form four - gurubesar.my filejabatan pelajaran negeri melaka . tingkatan 4 [ 2011 ] additional...

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1 3472/2 Form 4 Additional Mathematics Paper 2 2011

22

1hours

PEPERIKSAAN SELARAS AKHIR TAHUN SEKOLAH-SEKOLAH MENENGAH NEGERI MELAKA

Kelolaan PEJABAT PELAJARAN DAERAH

JASIN * ALOR GAJAH * MELAKA TENGAH Dengan kerjasama :

JABATAN PELAJARAN NEGERI MELAKA TINGKATAN 4 [ 2011 ]

ADDITIONAL MATHEMATICS Paper 2

22

1hours

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

1. This questions paper consists of three sections : Section A, Section B and Section C. 2. Answer all questions in Section A , four questions from Section B and two questions from Section C. 3. Give only one answer / solution to each question.. 4. Show your working. It may help you to get marks. 5. The diagram in the questions provided are not drawn to scale unless stated. 6. The marks allocated for each question and sub-part of a question are shown in brackets.. 7. A list of formulae is provided on pages 2 to 3. 8. A booklet of four-figure mathematical tables is provided. 9. You may use a non-programmable scientific calculator.

Kertas soalan ini mengandungi 17 halaman bercetak

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The following formulae may be helpful in answering the questions.The symbols given are the ones commonly used. Rumus-rumus berikut boleh digunakan untuk membantu anda menjawab soalan. . Simbol-simbol yang diberi adalah yang biasa digunakan.

ALGEBRA

1. x =a

acbb

2

42

2 am an = a m + n 3 am an = a m - n

4 (am) n = a nm 5 loga mn = log am + loga n

6 loga n

m = log am - loga n

7 log a mn = n log a m

8 logab = a

b

c

c

log

log

9 Tn = a + (n-1)d

10 Sn = ])1(2[2

dnan

11 Tn = ar n-1

12 Sn = r

ra

r

ra nn

1

)1(

1

)1( , (r 1)

13 r

aS

1 , r <1

CALCULUS

1 y = uv , dx

duv

dx

dvu

dx

dy

2 v

uy ,

2v

dx

dvu

dx

duv

dx

dy

,

3 dx

du

du

dy

dx

dy

GEOMETRY

1 Distance = 2

21

2

21 )()( yyxx

2 Midpoint

(x , y) =

2

21 xx ,

2

21 yy

3 22 yxr

4 ř 22 yx

yjxi

4 Area under a curve

= b

a

y dx or

= b

a

x dy

5 Volume generated

= b

a

y 2 dx or b

a

x 2 dy

5 . A point dividing a segment of a line

( x,y) = ,21

nm

mxnx

nm

myny 21

6. Area of a triangle =

)()(2

1312312133221 1

yxyxyxyxyxyx


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STATISTICS

TRIGONOMETRY

1 x = N

x

2 x =

f

fx

3 = N

xx 2)( =

2_2

xN

x

4 =

f

xxf 2)( =

22

xf

fx

5 M = Cf

FN

Lm

2

1

6 1000

1 P

PI

1 x = N

x

2 x =

f

fx

3 = N

xx 2)( =

2_2

xN

x

4 =

f

xxf 2)( =

22

xf

fx

5 M = Cf

FN

Lm

2

1

6 1000

1 P

PI

7 1

11

w

IwI

8 )!(

!

rn

nPr

n

9. !)!(

!

rrn

nCr

n

10 P(AB)=P(A)+P(B)-P(AB)

11 p (X=r) = rnr

r

n qpC , p + q = 1

12 Min(mean) = np

13 npq

14 z =

x

1 Arc length , s = r

2 Area of a sector, L = 2

2

1r

3 sin 2A + cos

2A = 1

4 sec2A = 1 + tan

2A

5 cosek2 A = 1 + cot

2 A

6 sin2A = 2 sinAcosA

7 cos 2A = cos2A – sin

2 A

= 2 cos2A-1

= 1- 2 sin2A

8 tan2A = A

A2tan1

tan2

9 sin (A B) = sinAcosB cosAsinB

10 cos (A B) = cos AcosB sinAsinB

11 tan (A B) = BA

BA

tantan1

tantan

12 C

c

B

b

A

a

sinsinsin

13 a2 = b

2 +c

2 - 2bc cosA

14 Area of triangle = Cabsin2

1


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Section A

[ 40 Marks ]

Answer all questions from this section

Jawab semua soalan dari bahagian ini

1. The straight line 3 + x = 3y intersects the curve 1 + 2x2 = y2 + 3xy

at two points. Find the coordinates of the two points.

