Jawapan (1)

51 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan 16. (a) (i) x° = 35° x = 35 (ii) x° = 113° x = 113 (b) (i) x° = 180° – 132° = 48° x = 48 (ii) x° = 180° – 28° = 152° x = 152 17. (a) x° = 180° – 54° Sudut bersebelahan Adjacent angles = 126° x = 126 y° = ∠AFC Sudut bertentang bucu Vertically opposite angles = 126° – 90° = 36° y = 36 (b) x° = 55° + 30° y° = 180° – 85° = 85° = 95° x = 85 y = 95 (c) 4x° + 2x° = 180° 6x° = 180° x° = 30° x = 30 3y° = 4x° = 4 × 30° = 120° y° = 40° y = 40 18. (a) TU (b) EF Sudut bertentang bucu Vertically opposite angles Mahir Diri Sudut sepadan tidak sama → Bukan garis selari. Corresponding angles are not the same → Not parallel lines Sudut selang-seli tidak sama → Bukan garis selari Alternate angles are not the same → Not parallel lines 55° + 125° = 180° Hasil tambah sudut pedalaman = 180° → Garis selari The sum of interior angles = 180° → Parallel lines A B C D 140° 130° A B C D 125° 55° A B C D 85° 75° 19. (a) p = 60 (b) q = 125 (c) r = 180 − 75 = 105 20. (a) (i) Garis mengufuk: ST dan / and UV Horizontal line (ii) Sudut dongak: ∠TSU Angle of elevation (iii) Sudut tunduk: ∠VUS Angle of depression (b) (i) Garis mengufuk: PN Horizontal line (ii) Sudut dongak: ∠PNM Angle of elevation (iii) Sudut tunduk: ∠PNO Angle of depression (c) (i) Garis mengufuk: AB dan / and CD Horizontal line (ii) Sudut dongak: ∠DCB Angle of elevation

© Penerbitan Pelangi Sdn. Bhd. 52 Matematik Tingkatan 1 Jawapan (iii) Sudut tunduk: ∠ABC, ∠DCE Angle of depression 21. (a) 50° Sudut 80° selang-seli Alternate angle 80° x° D B C A 80° + x° + 50° = 180° x° + 130° = 180° x° = 180° – 130° = 50° x = 50 (b) 50° 110° x° y° B C A ∠ABC = 180° – 110° Sudut pedalaman Interior angles = 70° x° + 50° = 70° x° = 70° – 50° = 20° x = 20 y° + 50° = 180° Sudut pedalaman Interior angles y° = 180° – 50° = 130° y = 130 (c) 150° 3x° 2x° x° z° y° z° = 360° – 90° – 150° = 120° z = 120 y° = 180° – 120° = 60° y = 60 y° + 2x° + 3x° + x° = 360° 60° + 6x° = 360° 6x° = 300° x° = 50° x = 50 Mahir Diri Q R P K N L M 64° 64° 3x° x° ∠QLN = 64° Sudut sepadan / Corresponding angle x° + 3x° + 64° = 180° Hasil tambah sudut pada garis lurus = 180° The sum of angles on a straight line = 180° 4x° + 64° = 180° 4x° = 180° – 64° = 116° x° = 29° x = 29 22. (a) ∠SPO (b) ∠SVU = 45° ∠TVS = 27° x° = 27° + 45° = 72° x = 72 W U x° S P V Y T Q R O 45° 15 m 30 m 15 m 30 m 15 m 27° Sudut selang-seli Alternate angle Sudut selang-seli Alternate angle

53 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan (c) Sudut dongak R dari Q = Sudut Q dari P Angle of elevation R from Q = angle of Q from P = 34° Sudut dongak R dari Q ialah ∠YQR. Perhatikan bahawa titik P, Q dan R adalah segaris. PS dan QY ialah dua garis mengufuk yang selari. Maka, ∠YQR dan ∠SPQ ialah sudut sepadan. Angle of elevation R from Q is ∠YQR. Note that the points P, Q, and R are in a straight line. PS and QY are two parallel horizontal lines. Therefore, ∠YQR and ∠SPQ are corresponding angles. Praktis Masteri 8 Bahagian A 1. Sudut cakah ialah nilai sudut antara 90° dengan 180°. An obtuse angle is an angle between 90° and 180°. Jawapan / Answer: B 2. Jumlah nilai bagi pasangan sudut konjugat ialah 360°. The total angle of a pair of conjugate angles is 360°. Maka, nilai sudut konjugat bagi 145° Therefore, the conjugate angle of 145°. = 360° – 145° = 215° Jawapan / Answer: C 3. 3x + 9x = 180° 12x = 180° x = 15° Jawapan / Answer: A 4. 3x + 40° + 75° + 2x + 15° = 180° 5x + 130° = 180° 5x = 50° x = 10° Jawapan / Answer: B 5. Jawapan / Answer: B Bahagian B 1. (a) (i) sudut tirus / acute angle (ii) sudut refleks / reflex angle (b) (i) P 5 cm Q (ii) M 4.5 cm N 2. (a) (i) ∠HGL (ii) ∠GMK (b) (ii) 31° (iii) 120° Bahagian C 1. (a) (i) p = 90° − 74° = 16° (ii) p = 180° − 132° = 48° (iii) p = 360° − 105° − 90° = 165° (b) a° + b° + c° + 112° = 360° a° + b° + c° = 360° – 112° = 248° a + b + c = 248° (c) (i) 4 cm Q R P PR = 6.6 cm

© Penerbitan Pelangi Sdn. Bhd. 54 Matematik Tingkatan 1 Jawapan Poligon Asas Basic Polygons BAB 9 1. (a) (i) Bilangan bucu = 7 Number of vertices (ii) Bilangan pepenjuru Number of diagonals = 7(7 – 3) 2 = 14 (b) (i) Bilangan bucu = 10 Number of vertices (ii) Bilangan pepenjuru Number of diagonals = 10(10 – 3) 2 = 35 2. (a) (i) Sisi empat / Quadrilateral (ii) 4 (iii) 4 (iv) 2 (b) (i) Pentagon (ii) 5 (iii) 5 (iv) 5 (c) (i) Heksagon / Hexagon (ii) 6 (iii) 6 (iv) 6 3. (a) (i) Oktagon / Octagon (ii) 8 (iii) 8 (iv) 20 (b) (i) Nonagon (ii) 9 (iii) 9 (iv) 27 (c) (i) Dekagon / Decagon (ii) 10 (iii) 10 (iv) 35 4. (a) (b) • Terdapat 3 paksi simetri. There are 3 axes of symmetry. • Semua sisi adalah sama panjang. All sides are equal. • Semua sudut pedalaman ialah 60°. Each interior angle is 68°. • Terdapat 1 paksi simetri. There is one axis of symmetry. • Dua daripada sisinya sama panjang. Two of its sides are equal. • Dua daripada sudutnya adalah sama. Two of its angles are equal. • Tidak ada paksi simetri. There is no axis of symmetry. • Semua sisinya tidak sama panjang. All sides are not the same. • Semua sudut pedalaman adalah tidak sama. All the interior angles are not the same. 5. (a) Segi tiga bersudut cakah / Obtused-angled triangle (b) Segi tiga bersudut tegak / Right-angled triangle

55 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan 6. (a) (i) x° = 180° – 110° – 40° = 30° x = 30 (ii) x° = 180° – 34° – 36° = 110° x = 110 (b) (i) x° = 180° – 90° – 45° = 45° x = 45 (ii) x° = 180° – 90° – 74° = 16° x = 16 (c) (i) x° + x° + 100° = 180° 2x° + 100° = 180° 2x° = 180° – 100° = 80° x° = 40° x = 40 (ii) x° + 35° + 35° = 180° x° + 70° = 180° x° = 180° – 70° = 110° x = 110 7. (a) (i) x° = 74° + 15° = 89° x = 89 (ii) x° = 35° + 35° = 70° x = 70 (b) (i) x° + 35° = 90° x° = 90° – 35° = 55° x = 55 (ii) x°+ 60° = 125° x° = 125° – 60° = 65° x = 65 8. (a) ∠BFC = x° x° + 40° = 110° x° = 110° – 40° = 70° x = 70 40° + y° + 20° = 110° y° + 60° = 110° y° = 110° – 60° = 50° y = 50 x + y = 70 + 50 = 120 Mahir Diri Q x° x° P S 30° 64° T R Dalam / in ∆PQS, ∠PSQ = 180° – 90° – 30° = 60° x° + x° = 64° + 60° 2x° = 124° x° = 62° x = 62 Jumlah sudut dalam satu segi tiga ialah 180°. The sum of angles in a triangle is 180°. Sudut peluaran adalah sama dengan hasil tambah dua sudut pedalaman bertentangan. The exterior angle is the sum of the two interior opposite angles 9. (a) (i) Segi empat tepat / Rectangle (ii) 2 (b) (i) Segi empat selari / Parallelogram (ii) 0 (c) (i) Trapezium (ii) 0 (d) (i) Rombus / Rhombus (ii) 2 (e) (i) Lelayang / Kite (ii) 1 10. (a) (i) x° + 86° + 124° + 58° = 360° x° + 268° = 360° x° = 92° x = 92 x° 20° x° y° 40° B A 110° C D E F

