Matematik TambahanAdditional Mathematics

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1

Question 1

[3 marks] (a)Selesaikan persamaan kuadratik 3x² − 8x − 2 = 0 dengan menggunakan rumus. Beri jawapan betul kepada tiga tempat perpuluhan.Solve the quadratic equation 3x² − 8x − 2 = 0 using the formula. Give the answers correct to three decimal places.

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x = 2.897 or x = −0.230

• x = [8 ± √(64 + 24)] / 6, i.e. correct substitution into the formula1
• x = 2.8971
• x = −0.2301

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[2 marks] (b)Bentukkan persamaan kuadratik yang mempunyai punca 3 dan −½. Beri jawapan dalam bentuk ax² + bx + c = 0.Form the quadratic equation which has roots 3 and −½. Give the answer in the form ax² + bx + c = 0.

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2x² − 5x − 3 = 0

• Sum of roots = 5/2 and product of roots = −3/2, or (x − 3)(x + ½) = 01
• 2x² − 5x − 3 = 01

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[2 marks] (c)Cari julat nilai x bagi x² + 3x − 10 ≥ 0.Find the range of values of x for x² + 3x − 10 ≥ 0.

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x ≤ −5 or x ≥ 2

• (x + 5)(x − 2) ≥ 0 with critical values −5 and 21
• x ≤ −5 or x ≥ 21

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For the teacher: Applying, low difficulty, Chapter 2, standard 2.1.2

Textbook p. 40

2

Question 2

[2 marks] (a)Diberi fungsi f(x) = 5 − 3x. Cari f⁻¹(x).Given the function f(x) = 5 − 3x. Find f⁻¹(x).

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f⁻¹(x) = (5 − x)/3

• Let y = 5 − 3x, then x = (5 − y)/31
• f⁻¹(x) = (5 − x)/31

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[2 marks] (b)Diberi g(x) = x². Cari gf(x) dalam bentuk termudah.Given g(x) = x². Find gf(x) in its simplest form.

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gf(x) = 9x² − 30x + 25

• gf(x) = g(5 − 3x) = (5 − 3x)²1
• = 25 − 30x + 9x²1

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[2 marks] (c)Selesaikan persamaan gf(x) = 16.Solve the equation gf(x) = 16.

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x = 1/3 or x = 3

• (5 − 3x)² = 16, so 5 − 3x = 4 or 5 − 3x = −41
• x = 1/3 or x = 31

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For the teacher: Applying, low difficulty, Chapter 1, standard 1.3.2

Textbook p. 25

3

Question 3

[2 marks] (a)Diberi OP = 5i − 2j dan OQ = −i + 6j. Cari PQ dalam sebutan i dan j.Given OP = 5i − 2j and OQ = −i + 6j. Find PQ in terms of i and j.

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PQ = −6i + 8j

• PQ = OQ − OP = (−i + 6j) − (5i − 2j)1
• PQ = −6i + 8j1

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[2 marks] (b)Cari |PQ|.Find |PQ|.

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10

• |PQ| = √[(−6)² + 8²]1
• = 101

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[2 marks] (c)Cari vektor unit dalam arah PQ.Find the unit vector in the direction of PQ.

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(−3i + 4j)/5

• unit vector = PQ ÷ |PQ| = (−6i + 8j)/101
• = −(3/5)i + (4/5)j1

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For the teacher: Applying, low difficulty, Chapter 8, standard 8.3.2

Textbook p. 226

4

Question 4

[2 marks] (a)Selesaikan persamaan 5^2x−1 = 125^x−2.Solve the equation 5^2x−1 = 125^x−2.

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x = 5

• 5^(2x−1) = 5^(3x−6), same base1
• 2x − 1 = 3x − 6, so x = 51

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[2 marks] (b)Nyatakan 6√5 − √2 dalam bentuk surd termudah dengan menasabahkan penyebut.Express 6√5 − √2 in its simplest surd form by rationalising the denominator.