Garis lurus 3 + x = 3y menyilang lengkung 1 + 2x2 = y2 + 3xy pada

dua titik. Cari koordinat bagi dua titik tersebut.

[ 5 marks ]

2. The functions f and g are defined by f ( x ) = x1

6 , x ≠ 1 and

g ( x ) = kx + 10 , where k is a constant.

Fungsi f dan g ditakrifkan sebagai f ( x ) = x1

6 , x ≠ 1 dan

g ( x ) = kx + 10 , dengan k ialah pemalar.

a ) Find the value of k if f -1g ( -2 ) = - ½

Cari nilai k jika f -1g ( -2 ) = - ½

b ) Find g 2 ( x )

Cari g 2 ( x )

[ 6 marks ]

3. A set of scores p1, p2, p3, p4 and p5 has mean 6 and a standard deviation

of 2.3

Suatu set skor p1, p2, p3, p4 and p5 mempunyai min 6 dan sisihan piawai 2.3

(a) Find

Cari

(i) the sum of the score, p

hasil tambah skor, p


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(ii) the sum of the squares of the scores, 2p

hasil tambah kuasa dua skor, 2p

[ 3 marks ]

(b) Each score is multiplied by 3 and the 2 is added to it. Find, for the new

set of scores,

Setiap skor didarab 3 dan ditambah 2 , cari nilai baru bagi set skor

(i) the mean

min

(ii) the variance

varians .

[ 4 marks ]

4. (a) Given y = xx

1, find

2

2

dx

yd.

Diberi y = xx

1, cari

2

2

dx

yd.

(b) Given y = 16

19

8

1 x is the equation of normal to the curve

y = ( 4x 5 )2 at P. Find the coordinates of P.

Diberi y = 16

19

8

1 x adalah persamaan normal kepada lengkung

y = ( 4x 5 )2 di P. Cari koordinat- koordinat di P.

[ 6 marks ]


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5. The straight line kxy 4 is the normal to the curve 2122 xy at point A.

Garis lurus kxy 4 adalah normal kepada lengkok 2122 xy pada titik A

Find

Cari

a) the gradient of the tangent at point A

kecerunan bagi tangen pada titik A

[ 2 marks ]

b) the coordinates of point A and the value of k

koordinat tititk A dan nilai bagi k

[ 4 marks ]

c) the equation of tangent at point A

persamaan bagi tangen pada tititk A

[ 2 marks ]

6. a) Given that ax 10log and by 10log , express 210

100log

y

xin terms of a and b

Diberi ax 10log dan by 10log , bentukkan 210

100log

y

x dalam bentuk a dan b

[ 3 marks ]

b) Solve the equation 622 105.2 xxx

Selesaikan persamaan 622 105.2 xxx

[ 2 marks ]

c) Solve the equation 4loglog 327 xx

Selesaikan persamaan 4loglog 327 xx

[ 3 marks ]


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Section B

Bahagian B

[ 40 marks ]

Answer four questions from this section

Jawab empat soalan dari bahagian ini

7. Given A(5,-2) and B(2,1) are two fixed points. Point Q moves such that the

ratio of AQ and QB is 2 : 1

Diberi A(5,-2) dan B(2,1) adalah dua titik tetap. Titik Q bergerak dengan nisbah bagi AQ

kepada QB adalah 2 : 1

a) Show that the equation of the locus of point Q is x2+y2-2x-4y-3=0.

Tunjukkan persamaan lokus bagi titik Q ialah x2+y2-2x-4y-3=0.

[ 2 marks ]

b) Show that point C(-1,0) lies on the locus of point Q.

Tunjukkan bahawa titik C(-1,0) melalui lokus titik Q.

[ 2 marks ]

c) Find the equation of the straight line AC

Cari persamaan bagi garis lurus AC

[ 3 marks ]

d) Given the straight line AC intersects the locus of point Q again at point D,

find the coordinates of point D.

