1 © Penerbitan Pelangi Sdn. Bhd. 1. (a) 45 = 4 × 4 × 4 × 4 × 4 (b) 1 2 2 2 = 1 2 × 1 2 (c) (0.01)3 = 0.01 × 0.01 × 0.01 (d) b4 = b × b × b × b (e) (uvw) 2 = uvw × uvw 2. (a) 4 × 4 × 4 × 4 × 4 × 4 × 4 = 47 (b) 1 2 2 4 = 1 2 × 1 2 × 1 2 × 1 2 (c) 10 × 10 × 10 = 103 (d) 2 5 × 2 5 × 2 5 = 2 5 2 3 (e) r × r × r × r × r = r5 3. (a) 36 = 3 × 3 × 3 × 3 × 3 × 3 = 729 (b) 1 7 2 2 = 1 7 × 1 7 = 49 (c) (–4)3 = –4 × –4 × –4 = –64 (d) 104 = 10 × 10 × 10 × 10 = 10 000 (e) 2 5 2 3 = 2 5 × 2 5 × 2 5 = 8 125 4. (a) 625 (asas 5) = 5 × 5 × 5 × 5 = 54 (b) 729 (asas 9) = 9 × 9 × 9 = 93 (c) 2 401 (asas 7) = 7 × 7 × 7 × 7 = 74 (d) 27 64 (asas 3 4 ) = 3 4 × 3 4 × 3 4 = 3 4 2 2 (e) – 64 125 (asas – 4 5 ) = – 4 5 × – 4 5 × – 4 5 = – 4 5 2 Strateg2 iPT3 21 (–1)5 5. (a) (i) 34 × 35 = 34 + 5 = 39 (ii) 45 × 46 = 45 + 6 = 411 (b) (i) p5 × p9 = p5 + 9 = p14 (ii) q7 × q5 = q7 + 5 = q12 6. (a) (i) 25 ÷ 22 = 25 – 2 = 23 Indeks Indices 1 Bidang Pembelajaran: Nombor dan Operasi
Matematik Tingkatan 3 Bab 1 Indeks © Penerbitan Pelangi Sdn. Bhd. 2 (ii) 55 × 53 = 55 – 3 = 52 (b) (i) u9 ÷ u7 = u9 – 7 = u2 (ii) v6 ÷ v2 = v6 – 2 = v4 7. (a) (i) 5k2 × 6k6 = 5 × 6 × k2 × k6 = 30 × k2 + 6 = 30k8 (ii) 2y8 × 3y4 = 2 × 3 × y8 × y4 = 6 × y8 + 4 = 6y12 (b) (i) 42y6 ÷ 6y2 = 42y6 6y2 = 426 × y6 – 2 = 7y4 (ii) 56z11 ÷ 7z4 = 56z11 7z4 = 567 × z11 – 4 = 8z Strateg 7 iPT3 2 a7 ÷ a5 = a7 – 5 = a2 8. (a) (i) (32)2 = 32 × 2 = 34 (ii) (42)3 = 42 × 3 = 66 (b) (i) (b4)3 = b4 × 3 = b12 (ii) (c5)4 = c5 × 4 = c20 9. (a) (i) (3 × 4)2 = 32 × 42 = 9 × 16 = 144 (ii) (2 × 5)3 = 23 × 53 = 8 × 125 = 1 000 (b) (i) (hk)8 = h8 × k8 = h8k8 (ii) (pq)11 = p11 × q11 = p11q11 10. (a) (i) (f 5g)3 = f 5 × 3g1 × 3 = f 15g3 (ii) (r2s7)4 = r2 × 4s7 × 4 = r8s28 (b) (i) (6j6)3 = 63 × j 6 × 3 = 216j 18 (ii) (4w3)3 = 43 × w3 × 3 = 64w9 11. (a) (–3)0 = 1 (b) 12 20 = 1 12. (a) 2–3 = 123 (b) 3–4 = 134 (c) 12–8 = 1 128
Matematik Tingkatan 3 Bab 1 Indeks 3 © Penerbitan Pelangi Sdn. Bhd. (d) k–5 = 1k5 (e) b–13 = 1b13 13. (a) 12 = 2–1 (b) 1 252 = 25–2 (c) 196 = 9–6 (d) 1n12 = n–12 (e) 1v4 = v–4 14. (a) 27 1—3 = 3√— 27 (b) 81 1—4 = 4√— 81 (c) b 1—2 = √—b (d) 2 187 1—7 = 7√— 2 187 (e) 25 32 2 1—3 = 3 25 32 15. (a) 3√— 12 = 12 1—3 (b) 5√— 4 = 4 1—5 (c) √—q = q 1—2 (d) 4√—m = m 1—4 (e) 3 64 125 = 64 125 2 1—3 16. (a) 64 3—4 = (4√— 64)3 (b) 125 2—3 = (3√— 125)2 (c) t 4—3 = (3√— t )4 (d) 1 000 6—5 = (5√— 1 000 )6 (e) 14 3—2 = (√— 14 ) Strateg 3 iPT3 23 m10 m5 m 3—2 1m2 m2 × m3 (m2)5 m–2 √—m3 17. (a) 27 2—3 = (33) 2—3 = 33 × 2—3 = 32 = 9 (b) 16 5—4 = (24) 5—4 = 24 × 5—4 = 25 = 32 (c) 216 2—3 = (63) 2—3 = 63 × 2—3 = 62 = 36 (d) 81 3—2 = (92) 3—2 = 92 × 3—2 = 93 = 729
Matematik Tingkatan 3 Bab 1 Indeks © Penerbitan Pelangi Sdn. Bhd. 4 (e) 125 4—3 = (53) 4—3 = 53 × 4—3 = 54 = 625 18. (a) 28 ÷ (24 × 3–2)2 = 28 ÷ (24 × 2 × 3–2 × 2) = 28 ÷ 28 ÷ 3–4 = 28 – 8 3–4 = 20 3–4 = 13–4 = 34 = 81 (b) (52 ÷ 22)2 ÷ (2–2)2 = 52 × 2 ÷ 22 × 2 ÷ 2–2 × 2 = 54 ÷ 24 ÷ 2–4 = 54 24 × 2–4 = 54 24 – 4 = 54 20 = 54 = 625 19. (a) 2 1—2 × 5 – 1—2 × 10 – 3—2 = 2 1—2 × 5 – 1—2 (2 × 5) – 3—2 = 2 1—2 – 3—2 × 5 – 1—2 – 3—2 = 2–1 × 5–2 = 12 × 152 = 1 50 (b) 2 1—2 × 3 1—2 × 6 1—2 = 2 1—2 × 3 1—2 × (2 × 3) 1—2 = 2 1—2 + 1—2 × 3 1—2 + 1—2 = 2 × 3 = 6 (c) 42n × 162n × 64–2n = (22)2n × (24)2n × (26)–2n = 24n × 28n × 2–12n = 24n + 8n – 12n = 20 = 1 (d) 27 2—3 × 25 3—2 = (3√— 27)2 × (√— 25)3 = 32 × 53 = 9 × 125 = 1 125 (e) (84 ÷ 27) 1—3 = ((23)4 ÷ 33) 1—3 = 23 × 4 ÷ 33 × 1—3 = 24 ÷ 3 = 16 Strateg 3 iPT3 24 22 × 32 23 × 81 = 22 – 3 × 32 – 4 = 2–1 × 3–2 = 12 × 19 = 1 18 20. (a) 3x4y × 4xy3 = 3 × 4 × x4 + 1 × y1 + 3 = 12x5y4 (b) 27m4n2 ÷ 9m2n = 279 × m4 – 2 × n2 – 1 = 3m2n (c) (2a2b–2)2 × a3b 2 –1 = 22 × a2 × 2 + 3 × –1 × b–2 × 2 + (–1 × –1) = 4ab–3 = 4ab3
Matematik Tingkatan 3 Bab 1 Indeks 5 © Penerbitan Pelangi Sdn. Bhd. 21. (a) 24a3b2 × 4b4 8ab–2 × b–4 = 24 × 4a3b2 + 4 8ab–2 – 4 = 96a3b6 8ab–6 = 12a3 – 1 b6 – (–6) = 12a2b12 (b) (p4)3 ÷ (p–2)4 p–4q5 × (p12q4) 1—4 = p4 × 3 ÷ p–2 × 4 p–4q5 × p 12 × 1—4 q 4 × 1—4 = (p12 ÷ p–8) p–4 + 3q5 + 1 = p12 – (–8) – (–1) q6 = q21 q6 (c) (2u–3v –2)2 ÷ 2u2v –6 2 1—2 u4v–2 ÷ (23u10v4) 1—2 = 22u–3 × 2v–2 × 2 ÷ 2u2v–6 2 1—2 u4v–2 ÷ 23 × 1—2 u 10 × 1—2 v 4 × 1—2 = 22u–6v–4 × 2 ÷ 2u2v–6 2 1—2 u4v–2 ÷ 2 3—2 u5v2 = 22 – 1u–6 – 2v–8 – (–6) 2 1—2 – 3—2 u4 – 5v –2 – 2 = 2u –8v2 2–1u–1v–4 = 21 – (–1)u–8 – (–1)v2 – (–4) = 22u–7v6 22. (a) 32n + 52 = 106 32n + 52 = 106 32n + 52 = 106 32n = 106 – 25 32n = 81 32n = 34 2n = 4 n = 2 (b) 3n × 3 = 35 × 33 3n + 1 = 35 + 3 n + 1 = 8 n = 7 (c) 16(3n ) = 81(2n) 32 2n = 81 16 32 2n = 32 24 n = 4 (d) 22n × 2 = 8 2—3 22n × 2 = (23) 2—3 22n + 1 = 22 2n + 1 = 2 2n = 1 n = 12 (e) 4n = 82n – 1 4n = 82n – 1 22n = (23)2n – 1 22n = 23 × (2n – 1) 22n = 26n – 3 2n = 6n – 3 4n = 3 n = 34 1. (a) Pernyataan Statements Benar / Palsu True / False 3√— 64 = 4 Benar True 32 3—5 = 8 Benar True (b) (i) a5 ÷ a3 = a5 – 3 = a–2 = 1a2
Matematik Tingkatan 3 Bab 1 Indeks © Penerbitan Pelangi Sdn. Bhd. 6 (ii) 2–2 × 33 4–2 × 81 = 2–2 × 33 (22 )–3 × 34 = 2–2 – (–6) × 33 – 4 = 24 × 3–1 = 16 × 1 3 = 5 1 3 (c) (i) k–2 × k3 = k–2 + 3 = k (ii) 22y – 1 = 16 22y – 1 = 24 2y – 1 = 4 2y = 5 y = 2 1 2
