Skema Jawapan - Maths Ting 3 Buku Latihan Mastery Kssm 2019 .pdf

Jawapan Bab 1 Indeks Indices

1.1 Tatatanda Indeks

Index Notation

1. base

2. (a) 8

7

indeks

an

asas



index

(b) 0.5

6



(c) (–h)

5

3 4 (b) (–9)6 (c) 152

3. (a) 75 4. (a) 81 = 3 × 3 × 3 × 3 = 34 3 3 3 3

81 27  9  3  1

(c) 32 = 2 × 2 × 2 × 2 × 2 = 25 2 2 2 2 2

32 16  8  4  2  1

5. (a) 27 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 128 3 (c) 1 4 2 = 4 × 4 × 4 5 5 5 5 64 = 125

(b) 256 = 4 × 4 × 4 × 4 = 44 4 4 4 4

256  64  16  4  1

(d) 625 = 5 × 5 × 5 × 5 = 54 5 5 5 5

625 125  25  5  1

(b) (–3)5 = (–3) × (–3) × (–3) × (–3) × (–3) = –243 (d) 0.64 = 0.6 × 0.6 × 0.6 × 0.6 = 0.1296

1

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 1

1.2 Hukum Indeks

Index Laws

6. (a) 3

2



6

3

×3

4



(b) p

4

p

7

×p

3



(c) a

m+n



7. 62 × 613 63 × 612

615

Any other possible answers

65 × 610

6 ×6 4

11

8. (a) 47 + 8 = 415 (d) 91 + 3 + 10 = 914

9. (a)

p

8

p

4

Mana-mana jawapan lain yang sesuai

(b) 65 + 5 = 610

(c) p6 + 12 = p18

(e) q2 + 4 + 1 = q7

(f )

(2 × 3) × (y3 × y × y4) = 6 × y3 + 1 + 4 = 6 × y8 = 6y8

6

3 m–n 2 a (b) = 2 (c) 3 2

10. 710 ÷ 75 78 ÷ 73

75

7 ÷7 9

11. (a) 86 – 2 = 84 (d) 615 – 5 = 610

2. (a) (49) 1

5



© Penerbitan Pelangi Sdn. Bhd.

Mana-mana jawapan lain yang sesuai

4

Any other possible answers

76 ÷ 7

(b) 411 – 5 = 46

(c) 318 – 9 = 39

(e) x16 – 4 = x12

7 (f ) 14y 2y = 7y7 – 1 = 7y6

(b) (2p)

7

mn

(c) a

2

Matematik  Tingkatan 3  Bab 1

13. (a) 34 × 3 = 312

(b) 64 × 5 = 620

(c) 92 × 8 = 916

(d) k3 × 6 = k18

(e) x 11 × 2 = x22

(f ) 21 × 4 × y3 × 4 = 16y12

14. (a)

(b) 3 (c) 3 (d)

4–1

15.



as

6–2

1 4

9–5

as

1 62

16. (a) 3 5

=1÷ 3 5 = 5 3

as

1 106

(b) 1–3 = 6

= 1 3 5



p–7

1 1 6

1 2

= 4–6

1 (2x)3 1 1 x

(d) 4 7

1 2

5

1 2



(c) 5 = 1 ÷ 9 9 5 = 1 9 5 –1 = 9 5

1 4 7

–2

1 2

5

=1÷ 1 x = x5

3

(b) 32 = 1 ÷ 12 3 = 1 1 32 = 1–2 3

1 2

(2x)–3

(c) 1–5 = x

3

1 2



as

1 p7

=1÷ 1 6 = 63



6 17. (a) 1 = 16 4 4



as

1 95

–1

1 2

10–6

=

2

1 2 = 1 ÷ 142 7

2



= 7 4

2

1 2

8 8 (d) p = p8 q q

1 2

1 2



8 = 1 ÷ q8 p



8 =1÷ q   p

1 2

=

1 q p

8

1 2

= q p

–8

1 2

18. 25

1 2

√25

as

64 3

1 3

√64

as

32 5

1 5

√32

as

256

1 4

h

as

4

7

√256

1 7

√h

3

as

k

1 10

10

√k

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 1

20. n

n

√am

( √a )m

3

√642

64

4

√813

81

√363

36

9

√x4 

x



2 3

3

(√64 )2

3 4

( √81 )3 4

3 2

(√36 )3

4 9

9

( √x  )4

21. (am)

1 n

1 n

(a )m

1

3 4

(16 )

(495)

1 2

16

49

(125 )

(y2)

1 7

125

y

(16 )

1 2

5 2

1

7 3

1 4 3

3 4

(49 )5

1 3 7

7 3

(125 )

1 7

2 7

(y )2



© Penerbitan Pelangi Sdn. Bhd.

4

Matematik  Tingkatan 3  Bab 1 1

1

22. (a) 27 3 = 3√27 =3

1

4

(b) 16 4 = √16 =2

3

3

(d) 9 2 = (√9 )3 = 33 = 27

5

4

(e) 81 4 = ( √81)3 = 33 = 27

4×3 1×3 23. (a) 3 × 4 12 = 3 × 43 4 5 (e) 63 7 4×5 = 63 × 5 7 20 = 615 7

1 2

5

(c) 243 5 = √243 =3 6

(f ) 729 6 = (√729 )5 = 35 = 243

(b) h–3 × 6 × k7 × 6 = h–18k42

(c) 95 × 2 × 83 × 2 × 4­6 × 2 = 910 × 86 × 412

3×4 (f ) 22 × 4 5 12 = 2 8 5

7×6 (g) x1 × 6 y 42 = x 6 y 

(d) p3 × 6 × q4 × 6 × r1 × 6 = p18q24r 6 (h)

m 1 2n 2 4

2

8

4×2 = m 2 8×2 (2 )n 8 = m16 4n

1

1

(b) 4p = 64 4p – 3 3 4p = 4p – 3 4 p 4 = 43 – (p – 3) p = 3 – p + 3 2p = 6 p = 3

24. (a) (32 5 × 81 4 )2 ÷ 3–1

= 32

1 ×2 5 2 5

× 81

1 ×2 4

÷ 3–1­

1 2

= 32 × 81 ÷ 3–1 5×

2 5

4×

1 2

=2 ×3 ÷ 3–1 = 22 × 32 ÷ 3–1 = 4 × 32 – (–1) = 4 × 33 = 108

(d) 1 × 625(6x – 13) = 1252x 25 1 × 54(6x – 13) = 53(2x) 52 5–2 × 524x – 52 = 56x 5–2 + 24x – 52 = 56x   24x – 54 = 6x 18x = 54 x = 3

(c) 40m17n14 ÷ 8m14n10 = 5mxny 40 m17 – 14n14 – 10 = 5mxny 8 5m3n4 = 5mxny Maka/ Hence, x = 3, y = 4

5

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 1

Praktis PT3 1. (a)

(ii) 3 = 81y 3 = (34)y 3 = 34 × y 31 = 34y 1 = 4y 1 y = 4

B

A =

(p3)4

p12

1

=

p3 × p4

=

4

√p3

(c) p7

1

3

p4

(b) (i) 63x9 ÷ 9x –4 63 = x9 – (–4) 9 = 7x13

3

3. (a) (i) q

(ii) 43 × 64–2 = 8x (2 ) × (26)–2 = (23)x 26 × 2–12 = 23x 26 + (–12) = 23x 6 – 12 = 3x x = –2

(ii)

2 3

8

1 16 2

5

0

(iii) (–10)

(b) (i) (m–2)3 × m7 = m–2 × 3 × m7 = m–6 × m7 = m–6 + 7 = m

1 4 ×44 2 = 4 096 x+1 3

2

(4

) = 46 (4 ) = 46 43x = 46 3x = 6 x = 2

1+x+1–2 3 x 3

3

49 2

(ii)

1

343 3 × 7–2 3

=

2. (a) (i) Benar

(72) 2 1

(73) 3 × 7–2

True

2×

(ii)

=

Palsu False



(iii) Benar True

2

(b) (i) 11–2 × 1 331 3 = 11–2 × (3√1 331)2 = 11–2 × 112 = 11–2 + 2 = 110 =1

© Penerbitan Pelangi Sdn. Bhd.

6×

2 –2 7 n = 4n s × 16 (23)n3s3 –2 7 3 2 s = 4n s ×3 4n 3 8n s = 2n–2 + 3 – 3 s7 + 2 – 3 = 2n–2 s6 6 = 2s2 n



(c)

4n–2s7 × (16n6s4) 2 (2ns)3

6

7 3×

1

3 2

7 3 × 7–2 = 73 – 1 – (–2)   = 74 = 2 401

1 1 4× 2 2

s

Matematik  Tingkatan 3  Bab 1 1

3 5)2 4. 15 × (32 × 2 2 400 153 × (32 × 5)2 = (15 × 32 × 5)2 153 × 322 × 52 = 152 × 322 × 52 = 153 – 2 × 322 – 2 × 52 – 2 = 15

–3­ 2 2 2 (c) (3m n ) × √36mn 3

27m 2 n–7 1

1

×2

2 –3 × 2­ 2 n × √36m 2 n = (3 )m 3 27m 2 n–7

2×

1 2

1

2 –6­ = 9m n ×3 6m n 27m 2 n–7 1

3

–6 + – = 9 × 6 m 2 2 n1 + 1 – (–7) 27

5. 16p6 × 4p ÷ p4 = 644 64p6 + 1 – 4 = 644 64p3 = 644 64p3 = 64(64)3

= 2m–7 n9 9 = 2n7 m

Secara perbandingan / By comparison, p = 64 = 82

Cabaran KBAT 1.

1 3

3 2

6. 41, 42, 43, 44, 45, 46, … 4, 16, 64, 256, 1 024, 4 096, …

27 ÷ 9 = x2 1

3

(33) 3 ÷ (32) 2 = x2 3×

1

2×

3

Pola yang berulang bagi digit terakhir ialah 4, 6. Indeks, 100, ialah satu daripada setiap ulangan untuk nombor yang kedua. Digit sa bagi 4100 ialah 6.

3 3 ÷ 3 2 = x2 3 ÷ 33 = x2 31 – 3 = x2 3–2 = x2 1 = x2 32 x = 1 , – 1 3 3

The repeating pattern for last digit is 4, 6. The index, 100, is one from each repeat for the second number. The units digit for 4100 is 6.

2. 53y – 1 = (52y)(25) 53y – 1 = (52y)(52) 53y – 1 = 52y + 2 3y – 1 = 2y + 2 y = 3

1

2×

3. 9 a = (9 1

8

4 5

1

× 92

×

2

4 5

7

) ÷ (9 2

×3

× 9–3 × 3)

21

9 a = (9 5 × 9 5 ) ÷ (9 2 × 9–9) 1

8

1

1

9a = 95

+

2 21 – (–9) – 5 2

9a = 92 1 = 1 a 2 a = 2

7

© Penerbitan Pelangi Sdn. Bhd.

Jawapan Bab 2 Bentuk Piawai

Standard Form

2.1 Angka Bererti

Significant Numbers

1. (a) 23.5 2.

(b) 23. 5

(c) 3

(a) 2 angka bererti

(b) 4 angka bererti

(c) 3 angka bererti

(d) 4 angka bererti

2 significant figures

4 significant figures

3 significant figures

4 significant figures

(e) 3 angka bererti

(f ) 4 angka bererti

(g) 5 angka bererti

(h) 4 angka bererti

3 significant figures

4 significant figures

5 significant figures

4 significant figures

(i) 1 angka bererti

(j) 1 angka bererti

(k) 2 angka bererti

(l) 5 angka bererti

1 significant figure

1 significant figure

2 significant figures

5 significant figures

(m) 1 angka bererti

(n) 3 angka bererti

(o) 2 angka bererti

(p) 3 angka bererti

1 significant figure

3 significant figures

2 significant figures

3 significant figures

(q) 3 angka bererti

(r) 5 angka bererti

(s) 2 angka bererti

(t) 5 angka bererti

3 significant figures

5 significant figures

2 significant figures

5 significant figures

3.

Angka bererti (a.b.) Significant figures (s.f.)

1 a.b.

2 a.b.

3 a.b.

4 a.b.

5 a.b.

200

0.0017

0.00670

56.33

33.300

0.1

380

1.08

3 007

56.012

6 000

4.0

5 090

80.00

22 506

0.04

0.012

0.210

9.045

540.09

1 s.f.

2 s.f.

3 s.f.

4 s.f.

5 s.f.



