MODULE AND MORE Matematik Tambahan Tingkatan 4

Kertas 1 1. Diberi bahawa A(6,−1) dan B(−2,5) terletak pada suatu satah cartes. It is given that A(6, −1) and A(−2, 5) lie on a cartesian plane. (a) Nyatakan jarak AB. State the distance AB. (b) Garis lurus AB dipanjangkan ke titik C dengan keadaan jaraknya dari titik B adalah dua kali jarak AB. Cari koordinat C. The straight line AB is extended to point C such that its distance from point B is twice the distance AB. Find the coordinates of C. (a) AB = √(6 + 2)2 + (–1 – 5)2 = √64 + 36 = 10 unit (b) A(6, –1) B(–2, 5) C(h, k) 1 2 (–2, 5) = 1 2(6) + 1(h) 3 , 2(–1) + 1(k) 3 2 12 + h 3 = –2 –2 + k 3 = 5 12 + h = –6 –2 + k = 15 h = –18 k = 17 \ C = (–18, 17) SPM 2019 2. Rajah menunjukkan tiga garis lurus, dengan keadaan k, p, q dan r ialah pemalar. Diagram shows three straight lines, such that k, p, q and r are constants. 0 –r y ky = 4x + 8 – + – = 1 x p y q x Ungkapkan / Express (a) k dalam sebutan q. k in terms of q. (b) r dalam sebutan k dan p. r in terms of k and p. (a) Pintasan-y / y-intercept = 8 k q = 8 k k = 8q (b) Kecerunan / Gradient = 4 k –r – 0 0 – p = 4 k r p = 4 k r = 4p k SPM 2019 PRAKTIS PRAKTIS SPM SPM 7 Matematik Tambahan Tingkatan 4 Bab 7 Geometri Koordinat 140 BAB 7 BAB 7

Kertas 2 1. Rajah menunjukkan kedudukan kilang K dan kilang L yang dilukis pada satah Cartes. Diagram shows the location of factory K and factory L drawn on a Cartesian plane. x 0 L(2, –1) K(–4, 1) y T S ST ialah jalan raya lurus dengan keadaan jarak dari kilang K dan kilang L ke mana-mana titik pada jalan raya adalah sentiasa sama. ST is a straight road such that the distance from factory K and factory L to any point on the road is always equal. (a) Cari persamaan bagi ST. Find the equation of ST. (b) Satu lagi jalan raya lurus, MN dengan persamaan y = 8 – 2x akan dibina. Another straight road, MN with an equation y = 8 – 2x is to be built. (i) Lampu isyarat akan dipasang di persimpangan kedua-dua jalan raya itu. Cari koordinat bagi lampu isyarat itu. A traffic light is to be installed at the cross roads of the two roads. Find the coordinates of the traffic light. (ii) Antara dua jalan raya itu, yang manakah melalui kilang R1– 4 3 , –12? Which of the two roads passes through factory R1– 4 3 , –12? (a) Katakan P(x, y) ialah titik pada ST Let P(x, y) is a point at ST. PK = PL   √[x – (–4)]2 + (y – 1)2 = √(x – 2)2 + [y – (–1)]2 (x + 4)4 + (y – 1)2 = (x – 2)2 + (y + 1)2 x2 + 8x + 16 + y2 – 2y + 1 = x2 – 4x + 4 + y2 + 2y + 1 12x – 4y + 12 = 0 3x – y + 3 = 0 SPM 2017 3. O y (km) x (km) Rumah Ali Ali’s House Rumah Siti Siti’s House Rajah di atas menunjukkan kedudukan rumah Ali dan rumah Siti. Koordinat bagi rumah Ali dan rumah Siti masing-masing ialah (14, 12) dan (−10, −8). Ali dan Siti berbasikal dari rumah ke arah satu sama lain pada sebatang jalan raya yang lurus dengan halaju berbeza. Diberi halaju Ali ialah tiga kali halaju Siti. Cari jarak antara rumah Ali dengan tempat mereka bertemu. The diagram above shows the positions of Ali’s house and Siti’s house. The coordinates of Ali’s house and Siti’s house are (14, 12) and (−10, −8) respectively. Ali and Siti cycle from their house towards each other on a straight road with different velocity. Given that the velocity of Ali is three times velocity of Siti. Find the distance between Ali’s house and the place where they meet. Kedudukan mereka bertemu The position where they meet = 1 3(–10) + 1(14) 3 + 1 , 3(–8) + 1(12) 3 + 1 2 = 1 –16 4 , –12 4 2 = (−4, –3) Jarak / Distance = [14 – (–4)]2  + [12 – (–3)]2 =  182 + 152 =  549 = 23.43 unit SPM 2015 1 3 A(14, 12) (x, y) B(–10, –8) Matematik Tambahan Tingkatan 4 Bab 7 Geometri Koordinat 141 BAB 7 BAB 7

(b) (i) y = 3x + 3 …… 1 y = 8 – 2x …… 2 Gantikan/Replace 1 ke dalam/ into 2: 3x + 3 = 8 – 2x 5x = 5 x = 1 Apabila/When x = 1, y = 3(1) + 3 = 6 \ koordinat bagi lampu isyarat = (1, 6) coordinates of the traffic light (ii) Gantikan koordinat kilang R1– 4 3 , –12 ke dalam kedua-dua persamaan jalan raya, ST dan MN. Replace the coordinates of factory R1– 4 3 , –12 into both equations of the road, ST and MN. ⇒ ST: y = 3x + 3 –1 = 31– 4 3 2 + 3 –1 = –4 + 3 –1 = –1 ⇒ MN: y = 8 – 2x –1 = 8 – 21– 4 3 2 –1 = 8 + 8 3 –1 ≠ 32 3 \ hanya jalan raya ST melalui kilang R. only the road ST passes through factory R. 2. Rajah menunjukkan laluan titik bergerak P(x, y). P sentiasa bergerak dengan jarak tetap dari titik A. Diagram shows the path of a moving point P(x, y). P is always moving at a constant distance from point A. 0 y x A(2, –1) P(x, y) B(1, 2) B(1, 2) dan R(3, q) terletak pada laluan titik P. Garis lurus BC ialah tangen kepada laluan itu dan bersilang dengan paksi-x pada titik C. Cari B(1, 2) and R(3, q) lie on the path of point P. The straight line BC is a tangent to the path and intersects the x-axis at point C. Find SPM 2019 (a) persamaan bagi laluan titik P. the equation of the path of point P. (b) nilai-nilai yang mungkin bagi q. the possible values of q. (c) luas ∆ABC. the area of ∆ABC. (a) Jejari/ Radius, AB = √(1 – 2)2 + (2 + 1)2 = √1 + 9 = √10 PA = √10 √(x – 2)2 + (y – (–1)2 = √10 (x − 2)2 + (y + 1)2 − 10 = 0 x2 − 4x + 4 + y2 + 2y + 1 − 10 = 0 x2 + y2 − 4x + 2y − 5 = 0 (b) R(3, q): x = 3, y = q 9 + q2 – 12 + 2q – 5 = 0 q2 + 2q – 8 = 0 q = 2, q = –4 (c) mAB × mBC = –1 1 –1 – 2 2 – 1 2 × mBC = –1 –3 × mBC = –1 mBC = 1 3 Persamaan BC / Equation of BC y – 2 = 1 3 (x – 1) 3y – 6 = x – 1 3y = x + 5 Pada C / At C, y = 0. x + 5 = 0 x = –5 \ C(–5, 0) Luas ∆ABC / Area of ∆ABC = 1 2 u 2 –1 1 2 –5 0 2 –1u = 1 2 z(4 + 0 + 5) – (–1 – 10 + 0)z = 1 2 z9 – (–11)z = 1 2 z20z = 10 unit2 Praktis SPM Ekstra Matematik Tambahan Tingkatan 4 Bab 7 Geometri Koordinat 142 BAB 7 BAB 7

Dalam rajah, SRT adalah selari kepada paksi-y. QR adalah berserenjang dengan SRT. Titik P(x, y) bergerak dengan keadaan jaraknya dari Q(6, 5) adalah dua kali jaraknya dari garis lurus SRT. In the diagram, SRT is parallel to the y-axis. QR is perpendicular to SRT. The point P(x, y) moves in such a way that its distance from Q(6, 5) is twice its distance from the straight line SRT. 0 R S y x T –4 Q(6, 5) (a) Cari persamaan lokus bagi titik P. Find the equation of the locus of point P. (b) Tentukan sama ada lokus itu menyilang paksi-x atau tidak. Determine whether the locus intersect the x-axis. (a) Katakan P(x, y) PQ = 2 × jarak SRT dengan P (x – 6)2 + (y– 5)2 = 2[x – (–4)]2 (x – 6)2 + (y – 5)2 = 4(x + 4)2 x2 – 12x + 36 + y2 – 10y + 25 = 4(x2 + 8x + 16) x2 + y2 – 12x – 10y + 61 – 4x2 – 32x – 64 = 0 –3x2 + y2 – 44x – 10y – 3 = 0 3x2 – y2 + 44x + 10y + 3 = 0 (b) Pada paksi-x, y = 0 At x-axis, y = 0 3x2 – (0)2 + 44x + 10(0) + 3 = 0 3x2 + 44x + 3 = 0 a = 3, b = 44, c = 3 b2 – 4ac = 442 – 4(3)(3) = 1 900 > 0 b2 – 4ac > 0 Lokus itu menyilang paksi-x pada dua titik. The locus intersects the x-axis at two points. Sudut Sudut KBAT KBAT KBAT Ekstra Kuiz 7 Matematik Tambahan Tingkatan 4 Bab 7 Geometri Koordinat 143 BAB 7 BAB 7

144 BAB 8 Vektor Vectors 8.1 Vektor Vectors NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN a ~ –2a ~ 2a ~ A B A B

BAB 8 8BAB 1. Nyatakan tembereng garis yang diberi dalam bentuk vektor. TP 2 State the line segments given in the vector form. CONTOH s ~ m ~ n ~ p ~ q ~ r ~ T S U P Q R Penyelesaian: →PQ = q ~ →QR = r ~ →SR = s ~ →ST = –→TS = –m~ →TU = –→UT = –n ~ →UP = p ~ (a) C c ~ O d ~ D b ~ B A a ~ →OA = a ~ →OB = – →BO = –b ~ →OD = d ~ →OC = – →CO = – c ~ (b) L s ~ M r ~ p ~ N K q ~ →KL = q ~ →KN = – →NK = –p ~ →MN = r ~ →LM = – →ML = – s ~ 2. Binakan vektor yang berikut. TP 2 Construct the following vectors. CONTOH Diberi →PQ = p ~ , bina Given that → PQ = p ~ , construct (i) →AB = 2p ~ (ii) →CD = – 1 2 p ~ (iii) →EF = – 3 2 p ~ Penyelesaian: P p ~ Q A 2p ~ – –p ~ 1 2 B C D E F – –p ~ 3 2 Matematik Tambahan Tingkatan 4 Bab 8 Vektor 145

BAB 8 (a) Diberi →OP = a ~, bina Given that → OP = a ~ , construct P a ~ Q A –a ~ 1 3 – –a ~ –2a ~ 5 3 –a ~ 2 3 B C D E H G F (i) →AB = 1 3 a ~ (iii) →EF = – 5 3 a ~ (ii) →CD = –2a ~ (iv) →GH = 2 3 a ~ 3. Tentukan magnitud vektor-vektor berikut. TP 3 Determine the magnitude of the following vectors. CONTOH r 3 unit / units 4 unit / units Penyelesaian: |r ~ | = 32 + 42 = 5 unit/units 12 unit / units t ~ 9 unit / units | t ~ | = 92 + 122 = 15 unit 4. Nyatakan pasangan vektor-vektor yang sama. TP 2 State the pairs of vectors that are equal. CONTOH p ~ s ~ t ~ u ~ v ~ x ~ m ~ w ~ Penyelesaian: p ~ = m~ u ~ = x ~ s ~ = v ~ Tip Vektor yang sama mempunyai magnitud dan arah yang sama. Vectors that are equal have equal magnitude and same direction. (a) a ~ s ~ t ~ b ~ v ~ r ~ c ~ w ~ s ~ = w~ t ~ = v ~ a ~ = c ~ b ~ = r ~ Matematik Tambahan Tingkatan 4 Bab 8 Vektor 146

BAB 8 5. Ungkapkan vektor berikut dalam sebutan a ~ atau b ~. TP 3 Express the following vectors in terms of a ~ or b ~. CONTOH a ~ Q B A C D E F P Penyelesaian: →AB = 1 2 a ~ →CD = –2a ~ →EF = 1 1 2 a ~ (a) a ~ A B V U P S R D C Q 3b ~ (i) →AB = – 1 2 a ~ (iii) →QS = 7b ~ (v) →SR = –2b ~ (ii) →CD = 1 1 2 a ~ (iv) →RQ = –5b ~ 6. Tentukan vektor-vektor yang selari dan nyatakan hubungan mereka. TP 3 Determine the vectors that are parallel and state their relationships. CONTOH A B F E G H P S R D C Q Penyelesaian: (i) →AB dan →RS adalah selari. → AB and → RS are parallel. →AB = 1 2 →RS (ii) →GH dan →CD adalah selari. → GH and → CD are parallel. →GH = −2 →CD (iii) →EF dan →PQ adalah selari. → EF and → PQ are parallel. →EF = −1 1 2 →PQ Matematik Tambahan Tingkatan 4 Bab 8 Vektor 147

