5 DWIBAHASA Tingkatan Matematik Tambahan Additional Mathematics Penulis Buku Teks Azizah binti Kamar (Guru Cemerlang) Dr. M. K. Wong Nurbaiti binti Ahmad Zaki Pan Asia Publications Sdn. Bhd. 199101016590 (226902-X) No. 2-16, Jalan SU 8, Taman Perindustrian Subang Utama, Seksyen 22, 40300 Shah Alam, Selangor Darul Ehsan, Malaysia. Tel: +603-5614 4168 Faks: +603-5614 4268 E-mel: [email protected] Laman web: www.panasiapub.com © Pan Asia Publications Sdn. Bhd. Hak cipta terpelihara. Sebarang bahagian dalam buku ini tidak boleh diterbitkan semula, disimpan dalam cara yang boleh dipergunakan lagi ataupun dipindahkan dalam sebarang bentuk atau cara, baik secara elektronik, mekanik, gambar, rakaman atau sebagainya, tanpa kebenaran daripada penerbit. Cetakan Pertama 2022 MODUL A+1 MATEMATIK TAMBAHAN Tingkatan 5 ISBN 978-967-466-631-6 Dicetak oleh Metrobay Industry Sdn. Bhd. (430102-W) • PDF Manual Guru • Rancangan Pengajaran Tahunan Bonus Guru
ii BAB 1 Sukatan Membulat Circular Measure 1.1 Radian Radian.......................................................................... 2 1.2 Panjang Lengkok Suatu Bulatan Arc Length of a Circle.................................................... 3 1.3 Luas Sektor Suatu Bulatan Area of Sector of a Circle ............................................... 8 1.4 Aplikasi Sukatan Membulat Application of Circular Measures.................................. 12 Soalan Berformat SPM.................................................... 13 BAB 2 Pembezaan Differentiation 2.1 Had dan Hubungannya dengan Pembezaan Limit and Its Relation to Differentiation......................... 19 2.2 Pembezaan Peringkat Pertama The First Derivative..................................................... 21 2.3 Pembezaan Peringkat Kedua The Second Derivative ................................................. 26 2.4 Aplikasi Pembezaan Application of Differentiation ....................................... 28 Soalan Berformat SPM.................................................... 37 BAB 3 Pengamiran Integration 3.1 Pengamiran sebagai Songsangan Pembezaan Integration as the Inverse of Differentiation................... 42 3.2 Kamiran Tak Tentu Indefinite Integral........................................................ 43 3.3 Kamiran Tentu Definite Integral .......................................................... 45 3.4 Aplikasi Pengamiran Application of Integration ............................................ 56 Soalan Berformat SPM.................................................... 57 BAB 4 Pilih Atur dan Gabungan Permutation and Combination 4.1 Pilih Atur Permutation................................................................ 62 4.2 Gabungan Combination ............................................................... 67 Soalan Berformat SPM.................................................... 69 BAB 5 Taburan Kebarangkalian Probability Distribution 5.1 Pemboleh Ubah Rawak Random Variable......................................................... 72 5.2 Taburan Binomial Binomial Distribution .................................................. 75 5.3 Taburan Normal Normal Distribution .................................................... 82 Soalan Berformat SPM.................................................... 88 BAB 6 Fungsi Trigonometri Trigonometric Functions 6.1 Sudut Positif dan Sudut Negatif Positive Angles and Negative Angles............................. 95 6.2 Nisbah Trigonometri bagi Sebarang Sudut Trigonometric Ratios of Any Angle................................ 97 6.3 Graf Fungsi Sinus, Kosinus dan Tangen Graphs of Sine, Cosine and Tangent Functions............. 100 6.4 Identiti Asas Basic Identities.......................................................... 103 6.5 Rumus Sudut Majmuk dan Rumus Sudut Berganda Addition Formulae and Double Angle Formulae .......... 105 6.6 Aplikasi Fungsi Trigonometri Application of Trigonometric Functions....................... 108 Soalan Berformat SPM...................................................110 BAB 7 Pengaturcaraan Linear Linear Programming 7.1 Model Pengaturcaraan Linear Linear Programming Model ....................................... 113 7.2 Aplikasi Pengaturcaraan Linear Application of Linear Programming............................ 119 Soalan Berformat SPM.................................................. 123 BAB 8 Kinematik Gerakan Linear Kinematics of Linear Motion 8.1 Sesaran, Halaju dan Pecutan sebagai Fungsi Masa Displacement, Velocity and Acceleration as a Function of Time ..................................................................... 126 8.2 Pembezaan dalam Kinematik Gerakan Linear Differentiation in Kinematics of Linear Motion ............ 129 8.3 Pengamiran dalam Kinematik Gerakan Linear Integration in Kinematics of Linear Motion ................. 133 8.4 Aplikasi Kinematik Gerakan Linear Application of Kinematics of Linear Motion................. 136 Soalan Berformat SPM.................................................. 138 Kertas Model SPM......................................................... 141 Jawapan ..................................................................... MG–1 Lembaran PBD........................................................ MG–14 00 Kand Modul A+ MateTam Tg5.indd 2 04/10/2021 2:11 PM