1 © Penerbitan Pelangi Sdn. Bhd. 1. (a) 21.1 3 (b) 308 3 (c) 2.001 4 (d) 2.345 4 (e) 0.0012 3 2. Nombor Numbers Angka Bererti Significant Figures 1 2 3 (a) 459 500 460 459 (b) 123 100 120 123 (c) 1 986 2 000 2 000 1 990 (d) 3 768 4 000 3 800 3 770 (e) 102 889 100 000 100 000 103 000 3. Nombor Numbers Angka Bererti Significant Figures 1 2 3 (a) 1.591 2.000 1.600 1.590 (b) 4.502 5.000 4.500 4.500 (c) 11.976 10.000 12.000 12.000 (d) 3.76 4.000 3.80 3.76 (e) 1.99 2.00 2.00 1.99 4. Nombor Numbers Angka Bererti Significant Figures 1 2 3 (a) 0.0591 0.06 0.059 0.0591 (b) 0.005028 0.005 0.0050 0.00503 (c) 0.11976 0.1 0.12 0.120 (d) 0.376 0.4 0.38 0.376 (e) 0.0199 0.02 0.020 0.0199 5. (a) 4.16 + 5.28 – 2.34 = 7.10 = 7.0 (b) 213 + 384 – 327 = 270 = 300 (c) 38 × 96 – 2 265 = 3 648 – 2 265 = 1 383 = 1 000 (d) 35 92 – 0.353 = 0.3804 – 0.353 = 0.0274 = 0.03 (e) 56.8 – 45.6 102 + 98 = 11.2 200 = 0.0056 = 0006 Bentuk Piawai 2 Standard Form Bidang Pembelajaran: Nombor dan Operasi
Matematik Tingkatan 3 Bab 2 Bentuk Piawai © Penerbitan Pelangi Sdn. Bhd. 2 6. (a) 880 – 5.13 ÷ 0.03 = 880 – 171 = 709 = 710 (b) 4.123 + 7.353 × 5.459 – 6.467 = 4.123 + 40.14 – 6.467 = 37.796 = 38.0 (c) 2.34 + 4.23 – 0.3 × 4.18 = 2.34 + 4.23 – 1.254 = 5.316 = 5.3 (d) 0.4765 + 0.9887 – 0.540 9 = 0.4765 + 0.9887 – 0.06 = 1.4052 = 1.40 (e) 54.27 – 18.08 – 75 46 = 54.27 – 18.08 – 1.63 = 34.56 = 35.0 7. (a) 12.3 × 8.12 ÷ 300 = 0.0033292 = 0.00333 (b) (11.2 + 0.8) × (1.5 – 0.9) = 12 × 0.6 = 7.2 = 7.20 (c) 44 × 26 0.4 + 23.08 = 44 × 65 + 23.08 = 2 860 + 23.08 = 2 883.08 = 2 880 (d) 41.625 × 5 73.214 × 8 = 20.8125 585.712 = 0.03553 = 0.0355 (e) 92.823 × 36.705 + 65.68 × 23 = 3 407.068 + 1 510.64 = 4 917.708 = 4 920 8. (a) 1 450 = 1.45 × 103 (b) 16 = 1.6 × 10 (c) 32.5 = 3.25 × 10 (d) 400.8 = 4.008 × 102 (e) 1 800 000 = 1.8 × 106 9. (a) 4.5 × 10 = 45 (b) 4.56 × 103 = 4 560 (c) 7 × 106 = 7 000 000 (d) 0.026 × 106 = 26 000 (e) 0.079 × 107 = 790 000 10. (a) 0.00045 = 4.5 × 10–4 (b) 0.000016 = 1.6 × 10–6 (c) 0.0000018 = 1.8 × 10–6 (d) 0.000036 = 3.6 × 10–5 (e) 0.00000423 = 4.23 × 10–6 11. (a) 4.5 × 10–2 = 0.045 (b) 5.6 × 10–4 = 0.000 56 (c) 1.67 × 10–7 = 0.000000167 (d) 1.4 × 10–4 = 0.00014 (e) 6 × 10–7 = 0.0000006 12. (a) 6.2 × 106 + 2.4 × 106 = (6.2 + 2.4) × 106 = 8.6 × 106 (b) 6.4 × 103 + 5.2 × 103 = (6.4 + 5.2) × 103 = 11.6 × 103 = 1.16 × 104 (c) 9.4 × 104 + 2.3 × 104 = (9.4 + 2.3) × 104 = 11.7 × 104 = 1.17 × 105 13. (a) 2.45 × 103 – 1.21 × 103 = (3.45 – 1.21) × 103 = 2.24 × 103
Matematik Tingkatan 3 Bab 2 Bentuk Piawai 3 © Penerbitan Pelangi Sdn. Bhd. (b) 4.3 × 104 – 2.5 × 104 = (4.3 – 2.5) × 104 = 0.8 × 104 = 8 × 103 (c) 6.8 × 106 – 3.2 × 106 = (6.8 – 3.2) × 106 = 3.6 × 106 14. (a) 3 × 5 × 106 = 15 × 106 = 1.5 × 107 (b) 26 × 4 × 105 = 10.4 × 105 = 1.04 × 106 (c) 6.3 × 5 × 104 = 31.5 × 104 = 3.15 × 105 15. (a) 3.6 × 106 6 = 0.6 × 106 = 6 × 105 (b) 4.5 × 106 6 = 0.75 × 106 = 7.5 × 105 (c) 8.0 × 103 5 = 1.6 × 103 16. (a) 2.6 × 107 + 5.2 × 108 = 0.26 × 108 + 5.2 × 108 = (0.26 + 5.2) × 108 = 5.46 × 108 (b) 7.4 × 106 + 4.5 × 105 = 7.4 × 106 + 0.45 × 106 = (7.4 + 0.45) × 106 = 7.85 × 106 (c) 8.3 × 103 + 1.6 × 104 = 0.83 × 104 + 1.6 × 104 = (0.83 + 1.6) × 104 = 2.43 × 104 17. (a) 3.5 × 105 – 5.2 × 104 = 3.5 × 105 – 0.52 × 105 = (3.5 – 0.52) × 105 = 3.02 × 105 (b) 9.7 × 107 – 8.8 × 106 = 9.7 × 107 – 0.88 × 107 = (9.7 – 0.88) × 107 = 8.82 × 107 (c) 5.5 × 107 – 5.5 × 106 = 5.5 × 107 – 0.55 × 107 = (5.5 – 0.55) × 107 = 4.95 × 107 18. (a) 7.0 × 106 × 4.6 × 107 = 7.0 × 4.6 × 106 + 7 = 32.2 × 1013 = 3.22 × 1014 (b) 0.0076 × 0.00245 = 7.6 × 10–3 + 2.45 × 10–3 = 7.6 × 2.45 × 10–3 + (–3) = 18.62 × 10–6 = 1.862 × 10–5 (c) 2.33 × 104 × 1.2 × 10–2 = 2.33 × 1.2 × 104 – (–2) = 2.796 × 106 19. (a) 3.4 × 10–10 40 000 = 3.4 × 1010 4 × 104 = 3.4 4 × 1010 – 4 = 8.5 × 106 (b) 0.225 × 400 4 × 104 = 0.225 × 4 × 102 4 × 104 = 0.225 × 4 4 × 102 – 4 = 0.225 × 10–2
Matematik Tingkatan 3 Bab 2 Bentuk Piawai © Penerbitan Pelangi Sdn. Bhd. 4 (c) 540 × 110 ÷ 1 000 = 59 400 ÷ 103 = 5.94 × 104 ÷ 103 = 5.94 × 104 – 3 = 5.94 × 10 20. (a) Wang simpanan = 3 000 × 100 – 10 – 50 – 10 100 = 3 000 × 30 100 = 900 = RM9 × 102 (b) Beza tinggi 1.66 – [1.66 – 1.5 ÷ 100] 1 m = 100 cm = 1.66 + [1.66 – 0.015] = 1.66 + 1.645 = 3.305 (c) Jisim purata = (43.5 + 57.8 + 88.7) ÷ 3 = 190 ÷ 3 = 63.33 = 6.33 × 10 kg 21. (a) (i) Luas permukaan = 2(20 × 6) + 2(20 × 10) + 2(6 × 10) = 240 + 400 + 120 = 760 = 7.6 × 102 cm2 (ii) Isi padu = 20 × 6 × 10 = 1200 = 1.2 × 103 cm3 1. (a) (i) 0.00004 (ii) 3.8 × 10–5 (b) (i) (3.1 – 1.43) ÷ 25 = 1.67 ÷ 25 = 0.0668 = 0.067 (ii) 8.8 – 0.8 × 0.88 = 8.8 – 0.00704 = 8.096 = 8.1 (c) (i) 12 000 (ii) 200 000 × 11.5 = 2 300 000 g 2 300 000 ÷ 1 000 = 2 300 kg = 2.3 × 103 kg
1 © Penerbitan Pelangi Sdn. Bhd. Matematik Pengguna: Simpanan dan Pelaburan, Kredit dan Hutang Consumer Mathematics: Savings and Investments, Credit and Debt 3 Bidang Pembelajaran: Nombor dan Operasi 1. (a) akaun simpanan tetap , akaun simpanan semasa (b) amanah saham , hartanah, saham, bon 2. Istilah kewangan Financial terms Definisi Definition Prinsipal Principal Pertukaran wang kepada asset dengan harapan asset itu menghasilkan keuntungan Exchange of money to asset with a hope to gain profits Faedah Interest Pegangan hak milik sesuatu syarikat Ownership of a company Pelaburan Investment Caj ke atas wang yang dipinjam atau dilabur Charge on the money borrowed or invested Saham Share Jumlah asal wang yang dipinjam atau dilabur Original amount of money borrowed or invested 3. Prinsipal Principal (RM) Kadar faedah mudah setahun Simple interest rate per annum Tempoh Duration Faedah mudah diperoleh Simple interest earned Jumlah simpanan Total savings (RM) (a) 4 000 2% 3 bulan 3 months I = 4 000 × 2 100 × 3 12 = 20.00 4 000 + 20.00 = 4 020.00 (b) 7 000 2.5% 2 tahun 2 years I = 7 000 × 2.5 100 × 2 = 350.00 7 000 + 350.00 = 7 350.00 (c) 2 000 1.5% 6 bulan 6 months I = 2 000 × 1.5 100 × 6 12 = 15.00 2 000 + 15.00 = 2 015.00 4. Saham . Hartanah . Akaun Simpanan