1

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 2

4. (a) 3 000,  2 600,  2 650 (c) 5 000,  4 500,  4 510 (e) 20 000,  22 000,  22 100

(b) 1 000,  1 200,  1 160 (d) 10 000,  12 000,  11 600 (f ) 30 000,  31 000,  30 800

(a) 30 (b) 20 (c) 32.9 (d) 70 020 5. (e) 0.002 (f ) 6.0 (g) 20 000 (h) 572 (i) 200 000 (j) 7.235 (k) 850 900 (l) 3.5 2.2 Bentuk Piawai

Standard Form

6. (a) 3 (b) 7 (c) 7 (d) 3 (e) 7 (f ) 7 7. (b) 5 × 1 000 = 5 × 103

(c) 2.3 × 10 000 = 2.3 × 104

(d) 5.623 × 1 000 = 5.623 × 103

(e) 7.819 × 100 000 = 7.819 × 105

(f ) 1.4 × 1 000 000 = 1.4 × 106

(h) 4.2 × 0.001 = 4.2 × 10–3

(i) 1.873 × 0.01 = 1.873 × 10–2

(j) 3.38 × 0.00001 = 3.38 × 10–5

(k) 5.2 × 0.0001 = 5.2 × 10–4

(l) 6.8 × 0.00001 = 6.8 × 10–5

8. (b) 8 × 100 000 = 800 000

(c) 9.3 × 1 000 000 = 9 300 000

(d) 1.82 × 10 000 000 = 18 200 000

(e) 7.265 × 100 000 000 = 726 500 000

(f ) 4.123 × 1 000 000 000 = 4 123 000 000

(h) 3 × 0.01 = 0.03

(i) 1.9 × 0.001 = 0.0019

(j) 2.7 × 0.0001 = 0.00027

(k) 3.023 × 0.00001 = 0.00003023

(b) 10–3 (e) 10–2

(c) 103 (f ) 1012

(l) 5.36 × 0.000001 = 0.00000536 9. (a) 106 (d) 10–9 10. (a) (3.4 + 6.1) × 104 = 9.5 × 104

(b) (8.9 – 1.2) × 103 = 7.7 × 103

(c) 0.93 × 10–4 + 2.13 × 10–4 = (0.93 + 2.13) × 10–4 = 3.06 × 10–4

(d) 4.05 × 10–7 – 0.22 × 10–7 = (4.05 – 0.22) × 10–7 = 3.83 × 10–7

(e) 3.5 × 1.6 × 104 + 4 = 5.6 × 108

(f ) 2.4 × 1.8 × 107 + (–3) = 4.32 × 104

(g) 7.2 × 106 × 1.5 × 10–4 = 7.2 × 1.5 × 106 + (–4) = 10.8 × 102 = 1.08 × 10 × 102 = 1.08 × 101 + 2 = 1.08 × 103

(h) 1.2 × 105 – 5 6 = 0.2 ×100 = 2 × 10–1

© Penerbitan Pelangi Sdn. Bhd.

2

100 = 1

Matematik  Tingkatan 3  Bab 2 –9 (i) 1.8 × 10 8 10 5 = 0.36 × 10–9 – 8 = 3.6 × 10–1 × 10–17 = 3.6 × 10–1 + (–17) = 3.6 × 10–18

4.8 × 103 2.5 × 106 3 = 4.8 × 106 2.5 10 = 1.92 × 103 – 6 = 1.92 × 10–3 (j)

11. (a) Ketebalan sehelai kertas

Thickness of a sheet of paper



= 55 ÷ 800 = 0.06875 = 6.875 × 10–2 mm

(b) 3 hari = 3 × 24 × 60 = 4 320 = 4.32 × 103 minit

3 days = 3 × 24 × 60 = 4 320 = 4.32 × 103 minutes

5 Jarak/ Distance = 3.84 × 103 4.32 × 10 = 3.84 × 105 – 3 4.32 = 0.889 × 102 = 8.89 × 10–1 × 102 = 8.89 × 10 km (c) (i) 2(3.2 × 106) + 2(5 × 105) = 6.4 × 106 + 10 × 105 = 6.4 × 106 + 1 × 101 + 5 = 6.4 × 106 + 1 × 106 = (6.4 + 1) × 106 = 7.4 × 106 m

(ii)

3.2 × 106 × 5 × 105 = 3.2 × 5 × 106 × 105 = 16 × 106 + 5 = 16 × 1011 = 1.6 × 10 × 1011 = 1.6 × 101 + 11 = 1.6 × 1012 m2

Praktis PT3 1. (a) Nombor Number

(i)

3 560

Underlined digit

Bererti

Tidak bererti

Bererti

Tidak bererti

Bererti

Tidak bererti

Significant

(ii)

0.270

Significant

(iii) (c)

8.401

(b) (i) 0.00069 = 6.9 × 10–4 Maka/ Hence, p = 6.9, q = –4

Digit bergaris

Significant

0.04 8 000 000 –2 = 4 × 10 6 8 × 10 = 0.5 × 10–2 – 6 = 5 × 10–1 × 10–8 = 5 × 10–9 (ii)

Not significant

Not significant

Not significant

6.48 × 1011 – 3.34 × 108 = 6.48 × 1011 – 0.00334 × 1011 = (6.48 – 0.00334) × 1011 = 6.47666 × 1011 = 6.48 × 1011 sejam

  (6.48 × 1011 per hour)

3

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 2

2. (a) 3  

3

(b) (i) Isi padu kotak / Volume of box = 0.00615 = 6.15 × 10–3 m3 (ii)

represent

10n

mega

mikro as

106



micro

as

10–6



Volume of water

= 400 × 200 × 600 = 4 × 102 × 2 × 102 × 6 × 102 = 4 × 2 × 6 × 102 + 2 + 2 = 48 × 106 = 4.8 ×10 × 106 = 4.8 × 101 + 6 = 4.8 ×107 cm3

2 2 = 27 × 10 × 21 × 10 30 × 30

27 × 21 × 102 + 2 30 × 30 = 0.63 × 104 = 6.3 × 10–1 × 104 = 6.3 × 103

=

tera 1012

as

nano 10–9 (ii)

P+Q = 1.3 × 103 + 4.5 × 102 = 1.3 × 103 + 0.45 × 10 × 102 = 1.3 × 103 + 0.45 × 101 + 2 = 1.3 × 103 + 0.45 × 103 = (1.3 + 0.45) × 103 = 1.75 × 103

(c) Isi padu air

Number of tiles required

0.00000036 – 2.4 × 10–8 = 3.6 × 10–7 – 2.4 × 10–8 = 3.6 × 10–7 – 0.24 × 10 × 10–8 = 3.6 × 10–7 – 0.24 × 101 + (–8) = 3.6 × 10–7 – 0.24 × 10–7 = (3.6 – 0.24) × 10–7 = 3.36 × 10–7

3. (a) mewakili

(b) (i)

(c) Bilangan jubin yang diperlukan



P–Q = 1.3 × 103 – 4.5 × 102 = 13 × 10–1 × 103 – 4.5 × 102 = 13 × 10–1 + 3 – 4.5 × 102 = 13 × 102 – 4.5 × 102 = (13 – 4.5) × 102 = 8.5 × 102

Kaedah Alternatif Isi padu air

Volume of water

= 400 × 200 × 600 = 48 000 000 = 4.8 ×107 cm3

Cabaran KBAT 1. (a) Diameter bakteria jenis B Diameter of bacteria species B

2.



= 1 × 10–6 × 1 × 10–3 = 1 × 10–6 + (–3) = 1 × 10–9 m

(b) 1 × 10–9 m = 1 nm Nilai yang diperoleh itu sama dengan 1 nanometer. The value obtained is equal to 1 nanometre.

Jarak = Laju × Masa

Distance = Speed × Time

= 100 × 717 × 365 × 24 = 628 092 000 = 6.28 × 108 km

© Penerbitan Pelangi Sdn. Bhd.

4

Jawapan Bab 3 Matematik Pengguna: Simpanan dan Pelaburan, Kredit dan Hutang

Consumer Mathematics: Savings and Investments, Credit and Debt

3.1 Simpanan dan Pelaburan

Savings and Investments

1. Saham Shares

Akaun semasa

Akaun simpanan

Bon

Hartanah

Current account

Amanah saham

Savings account

Bond

Unit trust

Real estate

2. (a) Faedah mudah Simple interest

(b) Faedah kompaun Compound interest

(c) Prinsipal Principal

(d) tempoh, kadar faedah, prinsipal, kekerapan pengkompaunan period, interest rate, principal, compounding frequency

3. Faedah dikompaun setiap • setengah tahun

•

MV = P 1 + r 4

2

Faedah dikompaun setiap • suku tahun

•

MV = P 1 + r 12

2

Faedah dikompaun bulanan

•

MV = P 1 + r 2

1

Interest compounded half yearly

Interest compounded quarterly

•

1

1

Interest compounded monthly

4. (a) Jumlah simpanan Total savings



= prinsipal + faedah





= 8 000 + 8 000 × 2 × 2 100 = RM8 320



  principal + interest



12t

2t

2

(b) Nilai matang



1

4t

2



1

Matured value

= 15 000 1 + 0.035 12 = RM17 864.14

1

12(5)

2

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 3

(d) Nilai matang

(c) Jumlah simpanan

Total savings





= prinsipal + faedah



= 5 000 + 5 000 × 2.5 × 10 100 12 = RM5 104.17



  principal + interest



1



2

5. (a) Tinggi

Matured value

= 20 000 1 + 0.04 1 = RM32 020.64

1

(b) Rendah

1(12)

2

(c) Tinggi

Higher Lower Higher

6. (a) ROI = (8 800 + 200) – 8 000 × 100  % 8 000 = 12.5%

3

(b) ROI =

4

350) – 10 000 3 (10 400 + 210×000

4

× 100  %

= 11% (c) (i) ROI =

– 4 000 × 100  % 1 4 8004 000 2

= 20% (ii) Kejadian pencemaran udara atau jerebu menyebabkan permintaan topeng muka bertambah. Jadi, harga topeng muka akan bertambah dan pulangan Encik Tan akan bertambah.

The occurrence of air pollution and haze increases masks’ demand. Consequently, mask’s price will hike and this will cause Encik Tan’s return to increase.

7. (a) ( 3 )

(b) ( 7 )

(c) ( 3 )

8. Rendah

Sederhana

Tinggi

akaun simpanan tetap

hartanah

saham

hartanah

saham

akaun simpanan tetap

Low

(a) Risiko Risk

(b) Kecairan Liquidity

9. (a) rendah

Medium

fixed deposit account

real estate

real estate

shares

(b) tinggi

(c) tinggi

High

shares

fixed deposit account

(d) rendah

lower higher higher lower

10. (a) Purata kos sesyer/ Average cost per share 7 000(2.10) + 5 000(2.04) = 7 000 + 5 000 = 2.075 8 000 Jumlah unit pembelian/ Total units acquired Jumlah unit pembelian = 8 000 3.20 Total units acquired = 2 500 (b)

© Penerbitan Pelangi Sdn. Bhd.

3.20 =

2

Matematik  Tingkatan 3  Bab 3

(c) (i) Purata kos sesyer/ Average cost per share 3 000(1.25) + 2 000(1.24) + 5 000(1.22) = 3 000 + 2 000 + 5 000 = 1.233 (ii) Strategi pemurataan kos ringgit digunakan. Strategi ini dapat mengurangkan risiko pelaburan dengan mengurangkan purata harga pembelian.



Ringgit cost averaging strategy is used. This strategy helps in reducing investment risk by bringing down average purchasing cost.

(d) (i) Purata kos sesyer

Average cost per share

= 12 × 200 3 105.62 = 0.7728 (ii) Jumlah unit yang dibeli

Total units purchased

= 2 400 0.85 = 2 823.53 (iii) Pembelian secara berasingan dan berturutan lebih bermanfaat kepada Jasraj. Hal ini demikian kerana Jasraj mendapat purata harga pembelian yang lebih rendah dan jumlah unit yang lebih banyak berbanding dengan pembelian sekali gus.



Separately and continually purchase is more beneficial for Jasraj. This is because Jasraj manages to get a lower average purchase price and hold more units when compared to lump sum purchase.

11. (a) (i) MV = 25 000 1 + 0.038 2 = RM25 959.03

1



Separuh simpanan Encik Subra telah dikeluarkan.

2(1)

2

Half of Encik Subra’s savings has uplifted.

Nilai pulangan pelaburan bagi simpanan tetap Return of investment of fixed deposit

= 25 959.03 – 25 000 × 100 % 25 000 = 3.84%

1

2

(ii) Purata kos sesyer



Average cost per share

25 000 6 000 + 5 000 = 2.2727

=

(iii) Nilai pulangan pelaburan bagi saham



Return of investment of share



11 000(2.45) – 25 000 × 100 % 25 000 = 7.8%



Pelaburan saham lebih menguntungkan kerana nilai pulangan pelaburannya lebih tinggi.

=



3

4

Jumlah unit dijual Total units sold

= 6 000 + 5 000 = 11 000

Share investment is more profitable because it has higher return of investment.