1 1.1 Radian 1. Satu radian ditakrifkan sebagai sudut yang dicangkum di pusat bulatan oleh satu lengkok yang sama panjang dengan jejari bulatan, j seperti yang ditunjukkan dalam rajah. One radian is defined as an angle subtended at the centre of the circle by an arc that has a length equal to the radius of the circle, j as shown in the diagram. 2. 360° = 2p rad q° (dalam darjah)/(in degrees) q (dalam radian)/(in radians) 1.2 Panjang Lengkok Suatu Bulatan/Arc Length of a Circle 1. Rumus berikut boleh digunakan: The following formula can be used: q ° 360° = q rad 2π rad = Panjang lengkok/Length of arc, s 2πj Jika diringkaskan, ia menjadi s = jq (q dalam radian/in radians) When simplified, it becomes 2. Petua kosinus digunakan untuk mencari panjang perentas AB. The cosine rule is used to find the length of the chord AB. AB2 = j 2 + j 2 – 2j 2 kos q AB2 = j 2 + j 2 – 2j 2 cos q dengan keadaan j ialah jejari dan q ialah sudut dalam darjah. where j is the radius and q is the angle in degrees. 1.3 Luas Sektor Suatu Bulatan/Area of Sector of a Circle 1. Hubungan di bawah boleh digunakan bergantung kepada maklumat yang diberi. The relationship below can be used depending on the information given. q ° 360° = q rad 2π rad = Panjang lengkok/Length of arc, s 2πj = Luas sektor/Area of sector, A πj 2 2. Jika sudut ialah dalam radian, maka Luas sektor/Area of a sector = 1 2 j 2 q If the angle is in radians, then Luas tembereng = 1 2 j 2 q – 1 2 j 2 sinq , dengan keadaan q ialah dalam radian bagi rumus 1 2 j 2 q dan q ialah dalam darjah bagi rumus 1 2 j 2 sin q. Area of a segment = 1 2 j 2 q – 1 2 j 2 sin q , where q is in radians in the formula 1 2 j 2 q and q is in degrees in the formula 1 2 j 2 sin q. × π 180° × 180° π s j θ B O j A θ j A B O θ θ A B O j Nota 1 rad O A B j j Sukatan Membulat Circular Measure BAB 1 01 Modul Series MateTam Tg5.indd 1 04/10/2021 2:15 PM
BAB 1 2 Radian Radian 1.1 1. Tukarkan setiap sudut dalam darjah yang berikut kepada radian. Convert each of the following angles in degrees into radians. TP 1 150° q = 150° × π 180° = 2.62 rad Contoh/Example (a) 75° q = 75° × π 180° = 1.31 rad (b) 230° q = 230° × π 180° = 4.01 rad (c) 315° 15′ q = 315.25° × π 180° = 5.50 rad (d) 41° q = 41° × π 180° = 0.72 rad (e) 175° 30′ q = 175° 30′ × π 180° = 3.06 rad (f) 85° 10′ q = 85.167° × π 180° = 1.49 rad (g) 178° 20′ q = 178.333° × π 180° = 3.11 rad (h) 228° 48′ q = 228.8° × π 180° = 3.99 rad 2. Tukarkan setiap sudut dalam radian yang berikut kepada darjah. Convert each of the following angles in radians into degrees. TP 1 2.1 rad q ° = 2.1 × 180° π = 120° 19′ Contoh/Example (a) 0.5 rad q° = 0.5 × 180° π = 28° 39′ (b) 1.3 rad q° = 1.3 × 180° π = 74° 29′ (c) 1.8π rad q° = 1.8π × 180° π = 324° (d) 2 5 π rad q° = 2 5 π × 180° π = 72° (e) 1 3 π rad q° = 1 3 π × 180° π = 60° (f) π 2 rad q° = π 2 × 180° π = 90° (g) 2.5 rad q° = 2.5 × 180° π = 143° 14′ (h) 4 5 π rad q° = 4 5 π × 180° π = 144° Buku Teks m.s. 2-4 01 Modul Series MateTam Tg5.indd 2 04/10/2021 2:15 PM
BAB 1 3 Uji Kendiri 1.1 Panjang Lengkok Suatu Bulatan Arc Length of a Circle 1.2 1. Hitung panjang lengkok bagi setiap yang berikut. Calculate the arc length for each of the following. TP 3 s = jq = 14 × 1.1 = 15.4 cm Contoh/Example B O A s 14 cm 1.1 rad (a) s = 7 × 1.8 = 12.6 cm (b) B A O s 240° 2.5 cm s = 2.5 × (360° – 240°) × p 180° = 5.24 cm (c) B A s O 1.5 cm 0.4 rad s = 1.5 × (π – 0.4) = 4.11 cm B A s O 7 cm 1.8 rad Buku Teks m.s. 5-11 1. Tukarkan 451° 15′ kepada radian. Berikan jawapan betul kepada tiga tempat perpuluhan. Convert 451° 15′ into radians. Give answer correct to three decimal places. q = 451° 15′ × π 180° = 7.876 rad 2. Tukarkan 4π 3 rad kepada darjah. Convert 4π 3 rad into degrees. q° = 4π 3 × 180° π = 240° 3. Rajah menunjukkan sebuah bulatan berpusat O dengan keadaan sudut minor AOB ialah 108° 42′. The diagram shows a circle with centre O where the minor angle AOB is 108° 42′. θ A B O 108° 42’ Cari nilai q, dalam radian. Find the value of q, in radians. 360° – 108° 42′ = 251° 18′ q = 251° 18′ × π 180° = 4.386 rad 4. Rajah menunjukkan sebuah sektor AOC berpusat O. AOB ialah segi tiga bersudut tegak di B. The diagram shows a sector AOC with centre O. AOB is a right-angled triangle at B. KBAT Mengaplikasi θ O B C A 4 5 Cari nilai q, dalam radian. Find the value of q, in radians. sin q = 4 5 q = 53.13° q = 53.13° × π 180° = 0.927 rad 01 Modul Series MateTam Tg5.indd 3 04/10/2021 2:15 PM
BAB 1 4 2. Hitung jejari, j bagi setiap yang berikut. Calculate the radius, j for each of the following. TP 2 Q O j P 1.5 rad 10.2 cm j = 10.2 1.5 = 6.8 cm Contoh/Example (a) P Q O j 24.5 cm 2.5 rad j = 24.5 2.5 = 9.8 cm (b) O Q P j 0.7 rad 30.5 cm j = 30.5 2π – 0.7 = 5.46 cm (c) O P Q j 15.7 cm 3.1 rad j = 15.7 2π – 3.1 = 4.93 cm 3. Hitung sudut, dalam radian, bagi setiap yang berikut. Calculate the angle, in radians, for each of the following. TP 2 14 cm θ 16 cm Gunakan/Use s = jq 16 = 14q q = 16 14 = 8 7 rad Contoh/Example (a) P Q O 12.3 cm 7.5 cm θ q = 12.3 7.5 = 1.64 rad (b) O P Q 12 cm 30.2 cm θ q = 30.2 12 = 2.52 rad (c) 2π – q = 35 11 q = 2π – 35 11 = 3.1 rad O P Q 11 cm 35 cm θ (d) Q O P 5 cm θ q = π 3 rad Tip SPM Simbol q dibaca sebagai “téta”. The symbol q is read as “théta”. 01 Modul Series MateTam Tg5.indd 4 04/10/2021 2:15 PM
