Maths

JABATAN PENDIDIKAN NEGERI SELANGOR Project Work for Additional Mathematics 2012 Sample Answers PART 1 For (a) and (b) , accept any relevant answers. (c) Suggested answers to calculate the area of a triangle :

1 × base × height 2 1 2. Area = × a × b × sin C 2 1. Area = 3. Area = s( s − a)( s − b)( s − c ) , where s = 4. Area = 1 x1 2 y1 x2 y2 x3 y3 x1 y1 1 (a + b + c) , or the semi-perimeter. 2

5. Using graphical method.

PART 2 (a) (b) Cost = RM 20 × 300 = RM 6000. First method

a m

θº

b m m mm 100 m c Diagram 1

Using cosine rule, cos θ ° =

a2 + b2 − c2 2ab

1 Area = ab × sin θ 2

p ( m)

q ( m)

θo

Area (m2 )

50 60 65 70 80 85 90 95 99 100 Second method : Using Microsoft Excel p (m) 50 55 60 65 70 75 80 85 90 95 100 105 110 115 (c)

150 140 135 130 120 115 110 105 101 100

0 38.2145 44.8137 49.5826 55.7711 57.6881 58.9924 59.7510 59.9901 60

0 2598.15 3092.33 3464.10 3968.63 4130.68 4242.64 4308.42 4329.26 4330.13

q (m) 150 145 140 135 130 125 120 115 110 105 100 95 90 85

θ (degree) 0.0000 28.2516 38.2132 44.8137 49.5826 53.1301 55.7711 57.6882 58.9924 59.7510 60.0000 59.7510 58.9924 57.6882

Area ( m2) 0.00 1887.46 2598.08 3092.33 3464.10 3750.00 3968.63 4130.68 4242.64 4308.42 4330.13 4308.42 4242.64 4130.68

The herb garden is an equilateral triangle of sides 100 m with a maximum area of 4330.13 m 2 .

(d)(i) 50 < p < 150, 50 < q < 150 (ii) p + q ˃ 100 (iii) The sum of the lengths of any two sides of a triangle is greater than the length of the third side. Triangle Inequality Theorem. PART 3

(a)

Quadrilateral (suggested answer) 2x + 2y = 300 m2 x + y = 150 m2 Area = x y y 140 130 120 110 100 90 80 75 Area = x y 1400 2600 3600 4400 5000 5400 5600 5625

ym xm x 10 20 30 40 50 60 70 75

The maximum area is 5625 m2. (b) Regular Pentagon 5a = 300 a = 60 t 30 t = 30 tan 54° = 41.2915 m 1 Area = ( 41.2915 × 60) × 5 = 6193.73 m 2 2 tan 54° = (c) A Semicircle a a

a 54o

72o t a 54o

a

rm

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