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2√5 + 2√2

• Multiply numerator and denominator by (√5 + √2): 6(√5 + √2)/(5 − 2)1
• = 2√5 + 2√21

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[3 marks] (c)Diberi log₂ 3 = p dan log₂ 5 = q. Ungkapkan setiap yang berikut dalam sebutan p dan q.
(i) log₂ 15
(ii) log₂ 758Given log₂ 3 = p and log₂ 5 = q. Express each of the following in terms of p and q.
(i) log₂ 15
(ii) log₂ 758

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(i) p + q; (ii) p + 2q − 3

• (i) log₂ 15 = log₂ 3 + log₂ 5 = p + q1
• (ii) log₂ 75 = log₂ 3 + 2 log₂ 5 = p + 2q1
• log₂ 8 = 3, so log₂(75/8) = p + 2q − 31

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For the teacher: Applying, low difficulty, Chapter 4, standard 4.2.3

Textbook p. 103

5

Question 5

[2 marks] (a)Mei Ling menyimpan RM12 pada minggu pertama, RM19 pada minggu kedua, RM26 pada minggu ketiga dan seterusnya. Simpanannya membentuk satu janjang aritmetik. Cari simpanannya pada minggu ke-15.Mei Ling saves RM12 in the first week, RM19 in the second week, RM26 in the third week and so on. Her savings form an arithmetic progression. Find her savings in the 15th week.

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RM110

• a = 12, d = 7, T₁₅ = 12 + 14(7)1
• RM1101

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[3 marks] (b)Cari bilangan minggu paling sedikit supaya jumlah simpanannya melebihi RM1 000.Find the least number of weeks for her total savings to exceed RM1 000.

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16 weeks

• S_n = n/2 [24 + 7(n − 1)] and set S_n > 1 000, i.e. 7n² + 17n − 2 000 > 01
• n > 15.7 (or S₁₅ = 915 and S₁₆ = 1 032)1
• 16 weeks1

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[3 marks] (c)Janjang geometri 160, 120, 90, … mempunyai hasil tambah ketakterhinggaan. (i) Nyatakan nisbah sepunya. (ii) Cari hasil tambah ketakterhinggaan.The geometric progression 160, 120, 90, … has a sum to infinity. (i) State the common ratio. (ii) Find the sum to infinity.

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(i) 3/4; (ii) 640

• (i) r = 120/160 = 3/41
• (ii) S∞ = a/(1 − r) = 160/(1 − 3/4)1
• = 6401

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For the teacher: Applying, medium difficulty, Chapter 5, standard 5.1.4

Textbook p. 138

6

Question 6

Rajah 1 menunjukkan segi tiga ABC pada satah Cartesan dengan A(−2, 1), B(6, 5) dan C(4, −3).Diagram 1 shows triangle ABC on a Cartesian plane with A(−2, 1), B(6, 5) and C(4, −3).

[2 marks] (a)Titik P membahagikan tembereng garis AB dengan nisbah AP : PB = 3 : 1. Cari koordinat P.Point P divides the line segment AB in the ratio AP : PB = 3 : 1. Find the coordinates of P.

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P(4, 4)

• P = ((1(−2) + 3(6))/4, (1(1) + 3(5))/4)1
• P(4, 4)1

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[3 marks] (b)Cari persamaan garis lurus yang melalui C dan berserenjang dengan AB.Find the equation of the straight line passing through C and perpendicular to AB.

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y = −2x + 5

• Gradient of AB = (5 − 1)/(6 − (−2)) = 1/21
• Gradient of the perpendicular line = −21
• y + 3 = −2(x − 4), i.e. y = −2x + 51

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[3 marks] (c)Hitung luas segi tiga ABC.Calculate the area of triangle ABC.

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28 unit²

• Area = ½ |(−2)(5) + 6(−3) + 4(1) − [1(6) + 5(4) + (−3)(−2)]|1
• = ½ |−24 − 32|1
• 28 unit²1

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For the teacher: Applying, medium difficulty, Chapter 7, standard 7.3.1

Textbook p. 196

7

Question 7

[2 marks] (a)Nilai sebuah mesin, dalam RM, selepas t tahun diberi oleh V = 80 000(0.85)^t. Cari nilai mesin itu selepas 3 tahun.The value of a machine, in RM, after t years is given by V = 80 000(0.85)^t. Find the value of the machine after 3 years.

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RM49 130

• V = 80 000(0.85)³1
• RM49 1301

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[3 marks] (b)Cari bilangan tahun paling sedikit supaya nilai mesin itu kurang daripada RM20 000.Find the least number of years for the value of the machine to fall below RM20 000.

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9 years

• 80 000(0.85)^t < 20 000, so 0.85^t < 0.251
• t log 0.85 < log 0.25, so t > log 0.25 / log 0.85 = 8.53 (the inequality sign reverses because log 0.85 is negative)1
• 9 years1

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[3 marks] (c)Sebuah segi empat tepat mempunyai luas (7 + 3√5) cm² dan lebar (2 + √5) cm. Cari panjangnya dalam bentuk a + b√5.A rectangle has area (7 + 3√5) cm² and width (2 + √5) cm. Find its length in the form a + b√5.