Matematik Tingkatan 3 Bab 3 Matematik Pengguna: Simpanan dan Pelaburan, Kredit dan Hutang © Penerbitan Pelangi Sdn. Bhd. 2 5. (a) Nilai matang = 5 000 1 + 0.03 1 2 1(6) = 5 970.26 (b) Nilai matang = 20 000 1 + 0.033 12 2 12(10) = 27 806.77 (c) Nilai matang = 14 000 1 + 0.032 2 2 2(5) = 16 408.36 6. (a) Jumlah pelaburan = 3 000(1.41) + 5 000(1.40) + 6 000(1.36) = RM19 390 Bilangan unit saham = 3 000 + 5 000 + 6 000 = 14 000 Kos purata seunit saham = RM19 390 14 000 = RM1.385 (b) Jumlah pelaburan = 3 000(1.05) + 5 000(1.03) + 4 000(1.00) + 2 000(1.06)) = RM14 420 Bilangan unit saham = 3 000 + 5 000 + 4 000 + 2 000 = 14 000 Kos purata seunit saham = RM14 420 14 000 = RM1.03 7. (a) Jumlah pulangan = (8 900 + 60(20)) – 8 000 = RM1 020 Nilai pulangan pelaburan = RM1 020 RM8 000 × 100% = 12.75% (b) (i) Jumlah pulangan = (11 400 + 150) – 10 000 = RM1 550 Nilai pulangan pelaburan = RM1 550 RM10 000 × 100% = 15.5% (ii) Puan Mariam mendapat 15.5% keuntungan melalui pelaburan ini. (c) Jumlah pulangan = 10 000 – 12 000 = –2 000 Nilai pulangan pelaburan = – RM2 000 RM12 000 × 100% = –16.67% Keadaan ekonomi dan politik yang tidak stabil boleh merugikan pelaburan. 8. (a) 6 157 = P1 + 0.032 4 2 4(5) 6 157 = P(1.008)20 P = 6 157 1.00820 P = 5 250 (b) (i) 1.02 = 7 650 jumlah unit pembelian Jumlah unit pembelian = 7 650 1.02 = 7 500 (ii) Biar x = harga seunit sesyer 12% = (7 500x + 100) – 7 650 7 650 × 100% 0.12(7 650) + 7 650 – 100 = 7 500x x = 1.129 \ RM1.129 seunit (c) (i) Kos purata seunit saham = 2 000(1.76) + 3 000(1.78) + 5 000(1.72) 1 000 + 3 000 + 5 000 = 1.746 (ii) Strategi Kok Liang lebih menguntungkan. Ini kerana purata kos sesyer lebih rendah berbanding dengan pembelian saham oleh Haris. (d) (i) Nilai pulangan pelaburan = 5 000(1.72) – 8 000 8 000 × 100% = 7.5% (ii) 0.80 = 8 000 Jumlah unit pembelian Jumlah unit pembelian = 8 000 0.80 = 10 000
Matematik Tingkatan 3 Bab 3 Matematik Pengguna: Simpanan dan Pelaburan, Kredit dan Hutang 3 © Penerbitan Pelangi Sdn. Bhd. Biar x = harga jualan unit amanah saham 7.5% – 2% = 10 000x – 8 000 8 000 × 100% 440 + 8 000 = 10 000x x = 0.844 9. Pernyataan Statement Tanda Mark (a) Siti membelanjakan lebih daripada kemampuan bayaran baliknya. Siti spends more than her repayment ability. (b) Aida menggunakan cara auto debit untuk membayar bil. Aida uses auto debit method to pay her bills. ✓ (c) Aiman membuat belanjawan setiap bulan. Aiman makes budget every month. ✓ (d) Hamid selalu lewat membuat pembayaran balik. Hamid always make late repayment. 10. Kelebihan Advantages A C Kelemahan Weaknesses B D Penggunaan kad kredit The use of credit card 11. Baki belum jelas Outstanding balance (RM) Bayaran balik dibuat dalam tempoh tanpa faedah Repayment done within interest free period Caj kewangan 18% setahun Finance charge 18% per annum Caj bayaran lewat 1% ke atas baki tertunggak Late payment charge 1% of the outstanding balance (a) 800 Tiada bayaran No payment 800 × 18 100 × 15 365 = 5.92 800 × 1% = RM8 (b) 1 500 Pembayaran minimum RM75 Minimum repayment of RM75 (1 500 – 75) × 18 100 × 15 365 = 10.54 0 (Ini kerana pembayaran minimum dibuat) (c) 2 000 Tiada bayaran No payment 2 000 × 18 100 × 15 365 = 14.97 2 000 × 1% = 20 12. (a) Caj kewangan = (RM1 200 – RM60) × 18 100 × 15 365 = RM8.43 Jumlah tunggakan pada bulan September = RM1 140 + RM8.43 = RM1 148.43
Matematik Tingkatan 3 Bab 3 Matematik Pengguna: Simpanan dan Pelaburan, Kredit dan Hutang © Penerbitan Pelangi Sdn. Bhd. 4 (b) (i) Caj kewangan = RM2 800 × 18 100 × 12 365 = RM16.57 Caj bayaran lewat = RM2 800 × 1% = RM28 Jumlah tunggakan pada bulan September = RM2 800 + RM16.57 + RM28 = RM2 844.57 (ii) Perbezaan = RM2 844.57 – RM2 800 = RM44.57 (c) (i) USD1.00 = RM3.80 Harga buku = USD52 Jumlah penyata kad kredit = USD52 × RM3.80 USD1.00 × 101% = RM199.58 (ii) Caj kewangan = RM199.58 × 18 100 × 15 365 = RM1.48 Caj bayaran lewat = RM199.58 × 1% = RM2 Jumlah tunggakan pada penyata bulan September = RM199.58 + RM1.48 + RM2 = RM203.06 13. Jumlah pinjaman Total loan (RM) Kadar faedah tahunan Interest rate per annum Tempoh pinjaman Loan period Jumlah bayaran balik Total repayment Bayaran ansuran (sebulan) (RM) Instalment (per month) (RM) (a) 60 000 7% 4 tahun 4 years 60 000 + (60 000 × 0.07 × 4) = 76 800 RM76 800 4 × 12 = 1 600.00 (b) 20 000 6% 3 tahun 3 years 20 000 + (20 000 × 0.06 × 3) = 23 600 RM161 200 3 × 12 = 655.56 (c) 80 000 5% 6 tahun 6 years 80 000 + (80 000 × 0.05 × 6) = 104 000 RM104 200 6 × 12 = 1 444.44 14. (a) (i) Jumlah bayaran balik = 54 000 + (54 000 × 0.07 × 6) = 76 680 Bayaran ansuran = RM76 680 6 × 12 = 1 065 (ii) Katakan x = tempoh bayaran balik baru. Jumlah bayaran balik = 54 000 + (54 000 × 0.07 × x) = 54000 + 3780x 877.50 = 54 000 + 3 780x x × 12 10 530x = 54 000 + 3 780x 6 750x = 54 000 x = 8 (b) Bulan pertama: Baki pinjaman = RM342 000 Faedah bulan pertama = RM342 000 × 4.9% 12 = RM1 368 Jumlah pinjaman pada akhir bulan pertama = RM342 000 + RM1 368 = RM342 368 Baki selepas bayaran ansuran = RM343 368 – RM1 697 = RM341 671