BAB 8 (a) M N S T K L P G H D C Q (i) →MN dan/ and →KL adalah selari/ are parallel. →MN = 1 3 →KL (ii) →CD dan/ and →GH adalah selari/ are parallel. →CD = −2 →GH (iii) →ST dan/and →PQ adalah selari/ are parallel. →ST = 3 →PQ 7. Tentukan sama ada titik P, Q dan R adalah segaris. TP 4 Determine whether points P, Q and R are collinear. CONTOH Diberi / Given →PQ = 6a ~ dan / and →QR = 2a ~. Penyelesaian: →PQ = 3(2a ~) →PQ = 3 →QR \ →PQ adalah selari dengan →QR, dengan Q ialah titik sepunya. Titik P, Q dan R adalah segaris. \ → PQ is parallel to → QR, with Q is common point. Points P, Q and R are collinear. (a) Diberi / Given →PQ = 10a ~ dan / and →PR = 4a ~. →PQ = 5 2 (4a ~) →PQ = 5 2 →PR \ →PQ adalah selari dengan →PR, dengan P ialah titik sepunya. \ → PQ is parallel to → PR , with P is common point. Titik P, Q dan R adalah segaris. Points P, Q and R are collinear. 8.2 Penambahan dan Penolakan Vektor Addition and Subtraction of Vectors NOTA IMBASAN NOTA IMBASAN 1. Hasil tambah dua vektor a ~ dan b ~, dikenali sebagai vektor paduan dan ditulis sebagai a ~ + b ~. The addition of two vectors a ~ and b ~, is known as resultant vector and is written as a ~ + b ~. 2. Apabila dua vektor selari ditambah, vektor paduan mempunyai magnitud tertentu dan selari dengan dua vektor tersebut. When two parallel vectors are added, the resultant vector has a certain magnitude and parallel with the two vectors. | a ~ + b ~ | = |a ~| + |b ~| 3. Hasil tambah vektor-vektor tidak selari / Addition of non-parallel vectors (a) Hukum Segi Tiga Triangle Law u + v ~ ~ A B C u ~ v ~ →AC = →AB + →BC (b) Hukum Segi Empat Selari Parallelogram Law A B D C u ~ v ~ u + v ~ ~ →AC = →AB + →BC = →AD + →DC (c) Hukum Poligon Poligon Law O A D B C →OD = →OA + →AB + →BC + →CD Matematik Tambahan Tingkatan 4 Bab 8 Vektor 148

BAB 8 NOTA IMBASAN NOTA IMBASAN 4. Penolakan bagi suatu vektor, a ~ – b ~, ialah penambahan vektor a ~ dengan vektor negatif b ~, iaitu Subtraction of vector, a ~ – b ~, is an addition of vector a ~ and a negative vector b ~, that is a ~ – b ~ = a ~ + (–b ~) (a) Penolakan vektor-vektor selari, a ~ dan b ~. Subtraction of parallel vectors, a ~ and b ~. a ~ b ~ a ~ –b ~ a – b ~ ~ (b) Penolakan vektor-vektor tidak selari, a ~ dan b ~. Subtraction of nonparallel vectors, a ~ and b ~. b ~ a ~ O B A –b ~ –b ~ a – b ~ ~ a ~ B O C A NOTA 8. Tentukan vektor paduan bagi vektor-vektor yang berikut. TP 3 Determine the resultant vector of the following vectors. CONTOH 1 3b ~ + 2a ~ – 3 2 b ~ + 4a ~ Penyelesaian: 13 – 3 2 2b ~ + (2 + 4)a ~ = 3 2 b ~ + 6a ~ Hanya vektor yang selari boleh ditambahkan. Only parallel vector can be added. CONTOH 2 5x ~ – 5 2 y ~ – 7x ~ + 8y ~ Penyelesaian: (5 – 7) x ~ + 1– 5 2 + 82 y ~ = –2x ~ + 11 2 y ~ (a) 6m~ – 3n ~ – 5m~ – 2n ~ = (6 – 5)m~ + (–3 – 2)n ~ = m~ – 5n ~ (b) 8p ~ – 3q ~ + p ~ – 2q ~ = (8 + 1)p ~ + (–3 – 2)q ~ = 9p ~ – 5q ~ (c) 2s ~ + 3 4 t ~ – 1 2 s ~ + 5 t ~ 12 – 1 2 2s ~ + 1 3 4 + 52 t ~ = 3 2 s ~ + 23 4 t ~ = 1 1 2 s ~ + 5 3 4 t ~ (d) 3x ~ – 4y ~ – 7x ~ – 5y ~ = (3 – 7)x ~ + (–4 – 5)y ~ = –4x ~ – 9y ~ Matematik Tambahan Tingkatan 4 Bab 8 Vektor 149

BAB 8 9. Tentukan vektor paduan bagi vektor-vektor tidak selari yang berikut. TP 3 Determine the resultant vectors of the following non-parallel vectors. CONTOH 1 P U T Q R S Penyelesaian: (i) →PQ + →QR = →PR (ii) →QS + →SU = →QU (iii) →PT + →TS = →PS (iv) →PS + →SQ = →PQ CONTOH 2 M K N J L Penyelesaian: (i) →JN – →MN = →JN – (– →NM) = →JM (ii) →JM – →KM = →JM – (– →MK) = →JK (iii) →KN – →LN = →KN – (– →NL) = →KL (a) PQRS ialah segi empat selari. PQR is a parallelogram P Q S R (i) →PQ + →RS = →PR (ii) →SP + →SR = →SQ (iii) →QP + →QR = →QS (iv) →SR + →RQ = →SQ (b) L K J N M y ~ x ~ z ~ (i) x ~ + y ~ = →JL (ii) y ~ + z ~ = →KM (iii) →JK + →KM = →JM (iv) →KN + →NJ = →KJ (v) →LN + →NK = →LK (c) LK J P N M (i) →JK – →MK = →JK – (– →KM) = →JM (ii) →KN – →PN = →KN – (– →NP) = →KP (iii) →JL – →NL – →JN = →JL + →LN + →NJ = 0 ~ (iv) →KM – →PM – →JP = →KM + →MP + →PJ = →KJ (d) T Q S P R a ~ b ~ c ~ (i) →QR – →SR = b ~ – c ~ / →QS (ii) →SP = →SR + →RQ + →QP = c ~ – b ~ – a ~ (iii) →PR = →PQ + →QR = a ~ + b ~ (iv) →QS = →QR + →RS = b ~ – c ~ Matematik Tambahan Tingkatan 4 Bab 8 Vektor 150

BAB 8 10. Selesaikan masalah yang melibatkan penambahan dan penolakan vektor. TP 4 Solve the problems involving addition and subtraction of vectors. CONTOH Rajah menunjukkan sebuah segi tiga PQR. Diberi bahawa →PS = 3x ~, →QR = 6y ~ , →QS = 6y ~ – 4x ~ dan →PT = →TQ. Cari dalam sebutan x ~ dan/ atau y ~ . The diagram shows a triangle PQR. Given that → PS = 3 x ~, → QR = 6y ~, → QS = 6y ~ – 4 x ~ and → PT = → TQ . Find in terms of x ~ and / or y ~. (i) →RS (ii) →PQ (iii) →TS Penyelesaian: (i) →RS = →RQ + →QS = −6y ~ + (6y ~ − 4x ~) = − 4x ~ (ii) →PQ = →PR + →RQ = [3x ~ – (–4x ~)] − 6y ~ = 7x ~ − 6y ~ (iii) →TS = →TP + →PS = – 1 2 (7x ~ − 6y ~ ) + 3x ~ = – 1 2 x ~ + 3y ~ T S R Q P (a) Rajah di bawah menunjukkan sebuah sisi empat OACD. The diagram below shows a quadrilateral OACD. B A 2a ~ b ~ O C D Diberi bahawa →OA = 2a ~, →OB = b ~, →CD = a ~ − 3b ~ dan 2 →AB = →BC. Cari vektor-vektor berikut dalam sebutan a ~ dan / atau b ~. Given that → OA = 2a ~, → OB = b ~, → CD = a ~ − 3b ~ and 2 → AB = → BC. Find the following vectors in terms of a ~ and / or b ~. (i) →AB (ii) →OC (iii) →DA (i) →AB = →AO + →OB = – →AO + →OB = −2a ~ + b ~ (ii) →OC = →OA + →AC = →OA + 3 →AB = 2a ~ + 3(−2a ~ + b ~) = −4a ~ + 3b ~ (iii) →DA = →DC + →CA = – →CD + (– →AC) = – →CD + (–3 →AB) = −(a ~ − 3b ~) − 3(−2a ~ + b ~) = 5a ~ (b) Rajah di bawah menunjukkan sebuah sisi empat PQRS. The diagram below shows a quadrilateral PQRS. x ~ y ~ S T U R P Q Diberi →PS = x ~, →PT = y ~ , →RU = 1 2 ( x ~ – 3y ~ ), →RU = 1 3 →RQ dan →PT = →TU. Cari vektor-vektor berikut dalam sebutan x ~ dan/atau y ~ . Given that → PS = x ~, → PT = y ~, → RU = 1 2 ( x ~ – 3y ~), → RU = 1 3 → RQ and → PT = → TU . Find the following vectors in terms of x ~ and / or y ~. (i) →ST (ii) →PQ (i) →ST = →SP + →PT = −x ~ + y ~ (ii) →PQ = →PU + →UQ = 2y ~ + 2 3 →RQ = 2y ~ + 2 3 (3 →RU) = 2y ~ + 21 1 2 ( x ~– 3y ~ ) 2 = x ~ − y ~ Matematik Tambahan Tingkatan 4 Bab 8 Vektor 151

BAB 8 8.3 Vektor dalam Satah Cartes Vectors in a Cartesian Plane NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN 1. Vektor unit ialah vektor dengan magnitud 1 unit. A unit vector is a vector with a magnitude of 1 unit. 2. Semua vektor dalam satah Cartes boleh diungkapkan dalam sebutan i ~ dan j ~ di mana i ~ ialah vektor unit dalam arah positif paksi-x, | i ~| = 1 unit dan j ~ ialah vektor unit dalam arah positif paksi-y, | j ~| = 1 unit. All vectors in a Cartesian plane can be expressed in terms of i ~ and j ~ where i ~ is a unit vector in the positive direction of the x-axis, | i ~ | = 1 unit and is a unit vector in the positive direction of the y-axis, | j ~| unit. 3. Vektor unit i ~ dinyatakan dalam bentuk vektor lajur, i ~ = 1 1 02 dan vektor unit j ~ dinyatakan dalam bentuk vektor lajur, j ~ = 1 0 12. Unit vector i ~ can be written as a column vector, i ~ = 1 1 0 2 and unit vector j ~ can be written as a column vector, j ~ = 1 0 1 2. 4. Vektor dalam satah cartes boleh diwakili dalam bentuk x i ~ + y j ~ atau 1 x y 2. Any vector on a Cartesian plane can be represented in the form x i ~ + y j ~ or 1 x y 2. y j ~ x i ~ P(x, y) y O x →OP = x i ~ + y j ~ = 1 x y 2 Magnitud bagi →OP / Magnitude of →OP | →OP| = x2 + y2 5. Vektor unit pada arah a ~ ditandakan sebagai ^ a ~ dengan / Unit vector in the direction of a ~ is denoted as ^ a ~ where ^ a ~ = a ~ |a ~| = (x i ~ + y j ~) x2 + y2 6. Penambahan dan penolakan dua atau lebih vektor: Addition of two or more vectors: Jika/if a ~ = x1 i ~ + y1 j ~ = 1 x1 y1 2 dan/ and b ~ = x2 i ~ + y2 j ~ = 1 x2 y2 2 maka,/then, a ~ + b ~ = 1 x1 y1 2 ± 1 x2 y2 2 = 1 x1 ± x2 y1 ± y2 2 7. Pendaraban vektor dengan skala Multiplication of vectors with a scalar Jika /if a ~ = x1 i ~ + y1 j ~ = 1 x1 y1 2 Maka / Then ka ~ = k 1 x y 2 = 1 kx ky2 k ialah pemalar. k is a constant. 11. Ungkapkan vektor-vektor berikut dalam bentuk 1 x y 2 dan x i ~ + y j ~ . TP 3 Express the following vectors in the form 1 x y2 and x i ~ + y j ~. CONTOH A 2 2 4 4 6 y O x Dari koordinat (x, y), tukar ke bentuk vektor lajur 1 x y2. From the coordinates (x, y), change to column vector 1 x y2. Penyelesaian: A(6, 3) →OA = 1 6 3 2 →OA = 6 i ~ + 3 j ~ Matematik Tambahan Tingkatan 4 Bab 8 Vektor 152

