–––––360°
× 2 × 22–––7
× 7
= 7 1––3
cm
Arc CD = 60°–––––360°
× 2 × 22–––7
× 14
= 14 2––3
cm
Perimeter of the whole diagram = FA + arc AB + BC + arc CD + DE + arc EF
= 14 + 7 1––3
+ 7 + 14 2––3
+ 7 + 7 1––3
= 57 1––3
cm
(b) Area of sector OCD
= 60°–––––360°
× 22–––7
× 142
= 102 2––3
cm2
Area of sector OBE = area of sector OAB = area of sector OEF
= 60°–––––360°
× 22–––7
× 72
= 25 2––3
cm2
Area of the shaded region = (area of sector OCD – area of sector OBE)
+ area of sector OAB + area of sector OEF
= 102 2––3
– 25 2––3
+ 25 2––3
+ 25 2––3
= 128 1––3
cm2
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