8/11/2019 201307041939343670_9789834506582-F4Ch8

1/5

24

1 LetOR= rcm

Perimeter of the sectorROS= 40 cm

r+ r+ 8 = 40

2r= 32

r= 16

=s

r

=8

16 = 0.5 rad.

2

46.64 cm 0.454 rad.O

A

B

r

Length of the major arcAB= 46.64 cm

r(2 0.454) = 46.64

r[(2 3.142) 0.454] = 46.64

r(5.83) = 46.64

r = 8

Radius = 8 cm

3 (a) Area of sector = 20 cm2

1

2r2(0.4) = 20

r2= 100

r= 10

Length of the arcAB

= 10 0.4 = 4 cm

(b) Reflex angleAOB

= 2 0.4

= (2 3.142) 0.4

= 5.884 rad.

= 5.884 180

3.142 = 337 5

4 AOB=8

12

=2

3rad.

Area of the shaded region

= Area of sectorOAB Area of secorAMN

=1

2(12)223

1

2(6)2(0.7)

= 48 12.6

= 35.4 cm2

5

1.287 rad.

5 cm

4 cm

5 cm5 cm

3 cm3 cm

5 cm

Q R

P S

O

(a) Length of arcQR

= 10 1.287

= 12.87 cm

(b) Area of the shaded region

= Area of sectorOQR Area of OPS

=1

2r2

1

2Base Height

=1

2(10)2(1.287)

1

26 4

= 64.35 12

= 52.35 cm2

Form 4: Chapter 8 (Circular Measure)

SPM Practice

Fully-Worked Solutions

Paper 1

8/11/2019 201307041939343670_9789834506582-F4Ch8

2/5

25

6

O

P

Q

R

12 cm6 cm

6 cm

(a) In OPR,

cosPOR=6

12 POR= 60

= 60 3.142

180 = 1.0473 rad.

(b) PR= 122 62

= 108

= 10.3923 cm

Area of shaded region

= Area of sectorOQR Area of OPR

=1

2 1221.0473

1

26 10.3923

= 75.4056 31.1769

= 44.23 cm2

8/11/2019 201307041939343670_9789834506582-F4Ch8

3/5

26

1

OP=

3

515

= 9 cm9 cm

6 cm

15 cm

12 cm

O

P

A

B

(a) From OPB,

cos =9

15=

3

5 = 53.13

= 53.13 3.142

180 = 0.9274 rad.

(b) Area of the shaded region

= Area of sectorOAB Area of OPB

= 12(15)2(0.9274) 129 12

= 104.3325 54 = 50.33 cm2

2 (a)

8 cm 60

60

60 60

6060

8 cm

O

P

R

J

K

QT L

Insert a line OQin the diagram.

SinceOPQRis a rhombus,

OP=PQ=QR=RO.

Since the radius of a circle is a constant,

OP=OQ=OR.

It can now be concluded that OPQis an

equilateral triangle because

OP=PQ=OQ.It can also be concluded that ORQis an

equilateral triangle because

OR=RQ=OQ.

Therefore,POQ=ROQ= 60

Hence,POR=

= 120

= 120

180

=2

3

rad.

(b) Based on JOQ,

cosJOQ=OQ

OJ

cos 60 =8

OJ

OJ= 8cos 60 = 16 cm

Length of the arcJLK

=OJJOK

= 16 2

3(3.142)

= 33.51 cm

(c) Area of the shaded segment

=1

2r2( sin)

=1

2(16)2233.142 sin 120

= 157.27 cm2

3 (a) In OQR,

cos 3r

=7

OR

cos 60 =7

OR

OR=7

cos 60 = 14 cm

In OQR,

tan 60 =QR

7 QR= 7 tan 60

= 12.1244 cm

rad.

7 cm

7 cm

7 cm

12.1244cm

3O

P

Q

A

S

R

B

Paper 2

8/11/2019 201307041939343670_9789834506582-F4Ch8

4/5

27

PR=OROP

= 14 7

= 7 cm

Length of arc PQ

= 7

3

= 7

3.142

3= 7.3313 cm

Hence, the perimeter of the shaded regionA

= PR+QR+ Length of arc PQ

= 7 + 12.1244 + 7.3313

= 26.46 cm

(b) Area of OQR

=1

2712.1244

= 42.4354 cm2

Area of the sectorORS

=1

2142

3.142

3= 102.6387 cm2

Hence, the area of the shaded regionB

= Area of sectorORS Area of OQR

= 102.6387 42.4354

= 60.20 cm2

4 (a)

10 cm

10 cm8 cm

2 cm

1.982 rad.

OA QP

R

C

10 cm

CAOis an isosceles triangle.CAO=COA

= 1.982

= 3.142 1.982

= 1.16 rad.

(b)

10 cm

10 cm

8 cm

4 cm 4 cm

1.982 rad.1.16 rad.

OA M QP

R

C

10 cm

Perimeter of the shaded region

=CR+ Length of arcCQ+ Length of arcRQ

= 8 + (10 1.982) + (18 1.16)

= 48.7 cm

(c) MC= CO2MO2

= 102 42

=

84cm

Area of shaded region

= Area of sectorARQ Area of ACO

Area of sector COQ

=1

2(18)2(1.16)

1

28 84

1

2(10)2(1.982)

= 52.16 cm2

5A

B

P30

30

Q

M

R

O

6 cm

6 cm

6 cm

60

60

(a) In PQO,

sin 30 =6

PO

1

2=

6

PO

PO= 12 cm

PM=PO+OM

= 12 + 6

= 18 cm

PA=PB=PM= 18 cm

8/11/2019 201307041939343670_9789834506582-F4Ch8

5/5

28

Length of arcAMB

= 18 60 3.142180= 18.85 cm

(b) In PQO, PQ= 122 62

= 10.3923 cm

Area of POQ= Area of POR

=1

2610.3923

= 31.1769 cm2

Area of sectorQMR

=1

262240 3.142180

= 75.408 cm2

Area of sector PAMB

=1

2182

60 3.142

180= 169.668 cm2Hence, the area of the shaded region

= Area of sector PAMB Area of POQ

Area of POR Area of sector QMR

= 169.668 31.1769 31.1769 75.408

= 31.91 cm2

6

O

R

C

P Q

N

6 cm

6cm

6 cm

10cm

4 cm

(a) In ONC,

cos = 416

= 75.52

= 75.52 3.142

180 = 1.318 rad.(correct to 4 s.f.)

(b) In ONC,

CON= 180 90 75.52

= 14.48

POR= 90 + 14.48

= 104.48

= 104.48 3.142

180

= 1.824 rad.Length of arc PR= 6 1.824

= 10.94 cm

(c) Area of sectorPOR

=1

2621.824

= 32.832 cm2

Area of sectorQCR

=1

21021.318

= 65.900 cm2

In ONC,

ON= 162 42

= 240

= 15.492 cm

Area of trapezium POCQ

=1

2(6 + 10) (15.492)

= 123.936 cm2

Area of the shaded region= Area of trapezium POCQ

Area of sector POR Area of sector QCR

= 123.936 32.832 65.900

= 25.20 cm2