BAB 8 B E D C 2 4 2 4 y O x 1 3 5 –2 –1 –3 –4 –7 –6 –5 –4 –3 –2 –1 1 3 5 B(5, 5) →OB = 1 5 5 2 →OB = 5i + 5j D(−6, −2) →OD = 1 –6 –22 →OD = −6i − 2j C(−3, 4) →OC = 1 –3 4 2 →OC = −3i + 4j E(3, −3) →OE = 1 3 –32 →OE = 3i − 3j 12. Ungkapkan vektor-vektor berikut dalam bentuk: Express the following vectors in the form: (i) x i ~ + y j ~ (ii) 1 x y 2 Seterusnya, cari magnitud. TP 4 Hence, find the magnitude. CONTOH B A D C 2 4 2 4 y O x –2 –6 –4 –2 Tip Guna titik tamat menolak titik mula. Use ending point subtracting starting point. Penyelesaian: (i) A(−4, 1), B(−2, 4) →AB = (−2 − (−4)) i ~ + (4 − 1) j ~ = 2 i ~ + 3 j ~ C(−1, 3), D(4, −2) →CD = (4 − (−1)) i ~ + (−2 − 3) j ~ = 5 i ~ − 5 j ~ (ii) →AB = 1 2 3 2 , →CD = 1 5 –52  →AB = 22 + 32 = 13 ,  →CD = 52 + (–5)2 = 50 (a) a ~ c ~ b ~ 2 4 2 4 y O x –2 –4 –6 –4 –2 (i) a ~ = 3 i ~ + 2 j ~ , b ~ = −3 i ~ − 6 j ~ , c ~ = 4 i ~ − 3 j ~ (ii) a ~ = 1 3 2 2 , b ~ = 1 –3 –62 , c ~ = 1 4 –32 a ~ = 32 + 22 = 13 b ~ = (–3)2 + (–6)2 = 45  c ~ = 42 + (–3)2 = 5 Matematik Tambahan Tingkatan 4 Bab 8 Vektor 153

BAB 8 13. Berdasarkan vektor-vektor a ~ dan b ~, yang diberikan, cari vektor yang berikut. TP 4 Based on the vectors a ~ and b ~ given, find the following vectors. CONTOH a ~ = 2 i ~ – 3 j ~, b ~ = 5 i ~ + 2 j ~ (i) a ~ + b ~ (ii) 3a ~ – 2b ~ (iii)2a ~ – 5b ~ Penyelesaian: (i) a ~ + b ~ = 1 2 –32 + 1 5 2 2 = 1 7 –12 = 7 i ~ – j ~ (ii) 3a ~ – 2b ~ = 31 2 –32 – 21 5 22 = 1 6 –92 – 1 10 4 2 = 1 –4 –132 = –4 i ~ – 13 j ~ (iii)2a ~ – 5b ~ = 21 2 –32 – 51 5 22 = 1 4 –62 – 1 25 102 = 1 –21 –162 = –21 i ~ – 16 j ~ (a) a ~ = 3 i ~ – 4 j ~, b ~ = –2 i ~ + 5 j ~ (i) a ~ + 3b ~ (ii) 2a ~ – 5b ~ (iii)4a ~ + 2b ~ (i) a ~ + 3b ~ = 1 3 –42 + 31 –2 5 2 = 1 3 –42 + 1 –6 152 = 1 –3 112 = –3 i ~ + 11 j ~ (ii) 2a ~ – 5b ~ = 21 3 –42 – 51 –2 5 2 = 1 6 –82 – 1 –10 25 2 = 1 16 –332 = 16 i ~ – 33 j ~ (iii)4a ~ + 2b ~ = 41 3 –42 + 21 –2 5 2 = 1 12 –162 + 1 –4 102 = 1 8 –62 = 8 i ~ – 6 j ~ 14. Tentukan vektor unit pada arah vektor yang diberi. TP 4 Determine the unit vector in the direction of given vector. CONTOH 1 →AB = –5 i ~ + 12 j ~ Penyelesaian:  →AB = (–5)2 + 122 = 13 Vektor unit pada arah →AB Unit vector in the direction of → AB = 1 13 (–5 i ~ + 12 j ~ ) = – 5 13 i ~ + 12 13 j ~ Cari magnitud vektor dan gantikan ke rumus. Find the magnitude of the vector and substitute to the formula. Vektor unit / Unit vector = a ~ |a ~| = (x i ~ + y j ~) x2 + y2 CONTOH 2 →PQ = 1 5 –82 Penyelesaian:  →PQ = 52 + (–8)2 = 89 Vektor unit pada arah →PQ Unit vector in the direction of → PQ = 1 89 1 5 –82 = 5 89 – 8 89 Matematik Tambahan Tingkatan 4 Bab 8 Vektor 154

BAB 8 (a) →CD = 6 i ~ – 8 j ~  →CD = 62 + (–8)2 = 10 Vektor unit / Unit vector = 1 10 (6 i ~ – 8 j ~ ) = 3 5 i ~ – 4 5 j ~ (b) →EF = 5 i ~ + 6 j ~  →EF = 52 + 62 = 61 Vektor unit pada arah →EF Unit vector in the direction of → EF = 1 61 (5 i ~ + 6 j ~ ) = 5 61 i ~ + 6 61 j ~ (c) →GH = 1 8 –152  →GH = 82 + (–15)2 = 17 Vektor unit pada arah →GH Unit vector in the direction of → GH = 1 17 1 8 –152 = 8 17 – 15 17 = 8 17 i ~ – 15 17 j ~ 15. Selesaikan masalah-masalah berikut melibatkan vektor. TP 5 Solve the following problems involving vectors. CONTOH 1 Diberi bahawa →AB = 2 i ~ − 5 j ~ dan →CD = i ~ + (4 − m) j ~ . Cari nilai m jika →AB adalah selari dengan →CD. Given that → AB = 2 i ~ − 5 j ~ and → CD = i ~ + (4 − m) j ~. Find the value of m if → AB is parallel to → CD. Penyelesaian: →AB = λ →CD 2 i ~ − 5 j ~ = λ( i ~ + (4 − m) j ~ ) = λ i ~ + (4 − m)λ j ~ 2 = λ , −5 = (4 − m)λ = (4 − m)(2) −13 = −2m m = 6 1 2 1. Jika selari, maka If parallel, then →AB = λ →CD 2. Bandingkan nilainilai i ~ dan j ~ . Compare the values of i ~ and j ~ . CONTOH 2 Rajah di sebelah menunjukkan vektor-vektor →OS dan →OT. The diagram shows vectors → OS and → OT . Ungkapkan / Express (i) →ST dalam bentuk 1 x y 2, → ST in the form 1 x y2, (ii) vektor unit pada arah →ST . unit vector in the direction of → ST . Penyelesaian: (i) →ST = →SO + →OT = − →OS + →OT = − 1 –4 7 2 + 1 6 5 2 = 1 10 –22 (ii)  →ST  = 102 + (–2)2 = 104 Vektor unit pada arah →ST / Unit vector → ST . = 1 104 1 10 –22 = 10 104 i ~ – 2 104 j ~ y S (–4, 7) T (6, 5) x O Matematik Tambahan Tingkatan 4 Bab 8 Vektor 155

BAB 8 CONTOH 3 Dalam rajah, diberi In the diagram, given that →PQ = 2x ~, →PS = 5y ~ dan/and →RS = 3x ~. (a) Ungkapkan dalam sebutan x ~ dan / atau y ~ , →PR dan →SQ. Express in terms of x ~ and / or y ~, → PR and → SQ. (b) Diberi →PT = h →PR dan →QT = k →QS, nyatakan →PT Given → PT = h → PR and → QT = k → QS, state → PT (i) dalam sebutan h, x ~ dan y ~ , in terms of h, x ~ and y ~, (ii) dalam sebutan k, x ~ dan y ~ . in terms of k, x ~ and y ~. Penyelesaian: (a) →PR = →PS + →SR = →PS + (– →RS ) = 5y ~ − 3x ~ →SQ = →SP + →PQ = (– →PS ) + →PQ = −5y ~ + 2x ~ (b) (i) →PT = h →PR = h(5y ~ − 3x ~) = 5hy ~ − 3hx ~ (ii) →PT = →PQ + →QT = 2x ~ + k →QS = 2x ~ − k(−5y ~ + 2x ~) = (2 − 2k) x ~ + 5k y ~ P Q R S T (a) Diberi bahawa →GH = i ~ + 3 j ~ dan →KL = (2p + 1) i ~ − 6 j ~ . Cari nilai p jika →GH adalah selari dengan →KL. Given that → GH = i ~ + 3 j ~ and → KL = (2p + 1) i ~ − 6 j ~. Find the value of p if → GH is parallel to → KL . →GH = λ →KL i ~ + 3 j ~ = λ((2p + 1) i ~ − 6 j ~ ) = (2p + 1)λ i ~ − 6λ j ~ 3 = − 6λ , 1 = (2p + 1)λ λ = − 1 2 1 = (2p + 1)1− 1 2 2 −2 = 2p + 1 p = − 3 2 (b) Diberi bahawa →PQ = 12 i ~ − 20 j ~ dan →RS = 3 i ~ + h j ~ . Cari nilai h jika →PQ adalah selari dengan →RS . Given that → PQ = 12 i ~ − 20 j ~ and → RS = 3 i ~ + h j ~. Find the value of h if → PQ is parallel to → RS . →PQ = λ →RS 12i ~ − 20 j ~ = λ(3i ~ + h j ~ ) = 3λi ~ + hλ j ~ 12 = 3λ , −20 = hλ λ = 4 −20 = 4h h = −5 Matematik Tambahan Tingkatan 4 Bab 8 Vektor 156

BAB 8 (c) Rajah menunjukkan vektor-vektor →OP dan →QO. The diagram shows vectors → OP and → QO. Ungkapkan / Express (i) →QO dalam bentuk 1 x y 2, → QO in the form 1 x y2, (ii) vektor unit pada arah →QP. unit vector in the direction of → QP. (i) →QO = − →OQ = − 1 4 2 2 = 1 –4 –22 (ii) →QP = →QO + →OP = 1 –4 –22 + 1 –8 6 2 = 1 –12 4 2  →QP = √(–12)2 + 42 = √160 Vektor unit pada arah →QP Unit vector in the direction of → QP. = 1 √160 1 –12 4 2 = 1 – 12 √160 4 √160 2 (d) Rajah menunjukkan vektor-vektor →OR dan →OS. The diagram shows vectors → OR and → OS. Ungkapkan / Express (i) →RO dalam bentuk 1 x y 2, → RO in the form 1 x y 2, (ii) vektor unit pada arah →RS . unit vector in the direction of → RS . (i) →RO = − →OR = − 1 –3 5 2 = 1 3 –52 (ii) →RS = →RO + →OS = 1 3 –52 + 1 –8 –72 = 1 –5 –122  →RS  = (–5)2 + (–12)2 = 13 Vektor unit pada arah →RS Unit vector in the direction of → RS . = 1 13 1 –5 –122 = 1 – 5 13 – 12 13 2 (e) Diberi bahawa a ~ = 2 i ~ + 7 j ~ dan b ~ = i ~ − 3 j ~ . Given that a ~ = 2 i ~ + 7 j ~ and b ~ = i ~ − 3 j ~. Cari / Find (i) a ~ + 5b ~, (ii) vektor unit pada arah a ~ + 5b ~. unit vector in the direction of a ~ + 5b ~. (i) a ~ + 5b ~ = 2 i ~ + 7 j ~ + 5( i ~ − 3 j ~ ) = 7 i ~ − 8 j ~ (ii) 7 i ~ − 8 j ~  = 72 + (–8)2 = 113 Vektor unit pada arah 7 i ~ − 8 j ~ unit vector in the direction of 7 i ~ – 8 j ~. = 1 113 (7 i ~ − 8 j ~ ) (f) Diberi bahawa O(0, 0), C(8, −3), D(5, 4), cari dalam sebutan i ~ dan j ~ bagi Given that O(0, 0), C(8, −3), D(5, 4), find in terms of i ~ and j ~ for (i) →CD, (ii) vektor unit pada arah →CD. unit vector in the direction of → CD. (i) →CD = →CO + →OD = − →OC + →OD = −(8 i ~ − 3 j ~ ) + (5 i ~ + 4 j ~ ) = −3 i ~ + 7 j ~ (ii)  →CD = (–3)2 + 72 = 58 Vektor unit pada arah →CD Unit vector in the direction of → CD. = 1 58 (–3 i ~ + 7 j ~ ) y P (–8, 6) Q (4, 2) x O y S (–8, –7) R (–3, 5) x O Matematik Tambahan Tingkatan 4 Bab 8 Vektor 157