Diberi garis lurus AC bersilang dengan titik lokus Q pada titik D, cari titik bagi D

[ 3 marks ]


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8. In the Diagram 8 , 090ABC and the equation of the straight line BC

is 2y+x+8=0.

Rajah 8 menunjukkan 090ABC dan persamaan garis lurus BC ialah

2y+x+8=0.

a) Find

Cari

(i) The equation of the straight line AB,

Persamaan garis lurus AB

(ii) The coordinates of B

Koordinat B

[ 5 marks ]

b) The straight line AB is extended to a point D such that AB : BD = 2 : 3.

Find the coordinates of D.

Garis lurus AB memanjang ke titik D dengan AB : BD = 2 : 3. Cari titik D

[ 2 marks ]

c) A point Q moves such that it’s distance from point A is always 6 units.

Find the equations of the locus of Q.

Titik Q bergerak dengan jarak 6 unit dari titik A. Cari persamaan lokus bagi Q

[ 3 marks ]

B

x

y A(5,11)

0

C

2y+x+8=0

Diagram 8 / Rajah 8


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9. In Diagram 9, is a right-angled triangle. PR is an arc of a circle with centre

O and radius 8 cm. ORST is a quadrant of a circle with centre O. The

ratio of OP : PQ = 2 : 1

Rajah 9, sebuah segitiga bersudut tegak. PR adalah lengkok bulatan dengan

berpusat O dan berjejari 8 cm. ORST ialah sukuan bulatan berpusat O. Nisbah

bagi OP : PQ = 2 : 1

. Calculate

Kirakan

a) The value of in radians.

Nilai dalam radian,

[ 2 marks ]

b) The perimeter, in cm, of the shaded region,

Perimeter, dalam cm . bagi kawasan berlorek,

[ 5 marks ]

c) The area, in cm2 , of the shaded region.

Luas, dalam cm2, bagi kawasan berlorek.

[ 3 marks ]

Diagram 9/ Rajah 9


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10. The Diagram 10 shows a right-angled triangle. LMN described in a

semicircle with centre O. LM = 4 cm and MN = x cm.

Rajah 10 menunjukkan sebuah segitiga bersudut tegak. LMN adalah semibulatan

berpusat O. LM = 4 cm dan MN = x cm.

a) Show that the area of the shaded region, A cm2 is given by

A = xx 28

2 2

Tunjukkan bahawa kawasan berlorek, A cm2 diberi sebagai A= xx 28

2 2

.

[ 5 marks ]

b) Find the value of x such that A is a minimum. Hence, find the minimum of A

in terms of .

Cari nilai bagi x dimana A adalah minimum. Seterusnya, cari nilai bagi A dalam

sebutan

[ 5 marks ]

L

M

N O·

Diagram 10/Rajah 10


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11. Table 11 shows the frequency distribution of the height of 40 children in a

kindergarden.

Jadual 11 menunjukkan taburan kekerapan bagi tinggi 40 orang kanak-kanak di

sebuah tadika.

Height / Tinggi ( cm )

Number of children

Bilangan kanak-kanak

100 – 104 8

105 – 109 5

110 - 114 10

115 – 119 6

120 – 124 7

125 – 129 4

( a ) Based on Table 11 , construct a cumulative frequency

distribution table.

Berdasarkan Jadual 11 bina jadual kekerapan longgokan bagi

taburan itu.

[ 2 marks ]

( b ) Without using ogive, calculate the first quartile of the distribution.

Tanpa menggunakan ogif, hitung kuartil pertama bagi taburan itu.

[ 3 marks ]

( c ) Without using ogive calculate the median height of the children.

Tanpa menggunakan ogif Hitung tinggi median kanak-kanak itu.

[ 3 marks ]

Table 11 / Jadual 11


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Section C

Bahagian C

[ 20 marks ]

Answer two questions from this section.

Jawab dua soalan daripada bahagian ini.

12. Table12 shows the prices, price indices and percentages of usage of four

items, A, B, C and D,which are the main ingredients in the manufacturing of a

type of soap.

Jadual 12 menunjukkan harga, indeks harga dan peratus penggunaan empat

barangan A,B,C dan D yang merupakan bahan utama bagi penghasilan sejenis

sabun.