© Penerbitan Pelangi Sdn. Bhd. 56 Matematik Tingkatan 1 Jawapan (ii) x° + 45° + 120° + 95° = 360° x° + 260° = 360° x° = 100° x = 100 (b) (i) x° + 78° + 154° + x° = 360° 2x° + 232° = 360° 2x° = 128° x° = 64° x = 64 (ii) 2x° + 110° + 106° + x° = 360° 3x° + 216° = 360° 3x° = 144° x° = 48° x = 48 11. (a) (i) x° = 110° x = 110 x° + y° = 180° 110° + y° = 180° y° = 70° y = 70 (ii) 2x° + x° = 180° 3x° = 180° x° = 60° x = 60 y° = x° = 60° y = 60 (b) (i) x° + 120° = 180° x° = 60° x = 60 y° = 120° y = 120 (ii) x° + 100° = 180° x° = 80° x = 80 y° = 100° y = 100 Sudut selangseli Alternate angles 12. (a) (i) x° + 36° + 90° + 90° = 360° x° + 216° = 360° x° = 360° – 216° = 144° x = 144 (b) (c) (i) x° + 25° = 125° x° = 125° – 25° = 100° x = 100 (i) ∠QRO = ∠SRO = 38° x° + 90° + 38° = 180° x° + 128° = 180° x° = 180° – 128° = 52° x = 52 13. ∠CBE = 46°, ∠BED = 180° – 46° = 134° ∠EAB + ∠AEB = ∠CBE ∠AEB + ∠AEB = 46° 2∠AEB = 46° ∠AEB = 23° x° = 360° – 134° – 23° = 203° x = 203 14. (a) ∠FGE = 60° DEFG ialah segi tiga sama sisi. DEFG is an equilateral triangle ∠EGH = 180° – 100° 2 = 40° x° = 60° – 40° = 20° x = 20 (b) EFGH ialah sisi empat. / EFGH is a quadrilateral. (c) Garis lurus FH yang dipanjangkan merupakan paksi simetri bagi sisi empat EFGH. The extended straight line FH is axis of symmetry of the quadrilateral EFGH. (d) EFGJ ialah rombus / EFGH is a rhombus. Mahir Diri x° + 135° + 90° + 75° = 360° x° + 300° = 360° x° = 360° – 300° = 60° x = 60 DABE ialah segi tiga sama kaki ABE is an isosceles triangle. ∠EAB = ∠AEB E x° J G H F 100°

57 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan Praktis Masteri 9 Bahagian A 1. Jawapan / Answer: C 2. Heksagon mempunyai 6 sisi. Hexagon has 6 sides. Jawapan / Answer: D 3. Pentagon mempunyai 5 pepenjuru. Pentagon has 5 diagonals. Segi tiga tidak mempunyai pepenjuru. Triangle does not have diagonal. Heksagon mempunyai 9 pepenjuru. Pentagon has 9 diagonals. Jawapan / Answer: D 4. ∠DBC = (180° – 36°) ÷ 2 = 144° ÷ 2 = 72° p = 72° (sudut bertentangan bucu) (vertically opposite angle) Jawapan / Answer: C 5. ∠TQV = 180° – 62° – 62° = 56° ∠TQV = 68° (sudut selang-seli / alternate angle) k = 68° – 56° = 12° Jawapan / Answer: A Bahagian B 1. (a) Poligon Polygon Heksagon Hexagon Bilangan sisi Number of sides 6 Bilangan bucu Number of vertices 6 Bilangan pepenjuru Number of diagonals 9 (b) y = 180° – 108°  ÷ 2 = 72° ÷ 2 y = 36° 2. (a) (i) (ii) (iii) (b) (i) Pentagon PQRST (ii) Oktagon PQRSTUVW Octagon PQRSTUVW Q P R T S Q P R V T W S U Bahagian C 1. (a) 70° 180° – 70° –––––––– 2 = 55° L K M KML = KLM = 55° (i) (ii) (iii) (b) 2p + 2p + 2p = 90° p = 15° q = 180° – 90° – 2 15°  q = 60° (c) ∠RTQ = 30° x° + 30° = 125° x° = 125° – 30° = 95° x = 95 x – y = 95 – 75 = 20 y° + 30° = 105° y° = 105° – 30° = 75° y = 75 ∠PTU dan ∠RTQ ialah sudut bertentang bucu. ∠PTU and ∠RTQ are vertically opposite angles.

© Penerbitan Pelangi Sdn. Bhd. 58 Matematik Tingkatan 1 Jawapan Perimeter dan Luas Perimeter and Area BAB 10 1. (a) 4 cm 3 cm 2.4 cm 3.2 cm Perimeter = 4 cm + 3 cm + 2.4 cm + 3.2 cm = 12.6 cm (b) 3.3 cm 2.2 cm 2.3 cm 1.5 cm 2.5 cm Perimeter = 2.2 cm + 1.5 cm + 2.3 cm + 2.5 cm + 3.3 cm = 11.8 cm 2. (a) 28 cm (b) 22 cm 3. (a) 2.8 cm 2.1 cm 5.5 cm Perimeter = 2.8 cm + 2.1 cm + 5.5 cm = 10.4 cm (b) 1.9 cm 2.4 cm 1.6 cm 3.1 cm 4 cm Perimeter = 2.4 cm + 4 cm + 3.1 cm + 1.6 cm + 1.9 cm = 13 cm 4. (a) Perimeter = 6 × 4 cm + 5 cm = 24 cm + 5 cm = 29 cm (b) Perimeter = 12 cm + 2 × 3 cm + 5 × 4 cm = 12 cm + 6 cm + 20 cm = 38 cm (c) 6 cm 6 cm 6 cm 8 cm 10 cm A E C B D Perimeter = 6 cm + 6 cm + 6 cm + 10 cm + 8 cm = 36 cm 5. (a) Q X W U V T P R S 13 m 7 m 7 m 13 m 7 m + ST + TU + UV + VW + WX + XP = 25 m ST + TU + UV + VW + WX + XP = 25 m – 7 m = 18 m Perimeter kawasan yang tidak ditampal kertas hias dinding The perimeter of the area that is not pasted with wallpaper = (ST + TU + UV + VW + WX + XP) + PQ + QR + RS = 18 m + 13 m + 7 m + 13 m = 51 m

59 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan (b) K N L M x m x m 14 m 14 m 20 m 5 m x m Perimeter kawasan yang berlorek Perimeter of the shaded area = 684 ÷ 12 = 57 m 14 + 20 + 5 + x + x + x = 57 39 + 3x = 57 3x = 18 x = 6 Perimeter kawasan tidak berlorek Perimeter of the area that is not shaded = (20 m – 6 m) + 6 m + 6 m + 5 m + 14 m = 45 m 6. (a) (i) a (ii) b (iii) 1 2 × a × b (b) (i) a (ii) b (iii) a × b (c) (i) a (ii) b (iii) 1 2 × a × b (d) (i) a (ii) h (iii) 1 2 × (a × b) × h (e) (i) a (ii) a (iii) a × a 7. (a) (i) Luas / Area = 1 2 × 14 × 5 = 35 cm2 (ii) Luas / Area = 1 2 × 7 × 6 = 21 cm2 (b) (i) Luas / Area = 17 × 11 = 187 cm2 (ii) Luas / Area = 9 × 6 = 54 cm2 (c) (i) Luas / Area = 1 2 × 9 × 6 = 27 cm2 (ii) Luas / Area = 1 2 × 7 × 6 = 21 cm2 (d) (i) Luas / Area = 1 2 × (17 + 5) × 9 = 99 cm2 (ii) Luas / Area = 1 2 × (9 + 12) × 7 = 73.5 cm2 8. (a) (i) 1 2 × x × 8 = 48 4 x = 48 x = 12 (ii) 1 2 × 9 × x = 27 x = 27 × 2 9 x = 6 (b) (i) 7 × x = 84 x = 84 ÷ 7 = 12 (ii) 15 × x = 75 x = 75 ÷ 15 = 5 (c) (i) 1 2 × 10 × x = 40 5 x = 40 x = 8 (ii) 1 2 × 14 × x = 35 7x = 35 x = 5 (d) (i) 1 2 × ( 8 + 12 ) × x = 40 10 x = 40 x = 4

