a p t er circular measure h 8 8.pdfsukatan membulat 1. (a) 0.6 rad.= 0.6 × 180° ... area of sector...

© Penerbitan Pelangi Sdn. Bhd.79

CHAPTER

CHAPTER

8Circular MeasureSukatan Membulat

1. (a) 0.6 rad. = 0.6 × 180°p

= 34.38°

(b) 1.8 rad. = 1.8 × 180°p

= 103.1°

(c) 4p5

rad. = 4p5

× 180°p

= 144°

(d) 3p2

rad. = 3p2

× 180°p

= 270°

2. (a) 158° = 158 × p180

= 2.758 rad.

(b) 42.6° = 42.6 × p180

= 0.7435 rad.

(c) 122.8° = 122.8 × p180

= 2.143 rad.

(d) 252°129 = 252.2 × p180

= 4.402 rad.

3. (a) s = 6 × 2.5 = 15 cm

(b) s = 8 × (100 × p180

)

= 13.96 cm

(c) s = 8.5 × (2p – 1.8) = 38.11 cm

4. (a) 12r = 2

5 p

= 9.549 cm

(b) 26r = 220 × p

180

= 6.771 cm

5. (a) q = 208.5

= 2.353 rad.

(b) q = 28.45.5

= 5.164 rad.

6. (a) In ΔOPM,P

R

OM

10 cm

0.6 rad.

PM sin 0.6 rad. = 10 PM = 10 × sin 0.6 rad. = 5.646 cmLength of arc PQR = 10 × 1.2 = 12 cmPerimeter of the shaded region= 12 + (2 × 5.646)= 23.29 cm

Additional Mathematics Form 4 Chapter 8 Circular Measure

© Penerbitan Pelangi Sdn. Bhd. 80

(b) In ΔOEN,

8 cm1.2 rad.

O

N GE

EN sin 1.2 = 8 EN = 8 × sin 1.2 rad. = 7.456 cmLength of arc EFG = 8 × 2.4 = 19.2 cm Perimeter of the shaded region = 19.2 + (2 × 7.456)= 34.11 cm

(c) In ΔOPN,

NP R

O

55°6 cm

PN sin 55° = 6 PN = 6 sin 55° = 4.915 cm

Length of arc PQR = 6 × 1110 × p180 2

= 11.52 cm Perimeter of the shaded region= 11.52 + (2 × 4.915)= 21.35 cm

7. (a) 80° = 1.396 rad. Length of arc PR = 5 × 1.396 = 6.98 cm

Length of arc QS = 12 × 1.396 = 16.75 cm

PQ = RS = 12 – 5 = 7 cm Therefore, perimeter of PQSR = 6.98 + 16.75 + 2(7) = 37.73 cm

(b) (i) OR = 15p – 1.8

Radius = 11.18 cm

(ii) Length of arc PQ = 11.18 × 1.8 = 20.12 cm

8. (a) A = 12

(8)2(1.4)

= 44.8 cm2

(b) A = 12

(10)2(0.86)

= 43 cm2

(c) A = 12

(7)2(110 × p180

)

= 47.04 cm2

(d) A = 12

(7.5)2(2p – 2p5

)

= 141.4 cm2

(e) A = 12

(9)2[(360 – 105) × p180

]

= 180.2 cm2

9. (a) r = 2(40)1.2

= 8.165 cm

(b) r = 2(60)0.8

= 12.25 cm

(c) r = 2(65)5p4

= 5.754 cm

(d) r = 2(20.5)30 × p

180 = 8.849 cm



(e) r = 2(88)120 × p

180 = 9.167 cm

10. (a) q = 2(43.2)62

= 2.4 rad.

(b) q = 2(48.5)62

= 2.694 rad.

(c) q = 2(256)102

= 5.12 rad.