Matematik Tingkatan 3 Bab 3 Matematik Pengguna: Simpanan dan Pelaburan, Kredit dan Hutang 5 © Penerbitan Pelangi Sdn. Bhd. Bulan kedua: Baki pinjaman = RM341 671 Faedah bulan kedua = RM341 671 × 4.9% 12 = RM1 366.68 Jumlah pinjaman pada akhir bulan kedua = RM341 671 + RM1 366.68 = RM343 037.68 Baki selepas bayaran ansuran = RM343 037.68 – RM1 697 = RM341 340.68 Jumlah faedah = 1 368 + 1 366.68 = 2 734.68 (c) (i) 5 100 × 1 5 = jumlah bayaran balik 7 × 12 1 020 = jumlah bayaran balik 84 Jumlah bayaran balik = 85 680 85 680 = 60 000 + (60 000 × r × 7) r = 0.0611 = 6.11% (ii) Jumlah bayaran balik = 60 000 + (60 000 × 0.037 × 6) = 73 320 Bayaran ansuran = 73 320 6 × 12 = 1 018.33 Boleh. Ini kerana bayaran ansuran bagi 6 tahun masih dalam 1 5 daripada gajinya. 1. Amanah Saham Unit trust ✓ Bon Bond ✓ Akaun Simpanan Semasa Current Saving Account Hartanah Real estates ✓ 2. Prinsipal Principal (RM) Kadar faedah mudah setahun Simple interest rate per annum Tempoh Duration Faedah mudah diperoleh Simple interest earned Jumlah simpanan Total savings (RM) (i) 10 000 1.8% 9 bulan I = 135 10 135 (ii) 4 000 3.3% 3 bulan 3 months I = 33 4 033 (iii) 70 500 2.5% 1 tahun I = 1 762.50 72 262.50 (iv) 55 000 1.5% 6 bulan 6 months I = 412.5 55 412.50 3. (i) Jumlah bayaran balik = 100 000 + (100 000 × 0.025 × 10) = RM125 000 Bayaran ansuran = 125 000 10 × 12 = RM1 041.67 (ii) Pinjaman bank = RM80 000 Kadar faedah = 2.5% – 0.5% = 2.0% Jumlah bayaran balik = 80 000 + (80 000 × 0.02 × 10) = 96 000 Bayaran ansuran = 96 000 10 × 12 = RM800 Jumlah faedah = RM96 000 – RM80 000 = RM16 000
1 © Penerbitan Pelangi Sdn. Bhd. 1. (a) III 2. (a) IV 3. (a) IV 4. (a) Skala = panjang sisi lukisan yang diberi panjang sisi objek sepadan = 4 2 = 2 1 = 1 1 2 Skala yang digunakan ialah 1 : 1 2 Lukisan Berskala 4 Scale Drawings Bidang Pembelajaran: Sukatan dan Geometri (b) Skala = panjang sisi lukisan yang diberi panjang sisi objek sepadan = 1 3 Skala yang digunakan ialah 1 : 3 (c) Skala = panjang sisi lukisan yang diberi panjang sisi objek sepadan = 3.4 5.1 = 2 3 Skala yang digunakan ialah 2 : 3
Matematik Tingkatan 3 Bab 4 Lukisan Berskala © Penerbitan Pelangi Sdn. Bhd. 2 StrategiPT3 21 Lukisan Drawings Panjang sisi rombus Length of sides of rhombus Skala Scale 2 cm Skala = panjang sisi lukisan panjang sisi objek Scale = length of side of drawing length of side of object = 2 4 = 1 2 Skala yang digunakan ialah 1 : 2 The scale used is 1 : 2 4 cm Lukisan adalah sama saiz dengan objek, maka skala ialah 1 : 1 The drawing is same size as the object, thus the scale is 1 : 1 12 cm Skala = panjang sisi lukisan panjang sisi objek Scale = length of side of drawing length of side of object = 12 4 = 3 1 = 1 1 3 Skala yang digunakan ialah 1 : 1 3 The scale used is 1 : 1 3 5. (a) Skala = panjang sungai pada lukisan panjang sebenar sungai = 4 8 000 = 1 2 000 Skala yang digunakan ialah 1 : 2 000 (b) Skala = tinggi pokok pada lukisan tinggi sebenar pokok = 9 90 × 100 = 1 1 000 = 1 : 1 000
Matematik Tingkatan 3 Bab 4 Lukisan Berskala 3 © Penerbitan Pelangi Sdn. Bhd. (c) Skala = tinggi bangunan pada lukisan tinggi sebenar bangunan 40 800 × 100 = 1 n n = 800 × 100 40 = 2 000 6. (a) Skala = 1 20 000 = 5 panjang jalan sebenar Panjang jalan sebenar = 5 × 20 000 ÷ 100 000 = 1 km (b) Skala = panjang lukisan panjang sebenar = 1 2 000 Panjang sebenar = (panjang lukisan × 2 000) ÷ 100 = 100 m (c) Skala = panjang lukisan panjang sebenar = 1 10 000 Panjang sebenar = (panjang lukisan × 100 000) ÷ 100 000 = 4.5 km 7. (a) Skala = tinggi pada lukisan tinggi sebenar = 1 200 Maka tinggi pada lukisan = tinggi sebenar × 100 ÷ 200 = 6 × 100 ÷ 200 = 60 m (b) Skala = jarak pada lukisan jarak sebenar = 1 3 000 000 Jarak pada lukisan = 300 × 100 000 3 000 000 = 10 cm (c) Skala = 1 100 000 = panjang sungai pada peta 400 000 Panjang sungai pada peta = 400 000 100 000 Strateg = 4 cm iPT3 2 Skala = 1 400 = panjang sisi lukisan pentagon panjang sisi sebenar pentagon Panjang sisi pentagon pada lukisan = 20 × 100 400 = 5 cm 8. (a) (b) (c)
Matematik Tingkatan 3 Bab 4 Lukisan Berskala © Penerbitan Pelangi Sdn. Bhd. 4 9. (a)
Matematik Tingkatan 3 Bab 4 Lukisan Berskala 5 © Penerbitan Pelangi Sdn. Bhd. 10. (a) Skala 1 : 400 bererti 1 cm pada lukisan mewakili 400 cm atau 4 m untuk padang sebenar. Maka luas sebenar padang = (4 × 4) × (6 × 4) = 384 m2 (b) Skala = panjang sisi lukisan panjang sisi sebenar = 1 600 Panjang sisi sebenar = 10 cm × 600 = 6 000 cm = 60 m Maka luas sebenar = 60 × 60 = 3 600 m2 (c) Panjang sisi lukisan berskala segi empat sama = 64 = 8 cm Maka panjang sebenar = 4 × panjang sisi lukisan = 4 × 8 = 32 cm 11. (a) (i) Skala = 20 : 4 000 = 1 : 200 (ii) Luas = 20 × 200 × 12 × 200 ÷ 10 000 = 9 600 000 ÷ 10 000 = 960 m2 (iii) Jumlah kos = 960 × 16 = RM15 360 1. (a) 20 cm : 1 km = 20 cm : 100 000 cm = 1 : 5 000 (b) (i) Skala = 4 10 × 100 = 1 250 = 1 : 250 (ii) Katakan lebar permaidani = x Maka perimeter permaidani = 2(x + 19) = 40 x = 40 2 – 19 = 1 m Luas lantai dewan = (24 – 5) × 10 = 190 Luas permaidani = 1 × 19 = 19 Luas lantai yang tidak ditutupi oleh permaidani = 190 – 19 = 171 m2 (c) (i) Skala = 1 400 1 400 = AB 28 × 100 AB = 28 × 100 4 = 7 cm