© Penerbitan Pelangi Sdn. Bhd. 60 Matematik Tingkatan 1 Jawapan (ii) 1 2 × (4 + 12) × x = 48 8x = 48 x = 6 9. (a) Luas kawasan yang berlorek Area of shaded region = Luas segi tiga + Luas segi empat selari Area of triangle + Area of parallelogram = 1 2 × 4 × 10 + 10 × 7 = 20 + 70 = 90 cm2 (b) Luas kawasan yang berlorek Area of shaded region = Luas segi tiga + Luas lelayang Area of triangle + Area of kite = 1 2 × 9 × 6 + 1 2 × 12 × 8 = 27 + 48 = 75 cm2 (c) Luas kawasan yang berlorek Area of shaded region = Luas segi empat selari – Luas segi tiga Area of parallelogram – Area of triangle = 12 × 8 – 1 2 × 12 × 8 = 96 – 48 = 48 cm2 (d) Luas kawasan yang berlorek Area of shaded region = Luas trapezium – Luas lelayang Area of trapezium – Area of kite = 1 2 × (15 + 7) × 10 – 1 2 × 10 × 5 = 110 – 25 = 85 cm2 (e) Luas kawasan yang berlorek Area of shaded region = Luas trapezium – Luas segi tiga Area of trapezium – Area of triangle = 1 2 × (12 + 3 + 4) × 10 – 1 2 × 4 × 10 = 95 – 20 = 75 cm2 10. (a) WR = 3 4 PQ PV = 30 m = 3 4 × 20 RS = 60 – 36 2 = 15 m = 12 m Luas kawasan A / Area of regions A = Luas trapezium PQSV – Luas segi tiga WRS Area of trapezium PQSV – Area of triangle WRS = 1 2 × (36 + 12 + 30) × 20 – 1 2 × 12 × 15 = 780 – 90 = 690 m2 Mahir Diri Luas tanah yang digunakan untuk penternakan itik, ABDEFH The area of the land used to rear ducks ABDEFH = Luas segi empat tepat ACEG – Luas ∆BCD – Luas ∆FGH Area of rectangle ACEG – Area of ΔBCD – Area of ΔFGH = 80 × 50 – 1 2 × 25 × 25 – 1 2 × 25 × 30 = 4 000 – 312.5 – 375 = 3 312.5 m2 G R E A H C D P B Q F 25 m 25 m 25 m 25 m 80 m 30 m 50 m 11. (a) (i) E P Q B D C A F 12 m x m x m 13 m 20 m 31 m 5 m Luas taman kanak-kanak = 90 The area of the playground 1 2 × (x + 5) × 12 = 90 6(x + 5) = 90 x + 5 = 15 x = 10

61 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan Panjang laluan batu kelikir The length of the gravel path = Perimeter trapezium ABCF Perimeter of trapezium ABCF = 20 + 31 + 13 + 10 = 74 m (ii) Luas taman bunga / Area of the garden = Luas trapezium ABCF / Area of trapezium ABCF = 1 2 × (10 + 31) × 12 = 246 m2 Praktis Masteri 10 Bahagian A 1. Perimeter = 8 cm + 8 cm + 12 cm + 13 cm = 41 cm Jawapan / Answer: C 2. Perimeter = 5 cm × 12 = 60 cm Jawapan / Answer: C 3. Luas segi tiga = 1 2 × 14 cm × x cm Area of triangle 63 cm2 = 7 cm × x cm x = 63 cm2 7 cm = 9 cm Jawapan / Answer: B Bahagian B 1. (a) (i) Perimeter = PQ + QR + PR (ii) 6 cm (b) Perimeter =  4 × 7 cm +  4 × 4 cm = 44 cm 2. b a h ● ● 1 2 × a × b a b ● ● a × b a b ● ● 1 2 × (a + b) × h a a ● ● a × a Bahagian C 1. (a) Panjang Length (cm) Lebar Width (cm) Luas Area (cm2 ) ü / û (i) 25 8 200 ü (25 × 8 = 200) (ii) 12 5 60 ü (12 × 5 = 60) (iii) 20 21 630 û (20 × 21 = 420) (b) Luas lantai / Area of the floor = 18 × 6 + 18 × 9 + 1 2 × 18 × 5 + 1 2 × (5 + 7) × 9 = 369 m2 Jumlah kos / Total cost = 369 × RM12 = RM4 428

© Penerbitan Pelangi Sdn. Bhd. 62 Matematik Tingkatan 1 Jawapan (c) (i) Panjang segi empat sama= 24 ÷ 4 The length of the square = 6 cm Luas segi empat sama= 6 × 6 Area of the square = 36 cm2 Panjang segi empat tepat= (24 – 4 – 4) ÷ 2 Length of the rectangle = 8 cm Luas segi empat tepat= 8 × 4 Area of the rectangle = 32 cm2 Maka, luas segi empat sama adalah lebih besar. / Thus, the area of the square is bigger. (ii) Katakan lebar / Let the width = 2 cm Panjang = (24 – 2 – 2) ÷ 2 Length = 10 cm 2 cm Maka, dimensi yang mungkin ialah panjang 10 cm dan lebar 2 cm. Therefore, the possible dimensions are 10 cm long and 2 cm wide. (Jawapan lain yang sesuai diterima juga) (Other suitable answer are also acceptable)

63 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan Pengenalan Set Introduction of Set BAB 11 1. (a) Haiwan / Animal: unta, tikus, ular / camel, rat, snake Tumbuh-tumbuhan / Plant: bunga, pokok, rumput / flower, tree, grass (b) Sayur-sayuran / Vegetable: bendi, bayam, kalian / lady’s finger, spinach, kailan Buah-buahan / Fruit: ciku, tembikai, langsat / ciku, watermelon, langsat (c) Warna / Colour: hitam, putih, jingga / black, white, orange Bentuk / Shape: pentagon, trapezium, lelayang / pentagon, trapezium, kite 2. (a) (i) Perihalan / Description: P ialah set nombor ganjil yang kurang daripada 10. / P is a set of odd numbers which are less than 10. Penyenaraian / Listing: P = {1, 3, 5, 7, 9} Tatatanda pembina set / Set builder notation: P = {x : x ialah nombor ganjil dan x  10} P = {x : x is an odd number and x  10} (b) (i) Perihalan / Description: P ialah set yang terdiri daripada bulan yang mempunyai kurang daripada 31 hari. / P is a set of months that have less than 31 days. Penyenaraian / Listing: P = {Februari, April, Jun, September, November} P = {February, April, June, September, November} Tatatanda pembina set / Set builder notation: P = {x : x ialah bulan dalam setahun yang mempunyai kurang daripada 31 hari} P = {x : x is a month in a year that have less than 31 days} 3. (a) (i) A = {Hari yang bermula dengan huruf J} {Days that start with the letter F} (ii) B = {Hari yang bermula dengan huruf E} {Days that start with the letter E} ü Set yang terdiri daripada semua hari dalam satu minggu The set of all the days in a week = {Ahad, Isnin, Selasa, Rabu, Khamis, Jumaat, Sabtu} {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday} Maka / Hence, A ≠ φ dan / and B = φ. Mahir Diri (a) [P = {2, 3, 5, 7}] (b) (c) 4. (a) (i) 11  Q 16  Q (ii) E  R N  R (iii) Perak  Y Sabah  Y Y = {Sabah, Sarawak} (iii) A  Z E  Z Z = {P, U, T, R, A, J, Y}

© Penerbitan Pelangi Sdn. Bhd. 64 Matematik Tingkatan 1 Jawapan Mahir Diri (a) 6  {gandaan bagi 3} Benar 6 ∈ True {multiples of 3} (b) orkid  {bunga} Palsu orchid ∉ {flowers} False (c) 49  {kuasa dua sempurna} Benar 49 ∈ True {perfect squares} 6 = 3 × 2. Maka, 6  {gandaan bagi 3} Hence, 6  {multiples of 3} Orkid ialah sejenis bunga. Maka, orkid  {bunga} An orchid is a type of flower. Hence, orchid  {flower} 49 = 72 , maka 49  {kuasa dua sempurna} 49 = 72 , hence 49  {perfect squares} 5. Set Sets (b) b bukan unsur bagi A. b is not an element of A. Maka / Hence, b  A (c) a ialah unsur bagi A. a is an element of A. Maka / Hence, a  A (d) A ialah set kosong. A is an empty set. Maka / Hence, A = φ / { } (a) Bilangan unsur A Number of elements in A = n(A) 6. (a) (i) n(Q) = 6 (ii) n(R) = 3 (b) (i) n(Y) = 4 ; Y = {2, 4, 6, 8} (ii) n(Z) = 5; Z = {7, 11, 13, 17, 19} 7. (a) (i) P = {p, q, r, s} Q = {p, r, q, t} R = {a, e, i, o, u} S = {i, o, a, u, e} s  P tetapi / but s  Q. Maka / Therefore, P ≠ Q. Set R dan set S mempunyai unsur yang sama. Maka, R = S. Set R and set S have the same elements. Therefore, R = S ü (ii) P = {1, 2, 3, 4, 6, 12} Q = {1, 12, 2, 6, 3, 4} R = {6, 12, 18, 24} S = {3, 6, 9, 12} Set P dan set Q mempunyai unsur yang sama. Maka, P = Q. Set P and set Q have the same elements. Therefore, P = Q 18  R tetapi / but 18  S. Maka / Therefore, R ≠ S. (b) (i) P ialah set yang terdiri daripada bulan yang bermula dengan F. P is a set of the months that start with the letter F. Q ialah set yang terdiri daripada bulan yang mempunyai 28 hari. Q is a set of the months that have 28 days. ü R ialah set yang terdiri daripada bulan yang bermula dengan J. R is a set of the months that start with the letter J. S ialah set yang terdiri daripada bulan yang mempunyai 30 hari. S is a set of the months that have 30 days. P = {Februari}, Q = {Februari} Set P dan set Q mempunyai unsur yang sama. Maka, P = Q. P = {February}, Q = {February}, Set P and Set Q have the same elements. Therefore, P = Q. R = {Januari, Jun, Julai}, S = {April, Jun, September, November} R = {January, June, July} S = {April, June, September, November} Januari  R tetapi Januari  S. January  R but January  S. Maka / Therefore, R ≠ S. ü