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(1 + √5) cm

• Length = (7 + 3√5)/(2 + √5)1
• Multiply by (√5 − 2)/(√5 − 2): (7 + 3√5)(√5 − 2)/(5 − 4)1
• = (1 + √5) cm1

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For the teacher: Analysing, high difficulty, Chapter 4, standard 4.4

Textbook p. 120

8Choose 3

Question 8

Kayu balak disusun dalam satu longgokan. Baris paling bawah mempunyai 25 batang kayu balak dan setiap baris di atasnya mempunyai 2 batang kurang daripada baris di bawahnya. Baris paling atas mempunyai 1 batang kayu balak.Logs are stacked in a pile. The bottom row has 25 logs and every row above has 2 logs fewer than the row below it. The top row has 1 log.

[2 marks] (a)Cari bilangan kayu balak pada baris ke-9.Find the number of logs in the 9th row.

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9

• a = 25, d = −2, T₉ = 25 + 8(−2)1
• 9 logs1

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[2 marks] (b)Cari bilangan baris dalam longgokan itu.Find the number of rows in the pile.

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13 rows

• 25 + (n − 1)(−2) = 11
• n = 131

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[3 marks] (c)Tunjukkan bahawa jumlah kayu balak dalam n baris pertama ialah S_n = n(26 − n).Show that the total number of logs in the first n rows is S_n = n(26 − n).

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S_n = n(26 − n)

• S_n = n/2 [2(25) + (n − 1)(−2)]1
• = n/2 (52 − 2n)1
• = n(26 − n)1

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[3 marks] (d)Satu longgokan yang disusun dengan cara yang sama mengandungi 120 batang kayu balak. Cari bilangan baris dalam longgokan itu. Terangkan mengapa satu daripada dua nilai n yang diperoleh tidak diterima.A pile built in the same way contains 120 logs. Find the number of rows in the pile. Explain why one of the two values of n obtained is not accepted.

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n = 6

• n(26 − n) = 120, so n² − 26n + 120 = 01
• (n − 6)(n − 20) = 0, so n = 6 or n = 201
• n = 6. n = 20 is rejected because a pile built this way can have at most 13 rows (the 20th row would have −13 logs, which is impossible)1

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For the teacher: Applying, low difficulty, Chapter 5, standard 5.1.4

Textbook p. 138

9Choose 3

Question 9

Ketinggian, h meter, air daripada satu pancutan taman pada jarak mengufuk x meter daripada muncung diberi oleh fungsi f(x) = −2x² + 8x + 10.The height, h metres, of water from a garden fountain at a horizontal distance of x metres from the nozzle is given by the function f(x) = −2x² + 8x + 10.

[2 marks] (a)Ungkapkan f(x) dalam bentuk a(x + p)² + q.Express f(x) in the form a(x + p)² + q.

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f(x) = −2(x − 2)² + 18

• f(x) = −2(x² − 4x) + 10 = −2[(x − 2)² − 4] + 101
• f(x) = −2(x − 2)² + 181

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[2 marks] (b)Nyatakan ketinggian maksimum pancutan itu dan jarak mengufuk pada ketinggian itu.State the maximum height of the jet and the horizontal distance at which it occurs.

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18 m at x = 2 m

• Maximum height = 18 m1
• at x = 2 m1

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[2 marks] (c)Pancutan itu jatuh ke tanah pada jarak mengufuk d meter. Cari d.The jet lands on the ground at a horizontal distance d metres. Find d.

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d = 5

• −2x² + 8x + 10 = 0, so (x − 5)(x + 1) = 01
• d = 5 (x = −1 is rejected because the distance cannot be negative)1

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[4 marks] (d)Seekor burung terbang secara mengufuk pada ketinggian 15 m dan memintas laluan pancutan di dua titik. Cari jarak mengufuk antara dua titik itu, betul kepada dua tempat perpuluhan.A bird flies horizontally at a height of 15 m and crosses the path of the jet at two points. Find the horizontal distance between the two points, correct to two decimal places.