Matematik Tingkatan 3 Bab 4 Lukisan Berskala © Penerbitan Pelangi Sdn. Bhd. 6 (ii)
1 © Penerbitan Pelangi Sdn. Bhd. 1. (a) Hipotenus Hypotenuse Sisi bertentangan Opposite side Sisi bersebelahan Adjacent side θ 2. kos q cos q sin q sin q tan q tan q 3 4 3 5 4 5 3. (a) kos q = 6 12 = 0.5 q = kos–1 0.5 = 60° (b) sin q = 5 13 q = sin–1 5 13 = 22.62° (c) tan q = 8 7 q = tan–1 8 7 = 48.81° 4. (a) sin 55° = 6 a a = 6 sin 55° = 7.32 cm (b) kos 70° = a 8 a = 8 kos 70° = 2.74 cm (c) tan 65° = a 8 a = 8 tan 65° = 17.16 cm 5. (a) sin q = O H = 3 3 6 = 3 2 q = sin–1 3 2 = 60° (b) tan q = O A = 6 6 = 1 q = tan–1 1 = 45° (c) kos q = A H = 3 6 = 1 2 q = kos–1 1 2 = 60° Nisbah Trigonometri 5 Trigonometric Ratios Bidang Pembelajaran: Sukatan dan Geometri
Matematik Tingkatan 3 Bab 5 Nisbah Trigonometri © Penerbitan Pelangi Sdn. Bhd. 2 6. (a) sin 30° = OH = 3x 12 = 3x x = 6 cm (b) tan 30° = OA = 3x 13 = 3x x = 3 3 cm (c) kos 45° = AH = x 10 12 = x 10 x = 10 × 12 x = 102 = 102 × 22 = 10 2 2 = 5 2 cm 7. (a) QR = 172 – 152 = 8 cm tan y = 8 10 = 45 (b) AB = 172 – 82 = 15 cm BD = 252 – 152 = 20 cm tan x = 15 20 = 34 x = tan–1 34 = 36.87° (c) Diberi tan x = 5 12 , maka PR = 12 cm y = sin–1 12 15 y = 53.13° (d) RT = 52 – 32 = 4 cm RQ = 2RT = 2 × 4 = 8 cm sin y = PR RQ = 58 y = sin–1 58 = 38.68° (e) (i) tan x = 43 BC AB = 43 BC 12 = 43 BC = 43 × 12 = 9 cm (ii) y = tan–1 AB BD = 12 27 = 23.96° 8. (a) DA = BC = 12 cm Diberi tan /DAE = 5 12 , maka ED = 5 cm dan EA = 13 cm EC = ED + DC = 5 + 5 = 9 cm EB = 92 + 122 = 15 cm sin x = OH = EC EB = 9 15 x = sin–1 35 = 36.87°
Matematik Tingkatan 3 Bab 5 Nisbah Trigonometri 3 © Penerbitan Pelangi Sdn. Bhd. StrategiPT3 21 (ii) sin y = 12 13 AC AE = 12 13 CE = 132 – 122 = 5 tan y = AC AE = 12 5 (ii) kos z = 4 5 CD BD = 4 5 CD 10 = 4 5 CD = 4 5 × 10 = 8 DE = CD + CE = 8 + 5 = 13 cm 9. (a) 5 m t 15° Katakan tinggi murid = t + 20 tan 15° = t + 20 5 t + 20 = 5 × tan 15° = 1.34 t = 1.34 + 0.2 = 1.54 m (b) 25° 1.65m 25° j Katakan jarak = j tan 25° = 1.65 j j = 1.65 tan 25° = 3.54 m 10. (a) Sudut di antara garis FC dan satah ABCD ialah /FCA. F A 3 cm C A B D C 8 cm AC = 82 + 42 = 80 cm tan /FCA = FA AC = 3 80 /FCA = tan–1 3 80 = 18.54° (b) M, N dan P masing-masing ialah titik tengah bagi BC, AD dan EF. Sudut di antara satah EFBC dan satah ABCD ialah /PMN. 3 cm N 4 cm M P tan /PMN = PN NM = 3 4 /PMA = tan–1 3 4 = 36.87°
Matematik Tingkatan 3 Bab 5 Nisbah Trigonometri © Penerbitan Pelangi Sdn. Bhd. 4 1. (a) tan q = PQ QR ✗ sin q = RQ PQ ✓ kos q = PQ QR cos q = PQ QR ✗ (b) (i) Diberi BC = CD = 5 cm AC = 13 cm AD = 5 + 13 = 18 cm AE = 182 + 62 = 6 10 cm (ii) tan EAD = 6 18 /EAD = tan–1 1 3 = 18.4° (c) (i) tan 60° = OP OA OP = OA × tan 60° = 4.85 m (ii) tan x = OP OB x = tan–1 4.85 5 = 44.13°
1 © Penerbitan Pelangi Sdn. Bhd. 1. (a) /AQB (b) /CQD (c) /ARB 2. (a) /POB = 180° – 2(30°) = 120° /QOB = 20° + 120° = 140° Maka q = 140° ÷ 2 = 70° (b) Sudut refleks /COA = 360° – 120° = 240° q = 240° ÷ 2 = 120° (c) Sudut refleks /DOC = 360° – 150° = 210° /DBC = 210° ÷ 2 = 105° q = 180° – 105° = 75° 3. (a) Lengkok AD = lengkok CD, maka /ABD = /DBC = 30° dan /ABC = 30° + 30° = 60° /ABC = 90° maka, q = 90° – 60° = 30° (b) /APO = (180° – 70°) ÷ 2 = 55° /OPB = 90° – 55° = 35° q = /OPB = 35° (c) /ODB = /OBD = 52° /ODA = 90° – 52° = 38° /OAD = /ODA = 38° q = 180° – 38° Strateg = 142° iPT3 21 a = /ACB = 30° /DBC = /CAD = 25° b = 180° – 120° – 25° = 35° 4. (a) /ADC = 180° – 35° – 30° = 115° x = 180° – 115° = 65° y = /ADC = 115° (b) x = 130° ÷ 2 = 65° y = 180° – 65° = 115° (c) /FDA= 180° – 110 = 70° /ADC + /ABC = 180° /ADC = x – 70° Maka (x – 70°) + y = 180° x + y = 180° + 70° = 250° Sudut dan Tangen bagi Bulatan Angles and Tangents of Circles 6 Bidang Pembelajaran: Sukatan dan Geometri
Matematik Tingkatan 3 Bab 6 Sudut dan Tangen bagi Bulatan © Penerbitan Pelangi Sdn. Bhd. 2 StrategiPT3 2 x = 120° /PQT= 180° – 105° = 75° /PTQ = 180° – 120° = 60° y = 180° – 75° – 60° = 65° 5. (a) /ORQ = /OQR = (180° – 80°) ÷ 2 = 50° x = 180° – (50° + 25°) = 105° /RQP = 180° – 70° = 110° y = 110° – 50° = 60° (b) /QSP = 180° – 110° = 70° /PSU = 130° – 70° = 60° /SUP = 90° Maka x = 90° – 60° = 30° dan y = 180° – 30° = 150° (c) /SOR = 60° Mak x = 60° ÷ 2 = 30° /SPQ = 90° – 35° = 55° /SRQ = 180° – 55° = 125° Maka y = 125° – 60° = 65° 6. Tangen Tangent Titik ketangenan Point of tangency MN I PQ C 7. (a) /BSP (b) /BTP (c) /BUP 8. (a) /OQA = 90° /OPA = 90° /PAO = 90° – 50° = 40° /QAO = /PAO = 40° /QOA = /POA = 50° AP = QA = 5 cm 9. (a) a = 70° /DBC = 40° b = 180° – 70° – 40° = 70° (b) a = 50° /BED = 80° b = 180° – 80° = 100° (c) /EDB = 60° /EDF = 90° a = 90° – 60° = 30° /DBA = 105° b = 105° – 60° = 45° 10. (a) /APB = 90° /BPQ = /BAP = x /PAQ + /APQ + /AQP = 180° x + (90° + x) + 40° = 180° 2x + 130° = 180° 2x = 50° x = 25°
