Page 22 - Modul A+1 Matematik Tambahan Tingkatan 5

π
BAB 1 RQ = (8) = 4π
2
Perimeter kawasan berlorek
Kertas 2/Paper 2
Perimeter of the shaded region
Bahagian A/Section A = 12 – ! 80 + 13.32 + 4π
1. (a) 3.4 = 8.6q = 28.94 cm
q = 3.4 (c) Luas sektor STQ/Area of sector STQ
8.6 1
2
= 0.4 rad = 2 (12) (1.11)
180° ©PAN ASIA PUBLICATIONS
∴∠POQ = 0.4 rad = 79.92 cm 2
1
(b) tan 50° = SR Luas/Area ∆ = (4)(8) = 16 cm 2
OR 2
SR = 8.6 tan 50° 1 2 2
= 10.25 cm Luas/Area ROQ = π(8) = 16π cm
4
Luas kawasan berlorek Luas kawasan berlorek
Area of the shaded region Area of the shaded region
1
1
= 1 (8.6)(10.25) – (8.6) sin 50° + (8.6) (0.4) = 79.92 – 16 – 16π
2
2
2 2 2 = 13.65 cm 2
= 30.54 cm 2 7.5
2. (a) 1.2 rad = 68.75° 5. (a) kos/cos q = 9
tan 68.75° = QR q = 33.56°
2 q = 0.586 rad
QR = 2 tan 68.75° (b) Perimeter kawasan berlorek
QR = 5.14 m Perimeter of the shaded region
kos/cos 68.75° = 2 = 9 + 6 + 15(0.586)
OP = 23.79 cm
OP = 2
kos/cos 68.75° (c) Luas kawasan berlorek
OP = 5.52 m Area of the shaded region
1
Perimeter pagar = 1 (15) (0.586) – (15)(9) sin 33.56°
2
Perimeter of the fence 2 2
= 5.14 + 5.52 + 10 + 8 = 28.61 cm 2
= 28.66 m 3
(b) Luas kawasan yang tidak ditanam dengan sayur 6. (a) sin ∠ROP = 6
Area of the area not planted with vegetables ∠ROP = 30°
1
= 1 (12) (1.2) – (10 + 8)(5.14) 30° × π = π
2
2 2 180° 6
= 40.14 m 2 ∠AOR = q
q = π – π
Bahagian B/Section B 6
q = 2.62 rad
3. (a) q = 360° – 2 × 60° 2 2
q = 240° (b) OP = ! 6 – 3
π
q = 240° × 180° = 3! 3 cm
q = 4π rad Perimeter kawasan berlorek
Perimeter of the shaded region
3
π
π
( )
+ 3
j
(b) 1 2 4π = 200π = 6 ( ) ( ) + 3 – (6 – 3! 3)
6
2
2 3 3
2
j = 100 = 10.05 cm
j = 10 cm (c) Luas sektor ROP/Area of sector ROP
( )
(c) Perimeter sektor berlorek = 1 (6) 2 π = 3π cm 2
Perimeter of the shaded sector 2 6
4π
1
= 10 + 10 + 10 ( ) Luas/Area ∆ROP = (3! 3)(3)
3
2
= 61.89 cm 9! 3
(d) ∠AOB = 120° = 2 cm 2
120° × π = 2π Luas sektor RPQ/Area of sector RPQ
3
( )
Luas tembereng ADBC/Area of the segment ADBC = 1 (3) 2 π
( )
1
= 1 (10) 2 2π – (10) sin 120° 2 2
2
2 3 2 = 9 π cm 2
4
= 61.42 cm 2 Luas kawasan berlorek
8
4. (a) tan q = Area of the shaded region

4
9! 3
q = 63.43° = 9 π – (3π – )
q = 1.11 rad 4 2
= 5.44 cm 2
2
2
(b) TR = ! 8 + 4 = ! 80
SQ = 12(1.11) = 13.32
MG-1
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