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2.45 m

• −2x² + 8x + 10 = 15, so 2x² − 8x + 5 = 01
• x = [8 ± √(64 − 40)]/41
• x = 3.22 and x = 0.78 (3.225 and 0.775)1
• Distance = 2.45 m1

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For the teacher: Applying, low difficulty, Chapter 2, standard 2.3.6

Textbook p. 61

10Choose 3

Question 10

Di kantin sekolah, dua nasi lemak, tiga keping kuih dan segelas minuman berharga RM17. Satu nasi lemak, dua keping kuih dan dua gelas minuman berharga RM14. Tiga nasi lemak, sekeping kuih dan dua gelas minuman berharga RM20. Harga sebungkus nasi lemak, sekeping kuih dan segelas minuman masing-masing ialah RM x, RM y dan RM z.At a school canteen, two packets of nasi lemak, three pieces of kuih and one glass of drink cost RM17. One packet of nasi lemak, two pieces of kuih and two glasses of drink cost RM14. Three packets of nasi lemak, one piece of kuih and two glasses of drink cost RM20. The prices of a packet of nasi lemak, a piece of kuih and a glass of drink are RM x, RM y and RM z respectively.

[2 marks] (a)Tulis tiga persamaan linear dalam x, y dan z untuk mewakili maklumat itu.Write three linear equations in x, y and z to represent the information.

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2x + 3y + z = 17; x + 2y + 2z = 14; 3x + y + 2z = 20

• 2x + 3y + z = 171
• x + 2y + 2z = 14 and 3x + y + 2z = 201

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[4 marks] (b)Dengan menggunakan kaedah penghapusan, cari harga sebungkus nasi lemak, sekeping kuih dan segelas minuman.Using the elimination method, find the price of a packet of nasi lemak, a piece of kuih and a glass of drink.

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x = 4, y = 2, z = 3

• Eliminate z: (1)×2 − (2): 3x + 4y = 20, and (3) − (2): 2x − y = 61
• Solve: y = 2x − 6 gives 3x + 8x − 24 = 20, so x = 41
• y = 2, substitute into (1): z = 31
• Nasi lemak RM4, kuih RM2, drink RM31

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[4 marks] (c)Papan menu berbentuk segi empat tepat mempunyai panjang l meter dan lebar w meter dengan l + 2w = 7 dan lw = 3, dengan keadaan l > w. Cari nilai l dan nilai w.A rectangular menu board has length l metres and width w metres, with l + 2w = 7 and lw = 3, where l > w. Find the value of l and of w.

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l = 6, w = 0.5

• l = 7 − 2w, so (7 − 2w)w = 31
• 2w² − 7w + 3 = 0, so (2w − 1)(w − 3) = 01
• w = 1/2, l = 6 or w = 3, l = 11
• l > w, so l = 6 and w = 0.51

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For the teacher: Applying, medium difficulty, Chapter 3, standard 3.1.3

Textbook p. 77

11Choose 3

Question 11

Pemboleh ubah x dan y dihubungkan oleh persamaan y = axⁿ, dengan a dan n ialah pemalar. Rajah 2 menunjukkan garis lurus penyuaian terbaik bagi graf log₁₀ y melawan log₁₀ x.The variables x and y are related by the equation y = axⁿ, where a and n are constants. Diagram 2 shows the line of best fit for the graph of log₁₀ y against log₁₀ x.

[2 marks] (a)Tunjukkan bahawa hubungan y = axⁿ boleh ditukarkan kepada bentuk linear Y = mX + c. Nyatakan Y, X, m dan c.Show that the relation y = axⁿ can be reduced to the linear form Y = mX + c. State Y, X, m and c.

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Y = log y, X = log x, m = n, c = log a

• log₁₀ y = n log₁₀ x + log₁₀ a1
• Y = log₁₀ y, X = log₁₀ x, m = n, c = log₁₀ a1

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[4 marks] (b)Menggunakan Rajah 2, cari nilai n dan nilai a.Using Diagram 2, find the value of n and of a.

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n = 1.5, a = 3.98

• Gradient = (2.1 − 1.2)/(1.0 − 0.4)1
• n = 1.51
• Y-intercept: 1.2 = 1.5(0.4) + c, so c = 0.61
• a = 10^0.6 = 3.981

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[2 marks] (c)Anggarkan nilai y apabila x = 5.Estimate the value of y when x = 5.

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y = 44.5

• log y = 0.6 + 1.5 log 5 = 1.6491
• y = 44.5 (accept 44 to 45)1

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[2 marks] (d)Rahim berkata, “Apabila nilai x digandakan, nilai y juga digandakan.” Adakah Rahim betul? Jelaskan jawapan anda.Rahim says, “When the value of x is doubled, the value of y is also doubled.” Is Rahim correct? Explain your answer.