BAB 1 5 4. Hitung perimeter bagi setiap kawasan berlorek. Calculate the perimeter for each of the shaded region. TP 3 D C 5 cm 4 cm O A B 1.6 rad Panjang lengkok AB/Length of arc AB = 5(1.6) = 8 cm Panjang lengkok CD/Length of arc CD = 9(1.6) = 14.4 cm Perimeter kawasan berlorek/Perimeter of the shaded region = AB + BC + CD + AD = 8 + 4 + 14.4 + 4 = 30.4 cm Contoh/Example (a) P 6.5 cm 0.45 rad Q O Perimeter = 6.5 + 6.5 + 6.5(0.45) = 13 + 2.93 = 15.93 cm (b) S R Q P 6.2 cm 1.8 rad O Panjang lengkok SR = 12.4(1.8) = 22.32 cm Panjang lengkok PQ = 6.2(1.8) = 11.16 cm Perimeter = 11.16 + 6.2 + 22.32 + 6.2 = 45.88 cm (c) 4 cm 3 cm 0.5 rad P T S Q R O OP = 4 0.5 = 8 cm Panjang lengkok SR = 11(0.5) = 5.5 cm Perimeter = 3 + 5.5 + 3 + 4 + 4 + 8 + 8 = 35.5 cm 5. Hitung jejari, j bagi setiap yang berikut. Calculate the radius, j for each of the following. TP 4 — 1 5 rad O V T U W j Perimeter rantau berlorek = 20 cm Perimeter of the shaded region = 20 cm VT = WU = 5 cm 5 + 1 5 j + 1 5 ( j + 5) + 5 = 20 2 5 j = 9 j = 45 2 = 22.5 cm Contoh/Example (a) 8 cm 13 cm 1.2 rad P Q O j Perimeter rantau berlorek = 22.4 cm Perimeter of the shaded region = 22.4 cm 1.2j + (13 – j) + 8 = 22.4 1.2j – j + 21 = 22.4 0.2j = 1.4 j = 7 cm 01 Modul Series MateTam Tg5.indd 5 04/10/2021 2:15 PM
BAB 1 6 (b) 1.5 rad j O R Q P S 1.5 rad OR = 2OP Perimeter rantau berlorek = 21 cm Perimeter of the shaded region = 21 cm 2j + 2j + 1.5(2j) + j + j + 1.5j = 21 10.5j = 21 j = 2 cm (c) j Q R O P θ Perimeter rantau berlorek = 16π + 24 Perimeter of the shaded region = 16π + 24 OP = OQ = QP º q = 60° = π 3 j + 2 3 πj + π 3 j + π 3 j + j = 16π + 24 2j + 4 3 πj = 4(4π + 6) 6j + 4πj = 12(4π + 6) j(6 + 4π) = 12(4π + 6) j = 12 cm 6. Tentukan perimeter bagi tembereng berlorek berdasarkan maklumat yang diberi. Determine the perimeter of the shaded segment based on the given informations. TP 4 8.2 cm A B O 1.2 rad 1.2 rad = 1.2 × 180° π = 68.75° Panjang perentas AB Length of chord AB = ! 8.22 + 8.22 – 2(8.2)2 kos/cos 68.75° = ! 85.74 = 9.26 cm Panjang lengkok AB Length of arc AB = 8.2 × 1.2 = 9.84 cm Perimeter = 9.84 + 9.26 = 19.1 cm Tip SPM Gunakan petua kosinus. Use the cosine rule. a2 = b2 + c2 – 2bc kos q a2 = b2 + c2 – 2bc cos q Contoh/Example (a) O 2.3 rad 11.2 cm P Q 2.3 rad = 2.3 × 180° π = 131.78° Panjang perentas PQ = ! 11.22 + 11.22 – 2(11.2)2 kos 131.78° = ! 418.034 = 20.45 cm Panjang lengkok PQ = 11.2 × 2.3 = 25.76 cm Perimeter = 25.76 + 20.45 = 46.21 cm (b) 1.5 rad 8.5 cm L K j O j(1.5) = 8.5 j = 5.67 cm 1.5 rad = 1.5 × 180° π = 85.94° Panjang perentas KL = ! 5.672 + 5.672 – 2(5.67)2 kos 85.94° = ! 59.745 = 7.73 cm Perimeter = 7.73 + 8.5 = 16.23 cm 01 Modul Series MateTam Tg5.indd 6 04/10/2021 2:15 PM
BAB 1 7 7. Selesaikan masalah yang melibatkan panjang lengkok berikut. Solve the following problems which involve the length of arc. TP 5 (a) Dane ingin menghasilkan beberapa buah topi seperti yang ditunjukkan dalam rajah untuk digunakan semasa jamuan hari jadi. Dane wants to make some decorative hats as shown in the diagram to be used in a birthday party. 9 cm 25 cm O P Dia merancang untuk menggunakan sekeping kadbod yang berbentuk segi empat tepat dengan ukuran 25 cm × 40 cm untuk menghasilkan setiap topi. Tentukan sama ada kadbod itu mencukupi untuk menghasilkan topi tersebut. Tunjukkan langkah kerja anda. He plans to use a piece of rectangular cardboard with a dimension of 25 cm × 40 cm to make each hat. Determine whether the cardboard is sufficient to make the hat. Show your working. S O P Q R 25 cm 25 cm 39.6° Panjang lilitan topi = 2π(9) Maka, panjang lengkok PQ = 18π cm 25 × ˙QOP × p 180° = 18π ∠QOP = 18 × 180° 25 = 129.6° ∠QOR = 129.6° – 90° = 39.6° Maka, OS = 25 sin 39.6° = 15.94 cm Panjang minimum kadbod = (25 + 15.94) cm = 40.94 cm ≈ 41 cm Saiz minimum kadbod = (25 × 41) cm Oleh kerana saiz minimum kadbod tersebut ialah (25 × 41) cm, maka saiz sebenar kadbod (25 × 40) cm itu tidak mencukupi untuk menghasilkan topi tersebut. Aminah sedang menjahit tampalan pada sehelai selimut. Dia menggunakan corak seperti yang ditunjukkan dalam rajah berikut. Ketiga-tiga kawasan berlorek adalah serupa dengan diameter bulatan itu ialah 12 cm. Jika dia perlu menjahit di sekeliling kawasan berlorek, hitung panjang benang, dalam m, yang diperlukan untuk menjahit 15 keping tampalan yang sama pada selimut tersebut. Aminah is sewing some patchwork on a blanket. She uses the pattern as shown in the following diagram. The three shaded regions are identical with the diameter of the circle is 12 cm. If she needs to sew around each shaded region, find the length of the thread, in m, needed to sew 15 pieces of same patchwork on the blanket. P R Q Jejari/Radius = 6 cm ∠POQ = 120° PQ = QR = PR Panjang PQ Length of PQ = ! 62 + 62 – 2(6)2 kos/cos 120° = 10.39 cm Panjang lengkok PQ Length of arc PQ = 6 × 120° × π 180° = 12.57 cm Jumlah panjang benang bagi setiap bulatan Total length of thread for each circle = 3(10.39 + 12.57) = 68.88 cm 15 tampalan/patches = 15 × 68.88 = 1 033.2 cm ÷ 100 = 10.332 m P Q R 120° 6 cm O Contoh/Example 01 Modul Series MateTam Tg5.indd 7 04/10/2021 2:15 PM