65 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan (ii) P = {x : x ialah nombor perdana dan x , 10} P = {x : x is a prime number and x , 10} Q = {x : x ialah nombor ganjil dan x , 9} Q = {x : x is an odd number and x , 9} R = {x : x ialah nombor genap dan x , 10} R = {x : x is an even number and x , 10} S = {x : x ialah gandaan 2 dan x , 10} S = {x : x is a multiple of 2 and x , 10} ü P = {2, 3, 5, 7}, Q = {1, 3, 5, 7} 2  P tetapi / but 2  Q. Maka / Therefore, P ≠ Q. R = {2, 4, 6, 8}, S = {2, 4, 6, 8} Set R dan set S mempunyai unsur yang sama. Maka, R = S. Set R and set S have the same elements. Therefore R = S. 8. (a) (i) P = {a, b, c, d, e, f}, Q = {a, e, i} Bukan / No i (salah satu unsur set Q) bukan unsur bagi set P, maka P bukan set semesta. i (one of the elements of set Q) is not an element of set P, therefore P is not the universal set. (ii) P = {i, j, k, l, m, n, o}, Q = {j, k, l} Ya / Yes j, k dan l (semua unsur set Q) ialah unsur bagi set P, maka P mungkin ialah set semesta. j, k and l (all elements of set Q ) are the the elements of set P, therefore P is probably the universal set. 9. (a) ξ a i e P h f g d c b P = {b, c, d, f, g, h} (b) ξ 2 8 4 P 6 7 9 5 3 1 10 x = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} P = {2, 4, 6, 8, 10} P = {1, 3, 5, 7, 9} 10. (a) (i)  (ii)  (b) (i)  (ii)  {nombor kuasa dua sempurna} = {1, 4, 9, 16, 25, 36, 49, …}, 25 ialah unsur dalam set nombor kuasa dua sempurna. {perfect squares} = {1, 4, 9, 16, 25, 36, 49, …}. 25 is an element in the set of perfect squares. 11. (a) Subset yang mungkin ialah { }, {1}. The possible subsets are { }, {1} (b) Subset yang mungkin ialah { }, {p}, {q}, {p, q}. The possible subsets are { }, {p}, {q}, {p, q} (c) Subset yang mungkin ialah { }, {2}, {4}, {6}, {2, 4}, {2, 6}, {4, 6}, {2, 4, 6}. The possible subsets are { }, {2}, {4}, {6}, {2, 4}, {2, 6}, {4, 6}, {2, 4, 6} 13. (a) P Q a e c f b d (b) P Q 1 6 5 3 2 4

© Penerbitan Pelangi Sdn. Bhd. 66 Matematik Tingkatan 1 Jawapan 12. (a) (i) P C Zul Kassim Kai Jian Kamal Billy B Elin Azlina Din Mei Mei Anita Chon Yao (ii) Set B ialah set semesta. Set B is the universal set. (iii) P  C C  B Praktis Masteri 11 Bahagian A 1. Jawapan / Answer: C 2.  = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11} P′ = {3, 5, 7, 9, 11} n(P′) = 5 Q′ = {3, 4, 5, 6, 7, 8, 9, 11} Subset Q = {2}, {10}, {2, 10} dan / and { }. Jawapan / Answer: A 3. P = {2, 4, 6} n(P) = 3 Jawapan / Answer: B 4.  = {kuning, hitam, hijau, ungu, merah, biru} {yellow, black, green, purple, red, blue} Q′ = {kuning, hitam, hijau, ungu} {yellow, black, green, purple} Subset Q = {merah / red}, {biru / blue}, {merah / red, biru / blue}, { } Jawapan / Answer: D 5. Jawapan / Answer: B Bahagian B 1. (a) (i)  (ii)  (b) (i) (ii) Bahagian C 1. (a) Peti sejuk Refrigerator • • Alat tulis Stationery Kertas Paper • Televisyen Television • • Peralatan elektrik Electrical appliance Dakwat Ink • (b) Zulkhairi dan / and Khairul (c) (i)  = {11, 12, 13, 14, 15, 16, 17, 18, 19, 20}  ={x : x ialah integer dan 11 < x < 20} ξ = {x : x is an integer and 11 < x < 20} G = {11, 13, 15, 17, 19} G = {x : x ialah nombor ganjil dan 11 < x < 20} G = {x : x is an odd number and 11 < x < 20} P = {11, 13, 17, 19} P = {x : x ialah nombor perdana dan 11 < x < 20} P = {x : x is a prime number and 11 < x < 20} (ii) G = {12, 14, 16, 18, 20}

67 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan Pengendalian Data Data Handling BAB 12 1. Kaedah Pengumpulan Data / Data Collection Method Temu bual / Interview Tinjauan / Survey Pemerhatian / Observation Eksperimen / Experiment 2. (a) Warna kegemaran murid Favourite colour of students (b) Jantina bayi yang dilahir Gender of babies (c) Tinggi pokok mangga di sebuah dusun Height of mango trees in an orchard (d) Jarak antara rumah dan sekolah Distance between house and school 3. (a) Integer Integer Gundalan Tally Kekerapan Frequency 6 ||| 3 7 || 2 8 |||| | 6 9 |||| 4 (b) Saiz Size Gundalan Tally Kekerapan Frequency S ||| 3 M |||| || 7 L |||| 5 XL ||| 3 (c) Gred Grade Gundalan Tally Kekerapan Frequency A |||| 5 B |||| 4 C |||| 4 D || 2 E | 1 4. Perwakilan Data Data Representation (b) Carta palang Bar chart (a) Carta pai Pie chart (c) Graf garis Line graph (d) Histogram (e) Plot titik Dot plot (f) Poligon kekerapan Frequency polygon Plot batangdan-daun Stem-and-leaf plot 5. (a) Jualan Kereta oleh Zhi Ming Sales of Cars by Zhi Ming Bulan / Month Bilangan kereta Number of cars Januari January 4 0 8 12 16 Februari February Mac March April April Mei May

© Penerbitan Pelangi Sdn. Bhd. 68 Matematik Tingkatan 1 Jawapan Bilangan kereta / Number of cars Bulan / Month Januari January Februari February Mac March April April Mei May 0 2 4 6 8 10 12 14 16 Jualan Kereta oleh Zhi Ming Sales of Cars by Zhi Ming Mahir Diri Bilangan Murid Tingkatan 1A Mengikut Rumah Sukan Number of Students in Form 1A According to Sport Teams Bilangan murid Number of students 2 0 4 6 8 10 Rumah sukan / Sport team Merah Red Kuning Yellow Hijau Green Biru Blue Bilangan murid rumah sukan merah The number of students in red sport team = 33 – 8 – 7 – 9 = 9 6. (a) Jualan Telur Gred A dan Gred B Sales of Grade A and Grade B Eggs Telur Gred A Grade A Egg Telur Gred B Grade B Egg Kuantiti telur (kotak) Quantity of eggs (boxes) 2 0 4 6 8 10 12 Kedai / Shop P Q R S 7. (a) Aiskrim Ice cream Bilangan murid Number of students Sudut sektor Angle of sector Coklat Chocolate 338 338 936 × 360° = 130° Vanila Vanilla 234 234 936 × 360° = 90° Oren Orange 221 221 936 × 360° = 85° Mangga Mango 143 143 936 × 360° = 55° Jumlah Total 936 360° Aiskrim Kegemaran bagi Sekumpulan Murid The favourite Ice cream of a Group of Students Vanila Vanilla Coklat Chocolate Mangga Mango Oren Orange 55° 85° 130° 8. (a) Tahun / Year Bilangan pekerja Number of workers 50 0 100 150 200 2012 2013 2014 2015 2016 Bilangan Pekerja di Syarikat Khidmat Maju Number of Workers in Syarikat Khidmat Maju

69 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan (b) Hari / Day Amaun (RM) Amount (RM) 500 0 1 000 1 500 2 000 Isnin Monday Selasa Tuesday Rabu Wednesday Khamis Thursday Jumaat Friday Hasil Jualan Restoran Ainin Sales of Restaurant Ainin 9. (a) 4 5 6 7 8 Bilangan Buku di dalam Beg Number of Books in Bag Bilangan buku / Number of books (b) 2.0 2.1 2.2 Masa (minit) / Time (Minutes) 2.3 2.4 Masa untuk Menyiapkan Satu Soalan Matematik Time to Solve a Mathematics Problem 10. (a) Kekunci: 5 | 4 bermakna 54 markah Key : 5 I 4 means 54 marks Markah Ujian Sejarah History Test Marks Batang / Stem Daun / Leaf 5 4 6 8 9 9 7 1 4 2 0 1 5 8 4 6 4 5 9 8 5 Markah Ujian Sejarah History Test Marks Batang / Stem Daun / Leaf 5 4 6 8 9 9 7 0 1 1 2 4 5 8 4 4 5 6 9 5 8 (b) Kekunci: 3 | 8 bermakna 38 kg Key : 3 I 8 means 38 kg Jisim Murid Mass of Students Batang / Stem Daun / Leaf 3 8 9 9 4 4 7 1 0 7 8 5 1 2 2 6 5 6 6 6 5 7 2 Jisim Murid Mass of Students Batang / Stem Daun / Leaf 3 8 9 9 4 0 1 4 7 7 8 5 1 2 2 5 6 6 6 5 6 7 2 11. (a) Kelas / Class Bilangan murid Number of students 10 0 20 30 40 1A 1B 1C 1D 1E Bilangan Murid di Lima Kelas Number of Students in Five Classes