Item

Barangan

Price per unit

( RM )

Harga seunit

( RM )

Price index in 2008

based on 2004

Indeks harga pada

2008 berasaskan

2004

Percentage of usage

( % )

Peratus penggunaan

( % )

2004 2008

A p 54 120 3k

B 36 q 125 2k

C 40 32 80 20

D 40 42 r 5k

Table 12 / Jadual 12

a) Find the values of p, q and r

Cari nilai-nilai p, q dan r

[ 3 marks ]


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b) State the value of k. Hence, calculate the composite index for the cost

of manufacturing of the soap in the year 2008 based on the year 2004

Nyatakan nilai k. Seterusnya, hitung indeks gubahan bagi kos penghasilan

sabun itu pada tahun 2008 berasaskan tahun 2004.

[ 3 marks ]

c) Calculate the price of a box of soap in the year 2004 if the

corresponding price in the year 2008 is RM250.

Hitung harga sekotak sabun pada tahun 2004 jika harga yang sepadannya

pada tahun 2008 ialah RM250

[ 2 marks ]

d) The cost of manufacturing of the soap is expected to increase by 30%

from the year 2008 to the year 2012. Find the expected composite

index in the year 2012 based on the year 2004.

Kos penghasilan sabun itu dijangka meningkat sebanyak 30% dari tahun

2008 kepada tahun 2012. Cari indeks gubahan yang dijangka pada tahun

2012 berasaskan tahun 2004.

[ 2 marks ]


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13 The Diagram 13 shows a bar chart indicating the weekly cost of the

iterms J, K, L, M and N for the year 2001. The table shows the prices

and the price indices for the items.

Diagram 13 menunjukkan carta bar merujuk kepada perbelanjaan mingguan

bagi iterm J, K, L, M and N pada tahun 2001. Jadual pula menunjukkan harga

dan indeks harga barang-barang tersebut.

Item

(Barangan)

Price in 2001

(Harga pada

2001)

Price in 2004

(Harga pada

2004)

Price index in 2004 based on2001

(Harga pada 2004 berasaskan

pada 2001)

J x RM2.40 120

K RM3.20 RM4.00 125

L RM4.00 RM5.20 y

M RM6.00 RM9.00 150

N RM2.40 z 110

Weekly cost (RM)

50

36

30

24

10

Items

Diagram 13 / Rajah 13

J

L

K

M

N


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a) Find the value of

Cari nilai bagi

i) x

ii) y

iii) z

[ 3 marks ]

b) Calculate the composite index for the items in the year 2004 based on the

year 2001.

Kirakan indeks gubahan bagi barang-barang itu bagi tahun 2004 berdasarkan

tahun 2001.

[ 2 marks ]

c) The total weekly cost of the items in the year 2001 was RM800. Calculate the

corresponding total weekly cost for tha year 2004.

Jumlah perbelanjaan mingguan barang-barang itu pada tahun 2001 ialah RM800.

Kirakan jumlah perbelanjaan mingguan yang sepadan bagi barang-barang itu pada

tahun 2004.

[ 2 marks ]

d) The cost of the items increases by 25% from the year 2004 to the year 2007.

Find the composite index for the year 2007 based on the year 2001.

Perbelanjaan barang-barang itu telah meningkat 25% dari tahun 2004 kepada tahun

2007. Cari indeks gubahan pada tahun 2007 berasaskan tahun 2001.

[ 3 marks ]


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14. Calculate

Hitung

(i) the length of AB

panjang AB

[ 3 marks ]

(ii) the angle of BAC

Sudut BAC

[ 3 marks ]

(iii) the area of a new triangle of ABC if AC is extended while the length

of AB , the length of BC and the angle of BAC are maintained

luas bagi segitiga baru ABC jika AC dipanjangkan manakala panjang AB,

BC dan sudut BAC tidak berubah.

[ 4 marks ]

Diagram 14 / Rajah 14

Figure 14 shows a triangle ABC Rajah 14 menunjukkan segitiga ABC


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15. Diagram 15 shows a quadrilateral PQRS with a triangle ABC. D, E and F are

points the sight AB, BC and AC respectively such that AB:DB= 1:2 and BE = 8

cm. It also given that AB = 12 cm and AC = 7 cm.

Rajah 15 menunjukkan sebuah segi tiga ABC. D, E dan F masing-masing ialah

titik ada sisi AB, BC dan AC dengan keadaan AD:DB=1:2 dan BE = 8 cm. Diberi

juga bahawa AB = 12 cm dan AC = 7 cm.