BAB 1 8 Luas Sektor Suatu Bulatan Area of Sector of a Circle 1.3 Uji Kendiri 1.2 1. Rajah menunjukkan sebuah sektor AOB berpusat O dan berjejari 8 cm. The diagram shows a sector AOB with centre O and a radius of 8 cm. KBAT Mengaplikasi 12 cm B A O θ 8 cm Diberi panjang lengkok AB ialah 12 cm, cari Given the length of arc AB is 12 cm, find (a) nilai q, dalam radian, the value of q, in radians, (b) panjang perentas AB. the length of the chord AB. (a) 8q = 12 q = 12 8 = 1.5 rad (b) 1.5 rad = 1.5 × 180° π = 85.94° Panjang perentas AB = ! 82 + 82 – 2(8)2 kos 85.94° = ! 118.937 = 10.91 cm 2. Rajah menunjukkan sebuah semibulatan berpusat O dan sebuah sektor APB dengan pusat P yang berjejari 10 cm. The diagram shows a semicircle with centre O and a sector APB with centre P and a radius of 10 cm. KBAT Mengaplikasi A B P 1.4 rad 10 cm O Cari Find (a) jejari semibulatan itu, the radius of the semicircle, (b) perimeter kawasan berlorek itu. the perimeter of the shaded region. (a) 1.4 rad = 1.4 × 180° π = 80.21° sin ( 80.21° 2 ) = OB 10 OB = 6.44 cm (b) Perimeter = 6.44π + 10(1.4) = 34.23 cm 1. Hitung luas kawasan berlorek yang berikut berdasarkan syarat-syarat yang diberi. Calculate the area of the following shaded regions based on the given conditions. TP 3 10 cm O A B 1.4 rad Gunakan/Use 1.4 2π = Luas/Area πj 2 Luas/Area = 1.4 2 (10)2 = 70 cm2 Contoh/Example (a) 3.5 cm 2.2 rad O Luas = 1 2 (3.5)2 (2.2) = 13.48 cm2 (b) 0.4 rad 7.2 cm Luas = 1 2 (7.2)2 (0.4) = 10.37 cm2 (c) 4.5 cm 3θ O θ q = π 4 3q = 3π 4 Luas = 1 2 (4.5)2 (3π 4 ) = 23.86 cm2 Buku Teks m.s. 12-19 01 Modul Series MateTam Tg5.indd 8 04/10/2021 2:15 PM
BAB 1 9 2. Hitung jejari, j dan/atau sudut q, dalam radian, bagi setiap yang berikut. Calculate the radius, j and/or the angle q, in radians, for each of the following. TP 3 7 cm θ Luas kawasan berlorek Area of the shaded region = 24 cm2 Luas/Area = 1 2 j 2 q 24 = 1 2 (7)2 q q = 48 49 = 0.98 rad Contoh/Example (a) 1.3 rad j Luas kawasan berlorek Area of the shaded region = 20 cm2 1 2 (1.3)j 2 = 20 j 2 = 20 × 2 1.3 j = 5.55 cm (b) O 10 cm θ Luas kawasan berlorek Area of the shaded region = 100 cm2 100 = 1 2 (10)2 (2π – q) 2 = 2π – q q = 2π – 2 = 4.28 rad (c) 10 cm j θ Luas kawasan berlorek Area of the shaded region = 30 cm2 1 2 j 2 q = 30.................a jq = 10.................b a ÷ b 1 2 j = 3 j = 6 cm ºq = 10 6 = 5 3 rad 3. Hitung luas bagi setiap kawasan berlorek yang berikut. Calculate the area for each of the following shaded regions. TP 4 C D O A B 1.6 rad 5 cm 3 cm Luas/Area OBC = 1 2 (8)2 (1.6) Luas/Area OAD = 1 2 (5)2 (1.6) Luas kawasan berlorek Area of the shaded region = 1 2 (8)2 (1.6) – 1 2 (5)2 (1.6) = 31.2 cm2 Contoh/Example (a) 1.4 rad 2.3 cm P Q O S R Luas kawasan berlorek = 1 2 (4.6)2 (2π – 1.4) – 1 2 (2.3)2 (2π – 1.4) = 38.75 cm2 01 Modul Series MateTam Tg5.indd 9 04/10/2021 2:15 PM
BAB 1 10 (b) 6 cm 7 cm B O A C θ tan q = 7 6 q = 49.4° Luas ∆OAB = 1 2 (6)(7) = 21 cm2 Luas sektor = 1 2 (6)2 (49.4° × π 180°) = 15.52 cm2 Luas kawasan berlorek = 21 – 15.52 = 5.48 cm2 (c) 8.2 cm B C O A ∠BOA = π 3 rad Luas semibulatan = 1 2 (8.2)2 π = 105.62 cm2 Luas sektor AOB = 1 2 (8.2)2 ( π 3 ) = 35.21 cm2 Luas kawasan berlorek = 105.62 – 35.21 = 70.41 cm2 4. Cari luas tembereng bagi setiap yang berikut. Find the area of segment for each of the following. TP 5 Luas tembereng Area of the segment = Luas sektor AOB – Luas segi tiga AOB Area of segment AOB – Area of triangle AOB = 1 2 j 2 q – 1 2 j 2 sin q° Jika q = 1.4 rad dan j = 4 cm, luas tembereng If q = 1.4 rad and j = 4 cm, the area of the segment = 1 2 (4)2 (1.4) – 1 2 (4)2 sin(1.4 × 180° π ) = 3.32 cm2 Contoh/Example 1.4 rad O 4 cm A B h C Tip SPM q hendaklah dalam darjah untuk trigonometri sinus. q must be in degrees for trigonometric sine. (a) 8 cm 1.1 rad O P Q 1.1 rad = 1.1 × 180° π = 63.02° Luas sektor = 1 2 (8)2 (1.1) = 35.2 cm2 Luas segi tiga = 1 2 (8)2 sin 63.02° = 28.52 cm2 Luas tembereng = 35.2 – 28.52 = 6.68 cm2 (b) O 9 cm P Q — π 3 ∠POQ = π – π 3 = 2π 3 rad Luas tembereng = 1 2 (9)2 ( 2π 3 ) – 1 2 (9)2 sin ( 2π 3 × 180° π ) = 49.75 cm2 Tip SPM Luas segi tiga/Area of triangle = 1 2 AB × h sin q 2 = AC j º AC = j sin q 2 , AB = 2j sin q 2 kos q 2 / cos q 2 = h j h = j kos q 2 /j cos q 2 Luas segi tiga/Area of triangle = 1 2 (2j 2 sin q 2 kos q 2 )/ 1 2 (2j 2 sin q 2 cos q 2 ) = 1 2 j 2 sin q 01 Modul Series MateTam Tg5.indd 10 04/10/2021 2:15 PM
BAB 1 11 Uji Kendiri 1.3 (c) 6.5 cm 30° O Q P ∠POQ = 180° – 60° = 120° 120° × π 180° = 2π 3 rad Luas tembereng = 1 2 (6.5)2 ( 2π 3 ) – 1 2 (6.5)2 sin 120° = 25.95 cm2 (d) 5 cm O A B C D ∠BOC = ∠AOB 3 = π 3 rad Luas tembereng = 3[ 1 2 (5)2 ( π 3 ) – 1 2 (5)2 sin 1 π 3 × 180° π 2] = 6.79 cm2 1. Rajah menunjukkan sebuah kipas tangan berbentuk sektor OABCD dan bahagian berlabel ABCD diperbuat daripada kain. The diagram shows a hand fan in a form of a sector OABCD and the area labelled ABCD is made of cloth. KBAT Mengaplikasi A B D O C Kain 14 cm Cloth 13 cm θ Diberi perimeter kain ialah ( 41 4 p + 26) cm, cari Given the perimeter of the cloth is ( 41 4 p + 26) cm, find (a) nilai q, dalam radian, the value of q, in radians, (b) luas kain itu, dalam sebutan π. the area of the cloth, in terms of π. (a) 14q + 27q = 41 4 p + 26 – 26 41q = 41 4 π q = 1 4 π rad (b) Luas kain = 1 2 (27)2 ( 1 4 π) – 1 2 (14)2 ( 1 4 π) = 91 1 8 π – 24 1 2 π = 66 5 8 π cm2 2. Rajah menunjukkan dua sektor bulatan, OACB dan CBOA, masing-masing dengan pusat O dan C. The diagram shows two sectors, OACB and CBOA, with centres O and C respectively. KBAT Mengaplikasi B A O C M Diberi bahawa OC adalah berserenjang dengan AB dan melalui titik tengah M. Jika OB = 8 cm, Given that OC is perpendicular to AB and passes through the midpoint M. If OB = 8 cm, (a) tunjukkan bahawa sudut AOB ialah 2 3 π, show that the angle AOB is 2 3 π, (b) cari luas kawasan berlorek itu. find the area of the shaded region. (a) OB = 8 cm, OM = 4 cm kos ∠MOB = 4 8 ∠MOB = 60° ∠AOB = 120° × p 180° = 2 3 π rad (b) Luas sektor AOB = 1 2 (8)2 ( 2 3 π) = 67.02 cm2 Luas ∆OAB = 1 2 (8)2 sin 120° = 27.71 cm2 Luas kawasan berlorek = 67.02 – 27.71 = 39.31 cm2 Luas sektor ACB = 2[ 1 2 (8)2 ( π 3 ) – 1 2 (8)2 sin 60°] = 11.60 cm2 Luas kawasan berlorek = 39.31 + 11.60 = 50.91 cm2 01 Modul Series MateTam Tg5.indd 11 04/10/2021 2:15 PM