© Penerbitan Pelangi Sdn. Bhd. 70 Matematik Tingkatan 1 Jawapan (b) Kawasan perumahan Residential area Bilangan murid Number of students P 45° 360° × 240 = 30 Q 90° 360° × 240 = 60 R 120° 360° × 240 = 80 S 75° 360° × 240 = 50 T 30° 360° × 240 = 20 (c) Markah Ujian Matematik bagi 15 orang Murid Mathematics Test Marks for 15 Students Markah / Marks 40 50 60 70 80 90 100 Mahir Diri (i) Bahasa Melayu / Malay language (ii) Mata pelajaran Subject Tinggi palang (bilangan petak) Bar Height (number of square) Sudut sektor Angle of sector Bahasa Melayu Malay language 9 9 9 + 5 + 6 × 360° = 162° Bahasa Inggeris English 5 90° Matematik Mathematics 6 6 9 + 5 + 6 × 360° = 108° Bahasa Melayu Malay language Bahasa Inggeris English Matematik Mathematics Mata Pelajaran Kegemaran Sekumpulan Murid Favourite Subjects of a Group of Students 108° 162° 12. (a) (i) Beza antara bilangan murid Difference in the number of students = 2 000 – 1 750 = 250 (ii) Tahun 2015 / Year 2015 (iii) Bilangan murid perempuan pada tahun 2016 The number of girls in 2016 = 55 100 × 2 000 = 1 100 (iv) 2 250 (a) (i) Peserta paling muda berumur 31 tahun. Peserta paling tua berumur 63 tahun. Maka, lingkungan umur peserta ialah (31 – 63) tahun. The youngest participant is 31 years old. The oldest participant is 63 years old. Therefore, the range of the participants’ age is (31 – 63) years old. (ii) Bilangan peserta yang umurnya 50 tahun dan ke atas The number of participants whose age is 50 years old and above = 8 + 3 = 11 Kawasan perumahan / Residential area Bilangan murid Number of students 20 0 40 60 80 P Q R S T Kedudukan Rumah bagi 240 orang Murid House Location of 240 Students

71 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan (iii) Beza antara bilangan peserta yang umurnya 40 – 49 dengan 50 – 59 Difference between the number of participants in the age group of 40 – 49 with 50 – 59 = 10 – 8 = 2 (c) (i) Sudut sektor untuk pinjaman kereta The angle of sector for car instalment = 360° – 125° – 100° – 40° – 50° = 45° Perbelanjaan terbesar keluarga Kamil adalah untuk pinjaman rumah (Sudut sektor untuk pinjaman rumah adalah paling besar) The largest expenditure for Kamil’s family is the house instalment (The angle of sector of the house instalment is the largest) (ii) Pecahan perbelanjaan untuk simpanan Fraction of the expenditure on savings = 40° 360° = 1 9 (iii) Jumlah peratusan perbelanjaan untuk pinjaman rumah dan pinjaman kereta Total percentage of expenditure on house instalment and car instalment. = 125° + 45° 360° × 100% = 47.22% (d) (i) Bungkusan terberat = 1.6 kg dan bungkusan teringan = 1.1 kg The heaviest parcel = 1.6 kg and the lightest parcels = 1.1 kg Maka, beza jisim antara bungkusan yang terberat dengan bungkusan yang teringan Thus, the difference in mass between the heaviest and the lightest parcel = 1.6 kg – 1.1 kg = 0.5 kg (ii) 40% daripada jumlah bilangan bungkusan 40% of the total number of parcels = 40 100 × (1 + 3 + 8 + 7 + 1) = 8 Daripada plot titik, jumlah bilangan bungkusan yang berjisim 1.5 kg dan 1.6 kg ialah 8. From the dot plot, the total number of parcels with the mass of 1.5 kg and 1.6 kg is 8. Maka / Thus, x = 1.4. (iii) Kebanyakan bungkusan yang diterima adalah berjisim 1.4 kg dan 1.5 kg. Most of the parcels received have a mass of 1.4 kg and 1.5 kg. 13. (a) (i) 35 biji mentol / light bulbs (ii) Bilangan mentol yang tempoh hayatnya kurang daripada 400 jam Number of light bulbs with a lifespan of less than 400 hours = 15 + 35 = 50 Bilangan mentol yang tempoh hayatnya melebihi 549 jam Number of light bulbs with a lifespan of more than 549 hours = 10 Beza antara bilangan mentol The difference in number of light bulbs = 50 – 10 = 40 (iii) Kebanyakan mentol mempunyai tempoh hayat 350 jam hingga 549 jam. Most of the light bulbs have a lifespan between 350 hours to 549 hours. (b) (i) Jumlah bilangan murid / Total number of students = 5 + 25 + 30 + 26 + 15 + 17 + 10 = 128 (ii) Bilangan murid yang wang sakunya melebihi RM3.50 The number of students whose pocket money is more than RM3.50 = 17 + 10 = 27 Peratusan murid yang wang sakunya melebihi RM3.50 Percentage of students whose pocket money is more than RM3.50 = 27 128 × 100% = 21.09% (iii) Kebanyakan murid mempunyai wang saku harian RM1.60 hingga RM3.00. Most of the students have daily pocket money between RM1.60 to RM3.00.

© Penerbitan Pelangi Sdn. Bhd. 72 Matematik Tingkatan 1 Jawapan Praktis Masteri 12 Bahagian A 1. Jumlah jualan ikan siakap di pasar A The total sale of sea bass at market A = RM780 – RM13(25) – RM13(15) = RM260 Bilangan ikan yang dijual The number of fish sold = RM260 ÷ RM13 = 20 Jawapan / Answer: A 2. Sudut yang diwakili pada carta pai The angle represents in pie chart = 30 40 + 50 + 30 + 50 × 360° = 30 170 × 360° = 63.5° Jawapan / Answer: B 3. Peratus kanak-kanak yang tidak dikenakan bayaran The percentage of kids who are not charged = 8 8 + 4 + 5 + 3 × 100% = 8 20 × 100% = 40% Jawapan / Answer: B Bahagian B 1. (a) Data yang dikumpulkan merupakan “Melayu, Cina, India dan sebagainya”. Maka, data yang dikumpulkan itu merupakan data kategori. The data collected are "Malay, Chinese, Indian and etcetera". Therefore, the data collected is categorical data. (b) Bilangan anak dalam beberapa buah keluarga The number of children in several families 1 2 3 4 5 • • • • • • • • Bilangan anak / Number of children Bahagian C 1. (a) Warna reben Colour of ribbons Sudut sektor Angle of sector Peratus Percentage Merah Red 72° 20% Kuning Yellow 162° 162° 360° × 100% = 45% Hijau Green 36° 10% Biru Blue 90° 90° 360° × 100% = 25% (b) Jumlah pelajar / Total number of students = 6 + 8 + 9 + 9 = 32 Peratus dalam kategori tersebut Percentage in the category = 15 32 × 100% = 46.88% Hari / Day Bilangan oren Number of oranges 0 100 200 300 400 500 Bilangan Oren yang Dijual oleh Puan Tasha Number of Oranges Sold by Puan Tasha Isnin Monday Selasa Tuesday Rabu Wednesday Khamis Thursday Jumaat Friday Bilangan oren yang dijual pada hari Rabu The number of oranges sold on Wednesday = 1 750 – 400 – 250 – 450 – 150 = 500 (ii) Hari Jumaat / Friday

73 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan (b) 5 cm 12 cm x cm Hipotenus Hypotenuse x2 = 52 + 122 = 25 + 144 = 169 x = 169 = 13 (c) 20 cm x cm 21 cm Hipotenus Hypotenuse x2 = 202 + 212 = 400 + 441 = 841 x = 841 = 29 4. (a) 8 cm x cm 10 cm Hipotenus Hypotenuse 102 = x2 + 82 x2 = 102 – 82 = 100 – 64 = 36 x = 36 = 6 (b) 30 cm x cm 34 cm Hipotenus Hypotenuse 342 = x2 + 302 x2 = 342 – 302 = 1 156 – 900 = 256 x = 256 = 16 (c) 41 cm x cm 40 cm Hipotenus Hypotenuse 412 = x2 + 402 x2 = 412 – 402 = 1 681 – 1 600 = 81 x = 81 = 9 5. (a) Dalam / In DBCD, 102 = BC2 + 62 BC2 = 102 – 62 = 100 – 36 = 64 BC = 64 = 8 Dalam / In DABC, x2 = 152 + 82 = 225 + 64 = 289 x = 289 = 17 Teorem Pythagoras The Pythagoras Theorem BAB 13 1. (a) x y z x (b) B A C AC 2. (a) x2 = y2 + x2 (b) AC2 = AB2 + BC2 Mahir Diri u v w Hipotenus ialah v. The hypotenuse is v. u w Hipotenus ialah w. The hypotenuse is w. v u w Hipotenus ialah w. The hypotenuse is w. v A B C Diberi / Given u2 = w2 – v2 . Maka, w ialah sisi terpanjang dalam segi tiga yang dilukis oleh Nadia, iaitu hipotenus ialah w. Therefore, w is the longest side in the triangle drawn by Nadia, i.e. the hipotenuse is w Antara segi tiga A, B dan C, hanya hipotenus segi tiga C ialah w. Jawapan ialah segi tiga C. Between triangles A, B and C only the hypotenuse of triangle C is w. The asnwer is triangle C. 3. (a) 3 cm 4 cm x cm Hipotenus Hypotenuse x ialah hipotenus x is the hypotenuse x2 = 3 2 + 4 2 = 9 + 16 = 25 x = 25 = 5

