06[a math cd]

7
1 © Penerbitan Pelangi Sdn. Bhd. Paper 1 1. E F G x ° 120° H J K JKE = (6 2) × 180° –––– 6 = 120° x° = 180° 120° x = 60 Answer: A 2. x° + x° + x° + 84° + 84° = 540° 3x° + 168° = 540° 3x° = 540° 168° = 372° x = 372 –––– 3 = 124 Answer: A 3. p° 30° 120° 30° H J G F K E JHG = (6 2) × 180° –––– 6 = 120° HJG = 180° 120° ––––––––––– 2 = 30° p° = 120° 30° 30° p = 60 Answer: C 4. 108° E F G H J K 108° EFG = (5 2) × 180° –––– 5 = 108° m° = 180° 108° ––––––––––– 2 m = 36 n° = 180° 108° n = 72 m + n = 36 + 72 = 108 Answer: A 5. x ° 125° 170° 100° V U T S R Q P y° y° = 180° 125° y = 55 x° + 100° + 170° + 90° + 125° + 90° = 720° x° + 575° = 720° x° = 720° 575° x = 145 x + y = 145 + 55 = 200 Answer: C CHAPTER 6 Polygons

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1 © Penerbitan Pelangi Sdn. Bhd.

Paper 1

1.

E F

G

120°

HJ

K

∠JKE = (6 − 2) × 180°––––6

= 120° ∴ x° = 180° − 120° x = 60

Answer: A

2. x° + x° + x° + 84° + 84° = 540° 3x° + 168° = 540° 3x° = 540° − 168° = 372°

x = 372––––3

= 124

Answer: A

3.

30°

120°30°

H

J G

FK

E

∠JHG = (6 − 2) × 180°––––6

= 120°

∠HJG = 180° − 120°–––––––––––2

= 30°

∴ p° = 120° − 30° − 30° p = 60

Answer: C

4.

108°

E

F

G H

J

K

108°

∠EFG = (5 − 2) × 180°––––5 = 108°

m° = 180° − 108°–––––––––––2

m = 36 n° = 180° − 108° n = 72∴ m + n = 36 + 72 = 108

Answer: A

5. x°

125°

170°

100°

V UT

S

RQ

P

y° = 180° − 125° y = 55

x° + 100° + 170° + 90° + 125° + 90° = 720° x° + 575° = 720° x° = 720° − 575° x = 145∴ x + y = 145 + 55 = 200

Answer: C

CHAPTER

6 PolygonsCHAPTER

2

Mathematics SPM Chapter 6

© Penerbitan Pelangi Sdn. Bhd.

6.

54°

P Q

R

S

T

∠TSR = (6 − 2) × 180°––––6

= 120° x° = 180° − 120° x = 60 y° = 180° − 90° − 54° y = 36

∴ x + y = 60 + 36 = 96

Answer: B

7.

46°

120°

120°

P Q R

S

UV

W

∠QRS = (6 − 2) × 180°––––6

= 120°

x° + 46° + 120° + 120° = 360° x° + 286° = 360° x° = 360° − 286° x = 74

Answer: B

Paper 1

1.

42°

H

MG

F

E

K

J

120°

30°69°

∠EFG = (6 – 2) × 180°6

= 120°

∠EGF = 180° – 120°2

= 30°

∠EGM = 180° – 42°2

= 69° x° = 120° – 30° – 69° x = 21

Answer: A

2.

K

E F

G

H

J

80°280°

2y°

2y°

2x° x°

Reflex ∠FGH = 360° – 80° = 280°

x° + x° + 2x° + 2y° + 2y° + 280° = 720° 4x° + 4y° = 720° – 280° 4(x° + y°) = 440° x + y = 110

Answer: B

3.

130°74°

82°

M

H G

FE

K

J

L

50°

286°

∠JHM = 180° – 130° = 50°Reflex ∠JKL = 360° – 74° = 286° p° + q° + 50° + 82° + 286° = 540° p° + q° + 418° = 540° p° + q° = 540° – 418° p + q = 122

Answer: A

3

Mathematics SPM Chapter 6

© Penerbitan Pelangi Sdn. Bhd.

4.

x°72°

72° y°W V U

T

S

RO

∠ROS = 360°5

= 72° x° = 2 × 72° = 144°

∠TVU = 360°5

= 72° y° = 180° – 72° – 72° = 36°

∴ x° + y° = 144° + 36° x + y = 180

Answer: D

5.

60°

G

H

J

KE

F74°

80°

46°

23°x°

∠GFJ = 180° – 60° – 74° = 46°Given ∠GFJ = 2∠EFJ

∠EFJ = 46°––––2

= 23° ∴ x° = 180° − 80° – 23° x = 77

Answer: C

6. y°

K L

G F

E

R

Q

P

N

M

H

J120°135°

∠EML = (8 – 2) × 180°8

= 135°

∠EMN = (6 – 2) × 180°6

= 120°

y° = 360° – 135° – 120° y = 105

Answer: B

7. ∠HJE = (5 – 2) × 180°5

= 108°∠LJE = 180° − 108° = 72° y° = 180° – 72° – 90° y = 18

Answer: A

8.

70°

150°

60°

30°

120°E

M

L

K J

H

GF

∠EFL = 360°––––6

= 60°∠LFG = 180° − 60° = 120°

∠FLG = 180° – 120°––––––––––2

= 30°∠FLM = 180° − 30° = 150°

x° = 360° – 70° – 150° – 60° x = 80

Answer: A

9.