Matematik Tingkatan 3 Bab 6 Sudut dan Tangen bagi Bulatan 3 © Penerbitan Pelangi Sdn. Bhd. (b) /STQ = 130° /OTS = (180° – 40°) ÷ 2 = 70° x = 130° – 70° = 60° y = 20° (c) /CBE = /CDE = 90° /CED = 180° – 142° = 38° x = 90° – 38° = 52° /ABE = y /ABC = /ABE + /CBE = y + 90° y + y + 90° + 34° = 180° 2y + 124° = 180° 2y = 56° y = 28° 11. (a) (i) kos /AOB = 5 13 , maka /AOB = 67.38° /AOC = 2 × /AOB = 134.76° (ii) AB = 132 – 52 = 12 cm (iii) Luas OABC = 2 × luas segi tiga OAB = 21 1 2 × 5 × 122 = 60 cm2 (b) (i) /POR = 1 2 /POQ = 60° tan 60° = 10 OP OP = 10 tan 60° = 5.77 cm (ii) lengkok PQ lilitan bulatan = 120° 360° Lengkok PQ = 1 2 × 2 × 22 7 × 5.77 = 12.09 cm Perimeter rantau berlorek = PR + QR + lengkok = 10 + 10 + 12.09 = 32.09 cm 1. (a) (i) SALAH (ii) BETUL (iii) BETUL (b) /ECB = /ABE = 45° /DEC = 180° – 160° = 20° /EDC = 90° /ECD = 180° – 90° – 20° = 70° x = /ECB + /ECD = 70° + 45° = 115° (c) (i) x = 360° – 80° = 280° (ii) /ACB = /ADE = 40° x = 180° – 20° – 40° = 120°
1 © Penerbitan Pelangi Sdn. Bhd. 1. (a) III ( ✓ ) (b) III ( ✓ ) (c) II ( ✓ ) 2. Objek Objects Pandangan dari atas Viewed from top Pandangan dari depan Viewed from front Pandangan dari sisi Viewed from side (a) (b) (c) Pelan dan Dongakan Plans and Elevations 7 Bidang Pembelajaran: Sukatan dan Geometri
Matematik Tingkatan 3 Bab 7 Pelan dan Dongakan © Penerbitan Pelangi Sdn. Bhd. 2 3. (a) 2 cm 4 cm E F D/A C/B Panjang sisi dan saiz sudut adalah sama kecuali bentuk ED dan FC. (b) 2 cm 4 cm F/B G/C E/A H/D Panjang sisi dan saiz sudut adalah sama kecuali bentuk EA, HD, FB dan GC. 4. (a) Sukuan I Pelan Sukuan II 3.2 cm 4.8 cm 2.4 cm 3.2 cm 2.4 cm 3.2 cm Dongakan depan Sukuan III Dongakan sisi Sukuan IV D C B A A B D C 4.8 cm 2.4 cm A B D C 45°
Matematik Tingkatan 3 Bab 7 Pelan dan Dongakan 3 © Penerbitan Pelangi Sdn. Bhd. (b) Sukuan I Pelan Sukuan II 6 cm Dongakan depan Sukuan III Dongakan sisi Sukuan IV A B C D E E E 45° 6 cm 5 cm 5 cm A/B 5 cm D/C D/A 5 cm C/B
Matematik Tingkatan 3 Bab 7 Pelan dan Dongakan © Penerbitan Pelangi Sdn. Bhd. 4 5. (a) Guna skala 1 : 30 Panjang sisi unjuran ortogon kuboid = 120 ÷ 30 = 4 cm. Lebar sisi unjuran ortogon kuboid = 150 ÷ 30 = 5 cm. Tinggi kuboid sisi unjuran ortogon kuboid = 90 ÷ 30 = 3 cm. Sukuan I Pelan 4 cm 5 cm 3 cm 4 cm 5 cm 3 cm Sukuan II Dongakan depan Sukuan III Dongakan sisi Sukuan IV C/B G/F J J A/B D/C D/A F/B G/C E/A I H/D E/F I/J H/G H/E I 45°
Matematik Tingkatan 3 Bab 7 Pelan dan Dongakan 5 © Penerbitan Pelangi Sdn. Bhd. (b) Guna skala 1 : 2 Sukuan I Pelan Sukuan II Dongakan depan 2 cm 1.5 cm 2 cm 1.5 cm 2 cm 2 cm 2 cm 2 cm 3 cm 3 cm Sukuan III Dongakan sisi Sukuan IV A/B F/C E/D E/F/A G/F E K/J/A L/I/B H/C D J/I K/L G/H G/J K D/C/B H/I L 45° 6. (a) A E C T S R Q P D B 5 cm 2 cm (b) A D C B 4 cm 6 cm
Matematik Tingkatan 3 Bab 7 Pelan dan Dongakan © Penerbitan Pelangi Sdn. Bhd. 6 7. (a) (i) 6 × 9.5 = 57 m2 (ii) 2.5 m × 3 m (iii) Luas ruang tamu = 500 cm × 300 cm =150 000 cm2 Luas sekeping jubin = 15 × 15 = 225 cm2 Bilangan jubin yang diperlukan = 150 000 ÷ 225 = 666.67 = 667 keping (iv) Jumlah kos jubin = 667 × RM3.00 = RM2 001.00 1. (a) ( ü ) ( ü ) ( ü ) ( ü ) (b) ( ü ) ( ü ) ( ü ) ( ü ) 2. ( III ) ( III ) ( I ) ( II ) ( ) ( )
Matematik Tingkatan 3 Bab 7 Pelan dan Dongakan 7 © Penerbitan Pelangi Sdn. Bhd. 3. Guna skala 1 : 2 Panjang unjuran ortogon EF = 12 ÷ 2 = 6 cm. Panjang unjuran ortogon AB = 6 ÷ 2 = 3 cm. Panjang unjuran ortogon AD = 12 ÷ 2 = 6 cm. Panjang unjuran ortogon bagi tinggi trapezium = 8 ÷ 2 = 4 cm Dongakan depan Sukuan III Dongakan sisi Sukuan IV D/H C G C/D G/H A/E A/D B/C B F B/A F/E Sukuan I Pelan Sukuan II 6 cm F/G 4 cm 5 cm 3 cm 3 cm 3 cm 4 cm 5 cm E/H 45°
1 © Penerbitan Pelangi Sdn. Bhd. 1. (a) Lokus Lokus bagi bandul yang berayun ialah lengkok suatu bulatan. (b) Lokus Lokus bagi suatu kincir angin yang berputar ialah bulatan. (c) Lokus syiling yang bergerak ialah berbentuk bulatan. 2. (a) O (b) O (c) O Lokus dalam Dua Dimensi Loci in Two Dimensions 8 Bidang Pembelajaran: Sukatan dan Geometri
Matematik Tingkatan 3 Bab 8 Lokus dalam Dua Dimensi © Penerbitan Pelangi Sdn. Bhd. 2 3. (a) B C A D Lokus Y BD (b) A B D C Lokus Y BD (c) A B D C Lokus Y BD 4. (a) 1 cm A B 1 cm Lokus Z (b) 1 cm A B 1 cm Lokus Z (c) 2 cm A B 2 cm Lokus Z 5. (a) 1 cm A C 1 cm B Lokus P D (b) 1 cm A C B D 1 cm Lokus P (c) 1 cm A C B 1 cm D Lokus P
Matematik Tingkatan 3 Bab 8 Lokus dalam Dua Dimensi 3 © Penerbitan Pelangi Sdn. Bhd. 6. (a) A D P Q B C R S Lokus ialah PQ dan RS (b) A B Q P R D C S Lokus ialah PR dan QS (c) C S B D A R P Q Lokus ialah PD dan DQ 7. (a) C Lokus S B D A (b) B R C A P Q Lokus S (c) B C A D Lokus R 1.5 cm (d) A Lokus T B C D 8. (a) A B C D Lokus Q Lokus P Lokus R
Matematik Tingkatan 3 Bab 8 Lokus dalam Dua Dimensi © Penerbitan Pelangi Sdn. Bhd. 4 (b) B C D A Lokus Q Lokus R Lokus P (c) B C A D E Lokus X Lokus Y Lokus Z 1.7 cm (d) Titik tengah Midpoint A C B D Lokus Y Lokus Z Lokus X
Matematik Tingkatan 3 Bab 8 Lokus dalam Dua Dimensi 5 © Penerbitan Pelangi Sdn. Bhd. StrategiPT3 21 2 cm C E B D A F Lokus Y Lokus Z Lokus X 1. (a) (i) P Q ( ü ) (ii) P 2. (i) Garis EB (ii) Lokus Y Lokus X Lokus Z A E B D C F 3. (i), (ii) Lokus Y Lokus X Lokus Z A B D C