BAB 8 (g) Rajah menunjukkan sebuah segi tiga ABC. Garis lurus AD bersilang dengan garis lurus BE pada titik F. Diberi bahawa The diagram shows a triangle ABC. The straight line AD intersects the straight line BE at point F. Given that →AE = 1 2 →AC, 2 →BD = →DC, →AE = 2x ~ dan / and →AB = 6y ~ (a) Ungkapkan dalam sebutan x ~ dan y ~ bagi: Express in terms of x ~ and y ~: (i) →BE, (ii) →AD. (b) Diberi →BF = h →BE dan →FD = k →AD, Given → BF = h → BE and → FD = k → AD, (i) ungkapkan →BF dalam sebutan h, x ~ dan y ~ , express → BF in terms of h, x ~ and y ~, (ii) ungkapkan →FD dalam sebutan k, x ~ dan y ~ . express → FD in terms of k, x ~ and y ~. (a) (i) →BE = →BA + →AE = −6y ~ + 2x ~ (ii) →AD = →AB + →BD = →AB + 1 3 →BC = 6y ~ + 1 3 ( →BA + →AC) = 6y ~ + 1 3 (−6y ~ + 4x ~) = 4y ~ + 4 3 x ~ (b) (i) →BF = h →BE = h(−6y ~ + 2x ~) = −6hy ~ + 2hx ~ (ii) →FD = k →AD = k14y ~ + 4 3 x ~2 = 4ky ~ + 4 3 k x ~ A D B E C F Matematik Tambahan Tingkatan 4 Bab 8 Vektor 158

BAB 8 Kertas 1 1. Rajah menunjukkan vektor-vektor →OA, →OB dan →OD yang dilukis pada grid segi empat sama bersisi 1 unit. Diagram shows vectors → OA, → OB and → OD drawn on a square grid with sides 1 unit. D B O A (a) Cari/ Find |– →BO| (b) Diberi →OA = a ~ dan →OB = b ~, ungkapkan dalam sebutan a ~ dan b ~ Given → OA = a ~ and → OB = b ~, express in terms of a ~ and b ~ (i) →AB (ii) →OD (a) |– →BO| = | →OB| = 32 + 42 = 5 unit (b) (i) →AB = – →OA + →OB = –a ~ + b ~ = b ~ – a ~ (ii) →OD = →OB + →BD = →OB + 2 →BA = →OB + 2(– →AB) = b ~ + 2(a ~ – b ~) = 2a ~ – b ~ SPM 2017 2. Rajah di bawah menunjukkan tiga vektor, →OP , →OQ dan →OR, dilukis pada grid segi empat sama yang sama besar bersisi 1 unit. The diagram below shows three vectors, → OP , → OQ and → OR, drawn on a grid of equal squares with sides of 1 unit. P R Q O y ~ x ~ Tentukan / Determine (a)  →OP, (b) →OP dalam sebutan x ~ dan y ~ . → OP in terms of x ~ and y ~ . (a) →OP = 1 –2 4 2  →OP = (–2)2 + 42 = 4.472 unit (b) x ~ = 1 1 2 2 , y ~ = 1 3 3 2 Biar →OP = hx ~ + ky ~ , 1 –2 4 2 = h1 1 2 2 + k1 3 3 2 Maka, −2 = h + 3k .............a 4 = 2h + 3k ..........b a − b ; −2 − 4 = −h h = 6 Gantikan dalam a, −2 = 6 + 3k k = – 8 3 \ →OP = 6x ~ – 8 3 y ~ PRAKTIS PRAKTIS SPM SPM 8 Matematik Tambahan Tingkatan 4 Bab 8 Vektor 159

BAB 8 3. Rajah menunjukkan sebuah segi empat selari ABCD. Diagram shows a parallelogram ABCD. A B D E C T Titik E terletak pada DC dengan keadaan DE : EC = 1 : 2. Diberi bahawa →DE = 6u ~, →AD = 5v ~ dan T ialah titik tengah AC. Ungkapkan dalam sebutan u ~ dan / atau v ~, Point E lies on DC such that DE : EC = 1 : 2. It is given that → DE = 6u~, → AD = 5v ~ and T is the midpoint of AC. Express in terms of u~ and / or v ~, (a) →EC (b) →ET (a) →DE →EC = 1 2 →EC = 2 →DE = 2(6u ~) = 12u ~ (b) →ET = →EC + →CT = 12u ~ + 1 2 →CA = 12u ~ + 1 2 ( →CD + →DA) = 12u ~ + 1 2 (18u ~ + (–5v ~)) = 12u ~ + 9u ~ – 5 2 v ~ = 21u ~ – 5 2 v ~ 4. Diberi bahawa P(m, 4), Q(10, h), v ~ = 3 i ~ − 2 j ~, w~ = 8 i ~ + 4 j ~ dan →PQ = 3v ~ + kw~, dengan keadaan m, h dan k ialah pemalar. Ungkapkan h dalam sebutan m. It is given that P(m, 4), Q(10, h), v ~ = 3 i ~ − 2 j ~, w~ = 8 i ~ + 4 j ~ and → PQ = 3v ~ + kw~ , where m, h and k are constants. Express h in terms of m. SPM 2019 SPM 2019 →PQ = 3v ~ + kw~ →OQ – →OP = 31 3 –22 + k1 8 4 2 1 10 h 2 – 1 m 4 2 = 1 9 + 8k –6 + 4k 2 1 10 – m h – 4 2 = 1 9 + 8k –6 + 4k 2 10 – m = 9 + 8k h – 4 = –6 + 4k 8k = 1 – m h = –2 + 4k k = 1 – m 8 h = –2 + 41 1 – m 8 2 h = –2 + 1 2 – m 2 h = –m – 3 2 5. Rajah menunjukkan vektor-vektor →OC, →OD dan →ON dilukis pada grid segi empat sama. Diagram shows the vectors → OC, → OD and → ON drawn on a square grid. D N O M C y ~ x ~ (a) Ungkapkan →ON dalam bentuk hx ~ + ky ~ , dengan keadaan h dan k ialah pemalar. Express → ON in the form of hx ~ + ky ~ , where h and k are constants. (b) Pada rajah, tanda dan label titik M dengan keadaan →NM + 2 →OC = →OD. On the diagram, mark and label the point M such that → NM + 2 → OC = → OD. (a) →ON = 2→OB + (–→OC) = 2y ~ – x ~ (b) →NM + 2→OC = →OD →NM = →OD – 2 →OC = y ~ – 2x ~ SPM 2018 Matematik Tambahan Tingkatan 4 Bab 8 Vektor 160

BAB 8 Kertas 2 1. Rajah menunjukkan segi tiga ABC. Garis lurus AE bersilang dengan garis lurus BC di titik D. Titik F terletak pada garis lurus AE. Diagram shows a triangle ABC. The straight line AE intersects with the straight line BC at point D. Point F lies on the straight line AE. E D F A C B Diberi bahawa →CD = 1 2 →CB, →AC = 6a ~ dan →AB = 4b ~. It is given that → CD = 1 2 → CB, → AC = 6a ~ and → AB = 4b ~. (a) Ungkapkan dalam sebutan a ~ dan/ atau b ~: Express in terms of a ~ and/ or b ~: (i) →BC, (ii) →AD. (b) Diberi bahawa →AF = m →AD dan →BF = n(a ~ – 4b ~), dengan keadaan m dan n ialah pemalar. Cari nilai m dan n. It is given → AF = m → AD and → BF = n(a ~ – 4b ~), where m and n are constants. Find the value of m and of n. (c) Diberi →AF = ka ~ + 4 7 b ~, dengan keadaan k ialah pemalar, cari nilai k. Given → AF = ka ~ + 4 7 b ~, such that k is a constants, find the value of k. (a) (i) →BC = →BA + →AC = – →AB + →AC = −4b ~ + 6a ~ (ii) →AD = →AC + →CD = 6a ~ + 1 2 →CB = 6a ~ – 1 2 →BC = 6a ~ – 1 2 (–4b ~ + 6a ~) = 3a ~ + 2b ~ SPM 2017 (b) →BF = n(a ~ – 4b ~) →BA + →AF = n(a ~ – 4b ~) – →AB + m →AD = n(a ~ – 4b ~) –4b ~ + m(3a ~ + 2b ~) = n(a ~ – 4b ~) –4b ~ + 3ma ~ + 2mb ~ = na ~ – 4nb ~ 3ma ~ + (2m – 4)b ~ = na ~ – 4nb ~ Bandingkan / Compare: 3m = n, 2m – 4 = –4n Gantikan n = 3m → 2m – 4 = –4n Replace 2m – 4 = –4(3m) 14m = 4 m = 4 14 = 2 7 n = 31 2 7 2 = 6 7 (c) →AF = ka ~ + 4 7 b ~ = m →AD ka ~ + 4 7 b ~ = 2 7 (3a ~ + 2b ~) ka ~ + 4 7 b ~ = 6 7 a ~ + 4 7 b ~ Bandingkan / Compare: k = 6 7 2. Rajah menunjukkan segi tiga OBQ dan OPA dengan keadaan titik A berada pada OQ dan titik B berada pada OP. Garis lurus BQ dan garis lurus AP bersilang pada titik C. Diagram shows triangles OBQ and OPA where point A lies on OQ and point B lies on OP. The straight lines BQ and AP intersect at point C. Q A C B P O Diberi bahawa →OQ = 12x ~, →OP = 8y ~, OA : AQ = 2 : 1, OB : BP = 3 : 1, →AC = h →AP dan →BC = k →BQ, dengan keadaan h dan k ialah pemalar. It is given that → OQ = 12x ~, → OP = 8y ~, OA : AQ = 2 : 1, OB : BP = 3 : 1, → AC = h → AP and → BC = k → BQ, where h and k are constants. SPM 2018 Matematik Tambahan Tingkatan 4 Bab 8 Vektor 161

BAB 8 (a) Ungkapkan →OC dalam sebutan Express → OC in terms of (i) h, x ~ dan/ and y ~, (ii) k, x ~ dan/ and y ~. (b) Seterusnya, cari nilai h dan nilai k. Hence, find the value of h and of k. (c) Diberi u x ~u = 1 unit, u y ~u = 2 unit dan OQ berserenjang kepada OP, hitung u →ACu. Given ux ~u = 1 unit, uy ~u = 2 units and OQ is perpendicular to OP, calculate u → ACu. (a) (i) →OC = →OA + →AC = 2 3 →OQ + h →AP = 2 3 (12x ~) + h( →OP – →OA) = 8x ~ + h(8y ~ – 8x ~) = 8x ~ + 8hy ~ – 8hx ~ = (8 – 8h) x ~ + 8hy ~ (ii) →OC = →OB + →BC = 3 4 →OP + k →BQ = 3 4 (8y ~) + k( →OQ – →OB) = 6y ~ + k(12x ~ – 6y ~) = 12k x ~ + (6 – 6k) y ~ (b) Bandingkan / Compare: 8 – 8h = 12k ; 8h = 6 – 6k 2 – 2h = 3k … 1 3 – 4h = 3k … 2 2 – 1: 1 – 2h = 0 h = 1 2 Gantikan/ Replace h = 1 2 ke dalam /into 1; 2 – 21 1 2 2 = 3k 1 = 3k k = 1 3 (c) u →OQu = u12x ~u = u12(1)u = 12 u →OPu = u8y ~u = u8(2)u = 16 u →OAu = u 2 3 →OQu = u 2 3 (12)u = 8 u →APu = √82 + 162 = √320 = 8√5 \ u →ACu = 1 2 u →APu = 1 2 (8√5) = 4√5 3. Rajah di bawah menunjukkan sebuah trapezium ABCD dan titik E terletak pada AC. The diagram below shows a trapezium ABCD and point E lies on AC. A E D B C Diberi bahawa →AB = 8a ~, →AD = 12b ~ dan →AD = 2 →BC. It is given that → AB = 8a ~, → AD = 12b ~ and → AD = 2 → BC. (a) Ungkapkan dalam sebutan a ~ dan b ~, Express in terms of a ~ and b ~, (i) →AC, (ii) →BD. (b) Diberi bahawa →AE = k →AC, dengan keadaan k ialah pemalar. Cari nilai k jika titik-titik B, E dan D adalah segaris. It is given that → AE = k → AC, where k is a constant. Find the value of k if the points B, E and D are collinear. (a) (i) →AC = →AB + →BC = →AB + 1 2 →AD = 8a ~ + 6b ~ (ii) →BD = →BA + →AD = – →AB + →AD = –8a ~ + 12b ~ (b) →AE = k →AC →AE = k(8a ~ + 6b ~) = 8ka ~ + 6kb ~ Katakan / Let →BE = l →BD, Maka / Hence →BE = l(–8a ~ + 12b ~) = –8la ~ + 12lb ~ →AE = →AB + →BE 8ka ~ + 6kb ~ = 8a ~ + (–8la ~ + 12lb ~) 8ka ~ + 6kb ~ = (8 – 8l)a ~ + 12lb ~ SPM 2014 Matematik Tambahan Tingkatan 4 Bab 8 Vektor 162