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Not correct: y is multiplied by 2^1.5 = 2.83

• y = a(2x)^1.5 = 2^1.5 × ax^1.5, so y is multiplied by 2^1.5 ≈ 2.831
• Rahim is not correct, because y is multiplied by about 2.83, not 21

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For the teacher: Analysing, high difficulty, Chapter 6, standard 6.3.1

Textbook p. 166

12Choose 2

Question 12

Jadual 1 menunjukkan harga tiga jenis alat tulis di sebuah kedai buku pada tahun 2022 dan 2024, serta pemberatnya.

Alat tulis	Harga 2022 (RM)	Harga 2024 (RM)	Pemberat
Buku latihan	2.50	3.00	5
Pen	1.20	1.50	3
Kalkulator	40.00	52.00	2

Jadual 1Table 1 shows the prices of three types of stationery at a bookshop in 2022 and 2024, and their weightages.

Stationery	Price 2022 (RM)	Price 2024 (RM)	Weightage
Exercise book	2.50	3.00	5
Pen	1.20	1.50	3
Calculator	40.00	52.00	2

Table 1

[2 marks] (a)Hitung indeks harga bagi pen pada tahun 2024 berasaskan tahun 2022.Calculate the price index of the pen in 2024 based on 2022.

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125

• I = 1.50/1.20 × 1001
• 1251

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[3 marks] (b)Hitung indeks gubahan bagi kos alat tulis itu pada tahun 2024 berasaskan tahun 2022.Calculate the composite index for the cost of the stationery in 2024 based on 2022.

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123.5

• Index of exercise book = 120 and calculator = 1301
• (120×5 + 125×3 + 130×2)/101
• 123.51

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[2 marks] (c)Satu set alat tulis berharga RM85 pada tahun 2022. Cari harganya pada tahun 2024.A set of stationery cost RM85 in 2022. Find its cost in 2024.

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RM104.98

• 123.5/100 × 851
• RM104.98 (RM105)1

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[3 marks] (d)Dijangka harga alat tulis meningkat sebanyak 8% daripada tahun 2024 ke tahun 2026. Cari harga set itu pada tahun 2026.Prices of stationery are expected to rise by 8% from 2024 to 2026. Find the cost of the set in 2026.

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RM113.37

• Index in 2026 based on 2024 = 1081
• Index in 2026 based on 2022 = 123.5 × 108/100 = 133.381
• 85 × 1.3338 = RM113.371

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For the teacher: Applying, low difficulty, Chapter 10, standard 10.2.2

Textbook p. 284

13Choose 2

Question 13

Jadual 2 menunjukkan indeks harga empat bahan yang digunakan untuk membuat kek pada tahun 2023 berasaskan tahun 2020, serta pemberatnya.

Bahan	Tepung	Gula	Mentega	Telur
Indeks harga	115	135	125	150
Pemberat	3	w	4	2

Jadual 2Table 2 shows the price indices of four ingredients used to make a cake in 2023 based on 2020, and their weightages.

Ingredient	Flour	Sugar	Butter	Eggs
Price index	115	135	125	150
Weightage	3	w	4	2

Table 2

[2 marks] (a)Harga gula pada tahun 2020 ialah RM3.20 sekilogram. Cari harganya pada tahun 2023.The price of sugar in 2020 was RM3.20 per kilogram. Find its price in 2023.

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RM4.32

• 135/100 × 3.201
• RM4.321

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[4 marks] (b)Indeks gubahan bagi kos membuat kek pada tahun 2023 berasaskan tahun 2020 ialah 130. Cari nilai w.The composite index for the cost of making the cake in 2023 based on 2020 is 130. Find the value of w.

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w = 5

• (115×3 + 135w + 125×4 + 150×2)/(3 + w + 4 + 2) = 1301
• 1 145 + 135w = 130(9 + w)1
• 1 145 + 135w = 1 170 + 130w1
• w = 51

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[2 marks] (c)Kos membuat satu kek pada tahun 2020 ialah RM18. Cari kosnya pada tahun 2023.The cost of making a cake in 2020 was RM18. Find its cost in 2023.