3

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 3

(b) (i) MV = 15 000 1 + 0.036 12 = RM15 549

1

12(1)

2

(ii) Nilai pulangan pelaburan bagi simpanan tetap Return of investment of fixed deposit



= 15 549 – 15 000 × 100 % 15 000 = 3.66%

1

2

Katakan x = harga jualan per unit bagi amanah saham

Let x = selling price per unit of the unit trust



2(3.66) = 20 000x – 15 000 × 100 15 000 1 098 = 20 000x − 15 000 20 000x = 16 098 x = 0.8049

3.2 Pengurusan Kredit dan Hutang

Credit and Debt Management

12. (a) B

(b) P

(c) B

(d) B

T F T T

13. A

C

Pengurusan kredit dan hutang yang bijaksana Smart management of credit and debt

E 14. Memudahkan proses pembelian dalam talian Ease online purchase process

Kelebihan Benefits

Mudah terbelanja melebihi kemampuan Easy to spend beyond affordability

Boleh dikenakan faedah dan caj-caj lain Interest and other charges could be imposed

Kelemahan Weaknesses

Memberi tempoh bayar balik tanpa faedah Provide interest free period for repayment



15. (a) baki belum jelas

(b) bayaran minimum

outstanding balance

© Penerbitan Pelangi Sdn. Bhd.

minimum payment

4

Matematik  Tingkatan 3  Bab 3

16. (a) Caj kewangan/ Finance charge = 1 500 × 18% 12 0.18 = 1 500 × 12 = RM22.50

1 tahun = 12 bulan 1 year = 12 months

Caj bayaran lewat/ Late payment charge = 1 500 × 1% = 1 500 × 0.01 = RM15.00 (b) Baki tertunggak/ Outstanding balance = 2 000 – 100 = RM1 900 Caj kewangan/ Finance charge = 1 900 × 1.5% = RM28.50

Caj bayaran lewat/ Late payment charge = 0



Jumlah baki tertunggak pada bulan depan



= 1 900 + 28.50 + 0 = RM1 928.50



Total outstanding balance on next month

(c) (i) Caj kewangan/ Finance charge = 3 500 × 18% 12 = RM52.50 Caj bayaran lewat/ Late payment charge = 3 500 × 1% = RM35.00

Jumlah baki tertunggak pada bulan depan



= 3 500 + 52.50 + 35 = RM3 587.50



Total oustanding balance on next month

(ii) Beza/ Difference = 3 587.50 – 3500 = RM87.50 (iii) Kebaikan/ Benefit: Sharmila tidak perlu membayar dengan tunai.

Sharmila does not need to pay in cash.

Kelemahan/ Weakness: Sharmila terpaksa menanggung faedah yang lebih tinggi apabila dia tidak dapat membuat bayaran balik.

Sharmila bears higher interest when she can’t make a repayment.

17. Pinjaman peribadi Personal loan

Pinjaman kereta Car loan

Faedah sama rata Flat interest

Pinjaman barangan pengguna Consumer goods loan

Pinjaman perumahan Housing loan

5

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 3

18. (a) Prinsipal, P/ Principal, P = 80 000 − 8 000 = RM72 000

(b) Jumlah bayaran balik/ Total repayment = 100 000 + (100 000 × 0.05 × 9) = RM145 000

Jumlah bayaran balik/ Total repayment = P + Prt = 72 000 + (72 000 × 0.04 × 8) = RM95 040

Bayaran ansuran/ Instalment = 145 000 9 × 12 = RM1 342.59

1 9. (a) (i) Jumlah bayaran balik/ Total repayment = 40 000 + (40 000 × 0.055 × 6) = RM53 200 53 200 = RM738.89 Bayaran ansuran/ Instalment = 6 × 12 (ii) Jumlah bayaran balik/ Total repayment = 40 000 + (40 000 × 0.055 × 5) = RM51 000 51 000 = RM850 Bayaran ansuran/ Instalment = 5 × 12

Wang yang perlu ditambah/ Money to be added = 850 – 738.89 = RM111.11

(b) (i) Jumlah bayaran balik/ Total repayment = 60 000 + (60 000 × 0.06 × 8) = RM88 800 88 800 Bayaran ansuran/ Instalment = 8 × 12 = RM925 (ii) Katakan x = tempoh bayaran balik baru dalam tahun

Let x = new loan repayment period in years

925 + 175 = 60 000 + (60 000 × 0.06 × x) x × 12 13 200x = 60 000 + 3 600x 9 600x = 60 000 x = 6.25 (c) (i) 1 × 5 400 = Jumlah bayaran balik/ Total repayment 4 8 × 12 Jumlah bayaran balik/ Total repayment 1 350 = 96 Jumlah bayaran balik/ Total repayment = RM129 600 Katakan x = kadar faedah tahunan



Let x = yearly interest rate

129 600 = 90 000 + (90 000 × x × 8) 720 000x = 39 600 x = 0.055 = 5.5%

© Penerbitan Pelangi Sdn. Bhd.

6

Matematik  Tingkatan 3  Bab 3

(ii) Jumlah bayaran balik/ Total repayment = 90 000 + (90 000 × 0.045 × 7) = RM118 350 Bayaran ansuran/ Instalment = 118 350 = RM1 408.93 7 × 12

Tidak boleh. Hal ini demikian kerana bayaran ansuran bulanan melebihi



Cannot. This is because monthly instalment exceeds 1 of his monthly income. 4

1 daripada gaji bulanannya. 4

Praktis PT3 1. (a) (i) Faedah mudah

Simple interest

(ii) Faedah sama rata Flat interest

(iii) risiko risks

(b) Katakan t = masa/ Let t = time, 2(5 000) = 5 000 + (5 000)(0.02)(t) 10 000 = 5 000 + 100t 100t = 5 000 t = 50 tahun/ 50 years (c) (i) Jumlah bayaran balik/ Total repayment = 80 000 + (80 000 × 0.028 × 8) = RM97 920

Bayaran ansuran/ Instalment 97 920 = 8 × 12 = RM1 020 (ii) 940 = Jumlah bayaran balik 7 × 12 940 =



Total repayment 7 × 12

Jumlah bayaran balik/ Total repayment = RM78 960

Katakan x = pinjaman yang boleh dibuat, Let x = the loan amount that can be made,



78 960 = x + (x × 0.028 × 7) 78 960 = 1.196x x = RM66 020.07

2. (a) (i) 3 (ii) 3 (iii) 7 (b) Jumlah bayaran balik/ Total repayment = 1 576.40 × 12 × 10 = RM189 168 Katakan r = kadar faedah tahunan, Let r =  yearly interest rate,



189 168 = 150 000 + 150 000(r)(10) 189 168 – 150 000 r = 150 000(10) = 0.026 = 2.6%

7

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 3

(c) (i) Caj kewangan/ Finance charge = 2 200 × 18% 12 = RM33

Caj bayaran lewat/ Late payment charge = 2 200 × 1% = RM22



Jumlah baki tertunggak



= 2 200 + 33 + 22 = RM2 255

Total outstanding balance

(ii) Beza/ Difference = 2 255 – 2 200 = RM55 Harga pembelian Daniel lebih murah. Daniel’s purchase price is cheaper.

3. (a) Mudah terbelanja melebihi kemampuan

Easy to spend exceeding ability

Mudah digunakan untuk pembelian tanpa tunai Easy to use for cashless payment

Kelemahan Weakness

Memberi perkhidmatan bayar balik tanpa faedah

Boleh dikenakan faedah dan caj-caj lain Could be imposed interest and other charges

Provide interest free repayment service



(b) Nilai matang/ Matured value 4(10) = 12 000 1 + 0.035 4 = RM17 002.91

1

2

(c) (i) Purata kos sesyer/ Average cost per share (4 000)(2.10) + (4 000)(2.06) + (6 000)(2.08) = 4 000 + 4 000 + 6 000 = 2.08 (ii) Jumlah unit dibeli/ Total of units purchased = 4 000 + 4 000 + 6 000 = 14 000

Kos pelaburan/ Cost of investment = 4 000(2.10) + 4 000(2.06) + 6 000(2.08) = 29 120

© Penerbitan Pelangi Sdn. Bhd.

8

Matematik  Tingkatan 3  Bab 3

Katakan x = harga jualan seunit, Let x = selling price per unit,

12.5 = (x × 14 000) – 29 120 × 100 29 120 14 000x − 29 120 = 3 640 14 000x = 32 760 x = 2.34

Cabaran KBAT

350 × 12 Jumlah bilangan unit / Total number of units Jumlah bilangan unit/ Total number of units = 5 340.11 x % 2(1) (ii) 4 352.56 = 4 200 1 + 2 x % 4 352.56 = 1 + 2 4 200 x % 1.018 = 1 + 2 x% = 0.036 = 3.6% Maka, / Hence, x = 3.6

1. (i) 0.7865 =

1

2

(iii) Nilai pulangan pelaburan bagi simpanan tetap Return of investment of the fixed deposit

= 4 352.56 – 4 200 × 100 % 4 200 = 3.63%

1

2

Katakan y = harga jualan seunit, / Let y = selling price per unit, 10 + 3.63 = 5 340.11y – 12(350) × 100 12(350) 572.46 = 5 340.11y − 4 200 5 340.11y = 4 772.46 y = 0.8937

2. (i) Caj kewangan/ Finance charge = 4 500 × 1.25% = RM56.25 Caj bayaran lewat/ Late payment charge = 4 500 × 1% = RM45 Jumlah tunggakan/ Outstanding amount = 4 500 + 56.25 + 45 = RM4 601.25 (ii) Nilai matang/ Matured value = 4 500 1 + 0.031 12 = RM4 511.63

1

12(

2

1 ) 12

(iii) Ya. Hal ini demikian kerana kadar faedah hutang kad kredit lebih tinggi. Yes. This is because interest rate charged on credit card debt is higher.

9

© Penerbitan Pelangi Sdn. Bhd.

Jawapan Bab 4 Lukisan Berskala

Scale Drawings

4.1 Lukisan Berskala

Scale Drawings

1. Ukuran pada lukisan

Ukuran sebenar objek

Measurement on drawing

Actual measurement of object

Tinggi

15 cm

1.5 m

15 cm : 1.5 m = 15 cm : 150 cm = 1 : 10

Lebar

18 cm

1.8 cm

18 cm : 1.8 m = 18 cm : 180 cm = 1 : 10

Height

Width

(a) sama

Ratio of the measurement on drawing to the actual measurement of object

(c) lukisan berskala

(b) berkadaran

equal



Nisbah ukuran pada lukisan kepada ukuran sebenar objek

proportional

scale drawings

2.

Objek Object

12 cm 6 cm 10 cm 5 cm 15 cm

9 cm

6 cm

4 cm

18 cm

(  3 )

(  3 )

3. 1 cm pada lukisan berskala mewakili 20 cm pada objek sebenar

1:1

1 cm on the scale drawing represents 20 cm on the real object

1 cm pada lukisan berskala mewakili 0.5 mm pada objek sebenar

1: 1 20

1 cm on the scale drawing represents 0.5 mm on the real object

1 cm pada lukisan berskala mewakili 10 mm pada objek sebenar

1 : 20

1 cm on the scale drawing represents 10 mm on the real object



1

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 4

4. Lukisan adalah lebih kecil daripada

1:2

objek.

smaller than

The drawings are the objects.

Lukisan adalah sama saiz dengan

1:1

objek.

same size as

The drawings are the objects.

Lukisan adalah lebih besar daripada

1 1: 2

objek.

larger than

The drawings are the objects.

5.

(a) Skala / Scale = 2 : 4 =1:2 (b) Skala / Scale = 84 : 21 1 =1: 4 1 14 cm = 100 000 Jarak sebenar / Actual distance Jarak sebenar / Actual distance = 14 × 100 000 = 1 400 000 cm = 14 km (c)

(d)

1 = Panjang sungai pada peta / Length of river on map 40 000 2 km



Panjang sungai pada peta / Length of river on map 2 = 40 000 = 0.00005 km = 5 cm 6. (a)

(b) 4 cm

1.5 cm



© Penerbitan Pelangi Sdn. Bhd.

2.5 cm



2

Matematik  Tingkatan 3  Bab 4

(c)



(d)



7. (a) (i)

(ii)

1 cm

0.5 cm 1 cm 0.5 cm

(b) (i)

(ii)

0.5 cm 0.5 cm

1 cm 1 cm





8. (a)





3

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 4

(b)

(i)

(ii)

(iii)

(c)

(d)

2 cm 1 cm

2 cm

1.5 cm

4 cm





10.

(a)

1 20 cm = 3 000 000 Jarak sebenar / Actual distance

Jarak sebenar / Actual distance = 20 × 3 000 000 = 60 000 000 cm = 600 km (b) Panjang sisi PQRS / Side length of PQRS = √36  = 6 cm Panjang sisi ABCD / Side length of ABCD = √81  = 9 cm

Skala / Scale = 6 : 9 3 =1: 2

(c) 2.5 cm dalam pelan mewakili 150 000 cm.



2.5 cm in the plan represents 150 000 cm.

4.6 cm dalam pelan mewakili 4.6 × 150 000 = 276 000 cm 2.5 4.6 cm in the plan represents = 2.76 km Maka, panjang sebenar lebuh raya PQ ialah 2.76 km.



Hence, the actual length of the highway PQ is 2.76 km.

Tinggi model / Height of model = 1 35 m 500 35 × 1 Tinggi model / Height of model = 500 = 0.07 m = 7 cm

(d) (i)

(ii)

20 cm = 1 Panjang sebenar / Actual length 500

Panjang sebenar / Actual length = 20 × 500 = 10 000 cm = 100 m

© Penerbitan Pelangi Sdn. Bhd.

4

Matematik  Tingkatan 3  Bab 4

(e) Panjang lukisan / Drawing's length = 1 30.5 m 300 30.5 × 1 Panjang lukisan / Drawing's length = 300 = 0.102 m = 10.2 cm Lebar lukisan / Drawing's width = 1 15.25 m 300 15.25 × 1   Lebar lukisan / Drawing's width = 300 = 0.051 m = 5.1 cm



Luas lukisan / Area of drawing = Panjang / Length × Lebar / Width = 10.2 × 5.1 = 52.02 cm2 (f ) 1 : 1 bermaksud lukisan berskala adalah empat kali objek. 4 means the scale drawing is four times the object.