BAB 8 8k = 8 – 8l dan / and 6k = 12l l = 1 2 k Gantikan l = 1 2 k ke dalam 8k = 8 – 8l , Replace into 8k = 8 – 81 1 2 k2 8k = 8 – 4k 12k = 8 k = 2 3 4. Rajah di bawah menunjukkan sebuah segi tiga PQR. The diagram below shows a triangle PQR. P B Q R C A Diberi PB : BQ = 3 : 2, PA : AR = 1 : 4, →PQ = a ~ dan →PA = 2b ~. It is given PB : BQ = 3 : 2, PA : AR = 1 : 4, → PQ = a ~ and → PA = 2b ~. (a) Ungkapkan dalam sebutan a ~ dan/atau b ~: Express in terms of a ~ and/or b ~: (i) →QA, (ii) →QB. (b) Diberi a ~ = 3 i ~ dan b ~ = i ~ – 3 j ~, cari  →QA. Given that a ~ = 3 i ~ and b ~ = i ~ – 3 j ~, find  → QA. (c) Diberi →QC = m →QA dan →CB = n →RB, dengan keadaan m dan n ialah pemalar, cari nilai m dan nilai n. Given that → QC = m → QA and → CB = n → RB, where m and n are constants, find the value of m and of n. SPM 2015 (a) (i) →QA = →QP + →PA = –a ~ + 2b ~ (ii) →QB = 2 5 →QP = 2 5 (–a) = – 2 5 a ~ (b) →QA = –a ~ + 2b ~ = –3 i ~ + 2(i ~ – 3 j ~) = – i ~ – 6 j ~  →QA = (–1)2 + (–6)2 = 37 = 6.08 unit (c) →QC = m →QA = m(–a ~ + 2b ~) = –ma ~ + 2mb ~ →CB = n( →RP + →PB) = n(–10b ~ + 3 5 a ~) = 3 5 na ~ – 10nb ~ →QC = →QB + →BC = – 2 5 a ~ – 3 5 na ~ + 10nb ~ = 1 5 (–2 – 3n)a ~ + 10nb ~ Oleh itu, 1 5 (–2 – 3n) = –m....................a 10n = 2m n = 1 5 m..................b Gantikan b ke dalam a, Replace b into a, 1 5 1–2 – 3 5 m2 = –m 22 25m = 2 5 m = 5 11 n = 1 11 Praktis SPM Ekstra Matematik Tambahan Tingkatan 4 Bab 8 Vektor 163

BAB 8 1. Vektor-vektor 1 a b 2 dan 1 1 –32 adalah selari. Diberi bahawa 1 a b 2 mempunyai magnitud √250 dan a . 0, cari nilai a dan nilai b. The vectors 1 a b2 and 1 1 –3 2 are parallel. Given that 1 a b2 has a magnitude of √250 and a . 0, find the value of a and of b. 1 a b 2 = k 1 1 –32 k = a, –3k = b b = –3a Magnitud = √a2 + b2 Magnitude √250 = √a2 + (–3a)2 √10a2 = √250 10a2 = 250 a2 = 25 a = ±5 Sudut Sudut KBAT KBAT Kuiz 8 KBAT Ekstra Apabila /When a = 5, b = –3(5) = –15 Apabila /When a = –5, b = –3(–5) = 15 Memandangkan a . 0, It is seen that a . 0, Maka/ Thus, a = 5 dan b = –15 Matematik Tambahan Tingkatan 4 Bab 8 Vektor 164

165 BAB 9 Penyelesaian Segi Tiga Solution of Triangles 9.1 Petua Sinus Sine Rule NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN Petua Sinus / Sine rule A C a B b A B1 C a b A B2 C a b a c A A a c A c a a fi c A a c A c a c A C a B b

Matematik Tambahan Tingkatan 4 Bab 9 Penyelesaian Segi Tiga 166 BAB 9 1. Hitung panjang x bagi setiap segi tiga yang berikut. TP 4 Calculate the length of x in each of the following triangles. CONTOH 9.6 cm x cm 118° 23° Penyelesaian: x sin 118° = 9.6 sin 23° x = 9.6 × sin 118 sin 23° = 21.69 Sudut Kalkulator Tekan 9 . 6 × Press sin 1 1 8 ÷ sin 2 3 = (a) x cm 8.7 cm 53° 82° x sin 82° = 8.7 sin 53° x = 8.7 × sin 82° sin 53° = 10.79 (b) x cm 15.9 cm 123° 39° Q = 180° − 123° − 39° = 18° x sin 18° = 15.9 sin 123° x = 15.9 × sin 18° sin 123° = 5.859 (c) x cm 16.7 cm 42° 109° Q = 180° − 109° − 42° = 29° x sin 29° = 16.7 sin 109° x = 16.7 × sin 29° sin 109° = 8.563 2. Hitung nilai y bagi setiap segi tiga yang berikut. TP 4 Calculate the value of y in each of the following triangles. CONTOH x y 65° 9.1 cm 10.3 cm Penyelesaian: sin x 9.1 = sin 65° 10.3 sin x = 9.1 × sin 65° 10.3 = 53°12’ y = 180° − 65° − 53°12’ y = 61°48’ (a) x y 72° 8.5 cm 12.1 cm sin y 8.5 = sin 72° 12.1 sin y = 8.5 × sin 72° 12.1 y = 41°55’ x = 180° – 72° – 41°55’ x = 66°59

Matematik Tambahan Tingkatan 4 Bab 9 Penyelesaian Segi Tiga 167 BAB 9 (b) 15.3 cm 110° 10.8 cm x y sin x 10.8 = sin 110° 15.3 sin x = 10.8 × sin 110° 15.3 x = 41°33’ y = 180° – 110° – 41°33’ y = 28°27’ (c) x y 11.5 cm 16.3 cm 41° sin x 16.3 = sin 41° 11.5 sin x = 16.3 × sin 41° 11.5 x = 180° – 68°25’ x = 111°35’ y = 180° – 41° – 111°359 y = 27°259 3. Selesaikan setiap yang berikut. TP 5 Solve each of the following. CONTOH 1 Cari dua nilai yang mungkin bagi ∠C dalam segi tiga ABC. Seterusnya, lakar segi tiga ABC’. Diberi C’ terletak pada BC dengan keadaan AC = AC’. Find two possible values of ∠C in triangle ABC. Hence, sketch triangle ABC’, given C’ lies on BC such that AC = AC’. 34° 8.2 cm 5.7 cm A B C 53°33ʹ Penyelesaian: sin C 8.2 = sin 34° 5.7 sin C = 8.2 × sin 34° 5.7 sin C = 0.8045 ∠C = 53°33’ atau 180° – 53°33’ = 126°27’ 34° 8.2 cm 5.7 cm A B C 126°27ʹ 53°33ʹ Cʹ (a) Cari dua nilai yang mungkin bagi ∠C dalam segi tiga ABC. Seterusnya, lakar segi tiga ABC9, diberikan C9 terletak pada BC dengan keadaan AC = AC9. Find two possible values of ∠C in triangle ABC. Hence, sketch triangle ABC9, given C9 lies on BC such that AC = AC9. 42° 8 cm 11 cm A C B sin C 11 = sin 42° 8 sin C = 11 × sin 42° 8 sin C = 0.9201 ∠C = 66°56’ atau 180 – 66°56’ = 113°4’ 42° 113°4’ 66°56’ 8 cm 11 cm A C B Cʹ

Matematik Tambahan Tingkatan 4 Bab 9 Penyelesaian Segi Tiga 168 BAB 9 CONTOH 2 Dalam rajah di sebelah, BCD ialah garis lurus. Hitung In the diagram, BCD is a straight line. Calculate (i) ∠ABC, (ii) panjang AD. the length of AD. Penyelesaian: 28° 73° 9 cm 10.5 cm 8 cm A D C B (i) Dalam/ In ∆ABC, sin ∠ABC 9 = sin 73° 10.5 sin ∠ABC = 9 × sin 73° 10.5 = 0.8197 ∠ABC = 55°3’ (ii) Dalam/ In ∆ABD, AD sin 55°3’ = 8 sin 28° AD = 8 × sin 55°3’ sin 28° = 13.97 cm (b) Dalam rajah di sebelah, ABC ialah garis lurus. Hitung In the diagram, ABC is a straight line. Calculate (i) panjang BD, the length of BD, (ii) ∠BCD. 85° 43° A C B D 13 cm 11.2 cm (i) Dalam ∆ABD, BD sin 43° = 11.2 sin 85° BD = 11.2 × sin 43° sin 85° = 7.668 cm (ii) Dalam ∆BCD, ∠DBC = 43° + 85° = 128° sin ∠BCD 7.668 = sin 128° 13 sin ∠BCD = 7.668 × sin 128° 13 = 0.4648 ∠BCD = 27°42’ (c) Rajah di sebelah menunjukkan sebuah sisi empat ABCD. Hitung The diagram shows a quadrilateral ABCD. Calculate (i) panjang AC, the length of AC, (ii) ∠ADC. 55° 138° 9° A C B D 22 cm 6 cm (i) Dalam ∆ABC, AC sin 138° = 6 sin 9° AC = 6 × sin 138° sin 9° = 25.66 cm (ii) Dalam ∆ACD, sin ∠ADC 25.66 = sin 55° 22 sin ∠ADC = 25.66 × sin 55° 22 = 0.9554 ∠ADC = 72°509

Matematik Tambahan Tingkatan 4 Bab 9 Penyelesaian Segi Tiga 169 BAB 9 9.2 Petua Kosinus Cosine Rule NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN Petua kosinus / Cosine rule A a B b c C C a b a c b 4. Hitung nilai x bagi setiap segi tiga yang berikut. TP 4 Calculate the value of x in each of the following triangles. CONTOH x cm 7.5 cm 119° 9.3 cm Penyelesaian: x2 = 7.52 + 9.32 − 2(7.5)(9.3) kos 119° 2(7.5)(9.3) cos 119° x = 14.50 cm Gunakan petua kosinus Use cosine rule a2 = b2 + c2 – 2bc kos A a2 = b2 + c2 – 2bc cos A (a) 8.3 cm x cm 11.6 cm 81° x2 = 8.32 + 11.62 – 2(8.3)(11.6) kos/cos 81° x = 13.17 cm

Matematik Tambahan Tingkatan 4 Bab 9 Penyelesaian Segi Tiga 170 BAB 9 (b) x cm 108° 9.5 cm 12.1 cm x2 = 9.52 + 12.12 – 2(9.5)(12.1) kos/cos 108° x = 17.54 cm (c) 123° 16.3 cm 10.5 cm x cm x2 = 10.52 + 16.32 – 2(10.5)(16.3) kos/cos 123° x = 23.71 cm 5. Selesaikan setiap yang berikut. TP 4 Solve each of the following. CONTOH 8.9 cm 17.6 cm Q 11.4 cm R P Hitung sudut terbesar dalam segi tiga PQR. Calculate the largest angle in the triangle PQR. Penyelesaian: ∠Q ialah sudut terbesar kerana ∠Q bertentangan dengan sisi terpanjang. ∠Q is the biggest angle because ∠Q is opposite to the longest side. 17.62 = 8.92 + 11.42 – 2(8.9)(11.4) kos/cos Q 309.76 = 209.17 – 202.92 kos/cos Q kos/cos Q = 209.17 – 309.76 202.92 = –0.4957 ∠Q = 119°43’ (a) 10 cm 9 cm 8 cm Q R P Hitung ∠PRQ dalam segi tiga PQR. Calculate ∠PRQ in the triangle PQR. 102 = 82 + 92 – 2(8)(9) kos/cos ∠PRQ 100 = 145 – 144 kos/cos ∠PRQ kos ∠PRQ = 145 – 100 cos ∠PRQ 144 = 0.3125 ∠PRQ = 71°479 (b) 15 cm 8.2 cm 10 cm Q P R Hitung sudut terbesar dalam segi tiga PQR. Calculate the largest angle in the triangle PQR. ∠Q ialah sudut terbesar kerana ∠Q bertentangan dengan sisi terpanjang ∠Q is the biggest angle because ∠Q is opposite to the longest side 152 = 8.22 + 102 – 2(8.2)(10) kos/cos Q 225 = 167.24 – 164 kos/cos Q kos Q = 167.24 – 225 cos Q 164 = –0.3522 ∠Q = 110°379 (c) 11.5 cm 9.8 cm 6.5 cm Q R P Hitung sudut terkecil dalam segi tiga PQR. Calculate the smallest angle in the triangle PQR. ∠Q ialah sudut terkecil kerana ∠Q bertentangan dengan sisi terpendek ∠Q is the smallest angle because ∠Q is opposite to the shortest side 6.52 = 9.82 + 11.52 – 2(9.8)(11.5) kos/cos Q 42.25 = 228.29 – 225.4 kos/cos Q kos Q = 228.29 – 42.25 cos Q 225.4 = 0.8254 ∠Q = 34°22’