108°36°

P

QR

S

T

U V

∠STP = (5 − 2) × 180°––––5

= 108°

∠PST = 180° − 108°–––––––––––2

= 36° ∴ y = 36

Answer: B

4

Mathematics SPM Chapter 6

© Penerbitan Pelangi Sdn. Bhd.

Paper 1

1.

35°

60°

90° x°

J

K

L

M

N

PQ

∠PJQ = 360°––––6

= 60°∠PJK = 180° − 60° = 120°∠JPM = ∠LMP

= 540° − 120° − 120° − 120°–––––––––––––––––––––––2 = 90°∠JPQ = 180° − 35° − 60° = 85° ∴ x° = 90° + 85° x = 175

Answer: C

2.

y°45°

45°135°K

J

L

ME

F

G

T

H

∠HGT = 360°––––8

= 45°∠JHG = 180° − 45° = 135°

y° + 45° + 45° + (360° − 135°) = 360° y° + 315° = 360° y° = 360° − 315° y = 45

Answer: A

3.

30°72°

36°

E

F

G

H J K

L

72°

∠EJK = 360°––––5

= 72°∠JKE = 180° − 72° − 72° = 36°

m° + n° + (36° + 30°) = 180° m° + n° = 180° − 66° m + n = 114

Answer: B

4.

125°

60°60°

G H

J

K

FE

L

∠LFE = 180––––3

= 60°

p° + p° + 60° + 125° + 90° = 540° 2p° + 275° = 540° 2p° = 540° − 275° p = 132.5

Answer: C

5.

50°S

T U

V

RQ

P

∠RST = (5 − 2) × 180°––––5

= 108°∠TSV = 180° − 50° = 130°∴ m° = 360° − 108° − 130° m = 122

Answer: A

5

Mathematics SPM Chapter 6

© Penerbitan Pelangi Sdn. Bhd.

6. (n − 2) × 180°––––n = 140°

180° − 360°––––n = 140°

180° − 140° = 360°––––n

40° = 360°––––n

n = 360°––––40°

= 9

The number of sides of the regular polygon = 9

Answer: B

7.

x° y°

KJ

H

G

FE

22.5°45°

M

L

∠JKL = (8 − 2) × 180°––––8

= 135° y = ∠JKG = 180° − 135° = 45°

∠MKL = 180° − 135°–––––––––––2

= 22.5° x° = 135° − 45° − 22.5° x = 67.5°∴ x + y = 67.5 + 45 = 112.5

Answer: B

8. 360°––––n = 36°

n = 10

The number of sides of the regular polygon = 10

Answer: B

9. y°

150°

60°

P

Q

RS T

U

Sum of angles of PQRSTU= (6 − 2) × 180°= 720°

∠QPU = 720° – 4(150°)–––––––––––––2 = 60°∴ y° + 60° = 150° y° = 150° − 60° y = 90

Answer: C

10.

70°

70°

110°

105°P

QR S

T

U

n°m°

∠TUQ = ∠PQU = 70°∠RQU = 180° − 70° = 110°

Sum of angles of QRSTU= (5 − 2) × 180°= 540°

m° + n° + 110° + 70° + 105° = 540° m° + n° + 285° = 540° m° + n° = 540° − 285° m + n = 255

Answer: C

6

Mathematics SPM Chapter 6

© Penerbitan Pelangi Sdn. Bhd.

11.

JK

L

M

N

P

Q

40°

108°36°

x°y°

x° = 180° − 40° x = 140

∠KPN = (5 − 2) × 180°––––5

= 108°

∠MPN = 180° − 108°–––––––––––2

= 36° y° = 108° − 36° y = 72∴ x + y = 140 + 72 = 212

Answer: D

12.

40°40°

100°

P

V

U

Q

R S

T

∠RQU = (5 − 2) × 180°––––5

= 108°∠PQU = 180° − 80° = 100° ∴ x° = 360° − 100° − 108° x = 152

Answer: D

13.

R

95°

108°

72°108°

S

T

UV

W

P

Q Z

∠PQZ = (5 − 2) × 180°––––5

= 108°∠RQZ = 180° − 108° = 72°∠QZS = 360° − 108° − 108° = 144°

y° + 95° + 72° + 144° = 360° y° + 311° = 360° y° = 360° − 311° y = 49Answer: A

14.

NQ

R

S

E

45°

60° 135°120°

FG

H

JK

L

M P

∠PML = (6 − 2) × 180°6

= 120°∠PMN = 180° − 120° = 60°

∠HPQ = (8 − 2) × 180°8

= 135°∠PQN = 180° − 135° = 45°∠MPQ = 360° − 120° − 135° = 105°

y° + 60° + 105° + 45° = 360° y° + 210° = 360° y° = 360° − 210° y = 150

Answer: C

7

Mathematics SPM Chapter 6

© Penerbitan Pelangi Sdn. Bhd.

15.

x°H

30°

60°J K

L

EF

G

∠FGJ = (6 − 2) × 180°6

= 120°

∠FGE = 180° − 120°2

= 30° ∠EGJ = 120° − 30° = 90° ∠JGH = 180° − 90° = 90° y° = 180° − 30° y = 150 x° = 180° − 60° − 90° x = 30∴ x + y = 30 + 150 = 180

Answer: B

16.

QR

X

S

T

P

∠XRS = ∠XSR

= 360°9

= 40° ∠RXS = 180° − 40° – 40° = 100°Answer: C

17. (n – 2) × 180°n

= 8 × 360°n

n – 2 = 16 n = 18The number of sides of the regular polygon = 18

Answer: B