Panjang tapak lukisan

Tinggi lukisan



= 4 × 24 = 96 cm

=4×9 = 36 cm



Luas lukisan berskala bagi segi tiga itu





Length of the drawing’s base

Height of the drawing

Area of the scale drawing of the triangle

= 1 × 96 × 36 2 = 1 728 cm² (g) (i) Skala / Scale = 4 cm : 2 cm 1 =1: 2



Lukisan berskala itu adalah dua kali objeknya. Maka, The scale drawing is twice its object. Hence,

2 × x = 8 x = 4

x + y = 2 × 7 4 + y = 14 y = 10

2 × z = 10 z = 5

(ii) Isi padu / Volume = 1 × (4 + 7) × 2 × 5 2 = 55 cm2

5

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 4

Praktis PT3



1. (a)

Objek

Lukisan

Object

Skala (1 : n) Scale (1 : n)

Drawing

4 cm : 12 cm = 1 : 3 atau / or 1 cm : 3 cm = 1 : 3

1 cm 3 cm

4 cm

12 cm

6 cm

6 cm

12 cm

12 cm

4 cm

6 cm

18 cm

27 cm

12 cm : 12 cm = 1 : 1 atau / or 6 cm : 6 cm = 1 : 1

27 cm : 18 cm = 3 : 2 2 =1: 3 atau / or 6 cm : 4 cm = 3 : 2 2 =1: 3

Tinggi pada katalog / Height on the catalogue = 1 48 m 800 48 × 1 Tinggi pada katalog / Height on the catalogue = 800 = 0.06 m = 6 cm (b)

(c) 1 : 1 bermaksud lukisan berskala adalah tiga kali objek. 3 means the scale drawing is three times the object. Panjang AB pada lukisan / Length of AB on the drawing = 3 × 5 = 15 cm Panjang BC pada lukisan / Length of BC on the drawing = 3 × 6 = 18 cm A



t = √152 – 92 = √144  = 12 cm

15 cm

t

B

C 18 cm

Luas segi tiga pada lukisan / Area of the triangle in the drawing = 1 × 18 × 12 2 = 108 cm2

© Penerbitan Pelangi Sdn. Bhd.

6

Matematik  Tingkatan 3  Bab 4

2. (a) (i) 

P'Q' = PQ

(ii) P’Q’ = PQ 3 = 6 3 = 2 cm

(b) (i)

7

P'Q' = 3PQ

P'Q' = PQ 3

7

3

Luas segi empat sama P’Q’R’S’

Area of the square P'Q'R'S'

=2×2 = 4 cm2

Sisi segi empat sama P’Q’R’S’ Side of square P’Q’R’S’

8 cm = 1 Panjang sebenar / Actual length 300

Panjang sebenar / Actual length = 8 × 300 = 2 400 cm = 24 m (ii)

6 cm = 1 Panjang sebenar / Actual length 300

Panjang sebenar / Actual length = 6 × 300 = 1 800 cm = 18 m

5.4 cm = 1 Lebar sebenar / Actual width 300

Lebar sebenar / Actual width = 5.4 × 300 = 1 620 cm = 16.2 m Luas tikar getah / Area of the linoleum = 18 × 16.2 = 291.6 m²

3. (a)

Lukisan berskala

Skala

Scale drawing

Scale

1: 1 2

30 mm

49.5 mm

Ukuran sebenar (mm) Actual measurement (mm)

30 × 1 2 = 15 mm

1:6

49.5 × 6 = 297 mm

1:1

40 × 1 = 40 mm

40 mm



7

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 4

(b) (i)

10 cm 1 = Jarak sebenar / Actual distance 80 000

Jarak sebenar / Actual distance = 10 × 80 000 = 800 000 cm = 8 km (ii) Masa yang diambil / Time taken = 8 × 30 = 240 minit / 240 minutes = 4 jam / 4 hours

Maka, Akmal akan sampai ke stesen Y pada pukul 11:00 a.m. Hence, Akmal will reach the station Y at 11:00 a.m.

(c)

Cabaran KBAT 1. (i) 1 : 150 bermaksud 1 cm pada pelan mewakili 150 cm pada lantai 1 : 150 means 1 cm on the plan representing 150 cm on the floor

Panjang sebenar



Lebar sebenar

Actual length

Actual width

= 2.4 × 150 = 360 cm = 3.6 m

= 1.6 × 150 = 240 cm = 2.4 m

Luas sebenar bilik A / Actual area of room A = 3.6 × 2.4 = 8.64 m2

(ii) Kos / Cost = 8.64 × 48 = RM414.72 2. Perimeter lukisan berskala / Perimeter of the scale drawing = (3 × 7) × 8 = 168 cm

© Penerbitan Pelangi Sdn. Bhd.

8

Jawapan Bab 5 Nisbah Trigonometri

Trigonometric Ratios

5.1 Sinus, Kosinus dan Tangen bagi Sudut Tirus dalam Segi Tiga Bersudut Tegak

Sine, Cosine and Tangent of an Acute Angle in a Right-angled Triangle

1. (a) BC,  AB (b) PR,  PQ (c) b,  a 2. (a)

(b) θ

θ

Sisi bersebelahan Adjacent side

Sisi bertentangan Opposite side

Sisi bertentangan Opposite side





3. (a)

Nisbah panjang Segi tiga

Triangle

Ratio of length

sisi bertentangan hipotenus

sisi bersebelahan hipotenus

sisi bertentangan sisi bersebelahan

opposite side hypotenuse

adjacent side hypotenuse

opposite side adjacent side

PQR

4 5

3 5

4 3

PST

8 =4 10 5

6 =3 10 5

8=4 6 3

PUV

12 = 4 15 5

9 =3 15 5

12 = 4 9 3

(b) (i) tetap

(ii) tidak berubah

constant

sin q



=

opposite side hypotenuse

(iii) nisbah trigonometri

do not change

4. sisi bertentangan sin q = hipotenus

5.

Sisi bersebelahan Adjacent side

trigonometric ratios

sisi bersebelahan kos q = hipotenus cos q



=

adjacent side hypotenuse

sisi bertentangan tan q = sisi bersebelahan tan q



=

opposite side adjacent side

RS sin ∠T ST ST RT

kos ∠T cos ∠T

RS tan ∠T RT

1

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 5

6.

BC

(a) sin y =

(b) kos y = cos y

AC 7. (a)

QR

(c) tan y =

PQ

KM KL

Nisbah panjang (kepada dua tempat perpuluhan) Segi tiga

Sudut tirus

Ratio of length (to two decimal places)

sisi bertentangan hipotenus

sisi bersebelahan hipotenus

sisi bertentangan sisi bersebelahan

opposite side hypotenuse

adjacent side hypotenuse

opposite side adjacent side

Triangle

Acute angle

ABC

30°

5 = 0.50 10

8.7 = 0.87 10

5 = 0.57 8.7

DEF

50°

7.7 = 0.77 10

6.4 = 0.64 10

7.7 = 1.20 6.4

GHJ

70°

9.4 = 0.94 10

3.4 = 0.34 10

9.4 = 2.76 3.4

(b) • bertambah increases  



• berkurang



• bertambah

decreases

increases  

8. (a) sin 16°

sin 28°

sin 45°

sin 53°

sin 90°

kos 77°

kos 64°

kos 21°

kos 8°

kos 0°

tan 0°

tan 34°

tan 59°

tan 75°

tan 88°

(b) cos 77°

cos 64°

cos 21°

cos 8°

cos 0°

(c)

9. (a) sin q = 8 10 4 = 5

© Penerbitan Pelangi Sdn. Bhd.

(b) sin q = 5 13

(c) QR = √172 – 82 = 15 cm 15 sin q = 17

2

(d) BC = √252 – 242 = 7 cm 7 sin q = 25

Matematik  Tingkatan 3  Bab 5

10. (a) kos q = 12 13 cos q

(b) kos q = 15 17 cos q

(c) KM = √152 + 82 = 17 cm 15 kos q = 17 cos q

(d) SQ = √132 – 122 = 5 cm 5 kos q = 10 cos q = 1 2

11. (a) tan q = 6 8 3 = 4

(b) tan q = 12 9 4 = 3

(c) QR = √132 – 52 = 12 cm 12 tan q = 5

(d) LM = √172 – 152 = 8 cm 8 tan q = 15

13. (a) s in x = KL 52 KL = 12 52 13 12 × 52 KL = 13 = 48 cm

(b)

k os x = 24   (cos x = 24 ) PR PR 24 0.96 = PR PR = 24 0.96 = 25 cm

(c)

QR = √252 – 242 = 7 cm

JK = √522 – 482 = 20 cm

(i) sin x =

48 20 12 = 5 (i) tan x =

SU = √162 + 122 = 20 cm

7 25

12 20 3 = 5

(i) sin x =

(ii) tan x = 7 24

(ii) kos x = 20   (cos x = 20 ) 52 52 = 5 13

(ii) kos x = 16   (cos x = 16 ) 20 20 = 4 5

14. (a) 0.2588 (d) 0.7923 15.

(b) 0.6921 (e) 0.4115

bersamaan dengan sin 30o is equal to

16. (a) 1 + 1 2 2

as

1 2



=1

kos 45° cos 45°

1 √2

tan x = TU 16 TU = 3 16 4 3 TU = × 16 4 = 12 cm

as

tan 60o

as

√3

(c) 6.7720 (f ) 1.1868

sin 60o √3 2

(b) 1 + 1 2 = 11 2

kos 30°

tan 45o

√3 2

1

as cos 30° as

(c) 3 √3 – √3 2 2

1 2

3



= 2 √3 2



= √3

1 2

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 5

17. (a) (i) BC = √102 – 62 = 8 cm 8 kos x =  (cos x = 8 ) 10 10 = 4 5

(b) (i) kos x = AC   (cos x = AC ) 17 17 AC = 15 17 17 AC = 15 cm AD = √172 – 152 = 8 cm 8 tan x = 15

(ii) AC = 2 × 8 = 16 cm 6 tan y = 16 = 3 8

(ii) BC = √152 – 122 = 9 cm 9 sin y = 15 3 = 5

(c) (i) kos y = QS   (cos y = QS ) 17 17 QS = 15 17 17 QS = 15 cm

(d) (i) sin y = BE 17 BE = 15 17 17 BE = 15 cm DE = 6 + 15 = 21 cm

QR = √172 – 152 = 8 cm PR = 2 × 8 = 16 cm

6 BC 6 = 3 BC 4 3BC = 6 × 4 BC = 8 cm (ii) tan x =

(ii) QT = 15 – 9 = 6 cm PT = √82 + 62 = 10 cm 6 sin x = 10 3 = 5

CD = √62 + 82 = 10 cm 6 sin x = 10 = 3 5 (iii) AB = √172 – 152 = 8 cm 8   (cos y = 8 ) kos y = 17 17

18. (a) q = tan–1 (1.44) = 55o13’

(b) q = sin–1 (0.69) = 43o38’

(d) q = sin–1 (0.83) = 56o6’

(e) q = kos–1 (0.17)

© Penerbitan Pelangi Sdn. Bhd.

cos–1 (0.17)

= 80o13’

4

(c) q = kos–1 (0.92)

cos–1 (0.92)

= 23o4’ (f ) q = tan–1 (3) = 71o34’

Matematik  Tingkatan 3  Bab 5

PQ 7 PQ = 1 7 √2 PQ = 1 × 7 √2 = 4.95 m Lebar lebuh raya ialah 4.95 m.

Kaedah Alternatif

19. (a) sin 45° =



Katakan / Let PQ = a a2 + a2 = 72 2a2 = 49 49 a = 2 = 4.95 m

The width of the highway is 4.95 m.

(b) tan θ = 1.5 2.6 θ = tan–1 (0.5769) = 30° (c) (i) tan x = 12 5

12 cm

S

(ii) QR = √52 + 122 = 13 cm 12 sin x = 13

R

12 cm

x P

5 cm

Q

Praktis PT3 1. (a) (i) ✓  (b) tan y = KN 8 KN = 1 8 KN = 8 cm MN = √172 – 82 = 15 cm tan ∠MKN = 15 8

2. (a) (i)

3 5

(ii) ✓ (iii) 7  (c) (i) ST = 12 cm 12 tan x = 24 = 1 2

'

(ii) RT = √242 + 122 = 26.83 cm sin y = 24 26.83 y = sin–1 (0.8945) = 63°27'

20 20 (ii) (iii) 29 21

(b) (i) tan x = MR = 12 MQ 5 Maka/  Hence, MR = 12, MQ = 5

(ii) PM = 2 × 5 = 10 kos y = LP (cos y = LP ) 10 10 LP 3 = 10 5 3 LP = × 10 5 = 6

QR = √52 + 122 = 13 sin x = 12 13

5

LM = √102 – 62 = 8 tan y = 8 6 = 4 3

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 5

(c) kos ∠X = XS (cos ∠X = XS ) 5.2 5.2 XS 5 = 5.2 13 5 × 5.2 XS = 13 = 2 m QX = 8 – 2 = 6 m

Cabaran KBAT 1. (a) BD = √82 + 62 = 10 cm 6 10 ∠DBE = tan–1 (0.6) = 30°58'

E

(b) tan ∠DBE =

6 cm D

10 cm

2. Katakan tinggi yang dicapai oleh tangga = x m Let the height that the ladder can reach = x m

sin 24° = x 6 x = 6 sin 24° = 2.44

xm

6m 24°

3. (a) sin x = AE BE 12 4 = BE 5 4BE = 12 × 5 BE = 15 cm AB = √152 – 122 = 9 cm (b) Luas kawasan berlorek Area of shaded region

= 1  (12 + 24) × 9 2 = 162 cm2

© Penerbitan Pelangi Sdn. Bhd.