11. (a)

M CA

O

30°9 cm

In ∆OAM, AM sin 30° = —– 9 AM = 4.5 cm OM cos 30° = —– 9 OM = 7.794 cm

1 Area of ∆OAC = —(AC)(OM) 2 1 = —(2 × 4.5)(7.794) 2 = 35.07 cm2

1 Area of sector OABC = —(9)2(60 × p180

) 2 = 42.41 cm2

Area of segment ABC = 42.41 – 35.07 = 7.34 cm2

(b)

10 cm

0.6 rad. O

G

E

M

In ∆OEM, EM sin 0.6 rad. = —– 10 EM = 5.646 cm OM cos 0.6 rad. = —– 10 OM = 8.253 cm

1 Area of ∆OEG = —(EG)(OM) 2 1 = —(2 × 5.646)(8.253) 2 = 46.60 cm2

1 Area of sector OEFG = — (10)2(1.2) 2 = 60 cm2

Area of segment EFG = 60 – 46.60 = 13.40 cm2

(c)

O M

A

C

9 cm3π8

rad.

In ∆OAM,

sin 3π8

rad. = AM9

AM = 8.315 cm

cos 3π8

rad. = OM9

OM = 3.444 cm 1 Area of ∆OAC = —(2 × 8.315)(3.444) 2 = 28.637 cm2

1 Area of sector OABC = —(9)2( 3p4

) 2 = 95.426 cm2

Area of segment ABC = 95.426 – 28.637 = 66.79 cm2



12. p(a) (i) q = 60 × —— 180 = 1.047 rad.

120 × p(ii) Length of arc AC = 6 × ––——— 180 = 12.57 cm Length of arc AB = 12 × 1.047 = 12.56 cm The perimeter of the shaded region = 12.56 + 12.57 + 6 = 31.13 cm

(iii) Area of sector DAB 1 = —(12)2(1.047) 2 = 75.38 cm2

Area of sector OAC 1 = —(6)2(2.094) 2 = 37.69 cm2

CN = !w62 – 32 = 5.196 cm

3 cm

6 cm

NDO

C

Area of ∆OCD 1 = —(6)(5.196) 2 = 15.59 cm2

Area of shaded region = 75.38 – 37.69 – 15.59 = 22.10 cm2

(b) R 5 cm

5 cm 5 cm 13 cm

4 cm 12 cm1 cm

S

OθQ P N

(i) sin q = 513

= 0.395 rad.

(ii) (a) Length of arc PS = 13 × 0.395 = 5.135 cm

Perimeter of the shaded region = 5.135 + 5 + 5 + 4 = 19.135 cm

(b) Area of OQRS = 12

(5 + 17) × 5

= 55 cm2

Area of sector OPS = 1

2 × 132 × 0.395

= 33.38 cm2

Area of the shaded region = Area of OQRS – Area of sector

OPS = 55 – 33.38 = 21.62 cm2

8PracticeSPM

Paper 1

1. Area of segment DEB= Area of sector ABD – Area of ∆ABD

= 1 12

× 92 × π2 2 – 1 1

2 × AB × AD2

= 63.6255 – 1 12

× 9 × 92 = 23.1255 m2

Area of the shaded region= 92 – 2(23.1255)= 34.749 m2

2. (a) 126° × π180°

= 2.1994 rad.

(b) Perimeter of the sector OEF = Arc EF + OE + OF = 9(2.1994) + 2(9) = 37.795 cm



3. (a) OG = 9 cm

EG = 23

(9)

= 6 cm

(b) Area of sector OGH = 12

(92)q

= 812

q

Area of sector EFG = 12

(62)q

= 18q

Area of shaded region = 29.25 cm2

812

q – 18q = 29.25

452

q = 29.25 q = 1.3 rad.

4. (a) s = rq 10.2 = 6q q = 10.2

6 = 1.7 rad

(b) Area of sector OAB

= 12

r2(π – 1.7)

= 12

× 62(3.142 – 1.7)

= 25.96 cm2

5.

