06[a math cd]
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1 © Penerbitan Pelangi Sdn. Bhd.
Paper 1
1.
E F
G
x°
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.
p°
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.
m°
n°
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°
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.
x°
54°
P Q
R
S
T
y°
∠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°
x°
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°
x°
∠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°
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.
q°
130°74°
82°
M
H G
FE
K
J
L
50°
286°
p°
∠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
x°
∠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.
y°
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°
n°
m°
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.
p°
p°
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.
m°
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°
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°
x°
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.
y°
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.
y°
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.
y°
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