Matematik Tambahan Tingkatan 4 Bab 9 Penyelesaian Segi Tiga 171 BAB 9 6. Selesaikan setiap yang berikut. TP 5 Solve each of the following. CONTOH 8 cm 7 cm 6 cm 14 cm D C A B Dalam rajah di atas, BCD ialah garis lurus. Hitung In the diagram, BCD is a straight line. Calculate (i) ∠ACB, (ii) ∠ADC. Penyelesaian: (i) Dalam/In ∆ABC, 82 = 62 + 72 – 2(6)(7) kos/cos ∠ACB 64 = 85 – 84 kos/cos ∠ACB kos ∠ACB = 85 – 64 84 cos ∠ACB = 0.25 ∠ACB = 75°319 (ii) Dalam/In ∆ACD, ∠ACD = 180° – 75°319 = 104°299 sin ∠ADC 6 = sin 104°299 14 sin ∠ADC = 6 × sin 104°299 14 ∠ADC = 24°319 (a) 8 cm 10 cm 7 cm 15 cm R Q S P Dalam rajah di atas, PQR ialah garis lurus. Hitung In the diagram, PQR is a straight line. Calculate (i) ∠RQS, (ii) panjang PS. the length of PS. (i) Dalam ∆QRS, 152 = 72 + 102 – 2(7)(10) kos/cos ∠RQS 225 = 149 – 140 kos/cos ∠RQS kos ∠RQS = 149 – 225 140 cos ∠RQS = –0.5429 ∠RQS = 122°539 (ii) Dalam/In ∆PQS, ∠PQS = 180° – 122°539 = 57°7’ PS2 = 72 + 82 – 2(7)(8) kos/cos 57°79 = 52.19 PS = 7.224 cm (b) 8 cm 5 cm 12 cm 130° D C A B Dalam rajah di atas, ABCD ialah sebuah trapezium. Hitung In the diagram, ABCD is a trapezium. Calculate (i) ∠BAC, (ii) panjang AD. the length of AD. (i) Dalam/In ∆ABC, 82 = 52 + 122 – 2(5)(12) kos/cos ∠BAC 64 = 169 – 120 kos/cos ∠BAC kos ∠BAC = 169 – 64 120 cos ∠BAC = 0.875 ∠BAC = 28°579 (ii) Dalam/In ∆ACD, ∠ACD = 28°579 AD sin 28°57’ = 12 sin 130° AD = 12 × sin 28°57’ sin 130° AD = 7.583 cm

Matematik Tambahan Tingkatan 4 Bab 9 Penyelesaian Segi Tiga 172 BAB 9 9.3 Luas Segi Tiga Area of a Triangle NOTA IMBASAN NOTA IMBASAN A a B b c C NOTA 7. Hitung luas bagi setiap segi tiga yang berikut. TP 4 Calculate the area of each of the following triangles. CONTOH 1 116° 14 cm 9 cm A B C Penyelesaian: Luas ∆ABC / Area of ∆ABC = 1 2 (9)(14) sin 116° = 56.62 cm2 (a) L J K 63° 9.5 cm 11.6 cm Luas ∆JKL / Area of ∆JKL = 1 2 (11.6)(9.5) sin 63° = 49.09 cm2

Matematik Tambahan Tingkatan 4 Bab 9 Penyelesaian Segi Tiga 173 BAB 9 (b) 13.5 cm 12.6 cm 42° D E F Luas ∆DEF / Area of ∆DEF = 1 2 (13.5)(12.6) sin 42° = 56.91 cm2 (c) 12.1 cm Q P R 18.7 cm 121° Luas ∆PQR / Area of ∆PQR = 1 2 (12.1)(18.7) sin 121° = 96.98 cm2 CONTOH 2 8 cm 10 cm 9 cm P R Q s = a + b + c 2 s = 9 + 8 + 10 2 = 13.5 Tip Guna Hukum Heron untuk mencari luas segi tiga. Use Heron law to find the area of triangle. Luas/Area ∆PQR = √s(s – a)(s – b)(s – c) = √13.5(13.5 – 8)(13.5 – 9)(13.5 – 10) = √13.5(5.5)(4.5)(3.5) = √1 169.44 = 34.20 cm2 (d) 7.5 cm 12 cm 5.7 cm B C A s = 5.7 + 7.5 + 12 2 = 12.6 Luas/Area ∆ABC = √12.6(12.6 – 5.7)(12.6 – 7.5)(12.6 – 12) = √12.6(6.9)(5.1)(0.6) = √266.04 = 16.31 cm2 (e) 4.5 cm 10.1 cm 6 cm Q R P s = 4.5 + 6 + 10.1 2 = 10.3 Luas/Area ∆ABC = √10.3(10.3 – 4.5)(10.3 – 6)(10.3 – 10.1) = √10.3(5.8)(4.3)(0.2) = √51.376 = 7.168 cm2

Matematik Tambahan Tingkatan 4 Bab 9 Penyelesaian Segi Tiga 174 BAB 9 9.4 Aplikasi Petua Sinus, Petua Kosinus dan Luas Segi Tiga Application of Sine Rule, Cosine Rule and Area of a Triangle 8. Selesaikan setiap yang berikut. TP 5 Solve each of the following. CONTOH 10 cm 8 cm 5 cm R P Q U V T W S Rajah di atas menunjukkan sebuah kuboid. Cari The diagram shows a cuboid. Find (i) ∠TQU, (ii) luas ∆TQU. the area of ∆TQU. Penyelesaian: (i) QU =  52 + 102 = 11.18 cm QS =  102 + 82 = 12.81 cm QT =  52 + 12.812 = 13.75 cm 82 = 11.182 + 13.752 – 2(11.18)(13.75) kos/cos∠TQU kos ∠TQU = 314.0549 – 64 307.45 cos ∠TQU = 0.8133 ∠TQU = 35°359 (ii) Luas/Area ∆TQU = 1 2 (11.18)(13.75) sin 35°359 = 44.73 cm2 Teoem Pythagoras Pythagorean Theorem a2 = b2 + c2 Petua kosinus Cosine rule (a) 12 cm 8 cm 6 cm R S V U T W Rajah di atas menunjukkan sebuah prisma tegak. Cari The diagram shows a right prism. Find (i) ∠WSU, (ii) luas segi tiga WSU. the area of triangle WSU. (i) SU =  122 + 62 = 13.42 cm WS =  122 + 82 = 14.42 cm WU =  82 + 62 = 10 cm 102 = 13.422 + 14.422 – 2(13.42)(14.42) kos/cos∠WSU kos ∠WSU = 388.0328 – 100 387.0328 cos ∠WSU = 0.7442 ∠WSU = 41°55’ (ii) Luas ∆WSU = 1 2 (13.42)(14.42) sin 41°55’ Area ∆WSU = 64.64 cm2 (b) PQR ialah sebuah segi tiga dengan PQ = 7.3 cm, QR = 9.6 cm dan PR = 14.7 cm. Hitung PQR is a triangle where PQ = 7.3 cm, QR = 9.6 cm and PR = 14.7 cm. Calculate (i) luas ∆PQR, the area of ∆PQR, (ii) tinggi P dari QR. the height of P from QR. (i) s = 7.3 + 9.6 + 14.7 2 = 15.8 Luas/Area ∆PQR = √15.8(15.8 – 7.3)(15.8 – 9.6)(15.8 – 14.7) = √15.8(8.5)(6.2)(1.1) = √915.926 = 30.264 cm2 (ii) Katakan h = tinggi P dari QR Let h = height of P from QR Luas/Area ∆PQR = 1 2 × QR × h = 30.264 1 2 × (9.60) × h = 30.264 h = 6.305 cm

Matematik Tambahan Tingkatan 4 Bab 9 Penyelesaian Segi Tiga 175 BAB 9 (c) Rajah menunjukkan sebuah prisma tegak. Segi tiga ABC ialah keratan rentas prisma itu dengan keadaan ∠ABC ialah sudut cakah. The diagram shows a right prism. Triangle ABC is the uniform cross-section of the prism where ∠ABC is an obtuse angle. 5 cm 8 cm 20 cm D C F A B E Diberi luas segi tiga ABC ialah 62 cm2 , hitung Given the area of triangle ABC is 62 cm2, calculate (i) ∠ABC, (ii) panjang AC. the length of AC. (i) Luas/Area ∆ABC = 62 cm2 1 2 (8)(20) sin ∠ABC = 62 sin ∠ABC = 62 × 2 8 × 20 = 0.775 ∠ABC = 180° – 50°489 (sudut cakah) = 129°129 (obtuse angle) (ii) AC2 = 82 + 202 – 2(8)(20) kos/cos 129°129 = 666.25 AC = 25.81 cm Kertas 2 1. P R T S Q 20° 50° 115° Rajah di atas menunjukkan dua buah segi tiga PQR dan RST. Diberi PR = 3.3 cm, RS = 5.5 cm dan ST = 4.5 cm. The diagram shows two triangles PQR and RST. It is given that PR = 3.3 cm, RS = 5.5 cm and ST = 4.5 cm. (a) Hitung Calculate (i) panjang, dalam cm, bagi QR, the length, in cm, of QR, (ii) panjang, dalam cm, bagi QT, the length, in cm, of QT, (iii)luas, dalam cm2 , bagi ∆PQR. the area, in cm2, of ∆PQR. SPM 2014 (b) (i) Lakar sebuah ∆P’Q’R’ yang bentuknya berbeza daripada ∆PQR dengan keadaan P’Q’ = PQ, P’R’ = PR dan ∠P’Q’R’ = ∠PQR. Sketch a ∆P9Q9R9 which is different in shape to ∆PQR such that P9Q9 = PQ, P9R9 = PR and ∠P9Q9R9 = ∠PQR. (ii) Seterusnya, nyatakan saiz ∠Q’P’R’. Hence, state the size of ∠Q9P9R9. (a) (i) QR sin 115° = 3.3 sin 20° QR = 8.745 cm (ii) RT2 = 5.52 + 4.52 – 2(5.5)(4.5) kos/cos 50° = 18.68 RT = 4.322 cm QT = QR – RT = 8.745 – 4.322 = 4.423 cm (iii) ∠PRQ = 180° – 115° – 20° = 45° Luas ∆PQR = 1 2 (3.3)(8.745) sin 45° Area ∆PQR = 10.20 cm2 PRAKTIS PRAKTIS SPM SPM 9

Matematik Tambahan Tingkatan 4 Bab 9 Penyelesaian Segi Tiga 176 BAB 9 (b) (i) Pfi Rfi Qfi R 20° (ii) ∠Q’P’R’ + 20° = 45° ∠Q’P’R’ = 45° – 20° = 25° 2. Rajah menunjukkan sisi empat kitaran ABCD. Diagram shows a cyclic quadrilateral ABCD. 105° 5 cm 11 cm 10 cm A D C B (a) Hitung Calculate (i) panjang, dalam cm, AC, the length, in cm, of AC, (ii) ∠ACB. (b) Cari Find (i) luas, dalam cm2 , bagi ∆ACD, the area, in cm2, of ∆ACD, (ii) jarak terdekat, dalam cm, dari titik D ke AC. the shortest distance, in cm, from point D to AC. SPM 2016 (a) (i) ∠ADC = 180° – 105° = 75° AC2 = 112 + 102 – 2(11)(10) kos 75° AC = 12.809 cm (ii) sin ∠BAC 5 = sin 105° 12.809 ∠BAC = 22°9’ ∠ACB = 180° – 105° – 22°9’ = 52°51’ (b) (i) Luas/Area DACD = 1 2 × 11 × 10 × sin 75° = 53.126 cm2 (ii) D A t C 12.809 cm Luas/Area DACD = 53.126 1 2 × 12.809 × t = 53.126 t = 53.126 × 2 12.809 = 8.295 cm 3. Rajah menunjukkan sebuah segi empat ABCD di mana AC dan BD adalah garis lurus. Diagram shows a quadrilateral ABCD such that AC and BD are straight lines. A B C D 5 cm 6 cm Diberi bahawa luas ∆BCD = 12.58 cm2 dan ∠BCD adalah sudut cakah. It is given that the area of ∆BCD = 12.58 cm2 and ∠BCD is obtuse angle. (a) Cari / Find (i) ∠BCD, (ii) panjang, dalam cm, bagi BD, the length, in cm, of BD, (iii) ∠CBD. SPM 2019