Diagram 15/ Rajah 15

Calculate

Kirakan

a) (i) the length, in cm, of EC,

Panjang dalam cm , bagi EC.

[ 2 marks ]

(ii) ACB

[ 2 marks ]

b) It is given that AF = AC.

Diberi bahawa AF = AC.

(i) Calculate the area, cm2, of ∆ DEF.

kirakan luas, dalam cm2, bagi ∆ DEF.

[ 3 marks ]

(ii) Hence, find the perpendicular distance, in cm , from D to EF.

Oleh yang demikian , cari jarak serenjang, dalam cm, dari D ke EF.

[ 3 marks ]

END OF QUESTION PAPER / KERTAS SOALAN TAMAT

110o

E

A

C

B

F

D


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1

3472/2 Form 4 Additional Mathematics Paper 2 2011

PEPERIKSAAN SELARAS AKHIR TAHUN

SEKOLAH-SEKOLAH MENENGAH NEGERI MELAKA

Kelolaan

PEJABAT PELAJARAN DAERAH JASIN * ALOR GAJAH * MELAKA TENGAH

Dengan kerjasama :

JABATAN PELAJARAN NEGERI MELAKA

TINGKATAN 4 2011

ADDITIONAL MATHEMATICS

Paper 2

MARKING SCHEME

This marking scheme consists of 12 printed pages.


2

Number

Solution and marking scheme

Submarks

Full

marks

1

x = 3y - 3

8 y ² - 27 y + 19 = 0

y = )8(2

)19)(8(4)27()27( 2

y = 2.375 , y = 1

x = 4.125 , x = 0

( 4.125, 2.375 ) , ( 0, 1 )

atau setara

1

1

1

1, 1

5

2

a) g (-2) = -2k + 10

f 1 (-2k + 10) = )102(

6)102(

k

k

2

1 =

)102(

)42(

k

k

k = 3

b) g 2 ( x ) = g ( 3x + 10 ) = 3(3x + 10) + 10 = 9x + 40

1

1

1

1

1

1

6 3. (a) (i) Mean = 6

65

p

p = 30

(ii) standard deviation, = 2.3

1

1


3

2

2

65

p

= 2.3

29.5365

2

p

p2 = 206.45

b) (i) New mean = 6(3) + 2 = 20

(ii) new variance = 2.3 x 32 = 20.7

1

1

1

1

1

7

4 (a) 12 x

dx

dy

2

2

dx

yd2x

-3

= 3

2

x

(b) The normal, m2 = 8

1

m1 = 8

Gradient of tangent = 8

Given y = ( 4x – 5 ) 2

dx

dy = 8 ( 4x – 5 )

x = 2

3, y = 1

P ( 2

3 , 1 )

1

1

1

1

1

1

6

5. a)

4

1Nm

4Tm

b) (-1,3)

)2)(12(2 xdx

dy

4)12(4 x

3y atau 2)1)1(2( 2 y

1

1

1

1

1

1


4

c) ))1((43 xy

14 xy

1

1

8

6.

a) y

x 2

1

10

10log

yx 101010 loglog2

110log

ba

21

b) x210

62 xx

6x

c) 27log

log

3

3 x atau 3log12 3

3log3 x

27x

1

1

1

1

1

1

1

1

8

7. a) AQ = 2QB

[( x - 5)2 + ( y + 2)2]2 = [2( x - 2)2 + ( y – 1)2]2

3 x2 + 3 y

2 – 6 x - 12 y - 9 = 0

x2 + y

2 – 2 x - 4 y - 3 = 0 - terbukti

b) subtitude (-1,0) into locus Q

(-1)2 + (0)2 – 2(-1) – 4(0) - 3 = 0 – terbukti

c) mAC = 15

02

= 3

1

y – 0 = 3

1( x – (-1))

1

1

1,1

1

1


5

y = 3

1x

3

1

d) y = 3

1x

3

1

3 y = - x – 1

x = -1 – 3 y .........(1)

x2 + y

2 – 2 x - 4 y – 3=0.....(2)

substituted (1) into (2)