6

B

Jawapan Bab 6 Sudut dan Tangen bagi Bulatan

Angles and Tangents of Circles

6.1 Sudut pada Lilitan dan Sudut Pusat yang Dicangkum oleh Suatu Lengkok

Angle at the Circumference and Central Angle Subtended by an Arc

1. (a) a ,  b (b) x ,  y (c) c ,  d (d) n ,  m 2. (i) (a) x = 50°,  y = 50° (b) x = 25°,  y = 25° (c) x = 30°,  y = 30° Sudut-sudut pada lilitan yang dicangkum oleh lengkok yang sama atau sama panjang adalah sama (x = y). The angles at the circumference subtended by the same arc or the arcs of the same length are equal (x = y).

(ii) (a) x = 20° (b) x = 90° (c) x = 100° y = 40° y = 180° y = 200° Sudut pada pusat adalah dua kali sudut pada lilitan yang dicangkum oleh lengkok yang sama (y = 2x). The angle at the centre is twice the angle at the circumference subtended by the same arc (y = 2x).

90° 3. (a) p = q r = s

(b) a = b = c d = e

(c) x = y

4. (a) y = 2x

(b) y = 2x

(c) x = 2y

5. (a) x = 28°

(b) x = 60°

(c) x = 90°

(e) 2x = 80°   x = 80° 2 = 40°

(f ) 2x = 210°   x = 210° 2 = 105°

(d) x = 2 × 30° = 60°

6.

(a)

x

(b)

(c) 60°

224° O a

a = 360° – 224° = 136° x = 136° 2 = 68°

40°

O a

a = 2 × 60° = 120° x = 180° – 120° 2 60° = 2 = 30°

1

x

a O x

40°

a = 2 × 40° = 80° x = 360° – 80° = 280°

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 6

(d)

(e) O 2x

(f ) 25°

x 3x

55°

x = 180° – 90° – 25° = 65°

2x + 3x = 90° 5x = 90° 90° x = 5 = 18° (g)

(h) 70° a

35° O 35° x

O

x = 90° – 35° = 55°

a = 90° – 70° = 20° x = 2 × 20° = 40°

7. (a) ∠PQS = ∠PRS = 55° ∠OQP = ∠OPQ = 30°

ΔOPQ ialah segi tiga sama kaki. ΔOPQ is an isosceles triangle.

∠SQT = ∠PQS − ∠OQP = 55° − 30° = 25° x = 180° − 90° − 25° = 65° (b) ∠ABD = ∠BAC = 32° ∠ODB = ∠OBD = 32°

ΔOBD ialah segi tiga sama kaki. ΔOBD is an isosceles triangle.

x = 90° − 32° = 58° y = 32° 

ΔOAC ialah segi tiga sama kaki. ΔOAC is an isosceles triangle.

x + y = 58° + 32° = 90°

(c) ∠OHG = 180° – 102° 2 78° = 2 = 39°

ΔOGH ialah segi tiga sama kaki. ΔOGH is an isosceles triangle.

2x + x + 39° = 90° 3x = 90° − 39° 3x = 51° x = 51° 3 = 17°

© Penerbitan Pelangi Sdn. Bhd.

O

O

2

x

x = 90° – 55° = 35°

x

Matematik  Tingkatan 3  Bab 6

(d) (i) ∠LNM = ∠KNL = 28° ∠KNM = 28° + 28° = 56°

x = 180° − 90° − 56° = 34°

(e) (i) ∠PTR = 180° − 62° − 20° = 98° (ii) ∠RQS = ∠RTS = 180° − 98° = 82°

(ii) KN = 2 × 6.5 = 13 cm sin x = MN KN sin 34° = MN 13 MN = 13 × sin 34° MN = 7.27 cm

ΔKMN ialah segi tiga bersudut tegak. ΔKMN is a rightangled triangle.

(iii) ∠RUS = ∠QUT = 360° − 62° − 98° − 98° = 102°

ΔAOD ialah segi tiga sama sisi.

(f ) ∠AOD = 60° x = 60° 2 = 30°

ΔAOD is an equilateral triangle.

6.2 Sisi Empat Kitaran

Cyclic Quadrilaterals

8. (a) sisi empat kitaran

(b) bucu  ;

cyclic quadrilateral

vertices

lilitan

circumference

9. a=

60°

b = 100°

c = 120°

d=

80°

e = 100°

f = 120°

(a) a + c = 180°, b + d = 180° (b) Hasil tambah sudut-sudut pedalaman yang bertentangan dalam sebuah sisi empat kitaran ialah 180°. The sum of the interior opposite angles in a cyclic quadrilateral is 180°.

(c) Sudut peluaran sebuah sisi empat kitaran bersamaan dengan sudut pedalaman bertentang yang sepadan. The exterior angle of a cyclic quadrilateral is equal to its corresponding interior opposite angle.

10. (a) a + c = 180° b + d = 180° a + b + c + d = 360° (c)

(b) p + r = 180° q + s = 180° p + q + r + s = 360°

j + n = 180° g + h + k + m = 180° h + j + k = 180° g + m + n = 180° g + h + j + k + m + n = 360°

(d) x = y

3

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 6

11. (a) m = 180° − 105° = 75°

(b) m = 112° n = 97°

(c) m = 108° n = 180° − 100° = 80°

n = 180° − 95° = 85° 12. (a) 2x = 180° − 92° 2x = 88°   x = 88° 2 = 44°

x = 180° − 50° − 95° = 35°

13. (a) ∠OPQ = 180° – 66° 2 114° = 2 = 57°

(b) ∠ACB = ∠ABC = ∠BCD = 38°

(c) ∠GHJ = ∠GKL = 100° ∠HJG = ∠HGJ = 180° – 100° 2 80° = 2 = 40°



y = 180° − 88° = 92°

p = ∠CAB = 180° − 38° − 38° = 104°

x = 180° − 20° − 57° = 103°

∠GJK = 180° − 62° − 80°  = 38°

(d) ∠OBC = ∠OCB = 35°

∠OBA = ∠OAB = 30°

∠ABC = 35° + 30° = 65° ∠GKJ = 180° – 100° = 80°

x = 180° − 65° = 115°

x = 40° + 38° = 78°

6.3 Tangen kepada Bulatan

Tangents to Circles

14. (a) Garis TVS ialah tangen kepada bulatan pada titik V. Line TVS is a tangent to the circle at point V.

Garis SWU ialah tangen kepada bulatan pada titik W. Line SWU is a tangent to the circle at point W.

(b) Garis ABC ialah tangen kepada bulatan pada titik B. Line ABC is a tangent to the circle at point B.

Garis CDF ialah tangen kepada bulatan pada titik D. Line CDF is a tangent to the circle at point D.

15. (i) (a) (i) 90°   (ii)  90°   (iii) 90° (b) berserenjang

perpendicular

© Penerbitan Pelangi Sdn. Bhd.

n = 80°

(b) a = 180° − 85° = 95°

4y + 2y = 180° 6y = 180° y = 180° 6 = 30°



(d) m = 180° – 85° = 95°

4

88° 85° 50° y

x

a

Matematik  Tingkatan 3  Bab 6

(ii) (a) (i) 2.35 cm (iv) 23°

(ii) 2.35 cm (v) 67°

(iii) 23° (vi) 67°

(b) (i) 23° + 23° (ii) 67° + 67° (iii) 46° + 134° = 46° = 134° = 180° (c) (i) sama

(ii) pembahagi dua sama sudut

the same



(iii) sudut penggenap

angle bisector

(iv) sama,  kongruen

supplementary angles

the same,  congruent

(iii) (a) (i) 70°   (ii)  50°   (iii)  70°    (iv)  50° (b) sama the same

16. (a) x + y = 180°

(b) p = r q=s

17. (a) BC = CD Maka / hence, m = 1.3 n = 90°

(b) m = 70° n = 52°

18. (a) x = 180° – 90° – 70° = 20°

(b) x = 360° – 67° – 67° = 226°

(c) ∠POR = 360° − 248° = 112°

(d) ∠POR = 180° − 70° = 110°



∠QOR = 110° ÷ 2 = 55°

∠QOR = 112° ÷ 2 = 56° x = 180° − 90° − 56° = 34°



19. (a) x = 64° y = 180° – 64° 2 116° = 2 = 58°

x = 180° – 55° 2 125° = 2 = 62.5°



z = y = 58° x + y + z = 64° + 58° + 58° = 180°

(b) GJ = GK, maka ∆JGK ialah segi tiga sama kaki.

GJ = GK, hence ∆JGK is an isosceles triangle.



∠GJK = 75°

x = 180° − 75° − 75° = 30°

y = 180° − 68° − 75° = 37°

5

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 6

(c) (i) x = 180° − 80° = 100° (ii) AF = AE − FE AC = AB + BC = 34 − 22 = 12 + 22 = 12 cm = 34 cm (d) ∠TQR = 85°

x = 85° − 65° = 20° (ii) sin ∠QAR = QR AR sin 50° = 12 AR AR = 15.7 cm

(e) (i) ∠QAV = 50° + 50° = 100°

∠QRV = 180° – 100° = 80°

∠WRB = 80° ÷ 2 = 40° kos ∠WRB = RW RB cos ∠WRB kos 40° = 17 RB    cos 40° RB = 22.2 cm

m = 180° − 80° = 100°



AB = AR + RB = 15.7 + 22.2 = 37.9 cm

6.4 Sudut dan Tangen bagi Bulatan

Angles and Tangents of Circles

20. (a) x = 42° y = 180° − 42° − 102° ∠URS = 180° − 78° = 36° = 102° (b) x = 84° 2 = 42°

y = 110° − 54° = 56°

∠ABE = ∠EFG = 110° ∠FBE = 108° 2 = 54°

y − x = 56° − 42° = 14°

(c) ∠STQ = ∠SQR = 50° ∠OTS = 50° − 35° = 15° ∠TOS = 180° − 15° − 15° = 150° ∠TQS = 150° 2 = 75°

∠TSQ = x = 55° ∠OST = ∠OTS = 15° y = 55° − 15° = 40° x + y = 55° + 40° = 95°

x = 180° − 50° − 75° = 55° © Penerbitan Pelangi Sdn. Bhd.

6

Matematik  Tingkatan 3  Bab 6

Praktis PT3

3. (a) (i)

1. (a) (i) PQ 

RS 

TU 

(ii) PQ 

RS 

TU 

RS 

ST 

(iii) PQ 

3

x

3

O

3

x

(iii)

(ii) p = 106° q + ∠KLJ = p q + (180° − 125°) = 106° q + 55° = 106° q = 106° − 55° = 51° (c) (i) w = 84° 2 = 42°

•

•

90°

x O

•

•

135°

(ii) ∠KNP = ∠KJN = 180° − 75° = 105° q = 180° − 60° − 105° = 15°

2. (a) (i) PALSU / FALSE (ii) BENAR / TRUE (iii) BENAR / TRUE

(c) tan ∠CAD = CD AC CD tan 65° = 21 CD = 21 × tan 65° = 45 cm

(b) (i) x = 2 × 24° = 48° (ii) y = 180° – 48° 2 = 132° 2 = 66°

∠DBE = ∠CAD = 65° tan ∠DBE = DE BE DE tan 65° = 14 DE = 14 × tan 65° = 30 cm

(iii) y + z = 97° 66° + z = 97° z = 97° − 66° = 31° (iv) z + y − x = 31° + 66° − 48° = 49°

CE = CD + DE = 45 + 30 = 75 cm

(c) x = 180° − 112° = 68°

x + y = 68° + 70° = 138°

45°

(b) (i) ∠AOC = 2 × 67° = 134° p = 360° − 134° = 226°

(iii) y + z = 360° − 84° − 90° = 186°



45°

90°

(ii) x = 90° − 42° = 48°

y = 180° − 68° − 42° = 70°

•

(ii)

(b) (i) ∠EBD = ∠EAD = 34° x = 180° − 65° − 34° = 81° y = 42°



•

7

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 6

Cabaran KBAT 1. ∠MLQ = ∠LPQ = 180° − 120° = 60° 2x = 180° − 74° − 60° 2x = 46°   x = 46° 2 = 23° 2. ∠SQR = ∠STP = 75° x = 180° − 75° − 75° = 30° 3. ∠OST = ∠OTS = 32° ∠RST = 90° x = 90° − 32° = 58°

© Penerbitan Pelangi Sdn. Bhd.