Matematik Tambahan Tingkatan 4 Bab 9 Penyelesaian Segi Tiga 177 BAB 9 (b) Diberi AC = 11.2 cm dan ∠ADC = 90°, hitung luas, dalam cm2 , bagi ∆ABD. Given AC = 11.2 cm and ∠ADC = 90°, calculate the area, in cm2, of ∆ABD. (a) (i) Luas / Area of ∆BCD = 12.58 cm2 1 2 × 5 × 6 × sin ∠BCD = 12.58 sin ∠BCD = 12.58 15 ∠BCD = 180° − 57° = 123° (ii) BD2 = 52 + 62 − 2(5)(6) kos/ cos 123° BD = 9.679 cm (iii) sin ∠CBD 6 = sin 123° 9.679 sin ∠CBD = 6 × sin 123° 9.679 ∠CBD = 31°19’ (b) ∠BDC = 180° − 123° − 31°19’ = 25°41’ ∠ADB = 90° − 25°41’ = 64°19’ AD2 = 11.22 – 62 AD = 9.457 cm Luas /Area of ∆ABD = 1 2 × 9.457 × 9.679 × sin 64°19’ = 41.25 cm2 4. Rajah menunjukkan sisi empat PQRS pada suatu satah mengufuk. Diagram shows a quadrilateral PQRS on a horizontal plane. 22 m 12 m 70° 20 m R Q V P S VQSP ialah sebuah piramid dengan keadaan PQ = 8 m dan V adalah 5 m tegak di atas P. VQSP is a pyramid such that PQ = 8 m and V is 5 m vertically above P. SPM 2017 Cari / Find (a) ∠QSR, (b) panjang, dalam m, bagi QS, the length, in m, of QS, (c) luas, dalam m2 , bagi satah condong QVS, the area, in m2, of inclined plane QVS. (a) sin ∠QSR 20 = sin 70° 22 sin ∠QSR = 20 × sin 70° 22 = 0.8543 ∠QSR = 58°419 (b) ∠QRS = 180° – 58°41’ – 70° = 51°19’ QS2 = 202 + 222 – 2(20)(22) kos 51°19’ = 333.986 QS = 18.28 m (c) VQ = √82 + 52 = 9.434 m VS = √122 + 52 = 13 m 9.434 m 13 m 18.28 m S V Q s = 9.434 + 18.28 + 13 2 = 20.36 Luas/Area ∆QVS = √20.36(20.36 – 9.434)(20.36 – 18.28)(20.36 – 13) = √20.36(10.926)(2.08)(7.36) = √3 405.49 = 58.36 m2 Praktis SPM Ekstra

Matematik Tambahan Tingkatan 4 Bab 9 Penyelesaian Segi Tiga 178 BAB 9 Sudut Sudut KBAT KBAT KBAT Ekstra Rajah menunjukkan sebuah kuboid. Diagram shows a cuboid. Hitung / Calculate (a) luas, dalam cm2 , bagi satah condong PRT. the area, in cm2, of the inclined plane PRT. (b) jarak serenjang, dalam cm, dari T ke PR. the perpendicular distance, in cm, from T to PR. (c) sudut di antara satah condong PRT dan tapak PQRS. the angle between the inclined plane PRT and the base PQRS. (a) T P R K 2900 1300 3400 PR2 = 302 + 202 PR = √1300 TR2 = 302 + 502 TR = √3400 TP2 = 202 + 502 TP = √2900 (√3400)2 = (√2900)2 + (√1300)2 – 2(√2900)(√1300) kos/cos P 2(√2900)(√1300) kos/cos P = 2900 + 1300 − 3400 kos/cos P = 800 2 (√2900)(√1300) ∠P = 78°7’ Luas / Area of ∆PRT = 1 2 (√2900)(√1300) sin 78°7’ = 950.02 cm2 (b) Luas / Area of ∆PRT = 950.02 cm2 1 2 × √1300 × t = 950.02, di mana / where t = Titik tengah / Midpoint PR t = 950.02 × 2 √1300 = 52.70 cm (c) 50 cm 52.70 L S K sin ∠TKS = 50 52.70 ∠TKS = 71°35’ 30 cm 20 cm 50 cm Q R P S T U V W Kuiz 9

179 1. Selesaikan. TP 3 Solve. CONTOH Kilang P menghasilkan 250 000 dan 287 500 buah telefon bimbit masing-masing pada tahun 2017 dan 2019. Hitung nombor indeks untuk menunjukkan perubahan pengeluaran pada tahun 2019 menggunakan tahun 2017 sebagai tahun asas. Factory P produced 250 000 and 287 500 mobile telephones in the year 2017 and 2019 respectively. Calculate the index number to show the change in the production in the year 2019 using 2017 as the base year. Penyelesaian: Katakan / Let Q17 = Bilangan telefon bimbit pada tahun 2017 Number of mobile telephone in the year 2017 Q19 = Bilangan telefon bimbit pada tahun 2019 Number of mobile telephone in the year 2019 Nombor indeks/ Index number I = Q19 Q17 × 100 = 287 500 250 000 × 100 = 115 BAB 10 Nombor Indeks Index Numbers 10.1 Nombor Indeks Index Numbers NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN Nombor Indeks Index Numbers 1. Nombor indeks / Index number, I = Q1 Q0 × 100 dengan keadaan / where Q1 = Kuantiti pada masa tertentu Quantity at a specific time Q0 = Kuantiti pada masa asas Quantity at base time 2. Indeks harga / Price index, I = P1 P0 × 100 dengan keadaan / where P1 = Harga barang pada masa tertentu Price of item at a specific time P0 = Harga barang pada masa asas Price of item at base time NOTA

Matematik Tambahan Tingkatan 4 Bab 10 Nombor Indeks 180 BAB 10 (a) Bilangan buku latihan yang dijual di sebuah koperasi sekolah pada tahun 2016 dan tahun 2018 masingmasing adalah 25000 dan 30500 buah. Hitung indeks jualan pada tahun 2018 berasaskan tahun 2016. The number of exercise books sold by a school cooperative store in the year 2016 and 2018 were 25000 and 30500 respectively. Calculate the index number in the year 2018 based on the year 2016. Q16 = 25000 buku / books Q18 = 30500 buku / books Nombor indeks / Index number, I = Q18 Q16 × 100 = 30 500 25 000 × 100 = 122 2. Hitung indeks harga bagi setiap yang berikut. TP 3 Calculate the price index for each of the following. CONTOH Harga sebotol sos tomato pada tahun 2007 dan tahun 2011 masing-masing ialah RM2.50 dan RM3.25. Cari indeks harga bagi sos tomato itu pada tahun 2011 dengan menggunakan tahun 2007 sebagai tahun asas. The prices of a bottle of tomato sauce in the year 2007 and the year 2011 were RM2.50 and RM3.25 respectively. Find the price index of the tomato sauce in the year 2011 by using the year 2007 as the base year. Penyelesaian: Katakan P07 = Harga sebotol sos tomato pada tahun 2007/ Price of a bottle of tomato sauce in the year 2007 Let P11 = Harga sebotol sos tomato pada tahun 2011/ Price of a bottle of tomato sauce in the year 2011 Indeks harga, I = P11 P07 × 100 Price index = 3.25 2.50 × 100 = 130 (a) Harga sebuah kereta pada tahun 2011 dan tahun 2013 masing-masing ialah RM80 000 dan RM60 000. Cari indeks harga bagi kereta itu pada tahun 2013 berasaskan tahun 2011. The prices of a car in the year 2011 and the year 2013 were RM80 000 and RM60 000 respectively. Find the price index of the car in the year 2013 based on the year 2011. Indeks harga, I = P13 P11 × 100 Price index = 60 000 80 000 × 100 = 75 (b) Barangan Item Harga / Price (RM) 2001 2003 A 3.50 4.90 B 6.00 8.10 Jadual di atas menunjukkan harga bagi dua barangan A dan B pada tahun 2001 dan 2003. Hitung indeks harga bagi A dan B pada tahun 2003 berasaskan tahun 2001. The table shows the prices of two items A and B in the years 2001 and 2003. Calculate the price indices of A and B in the year 2003 based on the year 2001. Indeks harga barangan A, IA = P03 P01 × 100 Price index item A = 4.90 3.50 × 100 = 140 Indeks harga barangan B, IB = P03 P01 × 100 Price index item B = 8.10 6.00 × 100 = 135

Matematik Tambahan Tingkatan 4 Bab 10 Nombor Indeks 181 BAB 10 3. Selesaikan setiap yang berikut. Solve each of the following. CONTOH 1 Indeks harga sejenis jam tangan pada tahun 2009 dengan menggunakan tahun 2006 sebagai tahun asas ialah 130. Jika harga jam tangan itu pada tahun 2006 ialah RM420, cari harganya pada tahun 2009. The price index of a watch in the year 2009 by using the year 2006 as the base year was 130. If the price of the watch in the year 2006 was RM420, find its price in the year 2009. TP 4 Penyelesaian: Katakan P06 = Harga jam tangan pada tahun 2006/ Price of the watch in the year 2006 Let P09 = Harga jam tangan pada tahun 2009/ Price of the watch in the year 2009 Indeks harga = 130 Price index P09 P06 × 100 = 130 P09 420 × 100 = 130 P09 = 130 × 420 100 = RM546 (a) Indeks harga bagi sejenis mentol LED pada tahun 2003 berasaskan tahun 2001 ialah 85. Jika harga bagi mentol itu pada tahun 2001 ialah RM22, cari harganya pada tahun 2003. The price index of a LED bulb in the year 2003 based on the year 2001 was 85. If the price of the bulb in the year 2001 was RM22, find its price in the year 2003. Indeks harga = 85 Price index P03 P01 × 100 = 85 P03 22 × 100 = 85 P03 = 85 × 22 100 = RM18.70 (b) Indeks harga bagi sebuah rumah pada tahun 2012 berasaskan pada tahun 2009 ialah 150. Jika harga rumah itu ialah RM240 000 pada tahun 2012, cari harganya pada tahun 2009. The price index of a house in the year 2012 based on the year 2009 was 150. If the price of the house was RM240 000 in the year 2012, find its price in the year 2009. Indeks harga = 150 Price index P12 P09 × 100 = 150 240 000 P09 × 100 = 150 P06 = 240 000 × 100 150 = RM160 000

Matematik Tambahan Tingkatan 4 Bab 10 Nombor Indeks 182 BAB 10 (c) Jadual menunjukkan harga satu kilogram barangan A, B dan C pada tahun 2005 dan 2006. Table shows the price per kilogram of items A, B and C in the year 2005 and 2006. Barangan Item Harga / Price (RM) Indeks harga pada tahun 2006 berasaskan tahun 2005 Price index in the year 2006 based on the year 2005 2005 2006 A RM1.20 RM1.60 z B x RM2.30 110 C RM0.60 y 102 Cari nilai z, x dan y. Find the value of z, x and y. z = P06 P05 × 100 = 1.60 1.20 × 100 = 133.33 Indeks harga B = 110 Price index B P06 P05 × 100 = 110 2.30 x × 100 = 110 x = 2.30 × 100 110 = RM2.09 Indeks harga C = 102 Price index C P06 P05 × 100 = 102 y 0.60 × 100 = 102 y = 102 × 0.60 100 = RM0.61 (d) Barangan Item Harga / Price (RM) Indeks harga pada tahun 2012 berasaskan tahun 2010 Price index in the year 2012 based on the year 2010 2010 2012 R x 360 125 S 125 y 140 Jadual di atas menunjukkan harga bagi barangan R dan S pada tahun 2010 dan 2012 dan indeks harga pada tahun 2012 berasaskan tahun 2010. Cari nilai x dan y. The table shows the prices of items R and S in the years 2010 and 2012 and the price indices for both items in the year 2012 based on the year 2010. Find the values of x and y. Indeks harga barangan R = 125 Price index of item R P12 P10 × 100 = 125 360 x × 100 = 125 x = 360 × 100 125 = 288 Indeks harga barangan S = 140 Price index of item S P12 P10 × 100 = 140 y 125 × 100 = 140 y = 140 × 125 100 = 175

Matematik Tambahan Tingkatan 4 Bab 10 Nombor Indeks 183 BAB 10 CONTOH 2 Barangan Item Indeks harga Price index 2010 (2005 = 100) 2012 (2005 = 100) 2012 (2010 = 100) M 120 150 x N 110 y 140 Jadual di atas menunjukkan indeks harga bagi dua barangan M dan N. Cari nilai x dan y. TP 5 The table shows the price indices of two items M and N. Find the values of x and y. Penyelesaian: Bagi barangan M For item M P10 P05 × 100 = 120 ⇒ P10 P05 = 120 100 P12 P05 × 100 = 150 ⇒ P12 P05 = 150 100 x = P12 P10 × 100 = P12 P05 × P05 P10 × 100 = 150 100 × 100 120 × 100 = 150 120 × 100 x = 125 Bagi barangan N For item N P10 P05 × 100 = 110 ⇒ P10 P05 = 110 100 P12 P10 × 100 = 140 ⇒ P12 P10 = 140 100 y = P12 P05 × 100 = P12 P10 × P10 P05 × 100 = 140 100 × 110 100 × 100 y = 154 (e) Barangan Item Indeks harga /Price index 2002 (2001 = 100) 2008 (2001 = 100) 2008 (2002 = 100) J 125 175 x K 120 y 110 L z 130 150 Jadual di atas menunjukkan indeks harga bagi tiga barangan J, K dan L. Cari nilai x, y dan z. The table shows the price indices of three items J, K and L. Find the values of x, y and z.