O

P

Q

θ

(a) ∠OQP = ∠OPQ = 30° q = 180° – 30° – 30° = 120°

= 120°180°

× π

= 120°180°

× 3.142

= 2.095 rad

(b)

O

P

Q

M60°

10 cm

In ∆OPM,

sin 60° = PM10

PM = 8.660 cm

cos 60° = OM10

OM = 5 cm

Area of ∆OPM = 12

(2 × 8.660)(5)

= 43.3 cm2

Area of sector OPQ = 12

(10)2(2.095)

= 104.75 cm2

Area of shaded region = 104.75 – 43.3 = 61.45 cm2

6. r(6α) + rp + r = 36 r(6α + p + 1) = 36 r = 36

(6α + p + 1)



7. R

8 cm

4 cmP N Q

∠RPQ = p3

rad

RN = 82 – 42 = 48 cm

Area of ∆ PNR = 12

× 4 × 48 = 13.86 cm2

Area of semicircle = (3.142)(4)2

2 = 50.27 cm2

Area of the shaded region = 13.86 + 50.27 = 64.13 cm2

Paper 2

1. (a) w = the circumference of the base of the cylinder

= 2πr

= 2 × 227

× 72

= 22 cm

The arc of the sector of circle P = w rpq = 22

rp1 π2 2 = 22

rpπ = 44

rp1227 2 = 44

rp = 14

l = rp + height of cylinder = 14 + 10 = 24 Therefore, l = 24 and w = 22

(b) Area of rectangular card = 22 × 24 = 528 cm2

Area of sector of circle P

= 12

(14)21 π2 2

= 49π = 49 × 22

7 = 154 cm2

Area of rectangle Q = 22 × 10 = 220 cm2

Area of card which is not used = 528 – 154 – 220 = 154 cm2

2. (a) ∠OPQ = 180° – 36°2

= 72° = 72 × π

180 = 1.257 rad.

(b) ∠POQ = 36° = 36 × π

180 = 0.6284 rad.

Length of arc PRQ = OP × ∠POQ = 15 × 0.6284 = 9.426 cm

Length of arc QS = PQ × ∠OPQ = 9.27 × 1.257 = 11.65 cm

Perimeter of shaded region = 9.426 + 11.65 + 9.27 = 30.35 cm



(c)

Q

15 cmP

NO 18°

cos 18° = ON15

ON = 14.27 cm

Area of ∆OPQ = 1

2 × 9.27 × 14.27

= 66.14 cm2

Area of segment PRQ = Area of sector OPQ – Area of ∆OPQ = ( 1

2 × 152 × 0.6284) – 66.14

= 4.555 cm2

Area of sector PQS = 1

2 × 9.272 × 1.257

= 54.01 cm2

Area of the shaded region = 4.555 + 54.01 = 58.565 cm2

3. (a) OQ = 9 + 4 = 13 cm OR = 132 – 122

= 5 cm

tan q = 125

q = 1.176 rad.

(b) RS = 5 – 4 = 1 cm

Length of arc PS = 4 × 1.176 = 4.704 cm

Perimeter of shaded region = PQ + QR + RS + arc PS = 9 + 12 + 1 + 4.704 = 26.70 cm

!

(c) Area of shaded region = Area of ∆ORQ – Area of sector OPS

= 1 12

× 12 × 52 – 1 12

× 42 × 1.1762 = 20.59 cm2

4. (a) cos ∠BOX = 36

∠BOX = 60° = 1.047 rad.

(b) sin ∠BOX = BXOB

BX = 6 (sin 60°) = 5.196 cm

BC = (BX)2 + (CX)2 = (5.196)2 + 92 = 108 = 10.392 cm

Length of arc BC = 6 (∠BOC)

= 61120° × π180° 2

= 12.568 cm

The perimeer of the shaded region = BC + Arc BC = 10.392 + 12.568 = 22.96 cm

(c) Area of sector OBC

= 12

× 62 × 1120° × π180° 2

= 37.704 cm2

Area of ∆OBC

= 12

(6)(6) sin 120°

= 15.5885 cm2

The area of the shaded region = Area of sector OBC – Area of ∆OBC = 37.704 – 15.5885 = 22.116 cm2



5. Q

P O

U

T R

S

8 cm 4 cm

(a) cos ∠QOT = OTOQ

= 48

∠QOT = 60° × π180°

= 1.047 rad.