BAB 1 12 Aplikasi Sukatan Membulat Application of Circular Measures 1.4 1. Selesaikan setiap masalah yang berikut. Solve each of the following problems. TP 6 Rajah menunjukkan sekeping jubin berbentuk segi empat sama ABCD. Diberi bahawa panjang sisi AB ialah 14 cm. Keempat-empat kawasan berlorek itu adalah serupa. OPQ dan BPQ adalah sektor masingmasing dengan pusat O dan B. The diagram shows a square tile ABCD. Given that the length of side AB is 14 cm. The 4 shaded regions are identical. OPQ and BPQ are sectors with centres O and B respectively. A B S Q R P O D C Jika terdapat 150 keping jubin yang sama pada suatu lantai, cari luas yang dilitupi oleh kawasan berlorek itu. If there are 150 same tiles on the floor, find the area covered by the shaded regions. O Q P H Luas sektor OPQ Area of the sector OPQ = 1 4 × π(7)2 Luas ∆OPQ Area of ∆OPQ = 1 2 × 7 × 7 = 1 2 (7)2 Luas tembereng PHQ Area of the segment PHQ = 1 4 × π(7)2 – 1 2 (7)2 Maka, luas kawasan berlorek setiap jubin Hence, the area of the shaded region of each tile = 8[ 1 4 × π(7)2 – 1 2 (7)2 ] = 8(13.98) = 111.88 cm2 Jumlah luas kawasan berlorek Total area of the shaded regions = 111.88 × 150 = 1.6782 × 104 cm2 Contoh/Example (a) Rajah menunjukkan sebuah sektor dengan pusat O dan berjejari 9.3 cm. Diberi bahawa Q terletak pada lilitan bulatan. The diagram shows a sector with centre O and a radius of 9.3 cm. Given that Q lies on the circumference. Q P R 140° 48° O Cari Find (i) perimeter kawasan berlorek, the perimeter of the shaded region, (ii) luas kawasan berlorek. the area of the shaded region. Q P R 140° 48° 48° 22° O (i) ∠POQ = 180° – 2(48°) = 84° Panjang perentas PQ = ! 9.32 + 9.32 – 2(9.3)2 kos 84° = 12.45 cm Panjang lengkok PQ = 9.3 × 84° × p 180° = 13.63 cm ∠QOR = 360° – 84° – 140° = 136° Panjang perentas QR = ! 9.32 + 9.32 – 2(9.3)2 kos 136° = 17.25 cm Panjang lengkok QR = 9.3 × 136° × p 180° = 22.07 cm Maka, perimeter kawasan berlorek = 12.45 + 17.25 + 13.63 + 22.07 = 65.4 cm (ii) Luas kawasan berlorek = [ 1 2 × 9.32 × 84° × p 180° – 1 2 (9.3)2 sin 84°] + [ 1 2 × 9.32 × 136° × p 180° – 1 2 (9.3)2 sin 136°] = 93 cm2 Buku Teks m.s. 20-22 01 Modul Series MateTam Tg5.indd 12 04/10/2021 2:15 PM
BAB 1 13 Uji Kendiri 1.4 1. Rajah menunjukkan sebuah bulatan berjejari 6.5 cm dan berpusat O. Diberi bahawa panjang AB adalah sama dengan panjang AC. The diagram shows a circle with a radius of 6.5 cm and centre O. Given that the length of AB is the same as AC. KBAT Menilai B A C 50° O Cari/Find (a) perimeter kawasan berlorek, the perimeter of the shaded region, (b) luas kawasan berlorek. the area of the shaded region. (a) ∠BAO = 25° Panjang AB = 2(6.5 kos 25°) = 11.78 cm Panjang lengkok AB = 6.5 × 130° × p 180° = 14.75 cm Perimeter kawasan berlorek = 2(11.78) + 2(14.75) = 53.06 cm (b) Luas kawasan berlorek = 2[ 1 2 × 6.52 × 130° × p 180° – 1 2 (6.5)2 sin 130°] = 63.5 cm2 2. Rajah menunjukkan sepotong tembikai dengan keratan rentas seragamnya berbentuk sektor berpusat O. The diagram shows a uniform cross-section of a piece of watermelon in the shape of a sector with centre O. KBAT Menilai B A C θ D O 10 cm Diberi bahawa panjang lengkok AB dan CD masingmasing ialah 20 cm dan 16 cm dengan bahagian ABCD tidak boleh dimakan. Jika AC = 2 cm dan panjang potongan tembikai itu ialah 10 cm, cari Given that the arc length AB and CD is 20 cm and 16 cm respectively with the portion ABCD is not edible. If AC = 2 cm and the length of a piece of the watermelon is 10 cm, find (a) nilai q, dalam radian, the value of q, in radians, (b) isi padu bahagian yang boleh dimakan. the volume of the area that can be eaten. (a) OC(q) = 16 ..............a (OC + 2)q = 20 OC(q) + 2q = 20 ..............b b – a: 2q = 20 – 16 q = 2 rad (b) OC = 16 2 = 8 cm Isi padu = 1 2 (8)2 (2) × 10 = 640 cm3 Bahagian A/Section A 1. Rajah menunjukkan sebuah bulatan berpusat O dengan jejari 5 cm. Diberi OQTC ialah sebuah segi empat tepat dengan luas 40 cm2 , cari The diagram shows a circle with centre O and a radius of 5 cm. Given OQTC is a rectangle with an area of 40 cm2 , find KBAT Menilai (a) nilai q, dalam radian, the value of q, in radians, [2 markah/marks] (b) luas sektor OAB, the area of the sector OAB, [2 markah/marks] (c) perimeter kawasan berlorek. the perimeter of the shaded region. [2 markah/marks] (a) OC × CT = 40 5 × CT = 40 º CT = 8 cm tan q = 8 5 q = 58° × π 180° = 1.01 rad (b) Luas sektor OAB = 1 2 (5)2 ( π 2 – 1.01) = 7.01 cm2 (c) Panjang lengkok AB = 5( π 2 – 1.01) = 2.80 cm Panjang BT = ! 82 + 52 – 5 = 4.43 cm Perimeter = 2.80 + 4.43 + 3 + 5 = 15.23 cm Soalan Berformat SPM Kertas 1 Paper 1 T Q A B O C θ 01 Modul Series MateTam Tg5.indd 13 04/10/2021 2:15 PM
