07[a math cd]

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




= 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


= 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



Area of triangle OTQ

= 1––2

× 6 × 8

= 24 cm2


= 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


= 205 1––3

− 77 + 25 2––3

= 154 cm2

5



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


= 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


= 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



(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


= 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


= 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


= 308 − 2 51 1––3

= 205 1––3

cm2

7



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


= 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



Area of sector OCD

= 60°–––––360°

× 22–––7

× 122

= 75 3––7

cm2


= 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


= 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


= 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



(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


= 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


= 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




= 1––2

× 22–––7

× 62

= 56 4––7

cm2


= 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


= 1––2

× 22–––7

× 72

= 77 cm2


= 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


= 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



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


= 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

07[a math cd]

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