© Penerbitan Pelangi Sdn. Bhd. 74 Matematik Tingkatan 1 Jawapan (b) Dalam / In DQST, 52 = QS2 + 32 QS2 = 52 – 32 = 25 – 9 = 16 QS = 16 = 4 QR = 2 × 4 = 8 cm Dalam / In DPQR, 172 = x2 + 82 x2 = 172 – 82 = 289 – 64 = 225 x = 225 = 15 6. (a) (i) BE = CD = 8 m Dalam / In ∆ABE, 172 = AE2 + 82 AE2 = 172 – 82 = 289 – 64 = 225 AE = 225 = 15 m Panjang pagar / Length of the fence = Perimeter trapezium ABCD Perimeter of trapeium ABCD = AB + BC + CD + DE + EA DE = BC = AB + BC + CD + BC + EA = 17 m + 4 m + 8 m + 4 m + 15 m = 48 m (ii) Luas tanah / Area of the land = 1 2 × (BC + AD) × CD = 1 2 × [4 + (4 + 15)] × 8 = 1 2 × 23 × 8 = 92 m2 (b) (i) Dalam / In DPTS, PS2 = 82 + 152 = 64 + 225 = 289 PS = 289 = 17 cm ST = RS = QP = 8 cm RQ = SP = 17 cm Perimeter seluruh rajah Perimeter of the whole diagram = PQ + QR + RS + ST + TP = 8 cm + 17 cm + 8 cm + 8 cm + 15 cm = 56 cm (ii) Luas segi empat selari PQRS Area of the parallelogram PQRS = RS × PT = 8 × 15 = 120 cm2 (c) (i) PQ = 289 = 17 cm PQ = QR = RS = SP = 17 cm PQRS ialah segi empat sama. PQRS is a square Dalam / In DRTS, 172 = RT2 + 152 RT2 = 172 – 152 = 289 – 225 = 64 RT = 64 = 8 cm Perimeter kadbod yang tinggal Perimeter of the remaining cardboard = PQ + QR + RT + TS + SP = 17 cm + 17 cm + 8 cm + 15 cm + 17 cm = 74 cm (ii) Luas kadbod yang tinggal Area of the remaining cardboard = Luas segi empat sama PQRS – Luas DRTS Area of square PQRS – Area of DRTS = 289 – 1 2 × 8 × 15 = 289 – 60 = 229 cm2 7. Dalam / In ∆PQR, PR2 = 202 + 152 = 400 + 225 = 625 PR = 625 = 25 cm 1 2 × PQ × QR = 1 2 × PR × QS 1 2 × 15 × 20 = 1 2 × 25 × QS 300 = 25 × QS QS = 12 cm

75 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan Dalam / In ∆PSQ, 152 = PS2 + 122 PS2 = 152 – 122 = 225 – 144 = 81 PS = 81 = 9 cm Mahir Diri (i) Luas / Area ∆STU = 60 1 2 × 8 × TU = 60 4TU = 60 TU = 15 cm PV = 8 cm + 7 cm = 15 cm Dalam / In ∆STU, SU2 = 82 + 152 = 64 + 225 = 289 SU = 289 = 17 cm Dalam / In ∆PVQ, 252 = QV2 + 152 QV2 = 252 – 152 = 625 – 225 = 400 QV = 400 = 20 cm PU = 50 – 20 – 15 = 15 cm Perimeter = PQ + QR + RS + SU + UP = 25 + 50 + 7 + 17 + 15 = 114 cm (ii) Luas kawasan yang berlorek Area of the shaded region = Luas trapezium – Luas segi tiga Area of trapezium – Area of triangle = 1 2 × (50 + 30) × 15 – 60 = 600 – 60 = 540 cm2 8. (a) 13 cm, 5 cm, 12 cm ü Panjang sisi terpanjang ialah 13 cm. The length of the longest side is 13 cm. 132 = 169 52 + 122 = 25 + 144 Hasil tambah kuasa dua bagi dua sisi yang lain Sum of square of the other two sides = 169 \ 132 = 52 + 122 Maka, segi tiga itu ialah segi tiga bersudut tegak. Therefore, the triangle is a right-angled triangle. (b) 8 cm, 16 cm, 17 cm Panjang sisi terpanjang ialah 17 cm. The length of the longest side is 17 cm. 172 = 289 82 + 162 = 64 + 256 Hasil tambah kuasa dua bagi dua sisi yang lain Sum of square of the other two sides = 320  289 \ 172 ≠ 82 + 162 Maka, segi tiga itu bukan segi tiga bersudut tegak. Therefore, the triangle is not a right-angled triangle. 9. (a) (i) Perimeter = 70 cm Maka / Thus, QR + RS + SP + PT + TQ= 70 20 + 13 + 20 + PT + 12 = 70 PT + 65 = 70 PT = 70 – 65 = 5 cm (ii) PQ = 13 cm PQ2 = 132 Kuasa dua sisi terpanjang Square of the longest side = 169 PT2 + TQ2 = 52 + 122 Hasil tambah kuasa dua bagi dua sisi yang lain Sum of square of the other two sides = 25 + 144 = 169 \ PQ2 = PT2 + TQ2 Maka, segi tiga PTQ ialah segi tiga bersudut tegak. Therefore, triangle PTQ is a right-angled triangle Praktis Masteri 13 Bahagian A 1. GH = 252 − 202 = 225 = 15 cm Kuasa dua sisi terpanjang Square of the longest side Kuasa dua sisi terpanjang Square of the longest side

© Penerbitan Pelangi Sdn. Bhd. 76 Matematik Tingkatan 1 Jawapan EJH = 212 + (15 + 5)2 = 212 + 202 = 841 = 29 cm Jawapan / Answer: D 2. RST = 302 − 242 = 324 = 18 cm RT = 18 ÷ 2 = 9 cm SU = 92 + 242 = 657 = 25.63 cm Jawapan / Answer: C 3. EF = 72 + 242 = 625 = 25 cm EK= 52 + 122 = 169 = 13 cm JK = 25 − 13 = 12 cm Jawapan / Answer: A 4. EF = 900 = 30 cm EK= 30 –6 = 24 cm JK = 72 + 242 = 625 = 25 cm Jawapan / Answer: B Bahagian B 1. (a) (i) Ya / Yes 92 + 402 = 1 681 QR = 1 681 = 41 (ii) Ya / Yes 202 + 212 = 841 PQ = 841 = 29 (b) (i) (ii) Bahagian C 1. (a) (i) PR (ii) PR ialah sisi yang bertentangan dengan sudut tegak dalam segi tiga PQR. PR is the opposite side to the right angle of triangle PQR. (b) D A B C E 8 cm 17 cm 24 cm Dalam / In ∆ACD, AC2 = 172 – 82 = 225 AC = 225 = 15 cm Panjang BE / Length of BE = 24 cm ÷ 2 = 12 cm Tinggi ∆ABC / Height of ΔABC, AE2 = 152 – 122 = 81 AE = 81 = 9 cm Luas / Area = 12 × 24 × 9 + 12 × 8 × 15 = 108 + 60 = 168 cm2 (c) (i) QS= 7.5 + 7.5 = 15 cm Luas PSQ / Area of PSQ = 60 12 × 15 × PS = 60 PS = 60 × 2 15 = 8 cm PQ2 = 152 + 82 = 225 + 64 = 289 PQ = 289 = 17 cm Perimeter / Perimeter = PQ + QR + RU + UT + TP = 17 + 7.5 + 3 + 7.5 + (8 – 3) = 40 cm (ii) Luas / Area= 60 – 7.5 × 3 = 60 – 22.5 = 37.5 cm2