BAB 1 14 2. Rajah menunjukkan sebuah sektor AOB berpusat O. The diagram shows a sector AOB with centre O. KBAT Menilai 7.5 cm 130° O A B Cari Find (a) ∠AOB, dalam radian, ∠AOB, in radians, [2 markah/marks] (b) perimeter, dalam cm, sektor AOB. the perimeter, in cm, of the sector AOB. [3 markah/marks] (a) 130° × π 180° = 2.27 rad (b) Perimeter = 7.5 + 7.5 + 7.5(2.27) = 32.03 cm 3. Rajah menunjukkan dua sektor, POQ dan SOR, berpusat O. The diagram shows two sectors, POQ and SOR, with centre O. KBAT Menganalisis O P S R Q θ Diberi bahawa OP = 6 cm, OP : OS = 2 : 3 dan luas kawasan berlorek ialah 33.75 cm2 , cari Given that OP = 6 cm, OP : OS = 2 : 3 and the area of the shaded region is 33.75 cm2 , find (a) panjang OS, the length of OS, [2 markah/marks] (b) nilai q, dalam radian. the value of q, in radians. [3 markah/marks] (a) OP OS = 2 3 = 6 OS = OS = 9 cm (b) 1 2 (9)2 q – 1 2 (6)2 q = 33.75 22.5q = 33.75 q = 1.5 rad Bahagian B/Section B 4. Rajah menunjukkan dua bulatan sepusat berpusat O. Sudut yang dicangkum pada pusat O oleh lengkok major PQ ialah 8p rad dan perimeter bagi seluruh rajah ialah 48 cm. The diagram shows two concentric circles with centre O. The angle subtended at the centre O by the major arc PQ is 8p rad and the perimeter for the whole diagram is 48 cm. KBAT Menganalisis A O B Q P Diberi OP = 3 2 OA, OA = j cm dan ∠AOB = 4p rad, ungkapkan j dalam sebutan p. Given OP = 3 2 OA, OA = j cm and ∠AOB = 4p rad, express j in terms of p. [8 markah/marks] OP = 3 2 OA = 3 2 j Panjang lengkok major PQ = 3 2 j(8p) = 12jp Panjang lengkok AB = j(4p) = 4jp Jumlah perimeter: 12jp + 4jp + 2( 3 2 j – j) = 48 16jp + j = 48 j(16p + 1) = 48 j = 48 16p + 1 5. Rajah menunjukkan sebuah segi tiga sama sisi OPQ dan titik S yang terletak di dalam segi tiga itu dengan keadaan OS = SP = SQ. The diagram shows an equilateral triangle OPQ and the point S lies inside the triangle such that OS = SP = SQ. KBAT Menganalisis 12 cm O P Q S X Diberi bahawa S ialah pusat bagi sektor SPXQ dan panjang lengkok PXQ ialah 12 cm. Cari perimeter kawasan berlorek, dalam sebutan π dan !3. Given that S is the centre of the sector SPXQ and the length of arc PXQ is 12 cm. Find the perimeter of the shaded region, in terms of π and !3 . [8 markah/marks] kos ∠OQS = 6 SQ kos 30° = 6 SQ SQ = 6 kos 30° = 6 ! 3 2 = 12 ! 3 = 4! 3 cm Perimeter kawasan berlorek = 4! 3 + 4! 3 + (4! 3)(120° × π 180° ) = 8! 3 + 8! 3 3 π = 8! 3(1 + π 3 ) cm KLON SPM 30° 120° 12 6 O S P Q 01 Modul Series MateTam Tg5.indd 14 04/10/2021 2:15 PM
BAB 1 15 Bahagian A/Section A 1. Rajah menunjukkan sektor POR dengan pusat O. Diberi bahawa panjang lengkok PQ ialah 3.4 cm. Hitung The diagram shows a sector POR with centre O. It is given that the length of arc PQ is 3.4 cm. Calculate KBAT Menganalisis (a) ∠POQ, dalam radian, ∠POQ, in radians, [3 markah/marks] (b) luas, dalam cm2 , kawasan berlorek. the area, in cm2 , of the shaded region. [3 markah/marks] 2. Rajah menunjukkan sebuah kebun yang berbentuk sektor OAB dengan pusat O dan berjejari 12 m. Diberi OPQR ialah kawasan tanaman sayur yang berbentuk trapezium dengan keadaan OR = 10 m, PQ = 8 m dan QR adalah berserenjang dengan OA. Diberi bahawa ∠AOB = 1.2 rad, cari The diagram shows a garden in the shape of a sector OAB with centre O and a radius of 12 m. Given OPQR is a trapezoidal vegetable growing area such that OR = 10 m, PQ = 8 m and QR is perpendicular to OA. Given that ∠AOB = 1.2 rad, find KBAT Mengaplikasi (a) panjang pagar yang digunakan untuk memagar kawasan tanaman sayur, the length of the fence used to fence around the vegetable growing area, [3 markah/marks] (b) luas kawasan yang tidak ditanam dengan sayur. the area not planted with vegetables. [3 markah/marks] Bahagian B/Section B 3. Rajah menunjukkan sebuah bulatan berpusat O. Diberi D ialah titik tengah bagi AB dan luas kawasan berlorek ialah 200π 3 cm2 , cari The diagram shows a circle with centre O. Given D is the midpoint of AB and the area of the shaded region is 200π 3 cm2 , find KBAT Mengaplikasi (a) nilai q, dalam radian, the value of q, in radians, [2 markah/marks] (b) jejari bulatan itu, the radius of the circle, [2 markah/marks] (c) perimeter sektor berlorek, the perimeter of the shaded sector, [3 markah/marks] (d) luas tembereng ADBC. the area of the segment ADBC. [3 markah/marks] 4. Rajah menunjukkan sebuah semibulatan berpusat O dengan jejari 8 cm. TSQ ialah sebuah sektor dengan pusat T dan berjejari 12 cm. Diberi bahawa OR berserenjang dengan POQ. Hitung The diagram shows a semicircle with centre O and a radius of 8 cm. TSQ is a sector with centre T and a radius of 12 cm. Given that OR is perpendicular to POQ. Calculate KBAT Menganalisis (a) nilai q, dalam radian, the value of q, in radians, [2 markah/marks] (b) perimeter kawasan berlorek, the perimeter of the shaded region, [4 markah/marks] (c) luas kawasan berlorek. the area of the shaded region. [4 markah/marks] KLON SPM 60° C B O A D θ Kertas 2 Paper 2 P T O Q R S θ 8.6 cm S O R Q P 50° O 1.2 rad B A R Q P 01 Modul Series MateTam Tg5.indd 15 04/10/2021 2:15 PM
BAB 1 16 5. Rajah menunjukkan sebuah sektor AOB berpusat O dengan jejari 15 cm. Titik C terletak pada OB dengan keadaan OC = CA = 9 cm. Hitung The diagram shows a sector AOB with centre O and a radius of 15 cm. Point C lies on OB such that OC = CA = 9 cm. Calculate KBAT Menganalisis (a) nilai q, dalam radian, the value of q, in radians, [2 markah/marks] (b) perimeter kawasan berlorek, the perimeter of the shaded region, [4 markah/marks] (c) luas kawasan berlorek. the area of the shaded region. [4 markah/marks] 6. Rajah menunjukkan sebuah semibulatan dengan pusat O dan sebuah sukuan bulatan RPQ berpusat P. Hitung The diagram shows a semicircle with centre O and a quadrant RPQ with centre P. Calculate KBAT Menilai (a) nilai q, dalam radian, the value of q, in radians, [2 markah/marks] (b) perimeter kawasan berlorek, the perimeter of the shaded region, [4 markah/marks] (c) luas kawasan berlorek. the area of the shaded region. [4 markah/marks] 7. Rajah menunjukkan