(b) Perimeter of the shaded region = Arc PQ + Arc OQ + OP

= 38 × 1120° × π180° 24 + (8 × 1.047) + 8

= 16.7573 – 8.376 + 8 = 33.133 cm

(c) Area of sector OPQ

= 12

× 82 × 1120° × π180° 2

= 67.0293 cm

Area of sector ROQ

= 12

× 82 × 1.047

= 33.504

Area of ∆ROQ

= 12

× OR × QT

= 12

× 8 × 82 – 42

= 27.7128

Area of segment OUQ = Area of sector ROQ – Area of ∆ROQ = 33.504 – 27.7128 = 5.7912 cm2

The area of the shaded region = Area of sector OPQ – Area of segment OUQ = 67.0293 – 5.7912 = 61.238 cm2

∠QRT = ∠QOT

6. (a) r = 3, q = 2.095 rad

(b) 18.85 = q(6 + r) .......... 6.28 = q(r)

r = 6.28q

..........

into

18.85 = q16 + 6.28q

2 18.85 = 6q + 6.28 6q = 12.57 q = 2.095 r = 6.28

2.095 = 3

Area of the chord

= 12 (92)(2.095) – 1

2 (92) sin 120°

= 84.85 – 35.07 = 49.78 cm2

Area of COD = 12 (3)2 2.095

= 9.43 cm2

Area of shaded reegion = 49.78 + 9.43 = 59.21 cm2

MRSM Cloned Questions

1.

1.05 rad.

5 cm

B

D

N

C

A

O

Length of arc AB = OA × ∠AOB 16.8 = OA × 1.05 OA = 16 cm

OD = 58

× 16

= 10 cm



tan 1.05 rad. = CN5

CN = 8.717 cm

Area of ∆OCD = 12

× 10 × 8.717

= 43.585 cm2

Area of sector OAB= 1

2 × 162 × 1.05

= 134.4 cm2

Area of the shaded region= 134.4 – 43.585= 90.815 cm2

Challenge

1. R

S

PT

U Q

Since the sides of ∆RPS are equal length,∠RPS = π

3.

Area of segment RUS = Area of sector PRS – Area of PRS

= 1 12

× 82 × π3 2 – 1 1

2 × 82 × sin 60°2

= 33.5145 – 27.7128= 5.802 cm2

The area of grey region= Area of RUS + Area of RTS + Area of two circle= 2(5.802) + 2(π × 0.72)= 14.683 cm2

2. E

G

F

D

B

N

MA

C

0.3 m1.3 m

0.5 m0.5 m

(a) cos ∠ABN = 0.31.3

∠ABN = 76°40' = 1.338 rad. ∠ABD = ∠ABN = 1.338 rad.

(b) The longer arc of the metal foil = Length of arc MF = 0.8 × (3.142 – 1.338) = 0.8 × 1.804 = 1.443 m

(c) AE = 0.8 + 0.3 = 1.1 EF = AN = 1.32 – 0.32 = 1.2649

Area of trapezium, ABFE = 1

2(AE + BF) × EF

= 12

(1.1 + 0.8) × 1.2649

= 1.2017 m2

Area of the shaded region = Area of ABFE – Area of sector AGM

– Area of sector BFM

= 1.2017 – 1 12

× 0.52 × 1.3382 – 1 1

2 × 0.82 × 1.8042

= 1.2017 – 0.1673 – 0.5773 = 0.457 m2 The area of the metal foil is 0.457 m2.

a p t er circular measure h 8 8.pdfsukatan membulat 1. (a) 0.6 rad.= 0.6 × 180° ... area of sector...

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