77 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan Ujian Akhir Sesi Akademik (UASA) Bahagian A / Section A 1. Jawapan / Answer: B 2. Faktor bagi 39/ Factor of 39 = 1, 3, 13, 39 Jawapan / Answer: C 3. GSTK bagi 15 dan 30 LCM of 15 and 20 2 15 20 2 15 10 3 15 5 5 5 5 1 1 = 2 × 2 × 3 × 5 = 60 Jawapan / Answer: C 4. (– 0.5)3 = – 0.125 = – 1 8 Jawapan / Answer: D 5. Panjang kotak hadiah / The length of the gift box = 196 = 14 cm Panjang reben yang diperlukan untuk 4 kotak hadiah The length of the ribbon needed for 4 gift boxes = 14 × 8 = 112 cm Jawapan / Answer: A 6. 18 : 24 = 18 6 : 24 6 = 3 : 4 Jawapan / Answer: A 7. Bilangan bakul diperlukan The number of basket needed = 174 ÷ 29 = 6 Jawapan / Answer: C 8. 6rt2 × 5rs2 3r2 s = 6 × 5 3 r1 + 1 − 2 s2 − 1 t 2 = 10st2 Jawapan / Answer: A 9. Pensel yang dimiliki oleh Ammar = p Pencil owned by Ammar Pensel yang dimiliki oleh Aqil = p − 3 Pencil owned by Aqil Jumlah pensel yang dimiliki: The total number of pencils p + p − 3 = 11 2p − 3 = 11 Jawapan / Answer: C 10. Diberi / Given 2x + y = 12……… x 4 + y 2 = 3 ……… Daripada / From , y = 12 − 2x ……… Gantikan / Substitute  dalam / into  x 4 + 12 – 2x 2 = 3 x + 2(12 − 2x) = 12 −3x = −12 x = 4 x = 12−2(4) = 4 Jawapan / Answer: D 11. Jawapan / Answer: C 12. y + 30° = 90° y = 60° 2x + x + 60° + 45° + 30° = 360° 3x = 225° x = 75° Jawapan / Answer: A 13. x = 180° − 180° − 180° – 2x 2 = 180° − 28° = 152° Jawapan / Answer: D 14. Bilangan pepenjuru bagi oktagon The number of diagonals of octagon = 8(8 − 3)2 = 20 Jawapan / Answer: C

© Penerbitan Pelangi Sdn. Bhd. 78 Matematik Tingkatan 1 Jawapan 15. Panjang sisi segi empat sama The length of a side of the square = 36 = 6 cm Perimeter = 6 × 8 = 48 cm Jawapan / Answer: B 16. Jawapan / Answer: D 17. P = {12, 15, 18, 21, 24, 27, 30, 33} Jawapan / Answer: B 18. Beza jumlah markah Bahasa Melayu dan jumlah markah Bahasa Inggeris The difference between the total of Bahasa Melayu score and the total of English score = (40 + 60 + 50) − (50 + 40 + 50) = 10 Jawapan / Answer: B 19. Bilangan murid yang menggemari genre pop The number of students who like the pop genre = 360° − 72° − 90° − 30° 360° × 300 = 168° 360° × 300 = 140 Jawapan / Answer: A 20. Tinggi segi tiga sama kaki The height of the isosceles triangle = 132 − 52 = 144 = 12 cm Jawapan / Answer: A Bahagian B / Section B 1. (a) – 3 4 –0.54 3.0 22 7 0 –353 1 2 3 5.5 (b) 1 34 4 3 2 18 11 9 36 2. 64 4 2 4 2 2 2 2 16 2 4 Kesimpulan / Conclusion: 64 ialah kuasa dua sempurna / 64 is a perfect square. 3. (a) 7 : 5 : 13 = 21 : 15 : 39 = 10.5 : 7.5 : 19.5 (b) 4. (a) (i) (ii) (b) (i) 3 4 p + 1 5 = 3 (ii) −7r + 12 = 3 + 5r 5. (a) (i) 135° (ii) 308° 6 10 33 100 33 1 000 0.32 0.33 81 100

79 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan (b) Syiling Coins Gundalan Tally Kekerapan Frequency 20 sen   6 50 sen   7 Bahagian C / Section C 1. (a) (i) p2 q2 atau / or (pq)2 (ii) 6a2 b × 4bc ÷ (–3ab) = 2 6a2 b × 4bc –3ab = –8abc (b) 72 + 15 = 49 + 15 = 64 = 8 (c) Pengiraan / Calculation = 1.2 + 3 4 + (–2.1) = 1.2 + 0.75 – 2.1 = –0.51 juta / million (Syarikat mengalami kerugian dalam tiga bulan itu.) (The company suffered a loss in the three months.) (d) 2 4 7 3 2 2 7 3 7 1 7 3 3 1 1 3 1 1 1 Gandaan sepunya terkecil (GSTK) Lowest common multiple (LCM) = 2 × 2 × 7 × 3 = 84 Maka, mereka akan bersenam bersama pada hari ke-84. Therefore, they will be exercise together on 84th day. 2. (a) (i) 3p – 2q + 6 = 3(4) – 2(–3) + 6 = 12 + 6 + 6 = 24 (ii) pq + (–2p) + 4q 3 = 4(–3) + [–2(4)] + 4(–3) 3 = –12 – 8 – 4 = –24 (b) x = 40° y = 40° + 90° = 130° (c) y – x = 10 y = 10 + x ……  2x – y 2 = 15 2x – y = 30 ……  Gantikan  ke dalam  / Substitute  into  2x – (10 + x) = 30 2x – 10 – x = 30 2x – x = 30 + 10 x = 40 ..….  Gantikan  ke dalam  / Substitute  into  y = 10 + 40 = 50 3. (a) Integer terbesar / Largest integer : 4, Integer kedua terbesar / Second largest integer : –3 Hasil darab / Product = 4 × (–3) = –12 (b) 4 cm P M O N (c) Burung Bird Pemandu Driver 65° 80° 40° 82° Mujahid x y z y = (90° – 40°) + (90° – 82°) = 58° z = 80° – 65° = 15° x = 180° – 58° – 15° = 107°

© Penerbitan Pelangi Sdn. Bhd. 80 Matematik Tingkatan 1 Jawapan 4. (a) –5 – 3x < 13 –3s < 13 + 5 –s < 18 3 –s < 6 s > –6 (b) (i) Set P = {2, 7, 9, 13} Set Q = {4, 5, 7, 9, 13} (ii) n(P) = 2 (iii) Q  P (c) Panjang sisi segi empat sama The length of the side of a square =  50 2 =  25 = 5 m Luas trapezium / Area of trapezium = 1 2 × (15 + 10) × (7 + 5) = 150 m2 Luas kawasan berlorek / Area of shaded region = 150 – 50 = 100 m2 5. (a) Kuasa dua sisi terpanjang Square of the longest side = 132 = 169 Hasil tambah kuasa dua bagi dua sisi yang lain Sum of the squares of the other two sides = 62 + 122 = 36 + 144 = 180 ≠ 169 Maka, segi tiga tersebut bukan segi tiga bersudut tegak. Therefore, the triangle is not a right-angled triangle. (b) (i) Panjang satu tali / Length of one rope =  62 + 82 =  100 = 10 m Jumlah panjang tali yang diperlukan Total length of rope required = 4 × 10 = 40 m (ii) x = 180° – 53° = 27° (c) Luas segi empat selari / Area of the parallelogram = 9 × 10 = 90 m2 Luas kawasan cili / Area of chili region = 1 2 ×  2 3 × 9 × 10 = 30 m2 Luas kawasan jagung / Area of corn region = 90 – 30 = 60 m2 6. (a) (i) 1 000 0 2 000 3 000 4 000 5 000 Jualan (RM) Sales (RM) Produk / Product A B C D E Jumlah jualan produk oleh Syarikat Maju Total product sakes by Syarikat Maju Jumlah jualan A, B, C dan D Total sales of A, B, C and D = RM2 500 + RM3 500 + RM2 000 + RM4 000 = RM12 000 Julan produk E / Sales of product E = 12.5 100 × 12 000 = RM1 500 (ii) Beza hasil julan / Difference of sales = RM4 000 – RM1 500 = RM2 500 (b) (i) Markah terendah / Lowest mark = 23 Markah tertinggi / Highest mark = 89

81 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan (ii) Peratusan pelajar yang gagal Percentage of student who failed = Bilangan pelajar gagal Number of students who failed Jumlah pelajar Total number of students × 100% = 7 30 × 100% = 23.33%

© Penerbitan Pelangi Sdn. Bhd. 82 Matematik Tingkatan 1 Jawapan Ujian Pertengahan Sesi Akademik (UPSA) Bahagian A / Section A 1. A: –9 × (–10) + 5 – (–2) = 97 B: –9 + (–10) × 5 – (–2) = –57 C: –9 – (–10) + 5 × (–2) = –9 D: –9 × (–10) – 5 + (–2) = 83 Jawapan / Answer: A 2. – 7 10 , 1 10, 3 5 , – 2 5 – 2 5 = – 4 10 3 5 = 6 10 Tertib menurun / Descending order: 3 5 , 1 10, – 2 5 , – 7 10 Jawapan / Answer: A 3. P = 3.779 + [(3.784 – 3.779) ÷ 5 × 2] = 3.779 + (0.005 ÷ 5 × 2) = 3.779 + 0.002 = 3.781 Jawapan / Answer: C 4. A: 2.2 = 2 1 5 B: 0.35 = 7 20 C: 0.28 = 7 25 D: 0.15 = 3 20 Jawapan / Answer: C 5. 2 24 48 60 2 12 24 30 3 6 12 15 2 4 5 Faktor sepunya terbesar Highest common factor = 2 × 2 × 3 = 12 Jawapan / Answer: D 6. 3 15 45 5 5 15 3 1 3 1 1 Gandaan sepunya terkecil Highest common factor = 3 × 5 × 3 = 45 15, 30, 45, 60, 75, 90, … 45, 90, … Jawapan / Answer: B 7. 3 15 24 30 5 5 8 10 2 1 8 2 4 1 4 1 1 1 1 Gandaan sepunya terkecil Lowest common multiple = 3 × 5 × 2 × 4 = 120 Jawapan / Answer: A 8. Jawapan / Answer: A 9. A: 1 49 = 1 7 B: 3 1 343 = 1 7 C: 2 14 ≠ 1 7 D: 1 + 3 216 343 = 1 – 6 7 = 1 7 Jawapan / Answer: C 10. (3  p –  4)2 = 4 3  p –  4 =  4 3  p – 2 = 2 3  p = 2 + 2 = 4 p = 43 = 64 Jawapan / Answer: D