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RM23.40

• 130/100 × 181
• RM23.401

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[2 marks] (d)Pada tahun 2025, indeks harga setiap bahan berasaskan tahun 2023 ialah 110. Puan Zainab berkata kos membuat kek pada tahun 2025 telah meningkat lebih daripada 40% berbanding tahun 2020. Adakah dia betul? Tunjukkan pengiraan anda.In 2025, the price index of every ingredient based on 2023 is 110. Puan Zainab says the cost of making a cake in 2025 has increased by more than 40% compared with 2020. Is she correct? Show your calculation.

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Correct: increase of 43%

• Index 2025 based on 2020 = 130 × 110/100 = 1431
• Increase = 43% which is more than 40%, so she is correct1

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For the teacher: Applying, low difficulty, Chapter 10, standard 10.2.1

Textbook p. 281

14Choose 2

Question 14

Rajah 3 menunjukkan segi tiga ABC dengan AB = 12 cm, AC = 10 cm dan ∠ABC = 48°. Titik C′ terletak pada garis BC.Diagram 3 shows triangle ABC with AB = 12 cm, AC = 10 cm and ∠ABC = 48°. Point C′ lies on the line BC.

[2 marks] (a)Cari ∠ACB, dengan keadaan ∠ACB ialah sudut tirus.Find ∠ACB, given that ∠ACB is an acute angle.

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63.1°

• sin C / 12 = sin 48° / 101
• ∠ACB = 63.1°1

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[2 marks] (b)Titik C′ terletak pada garis BC dengan keadaan AC′ = AC. Nyatakan ∠AC′B.Point C′ lies on the line BC such that AC′ = AC. State ∠AC′B.

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116.9°

• Triangle ACC′ is isosceles, so ∠AC′C = ∠ACB = 63.1°1
• ∠AC′B = 180° − 63.1° = 116.9°1

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[3 marks] (c)Hitung luas segi tiga ABC.Calculate the area of triangle ABC.

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55.98 cm²

• ∠BAC = 180° − 48° − 63.1° = 68.9°1
• Area = ½(12)(10) sin 68.9°1
• = 55.98 cm²1

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[3 marks] (d)Hitung panjang BC′.Calculate the length of BC′.

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3.50 cm

• ∠BAC′ = 180° − 48° − 116.9° = 15.1°1
• BC′/sin 15.1° = 10/sin 48°1
• BC′ = 3.50 cm1

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For the teacher: Applying, medium difficulty, Chapter 9, standard 9.1.2

Textbook p. 247

15Choose 2

Question 15

Rajah 4 menunjukkan sebidang tanah PQRS yang berbentuk sisi empat. PQ = 6 m, QR = 8 m, RS = 9 m, SP = 7 m dan ∠PQR = 105°.Diagram 4 shows a plot of land PQRS in the shape of a quadrilateral. PQ = 6 m, QR = 8 m, RS = 9 m, SP = 7 m and ∠PQR = 105°.

[2 marks] (a)Hitung panjang PR, betul kepada dua tempat perpuluhan.Calculate the length of PR, correct to two decimal places.

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11.17 m

• PR² = 6² + 8² − 2(6)(8) cos 105°1
• PR = 11.17 m1

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[3 marks] (b)Hitung ∠PSR, betul kepada satu tempat perpuluhan.Calculate ∠PSR, correct to one decimal place.

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87.7°

• cos S = (7² + 9² − PR²)/(2 × 7 × 9), with PR² = 124.851
• = 0.04091
• ∠PSR = 87.7°1

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[2 marks] (c)Hitung luas segi tiga PQR.Calculate the area of triangle PQR.

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23.18 m²

• ½(6)(8) sin 105°1
• 23.18 m²1

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[3 marks] (d)Encik Hafiz ingin menutup seluruh kawasan PQRS dengan rumput hamparan pada harga RM18 setiap m². Dia mempunyai RM1 000. Adakah wang itu mencukupi? Tunjukkan pengiraan anda.Encik Hafiz wants to cover the whole of PQRS with turf at RM18 per m². He has RM1 000. Is the money enough? Show your calculation.

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Enough: about RM983.8

• Area of PRS = ½(9)(7) sin 87.7° = 31.47 m²1
• Total area = 23.18 + 31.47 ≈ 54.65 m², cost ≈ 54.65 × 18 ≈ RM983.8 (accept RM983.7 to RM983.8)1
• RM983.8 < RM1 000, so the money is enough1

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For the teacher: Evaluating, high difficulty, Chapter 9, standard 9.4

Textbook p. 266

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