dua sektor, AOB dan APQ, masing-masing berpusat O dan P. Diberi bahawa OB selari dengan PQ. Jika luas sektor APQ sama dengan luas kawasan berlorek, cari The diagram shows two sectors, AOB and APQ, with centre O and P respectively. Given that OB is parallel to PQ. If the area of the sector APQ is the same as the area of the shaded region, find KBAT Menganalisis (a) nisbah j 1 kepada j2 , the ratio of j1 to j2 , [5 markah/marks] (b) nisbah perimeter sektor APQ kepada AOB. the ratio of the perimeter of the sector APQ to AOB. [5 markah/marks] 8. Rajah menunjukkan suatu corak berbentuk bulatan yang dilukis oleh Donny bagi projek lukisannya. Diberi bahawa lapan kawasan yang berlorek itu adalah serupa. Panjang lengkok ACB dan ADB adalah sama dan perimeter bagi kawasan berlorek itu ialah 28π cm. Cari The diagram shows the circular design drew by Donny for his art project. Given that the eight shaded regions are congruent. The lengths of arc ACB and ADB are the same and the perimeter of the shaded region is 28π cm. Find KBAT Mengaplikasi (a) jejari bulatan itu, the radius of the circle, [5 markah/marks] (b) luas kawasan berlorek itu. the area of the shaded region. [5 markah/marks] O A C B θ 6 cm 3 cm R A O P Q θ A B O P Q j 1 j 2 θ KLON SPM B A C D 01 Modul Series MateTam Tg5.indd 16 04/10/2021 2:15 PM
2.3 Pembezaan Peringkat Kedua The Second Derivative Buku Teks m.s. 49-50 1. Cari dy dx dan d2 y dx2 bagi setiap yang berikut. Find dy dx and d2 y dx2 for each of the following. TP 2 y = 4x3 + 2x – 5 x y = 4x3 + 2x – 5x–1 dy dx = 12x2 + 2 + 5x–2 d2 y dx2 = 24x – 10x–3 = 12x2 + 2 + 5 x2 = 24x – 10 x3 Contoh/Example (a) y = (1 – 2x2 )4 dy dx = 4(1 – 2x2 )3 (– 4x) = –16x(1 – 2x2 )3 d2 y dx2 = –16x(3)(1 – 2x2 )2 (– 4x) + (1 – 2x2 )3 (–16) = (1 – 2x2 ) 2 (192x2 + 32x2 – 16) = (1 – 2x2 )2 (224x2 – 16) (b) y = 2x3 – 5 x y = 2x3 – 5x–1 dy dx = 6x2 + 5 x2 = 6x2 + 5x–2 = 6x2 + 5 x2 d2 y dx2 = 12x – 10 x3 2. Selesaikan masalah berikut. Solve the following problems. TP 3 Diberi g(x) = 3x3 + 2x2 – 9x – 7, cari gʹ(4) dan gʺ(–1). Given g(x) = 3x3 + 2x2 – 9x – 7, find gʹ(4) and gʺ(–1). g(x) = 3x3 + 2x2 – 9x – 7 gʹ(x) = 9x2 + 4x – 9 gʹ(4) = 9(4)2 + 4(4) – 9 = 151 gʺ(x) = 18x + 4 gʺ(–1) = 18(–1) + 4 = –14 Contoh/Example (a) Diberi bahawa f(x) = (5x – 2)5 , cari f'(1) dan f ʺ(0). Given that f(x) = (5x – 2)5 , find f'(1) and f ʺ(0). f(x) = (5x – 2)5 fʹ(x) = 5(5x – 2)4 (5) = 25(5x – 2)4 fʹ(1) = 25(5(1) – 2)4 = 2 025 f ʺ(x) = 25(4)(5x – 2)3 (5) = 500(5x – 2)3 f ʺ(0) = 500(5(0) – 2)3 = – 4 000 Tip SPM • d2 y dx2 = d dx ( dy dx ) • f(x) = d dx (f(x)) • d2 y dx2 ≠ ( dy dx ) BAB 2 2 26 02 Modul Series MateTam Tg5.indd 26 04/10/2021 2:42 PM
Uji Kendiri 5.1 1. Tulis dalam tatatanda set bagi pemboleh ubah rawak X yang mewakili jisim sebiji durian dari kebun Ronny yang jisimnya di antara 0.8 kg dengan 3 kg. Write down in a set notation of the random variable X that represents the mass of a durian from Ronny’s farm which has a mass between 0.8 kg and 3 kg. X = {x : 0.8 , x , 3} 2. Berikut merupakan sebahagian daripada sistem merit dan demerit di sebuah sekolah: The following is a part of the merit and demerit system in a school: • Tolak 1 markah jika lewat datang ke sekolah. Minus 1 mark if late to school. • Tambah 2 markah bagi setiap kerja rumah yang selesai. Add 2 marks for each finished homework. Tulis dalam tatatanda set bagi pemboleh ubah rawak X yang mewakili sistem itu apabila Isah mempunyai kerja rumah bagi 5 subjek pada suatu hari. Write down in a set notation of the random variable X that represents the system when Isah has a homework for 5 subjects on a certain day. X = {–1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} 4. Nilai yang mungkin bagi X ditunjukkan dalam jadual berikut berserta dengan kebarangkalian yang sepadan. Cari nilai p dan seterusnya lukis graf taburan kebarangkaliannya. The possible values of X are shown in the following table together with the corresponding probabilities. Find the value of p and hence draw the probability distribution graph. TP 3 X = r 1 3 5 7 9 P(X = r) p 2p p 1 2 p 1 2 p p + 2p + p + 1 2 p + 1 2 p = 1 5p = 1 p = 1 5 X = r 1 3 5 7 9 P(X = r) 0.2 0.4 0.2 0.1 0.1 P(X = r) 0.1 0 1 3 5 7 9 0.2 0.3 0.4 x x x x x r Contoh/Example PAUTAN INTERAKTIF PAUTAN INTERAKTIF Graf taburan kebarangkalian The graph of the probability distribution (a) X = r 3 4 5 6 7 P(X = r) 1 2 p 0.25 p 2 3 p 0.1 1 2 p + 0.25 + p + 2 3 p + 0.1 = 1 13 6 p = 0.65 p = 0.3 X = r 3 4 5 6 7 P(X = r) 0.15 0.25 0.3 0.2 0.1 P(X = r) 0.1 0 3 4 5 6 7 0.2 0.3 0.4 x x x x x r BAB 5 74 05 Modul A+ MateTam Tg5.indd 74 05/10/2021 12:29 PM
MG-1 BAB 1 Kertas 2/Paper 2 Bahagian A/Section A 1. (a) 3.4 = 8.6q q = 3.4 8.6 = 0.4 rad ∴∠POQ = 0.4 rad (b) tan 50° = SR OR SR = 8.6 tan 50° = 10.25 cm Luas kawasan berlorek Area of the shaded region = 1 2 (8.6)(10.25) – 1 2 (8.6)2 sin 50° + 1 2 (8.6)2 (0.4) = 30.54 cm2 2. (a) 1.2 rad = 68.75° tan 68.75° = QR 2 QR = 2 tan 68.75° QR = 5.14 m kos/cos 68.75° = 2 OP OP = 2 kos/cos 68.75° OP = 5.52 m Perimeter pagar Perimeter of the fence = 5.14 + 5.52 + 10 + 8 = 28.66 m (b) Luas kawasan yang tidak ditanam dengan sayur Area of the area not planted with vegetables = 1 2 (12)2 (1.2) – 1 2 (10 + 8)(5.14) = 40.14 m2 Bahagian B/Section B 3. (a) q = 360° – 2 × 60° q = 240° q = 240° × π 180° q = 4π 3 rad (b) 1 2 j 2 ( 4π 3 ) = 200π 3 j 2 = 100 j = 10 cm (c) Perimeter sektor berlorek Perimeter of the shaded sector = 10 + 10 + 10( 4π 3 ) = 61.89 cm (d) ∠AOB = 120° 120° × π 180° = 2π 3 Luas tembereng ADBC/Area of the segment ADBC = 1 2 (10)2 ( 2π 3 ) – 1 2 (10)2 sin 120° = 61.42 cm2 4. (a) tan q = 8 4 q = 63.43° q = 1.11 rad (b) TR = ! 