8

Jawapan Bab 7 Pelan dan Dongakan

Plans And Elevations

7.1 Unjuran Ortogon

Orthogonal Projections

1. (a) Ya / Yes

(b) Bukan / No

(c) Ya / Yes

2. D 3. (a)

(b)

4 cm



5 cm

4 cm

(c)

(d)

E/D

T

S

U

H/C

2 cm

F/A

4 cm

4 cm

G/B

P 2.5 cm

(e) E/D

(f ) H

R/N

R

2.5 cm

Q

Q/P

J/C

3 cm

4 cm

V/S

U/T

2 cm



F/A 1 cm G

2 cm

K/B

J/M

K/L

2 cm

1

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 7

4. (a)

3

(b) (c) 3

(d)

3

(e) (f )

5. (a)

Perbandingan antara pepejal dengan unjuran ortogon

Lukisan unjuran ortogon

Drawing of orthogonal projection

Comparison between the solid and its orthogonal projection

(i)

Semua panjang sisi, saiz sudut dan bentuk ADE adalah sama dengan prisma itu.

A/B

All the edges, angles and shape ADE are the same as the prism.

3 cm

D/C

E/F

4 cm

(ii) B/C

Sama/Same: AB, EF, DC, CF, DE; semua sudut tegak;

F

all the right angles; bentuk/ shape DCFE

Berbeza/ Different: AE, BF; bentuk/ shape ABFE

6 cm

A/D

(b)

4 cm

E

(i) U

T

Sama/Same: QR, UT; semua sudut tegak all the right angles

4 cm

Q/P

© Penerbitan Pelangi Sdn. Bhd.

R/S

8 cm

2

Berbeza/ Different: QU, RT; bentuk/ shape QRTU

Matematik  Tingkatan 3  Bab 7

(ii) S

T

Sama/Same: PQ, RS, PS, QR, UT; semua sudut tegak;

R

all the right angles ; bentuk/ shape PQRS

Berbeza/ Different: PU, ST; bentuk/ shape PSTU 8 cm

P

3 cm

U

Q

3 cm

7.2 Pelan dan Dongakan

Plans and Elevations

6. (a)

(b) 4 cm

E/A

D/C B/I

4 cm



2 cm

C/H

4 cm

F/B

2 cm

G/L

3 cm

A/J

D/E

F/K

(c) T/S

U/R

5 cm



W/P

2 cm

V/Q

3

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 7

7. (a) (i)

(b) (i) F/E

2 cm

G/H

P/N

K/J

2 cm

2 cm 1 cm

Q/T A/D

5 cm

B/C 2 cm

J/M

(ii)

K/L

3 cm

R/S

2 cm

(ii) H/E

G/F

N

P

1 cm K

2 cm

J 1 cm

B/A

T

C/D

3 cm

Q 2 cm

M/L/S

8. (a) (i)

4 cm

J/K/R

(ii) F/E

E/D

H/I

J/C

4 cm G/H

F/A

4 cm

G/L

2 cm

8 cm

K/B

3 cm

L/I

2 cm

K/J

2 cm



A/D

6 cm

© Penerbitan Pelangi Sdn. Bhd.

4

B/C

Matematik  Tingkatan 3  Bab 7

(iii) F

E

3 cm

H

G

3 cm

K/L

J/I

2 cm



B/A

C/D

4 cm

(b) (i) A/L

H/I

D/E

4 cm

B/K

5 cm

G/J

2 cm

C/F

(ii) C/D

B/A

5 cm

7 cm G/H

2 cm

F/E

3 cm



K/L

5 cm

J/I

5

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 7

(iii)

C

D 1 cm

B

A

4 cm

F/G

E/H

3 cm

(c) (i)

J/K

E/D

J

4 cm

I/L

4 cm

F/C

I 6 cm

4 cm



K/A

2 cm

L/H

G/B

(ii) K

2 cm

L

1 cm J/E

H/I

G/F

2 cm



A/D

B/C

4 cm

(iii) L/K 1 cm G/H

I/J

2 cm

F/E

4 cm

2 cm



B/A

© Penerbitan Pelangi Sdn. Bhd.

6 cm

C/D

6

Matematik  Tingkatan 3  Bab 7

9. (a)

(b) H E

G F

H J

5 cm

3 cm

K C

D

6 cm



A

4 cm

F

4 cm D

3 cm

G

E

A



B

6 cm

C 4 cm B

10. (a) (i) L/E/D

F/C

6 cm

H/J/A



4 cm

G/B

3 cm

(ii) H

G

1 cm J/K/L

4 cm

6 cm E

F

3 cm



3 cm

A/D

B/C

4 cm

7

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 7

(iii) G/H 1 cm K

J

L

3 cm

6 cm

F/E

3 cm



B/A

6 cm

C/D

(b) (i) 3 cm

K

J

2 cm

L

M

1 cm F/C

E/D

3 cm

4 cm P/Q/G

N/H

2 cm



A

4 cm

B

(ii) P/L

N/M

1 cm E

F

6 cm 5 cm



A/D

© Penerbitan Pelangi Sdn. Bhd.

4 cm

B/G/C/K

3 cm

8

H/J

Matematik  Tingkatan 3  Bab 7

(iii) 3 cm

N/P

M/L

F/E

3.5 cm

Q 5 cm

2.5 cm



B/A

2 cm

H/G

2 cm

C/D

3 cm

J/K

(c) (i)

(ii)

E/D

H/C

3 cm

2 cm

F/E F

G/H

G

3 cm 3 cm

A/D



A

4 cm

4 cm

B/C



B

(iii) 3 cm

2 cm

G/F

H/E

3 cm



B/A

6 cm

C/D

Isi padu gabungan pepejal = Isi padu prisma + Isi padu separuh silinder Volume of composite solid



=  Volume of prism + Volume of half cylinder

= 1 (3 + 6) × 3 × 4 + 1 × 3.142 × 22 × 3 2 2 = 54 + 18.852 = 72.852 cm3 9

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 7

Praktis PT3 1. (a) (i) A (ii) C (iii) B (b) Isi padu sebuah kubus Volume of a cube



= 256 ÷ 4 = 64 cm3 Panjang sisi sebuah kubus Length of side of a cube

= 3√64  = 4 cm Jumlah luas permukaan pepejal Total surface area of the solid



Luas satu muka × Jumlah muka yang tidak bertindih

= 42 × 18 = 288 cm2

Area of a face × Total number of non-overlapping faces

(c) Panjang sisi kubus besar

Length of side of the bigger cube

= 3√1 728  = 12 cm 2. (a) •

•

•

•

•

•

(b) (i)

(ii) P/N

T



Q/K

T

S/M

2 cm

2 cm

R/L

3 cm

Q/P

R/S

2 cm

(c) Tidak/ No K/N

© Penerbitan Pelangi Sdn. Bhd.

10

2 cm

L/M

Matematik  Tingkatan 3  Bab 7

Cabaran KBAT (a)

8 cm

1.5 cm

1.5 cm

(b)

1.5 cm

1.5 cm

3 cm

5.6 cm

3 cm

3 cm

(c) 2.6 cm

0.4 cm

2.6 cm

8 cm

11

© Penerbitan Pelangi Sdn. Bhd.

Jawapan Bab 8 Lokus dalam Dua Dimensi

Loci in Two Dimensions

8.1 Lokus Loci

1.  

Kedudukan atau lokus titik M

Situasi

Situation

Position or locus of point M

Titik M Point M

•

•

•

•

Seorang budak sedang • berjalan menuju ke pokok renek

•

Bilah-bilah kipas angin sedang berpusing Fan blades are spinning Titik M Point M

Sebuah kereta sedang dipandu turun cerun bukit

A car is being driven down the hill slope Titik M Point M

A boy is walking towards a shrub

Titik M Point M

  Kulit pisang dibuang oleh seekor monyet dari atas pokok

•

•

Banana skin thrown by a monkey from a tree



titik

 ; 

syarat

points

 ; 

conditions

1

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 8

8.2 Lokus dalam Dua Dimensi

Loci in Two Dimensions

3. (a) EF (b) AC (c) KL (d) GH (e) AB, DC (f ) EF, GH 4. (a) bulatan ,  pusat ,  jejari

(b) pembahagi dua sama serenjang

circle ,  centre ,  radius

(c) garis lurus ,  selari

perpendicular bisector

(d) garis lurus ,  tengah-tengah ,  selari

straight lines ,  parallel

straight line ,  middle ,  parallel

(e) pembahagi dua sama sudut

angle bisectors

5. (a)

(b) Lokus

Lokus

Locus

Locus

K

L

3 cm

2.5 cm

(c)

(d)

P

Lokus

Q

Locus

Lokus

Locus

M

R

S 2 cm

N

(e)

B Lokus Locus

C

A

6. (a)

(b)

Lokus R

K K

Locus R

L

L

2 cm Lokus R Locus R

N

© Penerbitan Pelangi Sdn. Bhd.

N M

2

M

Matematik  Tingkatan 3  Bab 8

(c)

Lokus R Locus R

K

L

Lokus R Locus R

N

M

(d)

(e) N

K

L K

N

M Lokus R



L



Locus R

M

Lokus R Locus R

7. (a) Lokus P sentiasa berjarak 2 cm dari garis AD. Locus P is always 2 cm from line AD.

(b) Lokus P sentiasa berjarak 1 cm dari titik D. Locus P is always 1 cm from point D.

(c) Lokus P sentiasa berjarak sama dari dua garis bersilang, AD dan DC. (atau AB dan BC) Locus P is always equidistant from two intersecting lines, AD and DC. (or AB and BC)

(d) Lokus P sentiasa berjarak 1.5 cm dari garis AB. Locus P is always 1.5 cm from line AB.

(e) Lokus P sentiasa berjarak sama dari dua garis bersilang, AC dan BD. Locus P is always equidistant from two intersecting lines, AC and BD.

8. (a) P

(b)

Q

Lokus M Locus M

A

Lokus M

Lokus N

Locus M

Locus N

B

Lokus N

Locus N

D

(c) S

C

(d)

R

Lokus N

J

Locus N

Lokus M

B

Locus M

Lokus N Locus N

A K

C

L Lokus M Locus M

E



D

3

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 8

(e)

Praktis PT3

Lokus N Locus N

1. (a) (i) QS (ii) SQ (iii) VW

Q

P

(b) (i) BE (ii), (iii)

R

U Lokus M Locus M

A

C

S

T

B Lokus Z Locus Z

2 cm

Lokus X, BE Locus X, BE

Lokus L Locus L

F

9. (a)

Lokus Y

D

Locus Y

4m

P

Q

E

(c) 5m

(b)

S

R

1 cm

A

B Pekan Delima

1 cm

Kedudukan bunga Position of the flower

Pekan Mutiara C

F

1.5 cm

2. (a) (c)

D

E

E

5m

O P

F

S

5m

(b) K

L

N

M

Lokus C Locus C

G

H

(d)

Lokus A Locus A

6m

A

B

Lokus B Locus B

2 cm

(c) (i) 1 cm

10 m

1 cm Jalan batu kelikir Gravel pathway



© Penerbitan Pelangi Sdn. Bhd.

D

C

4

Pekan Intan

Matematik  Tingkatan 3  Bab 8

Cabaran KBAT

(ii) L

K

1. A

1 cm

B Pagar

Fences

4.5 cm

Kedudukan pokok mangga



N



Position of the mango tree

M

D

C

3. (a) B

D

2. A Lokus R

U

T

S

Lokus Q

Locus R

Locus Q

Lokus X Locus X

Lokus Y

Lokus P

Locus Y

Locus P

(b)

C

E P

Q

R

1unit 1unit

P

Jambatan

(c)

Bridge

S Lokus Y Locus Y

T

W

Lokus X Locus X



V

U

5

© Penerbitan Pelangi Sdn. Bhd.

Jawapan Bab 9 Garis Lurus

Straight Lines

9.1 Garis Lurus

Straight Lines

1. Kecerunan

Pintasan-y

(a)

∞

Tiada / None

(b)

0

2

(c)

– 3 = –1 3

3

(d)

6 =3 2

1

(e)

2 =2 1

−4

Gradient

y-intercept

kecerunan ,  pintasan-y gradient ,  y-intercept

2.

Persamaan

y = 2x + 3

Equation

Kecerunan Gradient

(a) 

y = –x + 11

Equation

y-intercept

3.

(b) 

2

Seperti

y = –7x – 2

Seperti as

(e) 

11

2 5

y = 1 x + 15 2 (f ) 

–2

(a) ax + by = c by =

y= 2x+2 5 (c) 

4

as

(d) 

Seperti as

as

Persamaan Pintasan-y

Seperti

y = 4x – 5

15

(b) x + y = 1 a b –a  x +

c

a  x + b

c b

y = –

bx + ay =

ab

y = –

1

b  x + a

b

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 9

4.

Persamaan garis lurus

Kecerunan

Pintasan-y

x+y=4

−1

4

Equation of straight line

(a)

Gradient

y-intercept

Kecerunan = – Gradient

Pintasan-y = y-intercept

(b)

5x − y = −7

5

7

(c)

x + 6y = −1

– 1 6

– 1 6

Persamaan garis lurus

Pintasan-x

Pintasan-y

Kecerunan

(a)

x + y =1 3 2

3

2

– 2 3

(b)

–x + y = 1 5

–1

5

5

(c)

x –y=1 6

6

–1

1 6

5.