Matematik Tambahan Tingkatan 4 Bab 10 Nombor Indeks 184 BAB 10 Bagi barangan J, For item J, P02 P01 × 100 = 125 ⇒ P02 P01 = 125 100 P08 P01 × 100 = 175 ⇒ P08 P01 = 175 100 x = P08 P02 × 100 x = P08 P01 × P01 P02 × 100 x = 175 100 × 100 125 × 100 x = 175 125 × 100 = 140 Bagi barangan K, For item K, P02 P01 × 100 = 120 ⇒ P02 P01 = 120 100 P08 P02 × 100 = 110 ⇒ P08 P02 = 110 100 y = P08 P01 × 100 y = P08 P02 × P02 P01 × 100 = 110 100 × 120 100 × 100 = 132 Bagi barangan L, For item L, P08 P01 × 100 = 130 ⇒ P08 P01 = 130 100 P08 P02 × 100 = 150 ⇒ P08 P02 = 150 100 z = P02 P01 × 100 z = P02 P08 × P08 P01 × 100 z = 100 150 × 130 100 × 100 z = 130 150 × 100 z = 86.67 (f) Indeks harga bagi dua barangan untuk tahun-tahun tertentu diberi dalam jadual berikut. Price indices of two items for certain years are given in the following table. Barangan Item Indeks harga /Price index 1995 (1990 = 100) 1999 (1990 = 100) 1999 (1995 = 100) P 125 140 x Q 135 y 120 Hitungkan nilai bagi x dan nilai bagi y. Calculate the value of x and of y. Bagi barangan P: For item P P95 P90 × 100 = 125 ⇒ P95 P90 = 125 100 P99 P90 × 100 = 140 ⇒ P99 P90 = 140 100 x = P99 P95 × 100 x = P99 P90 × P90 P95 × 100 x = 140 100 × 100 125 × 100 x = 112 Bagi barangan Q: For item Q P95 P90 × 100 = 135 ⇒ P95 P90 = 135 100 P99 P95 × 100 = 120 ⇒ P99 P95 = 120 100 y = P99 P90 × 100 y = P99 P95 × P95 P90 × 100 y = 120 100 × 135 100 × 100 y = 162

Matematik Tambahan Tingkatan 4 Bab 10 Nombor Indeks 185 BAB 10 (g) Indeks harga bagi Perodua Kancil bagi tahun 1996 berasaskan tahun 1990 ialah 140 dan indeks harga bagi Perodua Kancil bagi tahun 2000 berasaskan tahun 1996 ialah 105. Cari indeks harga bagi Perodua Kancil bagi tahun 2000 berasaskan tahun 1990. The price index for Perodua Kancil in the year 1996 based on the year 1990 is 140 and the price index for Perodua Kancil in the year 2000 based on the year 1996 is 105. Find the price index of Perodua Kancil in the year 2000 based on the year 1990. P96 P90 × 100 = 140 ⇒ P96 P90 = 140 100 P00 P96 × 100 = 105 ⇒ P00 P96 = 105 100 P00 P90 × 100 = P00 P96 × P96 P90 × 100 = 105 100 × 140 100 × 100 = 147 (h) Indeks harga bagi suatu barangan elektrik pada tahun 2000 dan 2002 berasaskan tahun 1998 masingmasing ialah 116 dan 125. Hitungkan indeks harga bagi tahun 1998 dan 2000 jika tahun 2002 digunakan sebagai tahun asas. Price indices for a certain electric product in the year 2000 and 2002 based on the year 1998 are 116 and 125 respectively. Calculate the price index in the year 1998 and 2000 if the year 2002 is used as based year. P00 P98 × 100 = 116 ⇒ P00 P98 = 116 100 P02 P98 × 100 = 125 ⇒ P02 P98 = 125 100 P98 P02 × 100 = 100 125 × 100 = 80 P00 P02 × 100 = P00 P98 × P98 P02 × 100 = 116 100 × 100 125 × 100 = 92.8 10.2 Indeks Gubahan Composite Index NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN NOTA IMBASAN Indeks Gubahan Composite Index 1. Indeks gubahan / Composite index I = I1w1 + I2w2 + …… + Inwn w1 + w2 + …… + wn = Ii wi wi dengan keadaan / where I = Nombor indeks / Index number w = Pemberat / Weightage 2. Pemberat ialah nilai atau kuantiti yang diberikan kepada setiap barangan untuk menunjukkan kepentingan relatif setiap barangan itu. Weightage is the value or quantity assigned to each item to show the relative importance of each item.

Matematik Tambahan Tingkatan 4 Bab 10 Nombor Indeks 186 BAB 10 4. Hitung indeks gubahan bagi setiap yang berikut. TP 3 Calculate the composite index for each of the following. CONTOH Komponen Component Indeks harga Price index Pemberat Weightage P 120 5 Q 115 6 R 130 7 S 145 2 Penyelesaian: Indeks gubahan,/ Composite index, I = Ii wi wi = 120(5) + 115(6) + 130(7) + 145(2) 5 + 6 + 7 + 2 = 2 490 20 = 124.5 (a) Bahan Ingredient Indeks harga Price index Pemberat Weightage J 125 4 K 135 2 L 130 1 M 140 5 Indeks gubahan,/ Composite index, I = Ii wi wi = 125(4) + 135(2) + 130(1) + 140(5) 4 + 2 + 1 + 5 = 1 600 12 = 133.3 (b) Makanan Food Indeks harga Price index Pemberat Weightage A 115 2 B 120 3 C 140 5 D 105 6 Indeks gubahan,/ Composite index, I = Ii wi wi = 115(2) + 120(3) + 140(5) + 105(6) 2 + 3 + 5 + 6 = 1 920 16 = 120 (c) Barangan Item Indeks harga Price index Pemberat Weightage E 123 30 F 145 20 G 151 10 H 115 40 Indeks gubahan,/ Composite index, I = Ii wi wi = 123(30) + 145(20) + 151(10) + 115(40) 30 + 20 + 10 + 40 = 12 700 100 = 127

Matematik Tambahan Tingkatan 4 Bab 10 Nombor Indeks 187 BAB 10 5. Selesaikan setiap yang berikut. Solve each of the following. CONTOH 1 Barangan Item Jumlah perbelanjaan sebulan (RM) Total expenditure per month (RM) Pemberat Weightage 2007 2009 Elektrik Electricity 200 270 4 Pendidikan Education 350 420 1 Pengangkutan Transportation 220 242 3 Makanan Food 340 493 2 Jadual di atas menunjukkan perbelanjaan bulanan sebuah keluarga pada tahun 2007 dan tahun 2009 dan pemberatnya masing-masing. Hitung indeks gubahan pada tahun 2009 dengan menggunakan tahun 2007 sebagai tahun asas. TP 4 The table shows the monthly expenditure of a family in the years 2007 and 2009 and their respective weightages. Calculate the composite index in the year 2009 using year 2007 as the base year. Penyelesaian: Indeks harga bagi elektrik = 270 200 × 100 Price index of electricity = 135 Indeks harga bagi pendidikan = 420 350 × 100 Price index of education = 120 Indeks harga bagi pengangkutan = 242 220 × 100 Price index of transportation = 110 Indeks harga bagi makanan = 493 340 × 100 Price index of food = 145 Indeks gubahan/Composite index = 135 × 4 + 120 × 1 + 110 × 3 + 145 × 2 4 + 1 + 3 + 2 = 1 280 10 = 128 (a) Barangan Item Harga / Price (RM) Pemberat Weightage 2010 2013 A 30.00 37.50 2 B 45.00 49.50 3 C 50.00 70.00 4 D 25.00 32.50 3 Jadual di atas menunjukkan harga bagi empat barangan pada tahun 2010 dan tahun 2013 dan pemberatnya masing-masing. Hitung indeks gubahan pada tahun 2013 berasaskan tahun 2010. The table shows the prices of four items in the years 2010 and 2013 and their respective weightages. Calculate the composite index in the year 2013 based on the year 2010. IA = 37.50 30.00 × 100 = 125 IB = 49.50 45.00 × 100 = 110 IC = 70.00 50.00 × 100 = 140 ID = 32.50 25.00 × 100 = 130 Indeks gubahan Composite index = 125 × 2 + 110 × 3 + 140 × 4 + 130 × 3 2 + 3 + 4 + 3 = 1 530 12 = 127.5

Matematik Tambahan Tingkatan 4 Bab 10 Nombor Indeks 188 BAB 10 (b) Barangan Item Harga / Price (RM) Pemberat Weightage 2012 2014 E 80 120 3 F 40 46 6 G 55 66 7 H 95 133 4 Jadual di atas menunjukkan harga bagi empat komponen pada tahun 2012 dan tahun 2014 dan pemberatnya masing-masing. Hitung indeks gubahan pada tahun 2014 berasaskan tahun 2012. The table shows the prices of four components in the years 2012 and 2014 and their respective weightages. Calculate the composite index in the year 2014 based on the year 2012. IE = 120 80 × 100 = 150 IF = 46 40 × 100 = 115 IG = 66 55 × 100 = 120 IH = 133 95 × 100 = 140 Indeks gubahan / Composite index I = 150 × 3 + 115 × 6 + 120 × 7 + 140 × 4 3 + 6 + 7 + 4 = 2 540 20 = 127 CONTOH 2 Barangan Item Indeks harga Price index Pemberat Weightage A 120 1 B 114 y C x 5 D 130 2 Jadual di atas menunjukkan indeks harga bagi empat barangan pada tahun 2008 berasaskan tahun 2004 dan pemberatnya masing-masing. Harga bagi barangan C pada tahun 2004 dan tahun 2008 masing-masing ialah RM30 dan RM42. Diberi indeks gubahan bagi tahun 2008 berasaskan tahun 2004 ialah 128, cari nilai x dan nilai y. TP 5 The table shows the price indices of four items in the year 2008 based on the year 2004 and their respective weightages. The prices of item C in the years 2004 and 2008 were RM30 and RM42 respectively. Given the composite index in the year 2008 based on the year 2004 is 128, find the value of x and of y. Penyelesaian: x = P08 P04 × 100 = 42 30 × 100 = 140 Indeks gubahan/Composite index = 128 120 × 1 + 114 × y + 140 × 5 + 130 × 2 1 + y + 5 + 2 = 128 114y + 1 080 y + 8 = 128 114y + 1 080 = 128y + 1 024 14y = 56 y = 4 (c) Barangan Item Indeks harga Price index Pemberat Weightage P 125 4 Q x 3 R 120 y S 150 3 Jadual di atas menunjukkan indeks harga bagi empat barangan pada tahun 2010 berasaskan tahun 2007 dan pemberatnya masing-masing. Harga bagi barangan Q pada tahun 2007 dan tahun 2010 masing-masing ialah RM8.00 dan RM10.40. Diberi indeks gubahan pada tahun 2010 berasaskan tahun 2007 ialah 128.75, cari nilai x dan nilai y. The table shows the price indices of four items in the year 2010 based on the year 2007 and their respective weightages. The prices of item Q in the years 2007 and 2010 were RM8.00 and RM10.40 respectively. Given the composite index in the year 2010 based on the year 2007 is 128.75, find the value of x and of y. x = P10 P07 × 100 = 10.40 8.00 × 100 = 130 Indeks gubahan/ Composite index = 128.75 125 × 4 + 130 × 3 + 120 × y + 150 × 3 4 + 3 + y + 3 = 128.75 120y + 1 340 y + 10 = 128.75 120y + 1 340 = 128.75y + 1 287.5 8.75y = 52.5 y = 6

Matematik Tambahan Tingkatan 4 Bab 10 Nombor Indeks 189 BAB 10 CONTOH 3 Rajah menunjukkan carta palang yang mewakili kos bagi tiga komponen A, B, C dan indeks harga bagi komponen-komponen bagi tahun 2003 berasaskan tahun 2001 diberi dalam jadual di bawah. Diberi indeks gubahan ialah 107. Cari nilai m. TP 5 Diagram shows a bar chart which represents the cost for three components A, B and C and price indices for the components in the year 2003 based on the year 2001 are given in table below. Given the composite index is 107. Find the value of m. A CB Kos / Cost Komponen Component 4 6 m Komponen Component Indeks harga Price index A 122 B 80 C 110 Penyelesaian: Indeks gubahan/Composite index = 107 122 × m + 80 × 6 + 110 × 4 m + 6 + 4 = 107 122m + 920 10 + m = 107 122m + 920 = 1 070 + 107m 15m = 150 m = 10 (d) Jadual menunjukkan indeks harga dan peratus penggunaan empat bahan P, Q, R dan S, yang digunakan dalam pengeluaran sejenis biskut. Table shows price indices and percentage of usage for four items P, Q, R and S, used in a production of a biscuit. Bahan Item Indeks harga pada tahun 1995 berasaskan tahun 1993 Price index in the year 1995 based on the year 1993 Peratus penggunaan Percentage of usage (%) P 135 40 Q x 30 R 105 10 S 130 20 Diberi indeks gubahan pada tahun 1995 berasaskan tahun 1993 ialah 128. Cari nilai x. Given the composite index in the year 1995 based on the year 1993 is 128. Find the value of x. Indeks gubahan/ Composite index = 128 135 × 40 + x × 30 + 105 × 10 + 130 × 20 40 + 30 + 10 + 20 = 128 9 050 + 30x 100 = 128 9 050 + 30x = 12 800 30x = 3 750 x = 125