82 + 42 = ! 80 SQ = 12(1.11) = 13.32 RQ = π 2 (8) = 4π Perimeter kawasan berlorek Perimeter of the shaded region = 12 – ! 80 + 13.32 + 4π = 28.94 cm (c) Luas sektor STQ/Area of sector STQ = 1 2 (12)2 (1.11) = 79.92 cm2 Luas/Area ∆ = 1 2 (4)(8) = 16 cm2 Luas/Area ROQ = 1 4 π(8)2 = 16π cm2 Luas kawasan berlorek Area of the shaded region = 79.92 – 16 – 16π = 13.65 cm2 5. (a) kos/cos q = 7.5 9 q = 33.56° q = 0.586 rad (b) Perimeter kawasan berlorek Perimeter of the shaded region = 9 + 6 + 15(0.586) = 23.79 cm (c) Luas kawasan berlorek Area of the shaded region = 1 2 (15)2 (0.586) – 1 2 (15)(9) sin 33.56° = 28.61 cm2 6. (a) sin ∠ROP = 3 6 ∠ROP = 30° 30° × π 180° = π 6 ∠AOR = q q = π – π 6 q = 2.62 rad (b) OP = ! 62 – 32 = 3! 3 cm Perimeter kawasan berlorek Perimeter of the shaded region = 6( π 6 ) + 3( π 2 ) + 3 – (6 – 3! 3 ) = 10.05 cm (c) Luas sektor ROP/Area of sector ROP = 1 2 (6)2 ( π 6 ) = 3π cm2 Luas/Area ∆ROP = 1 2 (3! 3 )(3) = 9! 3 2 cm2 Luas sektor RPQ/Area of sector RPQ = 1 2 (3)2 ( π 2 ) = 9 4 π cm2 Luas kawasan berlorek Area of the shaded region = 9 4 π – (3π – 9! 3 2 ) = 5.44 cm2 Jwpn_G Modul A+ MateTam Tg5.indd 1 27/10/2021 11:11 AM
MG-14 BAB 1 Sukatan Membulat Tahap Penguasaan Tafsiran 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang sukatan membulat dalam konteks penyelesaian masalah rutin yang kompleks. 6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang sukatan membulat dalam konteks penyelesaian masalah bukan rutin secara kreatif. 1. Rajah menunjukkan sebuah segi tiga bersudut tegak PQR. Diberi A ialah titik tengah PQ dengan keadaan PQ = 50 cm, ∠PRQ = π 6 rad, PBQ ialah satu lengkok daripada sebuah sektor berjejari r cm dan AB = 10 cm. The diagram shows a right-angled triangle PQR. Given that A is the midpoint of PQ where PQ = 50 cm, ∠PRQ = π 6 rad, PBQ is an arc from a sector with radius r cm and AB = 10 cm. TP 5 A Q R P B Cari Find (a) nilai r, the value of r, (b) panjang lengkok PBQ, the arc length PBQ, (c) luas kawasan berlorek. the area of the shaded region. (a) OA = r – 10 OP2 = OA2 + AP2 r2 = (r – 10)2 + 252 r2 = r2 – 20r + 100 + 625 20r = 725 r = 36.25 cm (b) tan ∠POA = 25 26.25 ∠POA = 43.6° = 0.7610 rad ∠POQ = 1.52 rad Panjang lengkok PBQ/Arc length PBQ = 36.25(1.52) = 55.1 cm (c) Luas tembereng PBQ/Area of the segment PBQ = 1 2 (36.25)2 (1.52) – 1 2 (36.25)2 sin 87.2° = 342.44 cm2 Luas ∆PQR/Area of ∆PQR = 1 2 (50)(100) sin 60° = 2 165.06 cm2 Luas kawasan berlorek/Area of the shaded region = 2 165.06 – 342.44 = 1 822.62 cm2 Lembaran Pentaksiran Bilik Darjah (PBD) A O r cm 25 cm 30° 10 cm θ Q R P B Lemb PBD Modul A+ Maths Tg5.indd 14 27/10/2021 11:13 AM
MG-21 Lembaran PBD BAB 4 Pilih Atur dan Gabungan Tahap Penguasaan Tafsiran 3 Mengaplikasikan kefahaman tentang pilih atur dan gabungan untuk melaksanakan tugasan mudah. 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pilih atur dan gabungan dalam konteks penyelesaian masalah rutin yang mudah. 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pilih atur dan gabungan dalam konteks penyelesaian masalah rutin yang kompleks. 6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pilih atur dan gabungan dalam konteks penyelesaian masalah bukan rutin secara kreatif. 1. (a) Diberi n C1 = 10, nyatakan nilai Given n C1 = 10, state the value of TP 3 (i) n (ii) n C9 (b) Jika n Cr = n Cs , ungkapkan n dalam sebutan r dan s. If n Cr = n Cs , express n in terms of r and s. TP 5 (a) (i) 10 (ii) Apabila/When n = 10, 10C9 = 10 (b) n Cr = n Cn – r = n Cs n – r = s n = s + r 2. Cari bilangan cara untuk membina kata laluan yang terdiri daripada 6 abjad jika menggunakan semua abjad dan Find the number of ways to form a password consisting of 6 alphabets chosen from all the alphabets if TP 3 (a) ulangan abjad tidak dibenarkan, repetition of alphabets is not allowed, (b) ulangan abjad dibenarkan, repetition of alphabets is allowed, (c) huruf-huruf vokal digunakan dan huruf A digunakan dua kali. the vowel letters are used and the letter A is used twice. (a) 26P6 atau/or 26 × 25 × 24 × 23 × 22 × 21 = 165 765 600 (b) 266 atau/or 26 × 26 × 26 × 26 × 26 × 26 = 308 915 776 (c) 6! 2! = 360 3. Seorang jurulatih ingin memilih 6 orang pemain yang terdiri daripada 3 orang lelaki dan 3 orang perempuan untuk membentuk satu pasukan badminton. 6 orang pemain itu dipilih daripada sekumpulan 4 orang lelaki dan 5 orang perempuan. A coach wants to choose 6 players consisting of 3 boys and 3 girls to form a badminton team. These 6 players are chosen from a group of 4 boys and 5 girls. TP 5 (a) Cari Find (i) bilangan cara pasukan itu dapat dibentuk, the number of ways the team can be formed, (ii) bilangan cara menyusun ahli pasukan itu dalam satu baris untuk satu sesi bergambar jika lelaki dan perempuan duduk berselang-seli. the number of ways the team members can be arranged in a row for a group photograph if the boys and the girls sit alternately. (b) Semasa taklimat, ahli pasukan duduk dalam bentuk bulatan. Cari bilangan pilih atur yang mungkin. During briefing, the team members sit in circular form. Find the possible number of permutations. (a) (i) 4 C3 × 5 C3 = 40 (ii) 3! × 3! × 2 = 72 (b) (6 – 1)! = 120 Lemb PBD Modul A+ Maths Tg5.indd 21 27/10/2021 11:13 AM
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