Equation of straight line

6. (a) x + y = 1 3 4 4x + 3y = 12 3y = –4x + 12 y = – 4 x + 4 3 (b) 3x y – + =1 2 2 –3x + y = 2 y = 3x + 2

(c) 4x + 2y = –1 2y = –4x – 1 y = –2x – 1 2 (d) 1 x + y = 2 3 y = – 1 x + 2 3 (e) 3 y + 5x – 2 = 0 3y = –5x + 2 5 2 y = – x + 3 3

© Penerbitan Pelangi Sdn. Bhd.

x-intercept

y-intercept

Gradient

Kaedah Alternatif x y (a) + = 1 3 4 Diberi/ Given a = 3, b = 4 Jadi/ So m=– b =– 4, a 3 c=4 Maka/ Hence y=– 4x+4 3 (c) 4x + 2y = –1 Diberi/ Given a = 4, b = 2, c = –1 Jadi/ So m = – a = – 4 = –2, b 2 c 1 c= =– b 2 Maka/ Hence 1 y = –2x – 2

2

c b

a b

Kecerunan = – Gradient

b a

Matematik  Tingkatan 3  Bab 9



Bentuk persamaan garis lurus

7.

Form of equation of straight line

y = mx + c

ax + by = c

x + y =1 a b

(a)

y = 3x – 4

3x – y = 4

3x – y = 1 4 4

(b)

y = 4x + 9

−4x + y = 9

– 4x + y = 1 9 9

(c)

y = 7x + 3

−7x + y = 3

– 7x + y = 1 3 3

(d)

y = 10 x + 5 3

−10x + 3y = 15

– 2x + y = 1 3 5

(e)

y=– 5x+ 8 3 3

5x + 3y = 8

5x + 3y = 1 8 8

(f )

y=– 7x+7 4

7x + 4y = 28

x + y =1 4 7

8. x ,  y ,  sama ,  terletak pada x ,  y ,  equals ,  on

Jawapan Alternatif x ; y; tidak sama; tidak terletak pada x ; y; not equal ; not on

9. (a) Sebelah kiri:

Left hand side:



y=5



Sebelah kanan:



(b) Sebelah kiri:

Left hand side:

2x + y = 2(3) + (−7) =6−7 = −1

Right hand side:

3x − 1 = 3(2) − 1 =6−1 =5



5=5 Titik P terletak pada garis lurus.

−1 ≠ 3 Titik P tidak terletak pada garis lurus.

(c) Sebelah kiri:

(d) Sebelah kiri:

x + y = –0.5 + 9 4 8 4 8 =1

– x + y = – 2 + 5 2 3 2 3 2 = 3





Left hand side:



Left hand side:



Sebelah kanan: Right hand side:



1



Sebelah kanan: Right hand side:

1 2 ≠1 3 Titik P tidak terletak pada garis lurus.

1=1 Titik P terletak pada garis lurus.

Point P is not on the straight line.







Right hand side:

3

Point P is on the straight line.



Sebelah kanan:



Point P is on the straight line.



3

Point P is not on the straight line.

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 9

10. (a) y = 5x + 1 k = 5(4) + 1 = 20 + 1 = 21

(c) y = kx + 7 3 = k(1) + 7 k = 3 – 7 = –4

(b) 2y = 3x + 1 2(5) = 3k + 1 10 = 3k + 1 3k = 10 − 1 3k = 9 k = 3

11. sama ,  selari equal ,  parallel

Jawapan Alternatif tidak sama; tidak selari not equal ; not parallel

12. (a) mPQ = 8 – 2 4–1 6 = 3 =2

(b) mPQ = 9 – 3 6–1 6 = 5 mRS = 11 – 5 12 – 7 6 = 5



mRS = 13 – 7 9–6 6 = 3 =2





mPQ = mRS PQ selari dengan RS.



mPQ = mRS PQ selari dengan RS. PQ is parallel to RS.

PQ is parallel to RS.

(c) mPQ = 7 – (–2) 2–5 9 = –3 = –3

(d) mPQ =

9–5 –3 – (–1) = 4 –2 = –2

mRS = 9 – (–1) 7–8 = 10 –1 = –10 mPQ ≠ mRS PQ tidak selari dengan RS.

mRS = 7 – 1 2–5 = 6 –3 = –2 mPQ = mRS PQ selari dengan RS.







PQ is not parallel to RS.

© Penerbitan Pelangi Sdn. Bhd.



4

PQ is parallel to RS.

(d) x + 2 3 + 2

y = 1 3 k = 1 3 k = 1 – 3 3 2 k 1 = – 3 2 k = – 3 2

Matematik  Tingkatan 3  Bab 9

13.

Persamaan garis lurus

Selari / Tidak selari

Equation of straight line

AB (a)

y = 3x – 1

Parallel / Not parallel

CD y = 3x + 4

mAB = 3 mCD = 3 mAB = mCD Selari

Parallel

(b)

y = 3 x – 12 2

2y = 3x – 2 3   y= x–1 2

mAB = 3 2 mCD = 3 2 mAB = mCD Selari

Parallel

(c)

y= 4x–5 5

5x + 4y = –4 4y = –5x – 4 5 y = – x – 1 4

mAB = 4 5

mCD = – 5 4 mAB ≠ mCD Tidak selari Not parallel

(d)

3y = x + 24 1   y= x+8 3

3y = x – 12   y= 1x–4 3

mAB = 1 3 mCD = 1 3 mAB = mCD Selari

Parallel

(e)

– x – y = 1 5 2 –2x – 5y = 10 –5y = 2x + 10 2 y = – x – 2 5

5y = –3x + 10 3   y=– x+2 5

mAB = – 2 5 mCD = – 3 5 mAB ≠ mCD Tidak selari Not parallel

14. (a) (b) (c) (d)

y = 2x + 3 y=5 x=2 Kecerunan, m = – 3 4 Gradient, m y=– 3x+3 4

5

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 9

15. (a) y = mx + c 2 = 3(1) + c c = 2 – 3 = –1 y = 3x – 1

(b) y = mx + c 3 = 5(–2) + c c = 3 + 10 = 13 y = 5x + 13

(c) y = mx + c –3 = 3(–1) + c c = –3 + 3 = 0 y = 3x

16. (a) y = mx + c −5 = m(3) − 1 3m = −5 + 1 3m = –4 4 m = – 3 y = – 4 x – 1 3

(b) y = mx + c −8 = m(1) + 1 3 1 m = −8 – 3 m = – 25 3 25 1 y = – x + 3 3

(c) y = mx + c –2 = m(–7) + 3 –7m = –2 – 3 –7m = –5 5 m = 7 y = 5 x + 3 7

17. (a) m = 4 – 3 2–1 1 = 1 = 1

(b) m = –1 – 0 2–5 –1 = –3 = 1 3

(c) m =



y=x+c Pada / At (1, 3) 3 = 1 + c c = 3 − 1 = 2 y = x + 2



y= 1x+c 3 Pada / At (5, 0) 1 0 = (5) + c 3 c = – 5 3 1x– 5 y = 3 3

–7 – 3 1–0 = –10



y = –10x + c Pada / At (0, 3) 3 = –10(0) + c c = 3



y = –10x + 3

(d) m = 6 – (–3) 6–8 = – 9 2

y = –   9 x + c 2 Pada / At (6, 6) 9 6 = – (6) + c 2 c = 6 + 27 = 33 y = – 9 x + 33 2

18. (a) x + y = 1  2 3

© Penerbitan Pelangi Sdn. Bhd.

4 – (–5) 1 – (–2) 9 = 3 = 3

(e) m =



y = 3x + c Pada / At (1, 4) 4 = 3(1) + c c = 4 − 3 = 1 y = 3x + 1

9–7 –4 – (–6) 2 = 2 = 1

(f ) m =



(b) –   x + y = 1 3 5

6

y=x+c Pada / At (–6, 7) 7 = –6 + c c = 7 + 6 = 13 y = x + 13

(c) x – y = 1 6 4

Matematik  Tingkatan 3  Bab 9

19. (a) m = 6 – (–2) 0–4 8 = –4 = –2 y = –2x + c Pada/ At (0, 6), 6 = –2(0) + c c = 6 y = –2x + 6

(d) x – y = 1 3 2

(g) mM = 8 – 6 0–4 = 2 –4 = – 1 2 mN = – 1 2 Pada/ At (0, 1), 1 1 = – (0) + c 2 c = 1 1 y = – x + 1 2

–2 – 1 5 – 11 = –3 –6 = 1 2

(b) m =

(c) x + y = 1 8 5

y = 1 x + c 2 Pada/ At (11, 1), 1 1 = (11) + c 2 c = 1 – 11 2 9 = – 2 1 9 y = x – 2 2 (e) m = 2 y = 2x + c Pada/ At (1, 4), 4 = 2(1) + c c = 4 – 2 = 2

(f ) m =

2 3

y = 2x + 2

2x+c 3 Pada/ At (2, 1), 2 1 = (2) + c 3 c = 1 – 4 3 1 = – 3 2 1 y = x – 3 3

(h) 3x + y = 2 y = –3x + 2 mB = –3 mA = –3

(i) mC = – 4 1 = –4 mD = –4

Pada/ At (0, 8), 8 = –3(0) + c c = 8 y = –3x + 8

Pada/ At (2, 0), 0 = –4(2) + c c = 8 y = –4x + 8

7

y =

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 9

20. (a) x – 5y + 17 = 0 …… 1 2x + y = –1 ……… 2

(b) 5x – 3y – 8 = 0 …… 1 –3y = –2x + 5 …… 2

Daripada/ From 1, x = 5y – 17 ……… 3

Daripada/ From 2, 2 5 y = x – ……… 3 3 3

Gantikan 3 ke dalam 2,

Gantikan 3 ke dalam 1,

Substitute 3 into 2,

Substitute 3 into 1,

2(5y – 17) + y = −1 10y – 34 + y = −1 11y = 33 y = 3

5x – 3( 2 x – 5 ) – 8 = 0 3 3 5x – 2x + 5 – 8 = 0 3x = 3 x = 1

Gantikan y = 3 ke dalam 3,



Substitute y = 3 into 3,

Gantikan x = 1 ke dalam 3,

x = 5(3) – 17 = 15 – 17 = –2



Substitute x = 1 into 3,

y = 2 (1) – 5 3 3 3 = – 3 = –1

Titik persilangan = (–2, 3) Point of intersection



Titik persilangan = (1, –1) Point of intersection

x + 5y = 1 …… 1 2 6 5x – y =10 ……… 2

(c) 4x – 7y = 0 ……… 1 8x – y – 26 = 0 …… 2

(d)

Daripada/ From 2, y = 8x – 26 ……… 3

1 × 6:  3x + 5y = 6 …… 3 Daripada/ From 2, y = 5x – 10 ……… 4

Gantikan 3 ke dalam 1, Substitute 3 into 1,

4x – 7(8x – 26) = 0 4x – 56x + 182 = 0 52x = 182 7 x = 2 7 Gantikan x = ke dalam 3, 2

Substitute x =

Gantikan 4 ke dalam 3, Substitute 4 into 3,

3x + 5(5x – 10) = 6 3x + 25x – 50 = 6 28x = 56 x = 2

7 into 3, 2

Gantikan x = 2 ke dalam 4,

y = 8( 7 ) – 26 2 = 28 – 26 = 2





7 Titik persilangan = ( , 2) 2 Point of intersection





© Penerbitan Pelangi Sdn. Bhd.

Substitute x = 2 into 4,

y = 5(2) – 10 = 10 – 10 = 0

8

Titik persilangan = (2, 0) Point of intersection

Matematik  Tingkatan 3  Bab 9

21. (a) 7x – 4y = –7 ……… 1 5x + y = 22 ……… 2

(b) x – 2y = 15 ……… 1 3y = –2x + 2 ……… 2

2 × 4: 20x + 4y = 88 …… 3

1 × 2: 2x – 4y = 30 ……… 3

1 + 3 : 27x = 81   x=3

Daripada/ From 2, 2x + 3y = 2 …… 4 4 – 3 : 7y = −28   y = –4

Gantikan x = 3 ke dalam 1,



Substitute x = 3 into 1,

7(3) – 4y = –7 4y = 21 + 7 4y = 28 y = 7

Gantikan y = –4 ke dalam 1,

x – 2(–4) = 15 x + 8 = 15 x = 7

Titik persilangan = (3, 7) Point of intersection



(c) x – 3y + 12 = 0 …… 1 2x + 4y = – 4 ……… 2

3 × 3: 3x – 15y = 48 ……. 4 2 + 4 : −8y = 40   y = −5

Gantikan y = 2 ke dalam 2, Substitute y = 2 into 2,

2x + 4(2) = – 4 2x = –4 – 8 2x = –12 x = –6

Point of intersection

1 × 5: 5y = x – 16 x − 5y = 16 …… 3

2 – 3 : 10y = 20   y=2



Titik persilangan = (7, –4)

(d) y = 1 x – 16 …… 1 5 5 –3x + 7y = –8 …… 2

1 × 2: 2x – 6y + 24 = 0 2x – 6y = −24 …… 3



Substitute y = –4 into 1,

Gantikan y = –5 ke dalam 3,

Substitute y = –5 into 3,

x − 5(–5) = 16 x + 25 = 16 x = −9

Titik persilangan = (–6, 2) Point of intersection





9

Titik persilangan = (−9, –5)

Point of intersection

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 9

22. (a) y = x + 4 x y

0 4

(b) 4x – y = –1 x 0 y 1

–4 0

y = –x + 2 x y

0 2

x–y=2 x y

2 0

0 –2

y

y = –x + 2

4

1 5 1 –1

y 4x – y = –1

4

y=x+4

2

x–y=2

2

–4

–2

O

–4

x

2

–2

O

2

x

–2

–2 –4





Titik persilangan = (−1, 3) Point of intersection



0 1

1 0

y 4

–

y 4

x=3

O

2

–2

x

4



y = – x +1





Titik persilangan = (3, –2) Point of intersection



23.