1 © Penerbitan Pelangi Sdn. Bhd. 1. (a) Kecerunan garis lurus BO = 2 – 0 4 – 0 = 1 2 maka persamaan garis lurus BO ialah y = 1 2 x (b) Kecerunan garis lurus CO = 2 – 0 1 – 0 = 2 maka persamaan garis lurus CO ialah y = 2x (c) Pintasan-y = –4 Kecerunan garis lurus IJ = – 1 – 4 6 2 = 2 3 Maka persamaan garis lurus IJ ialah y = 2 3 x – 4 (d) Pintasan-y = 2 Kecerunan garis lurus KL = – 1 2 1 2 2 = – 4 Maka persamaan garis lurus KL ialah y = – 4x +2 (e) Persamaan garis lurus PQ dengan c = 1 ialah y = – 1 4 x – 1 (f) Persamaan garis lurus RS dengan c = –2 ialah y = – 1 4 x – 2 2. (a) y = 2 3 x + 10 (b) y = –6x – 7 (c) y = 4x + c –2 = 4(5) + c c = –22 Maka, y = 4x – 22 (d) y = 1 2 x + c 3 = 1 2 (–2) + c c = 4 Maka, y = 1 2 x + 4 (e) m = 5 – (–3) 1 – (–2) = 8 3 y = 8 3 x + c 5 = 8 3 (1) + c c = 7 3 y = 8 3 x + 7 3 (f) m = 2 – (–3) 4 – (–2) = 5 6 y = 5 6 x + c –3 = 5 6 (–2) + c c = – 4 3 y = 5 6 x – 4 3 3. (a) m = – 3 –2 = 3 2 , c = 3 \ y = 3 2 x + 3 (b) m = 1 2 , c = –2 \ y = 1 2 x – 2 (c) m = 0 \ y = –8 (d) m = ∞ \ x = –4 (e) m = 6 – 0 5 – 0 = 6 5 , c = 0 \ y = 6 5 x Garis Lurus A Straight Line 9 Bidang Pembelajaran: Perkaitan dan Algebra
Matematik Tingkatan 3 Bab 9 Garis Lurus © Penerbitan Pelangi Sdn. Bhd. 2 (f) m = 3 – (–2) 4 – 0 = 5 4 , c = –2 \ y = 5 4 x – 2 4. (a) 3y = 4x +2 y = 4 3 x + 2 3 (b) –3y = –2x + 4 y = 2 3 x – 4 3 (c) y 2 = x 6 + 1 y = 1 3 x + 2 (d) – y 2 = – x 4 + 1 y = 1 2 x – 2 5. (a) 3x – y – 5 = 0 (b) 20 × 3 x 4 + y 5 4 = 1 × 20 5x + 4y – 20 = 0 6. (a) 2y = x + 6 x – 2y = –6 bahagi kedua-dua belah persamaan dengan –6 – x 6 + y 3 = 1 (b) 5y – 3x – 15 = 0 –3x + 5y = 15 bahagi kedua-dua belah persamaan dengan 15 – x 5 + y 3 = 1 (c) y = 1 2 x – 4 1 2 x – y = 4 bahagi kedua-dua belah persamaan dengan 4 x 8 – y 4 = 1 7. Titik Point Persamaan garis lurus The equation of straight line Sebelah kiri Left-hand side Sebelah kanan Right-hand side Kesimpulan (Ya atau Tidak) Conclusion (Yes or No) (a) (4, 1) 2y = x – 2 2 4 – 2 = 2 Ya (b) (–2, –3) 3y = 2x + 1 –9 2(–2) + 1 = –3 Tidak (c) (6, 0) y = 1 2 x – 4 0 1 2 (6) – 4 = –1 Tidak 8. (a) 2y = –4x + 5 y = –2x + 5 2 , kecerunan m1 = –2 –3y = 6x – 1 y = –2x + 1 3 , kecerunan m2 = –2 m1 = m2 , maka L1 selari dengan L2 . (b) 3y +1 = – 1 2 x + 6 y = – 1 6 x + 5, kecerunan m1 = – 1 6 2y = 8x – 3 y = 4x – 3 2 , kecerunan m2 = 4 m1 ≠ m2 , maka L1 tidak selari dengan L2 .
Matematik Tingkatan 3 Bab 9 Garis Lurus 3 © Penerbitan Pelangi Sdn. Bhd. (c) 2y + x = –7 y = – 1 2 x – 7 2 , kecerunan m1 = – 1 2 4 – 3y = 9x – 2 y = –3x + 2 , kecerunan m2 = –3 m1 ≠ m2 , maka L1 tidak selari dengan L2 . 9. (a) 2y = 6 – 5x y = –5 2 x +3 m = – 5 2 persamaan baharu y = – 5 2 x + c pada titik P(6, 0) 0 = – 5 2 (6) + c c = 15 Maka, y = – 5 2 x + 15 (b) 3x – 2y = 10 y = 3 2 x – 5 m = 3 2 persamaan baharu y = 3 2 x + c pada titik P(0, –8) c = –8 Maka, y = 3 2 x – 8 (c) 3y = 5x + 12 y = 5 3 x + 4 m = 5 3 persamaan baharu y = 5 3 x + c pada titik P(–1, –2) –2 = 5 3 (–1) + c c = – 1 3 Maka, y = 5 3 x – 1 3 10. (a) y x –4 –2 2 4 4 2 –2 2y – 3x = 3 0 2y – 3x = 3 Apabila x = 0, y = 1.5 Apabila y = 0, 3x + 3 = 0, x = –1 x y 0 1.5 –1 0 (b) x y –4 –2 2 4 4 2 –2 x —4 y —5 + = 1 0 x 4 + y 5 = 1 Apabila x = 0, y = 5 Apabila y = 0, x = 4 x y 0 5 4 0
Matematik Tingkatan 3 Bab 9 Garis Lurus © Penerbitan Pelangi Sdn. Bhd. 4 (c) y x –4 –2 2 4 4 2 –2 2x – 3y + 6 =0 0 2x – 3y + 6 = 0 Apabila x = 0, 3y = 6, y = 2 Apabila y = 0, 2x = –6, x = –3 x y 0 2 –3 0 11. (a) y = 2x – 1 y = 3 – x x y 0 –1 1 2 0 x y 0 3 3 0 y x –4 –2 2 4 4 2 –2 y = 3 – x y = 2x – 1 0 Dari graf, titik persilangan = (1.4, 1.6) (b) 3y = 2x – 1 y = 2x + 3 x y 2 1 1 2 0 x y 0 3 –1.5 0 y x –4 –2 2 4 4 2 –2 0 y = 2x + 3 3y = 2x – 1 Dari graf, titik persilangan = (–2.5, –1.9) (c) y = 1 2 x – 2 2y = 5 – 2x x y 0 –2 4 0 x y 0 2.5 2.5 0 y x –4 –2 2 4 4 2 –2 2y = 5 – 2x 0 y = x – 2 1 —2 Dari graf, titik persilangan = (3.1, 0.5) 12. (a) y = 2x – 1 ...........................a y = 3 – x .............................b a – b 3x – 4 = 0 x = 1 1 3 Gantikan x = 1 1 3 dalam b y = 3 – 1 1 3 = 1 2 3 Maka titik persilangan = 11 1 3 , 1 2 3 2
Matematik Tingkatan 3 Bab 9 Garis Lurus 5 © Penerbitan Pelangi Sdn. Bhd. (b) y = –x + 16 ........................a 2y = –x + 10 ......................b b – a y = –6 Gantikan y = –6 dalam b 2(–6) = –x + 10 x = 10 + 12 x = 22 Maka titik persilangan = (22, –6) (c) 2y = x – 5 ...........................a y = 5 – 2x ...........................b Gantikan b ke dalam a 2(5 – 2x) = x – 5 10 – 4x = x – 5 5x = 15 x = 3 .....................c Gantikan c ke dalam b y = 5 – 2(3) y = –1 Maka titik persilangan = (3, –1) (d) 3y = –4x + 6 ......................a 6y = x – 1 ...........................b b ÷ 6, y = 1 6 x – 2 .........................c Gantikan c ke dalam a 31 1 6 x – 22 = –4x + 6 1 2 x – 6 = – 4x + 6 9 2 x = 12 x = 8 3 ...................d Gantikan d ke dalam c y = 1 6 1 8 3 2 – 2 y = – 14 9 Maka titik persilangan = 1 8 3 , – 14 9 2 13. (a) 2y = 4x + 3 ⇔ y = 2x + 3 2 Kecerunan PQ = 2; Persamaan garis lurus PQ, y = 2x + c Gantikan x = 1 dan y = 1, c = 1 – 2 = –1. Maka persamaan garis lurus PQ ialah y = 2x – 1 (b) Kecerunan OC = 2 – 0 –6 – 0 = – 1 3 Kecerunan AB = 6 – y –7 – 0 = – 1 3 ; 6 – y = 7 3 ⇔ y = 3 2 3 Maka, A10, 3 2 3 2 Persamaan garis lurus AB; y = – 1 3 x + c 3 2 3 ialah pintasan-y, maka c = 3 2 3 dan y = – 1 3 x + 11 3 Persamaan garis lurus AB ialah x + 3y – 11 = 0 14. (a) 2y = 8 + x; Apabila x = 0, y = 4, maka D(0, 4). Koordinat C = (4, 0) Kecerunan DC = 4 – 0 0 – 4 = –1 = Kecerunan AB Persamaan garis lurus AB: y = –x + c Gantikan x = 4, y = 8 8 = –4 + c; c = 12, maka y = –x + 12 (b) Kecerunan OP = 1 6 = Kecerunan RQ Persamaan garis lurus RQ: y = 1 6 x + c Gantikan x = 8, y = 6 6 = 1 6 (8) + c; c = 14 3 y = 1 6 x + 14 3