83 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan 11. p : q = 6 : 5 = 12 : 10 q : r = 10 : 3 p : q : r = 12 : 10 : 3 Jawapan / Answer: A 12. m : n = 3 : 5 3m – n : 4m = 3(3) – 5 : 4(3) = 9 – 5 : 12 = 4 : 12 = 1 : 3 Jawapan / Answer: B 13. Jumlah dua bahagian yang lebih kecil The sum of two smaller parts = 2 + 3 = 5 3 bahagian / parts → RM90 1 bahagian / part → RM90 ÷ 3 = RM30 Nilai bagi bahagian yang paling kecil The value of the smallest part = 2 × RM30 = RM60 Jawapan / Answer: C 14. Nilai bagi 1 bahagian / The value of 1 part = RM640 ÷ (1 + 4 + 3) = RM640 ÷ 8 = RM80 Murid / Student K = RM80 Murid / Student I = RM80 × 4 = RM320 Murid / Student J = RM80 × 3 = RM240 Jawapan / Answer: D 15. Jawapan / Answer: C 16. (3p – q) – (p – 4q) = 3p – q – p + 4q = 2p + 3q Jawapan / Answer: B 17. 2gh × 5hr = 10gh2 r Jawapan / Answer: D 18. 2(x + 1) = 4 2x + 2 = 4 2x = 4 – 2 = 2 x = 1 Jawapan / Answer: C 19. 4m + 2n = 6 ........  2m – 4n = 18 ........   × 2: 8m + 4n = 12 ........   + : 2m + 8m – 4n + 4n = 18 + 12 10m = 30 m = 3 Jawapan / Answer: C 20. –2h – (–4h) = –2h + 4h = 2h Jawapan / Answer: C Bahagian B / Section B 1. (a) 1 4 (b) 1 1 2 = 1.5 (c) 2 (d) 1.5 – 1 4 = 1 1 4 atau / or 5 4 atau / or 1.25 2. (a) S = 4 1 4 (b) 3. (a) 18 ÷ 2 = 9 Faktor perdana bagi 72 ialah 2 dan 3. The prime factors of 72 are 2 and 3.

© Penerbitan Pelangi Sdn. Bhd. 84 Matematik Tingkatan 1 Jawapan (b) (i) 42 ÷ 3 = 14 42 ÷ 4 = 10.5 Palsu / False (ii) 136 ÷ 4 = 34 136 ÷ 8 = 17 136 ÷ 17 = 8 Benar / True 4. 3–161 64 ÷ 25 4  2 = 3– 125 64 ÷ 5 2  2 =  – 5 4 × 2 5  2 = 1 4 5. (a) (i) 32 : 40 = 32 × 1.5 : 40 × 1.5 = 48 : 60 () (ii) 10 : 26 = 10 × 2.5 : 26 × 2.5 = 25 : 65 () (b) (i) RM1.20 ÷ 3 × 8 = RM3.20 (ii) RM4.40 ÷ 8 × 3 = RM1.65 = RM3.20 Bahagian C / Section C 1. (a) (i) Penurunan lif 8 tingkat The lift going down 8 floors = –8 (ii) Kedudukan penyu 5 m di bawah aras laut The position of a turtle at 5 m below sea level = –5 (iii) Kerugian sebanyak RM3 000 The loss of RM3 000 = –3 000 (b) –0.5 ×  2 1 2 – 1.4 = – 1 2 ×  5 2 – 7 5  = – 1 2 ×  25 10 – 14 10  = – 1 2 × 11 10 = – 11 20 (c) (i) 6 1 5 – 4 1 2 = 31 5 – 9 2 = 62 10 – 45 10 = 17 10 = 1 7 10 m (ii) –6 1 5 + 45.6 = –6.2 + 45.6 = 39.4 m 39.4 m di atas aras laut. 39.4 m above sea level. 2. (a) (i) a = 3 ÷ 21 × 14 = 2 (ii) a = 2 ÷ 1.6 × 2.4 = 3 (iii) a = 18 ÷ 2.25 × 3.125 = 25 (b) (i) 2 6 18 30 3 3 9 15 3 1 3 5 5 1 1 5 1 1 1 Gandaan sepunya terterkecil Lowest common multiple = 2 × 3 × 3 × 5 = 90 (ii) Faktor sepunya terbesar Highest common factor = 2 × 3 = 6 (m + 4) = 6 m = 6 – 4 = 2 (c) Bilangan jubin yang diperlukan The number of tiles needed = (4.95 ÷ 0.45) × (4.5 ÷ 0.45) = 11 × 10 = 110

85 © Penerbitan Pelangi Sdn. Bhd. Matematik Tingkatan 1 Jawapan 3. (a) (i) Kuasa tiga sempurna yang terletak di antara 200 dengan 400 The perfect cubes between 200 and 400 = 216 dan / and 343 Beza / Difference = 343 – 216 = 127 (ii) 60 2 30 2 15 3 5 60 = 2 × 2 × 3 × 5 (b) (i) x × x × x = 729 x3 = 729 x = 3  729 = 9 (ii) Isi padu kubus kecil The volume of small cube = 729 ÷ 8 = 91.125 3  91.125 = 4.5 Maka, isi padu kubus kecil itu bukan kuasa tiga sempurna. Therefore, the volume of the small cube is not a perfect cube. (b) Nilai bagi 1 bahagian nisbah The value of 1 part of ratio = RM1 152 ÷ (7 + 2 + 3) = RM96 Pertambahan wang Lisa Lisa’s money increase by = 25 100 × (96 × 7) = 0.25 × 672 = RM168 Pertambahan wang Alvin Alvin’s money increase by = 50 100 × (96 × 2) = 0.5 × 192 = RM96 Peratus pertambahan wang Edward The percentage increase of Edward’s money = 300 – 158 – 96 26 × 3 × 100% = 36 288 × 100% = 12.5% 4. (a) (i) p5 = p × p × p × p × p (ii) (x – 2y) 4 = (x – 2y) × (x – 2y) × (x – 2y) × (x – 2y) (iii) 8(1 + k)3 = 2(1 + k) × 2(1 + k) × 2(1 + k) (b) Beza antara bilangan murid perempuan dengan murid lelaki The difference between the number of female and male students = (8xy – 3) – 4y = 8xy – 4y – 3 (c) (i) (3x + 2) + (3x + 2) + 2x = 2(4x – 3) + 2(x + 1) 8x + 4 = 8x – 6 + 2x + 2 8x + 4 = 10x – 4 10x – 8x = 4 + 4 2x = 8 x = 4 (ii) 5 = (4x – 3) × (x + 1) –  1 2 × 2x × y = [4(4) – 3] × (4 + 1) – 4y = 13 × 5 – 4y = 65 – 4y 4y = 65 – 5 = 60 y = 15 5. (a) Nilai SGD yang diterima oleh Ahmad The amount of SGD received by Ahmad = (3 400 – 60) ÷ 3.34 = 3 340 ÷ 3.34 = SGD 1 000 (b) (i) 23 × 3  –216 ÷ 16 25 = 8 × (–6) ÷ 4 5 = –60

© Penerbitan Pelangi Sdn. Bhd. 86 Matematik Tingkatan 1 Jawapan (ii) 3 – 1 64 ×  32 – 1 7 9  = – 1 4 ×  9 – 4 3  = – 23 12 (c) Jumlah luas permukaan kubus M The total surface of cube M = 6 × (6 × 6) = 216 Jumlah luas permukaan kubus N The total surface of cube N = 6 × (x × x) = 6x2 9 4 = 216 6x2 9(6x2 ) = 864 54x2 = 864 x2 = 16 x =  16 = 4 6. (a) (i) Biar x ialah nombor pertama, maka nombor kedua ialah (54 – x). Let x be the first number, thus the second number is (54 – x). Hasil tambah bagi dua nombor The sum of two numbers ∴ x + (54 – x) = 54 (ii) x – (54 – x) = 22 2x – 54 = 22 2x = 76 x = 38 (b) 2m + n = 11 ......... m – n = 4 .........  + : 3m = 15 m = 5 Gantikan m = 5 ke dalam . Substitute m = 5 into . 5 – n = 4 n = 5 – 4 = 1 (c) 3 45 60 90 5 15 20 30 3 3 4 6 2 1 4 2 2 1 2 1 1 1 1 Bilangan minimum kotak pen penyerlah yang perlu dibeli The minimum number of highlighters has to buy = (3 × 5 × 3 × 2 × 2) ÷ 90 = 180 ÷ 90 = 2