(a) 5x + 7y − 35 = 0 5x + 7y = 35 5x + 7y = 35 35 35 35 x + y = 1 7 5 pintasan-x/ x-intercept = 7, pintasan-y/ y-intercept = 5   Luas kawasan AOB = 1 × 7 × 5 2 The area of the region AOB = 17.5 unit2

© Penerbitan Pelangi Sdn. Bhd.

2

–4

O

–2

x y + =1 2 3

2

x

–4



–4



x y + =1 3 3

–2

2

–2

Point of intersection

(d)

(c) y = –x + 1 x y

Titik persilangan = (−1, –3)

10

Titik persilangan = (0, 3) Point of intersection

Matematik  Tingkatan 3  Bab 9

(b) (i) 25 cm (ii) x = 4, y = 1 (4) + 25 2 = 2 + 25 = 27



Ketinggian selepas 4 minggu ialah 27 cm. The height after 4 weeks is 27 cm.

(iii) y = 45, 1 x + 25 = 45 2 1 x = 20 2 x = 40

Bilangan minggu yang diperlukan ialah 40 minggu. The number of weeks needed is 40 weeks.

(c) (i)

Pilihan

Deposit

Harga sejam

Persamaan

A B

RM10 RM15

RM4 RM2

y = 4x + 10 y = 2x + 15

Option

(ii) 4x + 10 = 2x + 15 4x − 2x = 15 − 10 2x = 5 x = 2.5

Deposit

Price per hour

Equation

Pada 2.5 jam, kos bayaran kedua-dua pilihan itu adalah sama.

At 2.5 hours, the two options cost the same amount.

(iii) (a) Pilihan A/ Option A: y = 4(2) + 10 = 8 + 10 = 18

Pilihan B/ Option B: y = 2(2) + 15 = 4 + 15 = 19

Pilihan A lebih baik jika Halim ingin menyewa 2 jam kerana lebih murah.

Option A is better if Halim wants to rent for 2 hours because it is cheaper.

(b) Pilihan A/ Option A: Pilihan B/ Option B: y = 4(5) + 10 y = 2(5) + 15 = 20 + 10 = 10 + 15 = 30 = 25 Pilihan B lebih baik jika Halim ingin menyewa 5 jam kerana lebih murah.

Option B is better if Halim wants to rent for 5 hours because it is cheaper.

(d) Jalan A/ Road A: y = −x + 3 ……… 1 Jalan B/ Road B: 2y = 3x + 1 ……… 2 Gantikan 1 ke dalam 2,

Substitute 1 into 2,

2(−x + 3) = 3x + 1 −2x + 6 = 3x + 1 −2x − 3x = 1 – 6 −5x = −5 x = 1

Gantikan x = 1 ke dalam 1,



Substitute x = 1 into 1,

y = –1 + 3 = 2

Koordinat titik pada kolam air pancut = (1, 2) The coordinates of point at the fountain

11

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Bab 9

Praktis PT3 1. (a)

2x + y = –1

(b) (i) Daripada y = 2x + 3, kecerunan = 2 From y = 2x + 3, gradient = 2

•

–4x + y = –1 • 7 x + y =1 4 7

•

• •

– 7 4

(ii) Kecerunan garis lurus I Gradient of straight line I



–2

•

28

Straight lines I and II are not parallel. The gradients 1 of both the straight lines are not equal, 2 ≠ – . 2

(b) (i) 3 x – y = 0 4 3 y = x 4 4x + y = 1 (ii) 7 7 4x + y = 7 y = –4x + 7

(c) (i) mSR = mPQ = 7 – 5 1–9 = 2 –8 =– 1 4 1 Pada/ At (−7, −1), y = –   x + c 4 1 −1 = – (−7) + c 4 c = −1 − 7 4 11 = – 4

(c) Jalan Tenteram: y = 2x − 15 m = 2

=– 1 2 Garis lurus I dan II tidak selari. Kecerunan kedua-dua garis lurus itu tidak sama, 2≠– 1. 2

Jadi, kecerunan bagi Jalan Aman = 2 So, the gradient of Jalan Aman

Persamaan SR/ Equation of SR: 1 11 y=– x– 4 4

Diberi/ Given (4, 2), y = 2x + c 2 = 2(4) + c c = 2 − 8 = –6

(ii) R ialah titik persilangan SR dan QR. R is an intersection point of SR and QR.



Persamaan Jalan Aman:

Persamaan QR/ Equation of QR:

y = 2x – 6



Equation of Jalan Aman:

x = 9 

QR selari dengan garis mancancang TU, x = 6

QR is parallel to vertical line TU, x = 6

Gantikan x = 9 ke dalam y = – 1 x – 11 , 4 4

6 2. (a) mAB = – 3 = −2 mCD = −2

Substitute x = 9 into y = –

y = – 1 (9) – 11 4 4 9 11 = – – 4 4 = −5 Koordinat titik R = (9, −5)

y = –2x + c

Pada/ At (1, 4), 4 = −2(1) + c c = 4 + 2 = 6

Coordinates of point R

3. (a) (i) −1 (ii) 3  

Persamaan CD/ Equation of CD: y = −2x + 6

(iii) 3

© Penerbitan Pelangi Sdn. Bhd.

1 11 x– , 4 4

12

Pintasan-y/y-intercept =

c 9 = b 3

Matematik  Tingkatan 3  Bab 9

(c) (i) 40%

(b) (i) Sebelah kiri: Left hand side:



y=7



Sebelah kanan:

(ii) c = 40 Titik (20, 80) daripada graf, Point (20, 80) from the graph,

Right hand side:

80 = m(20) + 40 20m = 40 m = 2

−5x + 2 = −5(−1) + 2 = 5 + 2 = 7

7=7 \ Titik Q terletak pada garis lurus.

Point Q is on the straight line.



Persamaan garis lurus itu ialah



y = 2x + 40.

The equation of the straight line is

(ii) Sebelah kiri: Left hand side:

−4x + 2y = −4(0) + 2(6) = 0 + 12 = 12

Sebelah kanan:

Right hand side:

12

12 = 12 \ Titik Q terletak pada garis lurus.

Point Q is on the straight line.

Cabaran KBAT 1. (i)

Yuran(RM) Fee (RM) 140

Pakej A Package

Pakej B Package

120

100

80

60

40

20



O

2

4

6

8

10

12

14

Bilangan minggu Number of weeks

(ii) Minggu ke-4, iaitu RM80. / 4th week, that is RM80. (iii) Yuran pakej A/ Fee of package A = 40 + 10(10) = 40 + 100 = RM140 Yuran pakej B/ Fee of package B = 20(10) = RM200 Beza bayaran/ Difference of payment = 200 − 140 = RM60

13

© Penerbitan Pelangi Sdn. Bhd.

Jawapan Penilaian Akhir Tahun 1. (a) X = 59,  Y = 52,  Z = 38 (b) 0.00025 – 1.3 × 10–4 = 2.5 × 10

1

–4



=   2.5

–



= 1.2 × 10–4

– 1.3 × 10–4 1.3

2 × 10

–4

(c) Masa = 1 jam 40 minit = 1 40 j 60 2 =1 j 3

Time = 1 hour 40 minutes 40 h =1 60 2 =1 h 3

Laju/ Speed 160 = 2 1 3 = 96 km/j  (96 km/h) Pecutan/ Acceleration 105 – 96 = 0.5 = 18 km/j2   (18 km/h2)

2. (a)

Sisi bertentangan

Sisi bersebelahan

Hipotenus

AB

BC

AC

Opposite side

Adjacent side

Hypotenuse

1

2 n (b)  23n – 1 = 16 –2× 4 2 4 × 4n = –2 2­ 2 2n = 2 ×–22 2 = 22 + 2n – (–2) = 24 + 2n

3n – 1 = 4 + 2n n = 5

1

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Penilaian Akhir Tahun

(c)

P

S

4 cm

Q



R

3. (a)

1 9

–32

1– 13 2

9

3–2

–9

–2

(b) sin x =

15 17

tan y = 8 15   AD = √152 + 202  = 25 cm kos z = 20   (cos z = 20 ) 25 25 = 4 5 (c) (i) 3x + 2.5y (ii) 3(2) + 2.5(6) +2(5) = 31 Tidak boleh. Ini kerana jumlah pembelian melebihi wangnya. No. Because the total purchase is more than his money.

4. (a) 60 852

© Penerbitan Pelangi Sdn. Bhd.

61 703

61 024

2

60 593

60 123

Matematik  Tingkatan 3  Penilaian Akhir Tahun

(b) (i) C (ii) x2 – 8xy + 16y2 = (x – 4y)(x – 4y) Panjang kertas/ Length of paper (cm) = x – 4y (c) (i) Mod/ Mode = 37, median = 36 (ii) Saiz 37. Sebab saiz kasut ini mempunyai permintaan yang paling tinggi. Size 37. Because this size has the highest demand.

5. (a) 36, 49, 64, 81, 100 (mana-mana tiga/ any three of them) (b) (ii) 3 (iii) 3 (v) 3 (ii) 4.5x + 6(1.5) = 18 x = 2 Jisim betik ialah 2 kg.

(c) (i) 4.5x + 3(2y) = 18 4.5x + 6y = 18

The mass of papaya is 2 kg.

6. (a) (i) 7 (ii) 3  (b) (i) (ii)

7 6

(iii) 7 

14 9

P P

O

(c) 300 × 200 × 500 = 30 000 000 cm3 = 3 × 107 cm3 7. (a) Jejari

Radius

Sektor minor Minor sector

Lilitan bulatan Circumference

(b) Jumlah bayaran balik Total repayment

= 24 000 + (24 000 × 0.06 × 5) = RM31 200 Bayaran ansuran bulanan Monthly instalment

= 31 200 5 × 12 = RM520

3

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Penilaian Akhir Tahun

(c) (i) P(bola kuning) P(a yellow ball)

= 1 − 1 − 10 4 32 7 = 16

(ii) Bilangan bola hijau

P(bola hijau)

Number of green balls

P(a green ball)

= 1 × 32 4 =8

8. (a)

8 32 – 4 = 2 7

=

36

72 18

48

88

56



108

(b) 4x − 3 < 7 + 6x , 4x − 6x < 7 + 3 −2x < 10 x > −5

2x + 4 . 3x 4 . 3x – 2x 4 . x x , 4

Berdasarkan garis nombor/ Based on the number line, x > m, x , n m = −5, n = 4 (c) (i) K L

12

2 6

1

4

3

(ii) L' = {1, 3} 9. (a) (i) 70 (ii) 15 (iii) 5 (i) 0.06 : 0.15 : 2.1 = 6   : 15   : 210 ÷ 3 ÷ 3 ÷3 = 2  : 5   : 70 (ii) 1 : 3 = 1 × 20 : 3 × 20   5 4 5 4 = 4 : 15

© Penerbitan Pelangi Sdn. Bhd.

4

Matematik  Tingkatan 3  Penilaian Akhir Tahun

(iii) 1 1 : 8 = 5 : 8     atau/ or 4 4 = 5 × 4 : 8 × 4 4 = 5 : 32

1 : 8 = 1.25 : 8 4 = 125 : 800 =   5 : 32

1

(b) x = 35 ∠SQU = 180° – 83° = 97° y° = 180° – 35° – 97° y = 48 (c) (i)

K/F

L

G

5 cm

8 cm

P/Q

N/E

(ii)

M

R/S

2 cm

M/T

H

5 cm

5 cm

R/P

L/K

2 cm

T/N

S/Q

4 cm

H/E

G/F

8 cm

5

© Penerbitan Pelangi Sdn. Bhd.

Matematik  Tingkatan 3  Penilaian Akhir Tahun

10. (a) 15 ,  3 ,  10 (b) (i) y = –1 (ii) Kecerunan PQ Gradient of PQ

= 2 – (–1) 0 – (–2) = 3 2 Persamaan PQ Equation of PQ



y= 3x+2 2

(c) (i) Sudut dongak puncak pokok dari kereta Angle of elevation of top of tree from car

= 180° − 110° − 52° = 18° (ii) Sudut tunduk puncak pokok dari titik P Angle of depression of top of tree from point P



= 30° − 18° = 12°

© Penerbitan Pelangi Sdn. Bhd.

6