Matematik Tingkatan 3 Bab 9 Garis Lurus © Penerbitan Pelangi Sdn. Bhd. 6 (c) Kecerunan PR = 1 – 6 4 – 0 = – 5 4 Titik R(0, 6) merupakan koordinat pintasan-y, maka c = 6 Persamaan garis lurus PR; y = – 5 4 x + 6 PQ selari dengan paksi-y, persamaan garis lurus PQ; x = 4 QR selari dengan paksi-x, persamaan garis lurus QR; y = 6 (d) Koordinat titik R = (4, 0) Kecerunan SR = – 8 4 = –2 = Kecerunan PQ Persamaan garis lurus PQ: y = –2x + c Gantikan x = –5, y = 3 3 = –2(–5) + c c = –7 y = –2x – 7 (e) (i) Persamaan garis lurus AQ ialah y = 4 (ii) Kecerunan PR = –2. Persamaan garis lurus PR ialah y = –2x + c. Titik P(0, 10) ialah koordinat pintasan-y, maka c = 10. Maka, y = –2x + 10 (iii) y = 4 ..........................a y = –2x + 10 ..............b Gantikan a dalam b 4 = –2x +10 2x = 6 x = 3 Maka Q(3, 4) 1. (a) x + 2y – 3 = 0 2y = –x + 3 y = – 1 2 x + 3 2 Kecerunan, m = – 1 2 (b) 2x + 3y – 12 = 0 12 ÷ [2x + 3y] = 12 ÷ 12 x 6 + y 4 = 1 Maka, pintasan-x = 6 dan pintasan-y = 4 2. (a) PQ = 5 unit RQ = PQ = 5 unit OQ = 4 unit RO = √ — 52 – 42 = 3 unit Kecerunan RQ, mRQ = OQ OR = 4 3 Persamaan garis lurus RQ ialah y = 4 3 x +c titik Q(0, 4) ialah koordinat pintasan-y, maka c = 4. y = 4 3 x + 4 (b) /RQO= tan–11 4 3 2 = 53.13° mRP = 4 – 0 5 – (–3) = 4 8 = 1 2 Persamaan garis lurus RP y = 1 2 x + c c = 4 – 1 2 (5) = 3 2 /RSO = tan–11 3 3 2 2 = 63.43° /RQO + /RSO = 53.13° + 63.43° = 116.56° 3. (a) mQS = 5 – 0 0 – 3 = – 5 3 Persamaan garis lurus QS y = – 5 3 x + 5 ............................a
Matematik Tingkatan 3 Bab 9 Garis Lurus 7 © Penerbitan Pelangi Sdn. Bhd. (b) Titik R = (–1, 0) mPR = 5 – 0 4 – (–1) = 5 5 = 1 Persamaan garis lurus PR y = x + c Pada titik R(–1, 0), 0 = –1 + c, c = 1 \ y = x + 1 ..............................b a – b 0 = – 9 4 x + 4 –4 = – 9 4 x x = 16 9 ...................................c Gantikan c dalam b y = 16 9 + 1 = 25 9 Titik persilangan PR dan QS = 1 16 9 , 25 9 2
1 © Penerbitan Pelangi Sdn. Bhd. 1. (a) (i) ✗ (ii) ✓ (iii) ✗ (b) (i) 4(–8 + 5) × 2.5 ÷ 3 4 = 4(–3) × 2.5 ÷ 3 4 = –12 × 2.5 ÷ 3 4 = –30 × 4 3 = –40 (ii) 225 – 3 –0.064 = 15 – (–0.4) = 15.4 (c) Q R S P 2 cm Lokus X Lokus Y 2. (a) sin 30° cos 60° kos 60° tan 30° 1 2 1 2 1 3 (b) (i) (x – 2y)(3x + y) = 3x2 – 6xy + xy – 2y2 = 3x2 – 5xy – 2y (ii) 5p2 – 45 = 5(p2 – 9) = 5(p – 3)(p + 3) (c) Isi padu bekas = (28 × 20 × 9) + 1 22 7 × 72 × 102 = 5 040 + 1 540 = 6 580 cm3 Jumlah serbuk diperlukan = 6 580 × 10 1 400 = 47 peket = 47 × 20 g = 940 g 3. (a) 2 x3 x5 ÷ x2 9 4 1– 2 3 2 2 x3 2x-3 4 9 3 x2 (b) (i) Jumlah simpanan = 5 000 + 15 000 × 3.2 100 × 32 = 5 000 + 480 = RM5 480 (ii) Jumlah simpanan = 5 00011 + 0.032 1 2 1(3) = RM5 495.52 (c) 15 + 12 , x + 2x x . 9 Nilai minimum x = 10 biji 4. (a) (i) 128 400 129 000 (ii) 129 379 (iii) 127 650 128 000 (b) (i) C (ii) Jumlah peket biskut = 1.2 × 102 × (60 × 16) = 1.2 × 102 × 9.6 × 102 = 11.52 × 104 = 1.152 × 105 (c) (i) Laju = 6 × 400 m (16 × 60) s = 2.5 m s–1 UJIAN AKHIR TAHUN
Matematik Tingkatan 3 Ujian Akhir Tahun © Penerbitan Pelangi Sdn. Bhd. 2 (ii) Katakan x = masa untuk 4 pusingan 2.3 = 4 × 400 m x x = 635.65 saat atau 10.59 minit Laju purata = 10 × 400 m [(16 + 10.59) × 60] s = 2.51 m s–1 5. (a) (i) data numerik (ii) data kategori (iii) data numerik (b) (i) P Q fi (ii) {}, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3} (c) (i) panjang TU = 56 – 2(13) 2 = 15 cm (ii) SU = 20 – 8 = 12 cm TS = 152 – 122 = 9 cm Luas kawasan berlorek = 202 – 1 1 2 × 12 × 92 = 400 – 54 = 346 cm2 6. (a) (b) (i) z = 2x + 1.5y (ii) z = 2(3) + 1.5(4) = 12 (c) (i) 2.6 = 0(1) + 1(3) + 2(6) + 3x + 4(3) + 5(2) 1 + 3 + 6 + x + 3 + 2 2.6(15 + x) = 37 + 3x 39 + 2.6x = 37 + 3x 2 = 0.4x x = 5 (ii) 0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, … , 4, 4, 4, 5, 5 9 x 5 9 = x + 5 x = 4 7. (a) Q T R U P S V (b) x = 180° – 130° = 50° Hasil tambah sudut pedalaman pentagon = (5 – 2) × 180° = 540° 540° = 50° + 50°(2) + y + 45° + 285° y = 60° (c) (i) P(biru) = 1 5 Katakan x = bilangan pen biru x 12 + 8 + x = 1 5 5x = 20 + x 4x = 20 x = 5 (ii) P(bukan merah) = 8 + 5 12 + 8 + 5 = 13 23 8. (a) (i) Salah (ii) Betul (iii) Betul median
Matematik Tingkatan 3 Ujian Akhir Tahun 3 © Penerbitan Pelangi Sdn. Bhd. (b) (i) Panjang lengkok PQ = 90° 360° × 2 × 22 7 × 7 = 11 cm (ii) Perimeter = 11 + (7 – 4) + 2 + 4 + (7 – 2) = 25 cm (c) (i) 2, 4, 8, 16, 32 Bilangan sisi empat yang dibentuk selepas peserta kelima = 32 (ii) 16 – 2 = 14 9. (a) 12 4 32 3 1 8 6 12 2 32 16 (b) /PQT = /PST x = 180° – 40°(2) = 100° y = 180° – [90°– (180° – 105° – 40°)] = 125° (c) (i) E / A F G / B H / K F Q / L 3 cm 2 cm N / D 5 cm M / C 3 cm 2 cm (ii) E / A G / B P Q 2 cm A / B D / C K / L N / M 5 cm 2 cm 3 cm 2 cm 2 cm 10. (a) (i) sudut penggenap (ii) sudut pelengkap (iii) sudut konjugat (b) (i) Kecerunan MN = 2, titik (-2, 4) y = mx + c 4 = 2(–2) + c c = 8 y = 2x + 8 (ii) y = 2x + 8 , y = 0 0 = 2x + 8 x = -4 pintasan-x = -4