(-1-3 y )2 + y2 – 2(-1-3 y ) - 4 y – 3=0

10 y2 + 8 y =0

y (10 y + 8)=0

y = 0, y = -5

4

Substituted y = -5

4into (1)

x = - 1 – 3(-5

4)

= 5

7

D (5

7,-

5

4)

1

1

1

1

10

8. a) (i) m1m2 = -1


6

-2

1 m2 = -1

m2 = 2

y – 11 = 2( x – 5 )

y = 2x + 1

(ii)y = 2x + 1.....(1)

2y+x+8=0......(2)

Substituted (1) into (2)

2(2x+1)+x+8=0

x= -2

Substituted x= - 10 into (1)

y = 2(-2)+1

= -3

B(-2,-3)

b)

5

2)11(3,

5

152 yx= (-2,-3)

5

152 x= -2 ,

5

2)11(3 y= -3

x = 2

25, y = - 24

c)

01102210

36121222510

6)11()5(

22

22

22

yxyx

yyxx

yx

1

1

1

1

1

1

1

1

1

1

10


7

9. a)

= 0.8411rad

b) Arc PR = 8 x 0.8411 = 6.769 cm QR2 = 122 – 82 or RT2 = 82 + 82 = 8.944 cm = 11.31 cm

Arc RST= 8 x

= 12.57 cm Perimeter = 4+ 8.944+6.729+11.31+12.57 = 43.56 cm

c) Area of PQR =(

( )

= 8.858 cm2 or

Area of RST =(

= 18.27 cm2 Area of shaded region = 8.858 + 18.272

= 27.13 cm2

1

1

1

1

1

1

1

1

1 1

10

10. a) LN2 = 42 + x2

LN =

ON = 2

16 2x

Area of sector OLMN= 2

=

Area of

= 2x

Area of shaded region = 2x

=

1

1

1

1 1


8

b) ,

when A has a stationary value,

,

hence A is a minimum when ,

Amin=

= cm2.

1

1

1

1 1

10

11

a)

Height (cm)

Number of

children

Cumulative frequency

100-104

8

8

105-109

5

13

110-114

10

23

115-119

6

29

120-124

7

36

125-129

4

40

[ 2 m ]

Note: at least 1 mistake, give 1mark

2

b) Q1 = 104.5 + ( 5

8)40(4

1

) ( 5 )

= 104.5 + 2 = 106.5

1

1

1


9

b)Q2 = 109.5 + ( 10

13)40(2

1

) ( 5 )

= 109.5 + 3.5 = 113

1

1

1

8

12. a) p = 45

q = 45

r = 105

b) 3k + 2k +20 + 5k = 100

k = 8

100

)105(5)80(20)125(2)120(3 kkk

100

)105(40)80(20)125(16)120(24

= 106.8

c) RM 250 x 8.106

100

= RM 234.08

d)106.8 x 100

130

= 138.84

1

1

1

1

1

1

1

1

1

1

10

13. a)

64.2

130

00.2

RMz

y

RMx

1

1

1


10

b)

150

)30110()50150()36130()10125()24120(01/04

I

= 130.73

c)

84.1045

73.130100800

04

04

RMP

P

d)

41.163

100

12573.130

01/07

01/07

I

I

12504/07 I

1

1

1

1

1

1

1

10

14 (i) AB2 = 6

2 + 8.1

2 – 2(6)(8.1) cos 130

0

AB = 12.81

(ii) 0130sinsin

6 AB

BAC

Sin BAC = AB

Sin 01306

= 81.12

1306 0Sin

BAC = 2102’

(iii) The area = 2

1(6)(8.1)sin 130 +

2

1(6)(6)sin80

= 36.34 cm2

1,1

1

1

1

1

1,1,1

1

10


11

15 AD : DB = 1 : 2

AD = = 4 cm

BD = 12 – 4 = 8 cm. a) (i) Using the Cosine Rule in ∆ABC,

BC2= 122 + 72 – 2(12)(7)Cos 110o BC = 15.8259 EC = 15.8259 – 8 = 7.8259 = 7.526 cm (ii) Using Sin Rule in ∆ABC,

Sin <ACB =

<ACB = 45.44o

b) (i) <ABC = 180o – 110o – 45.44o = 24.56o

Area of ∆ ABC =

= 39.4671 cm2 Or

Area of ∆ ADF =

= 6.5778 cm2 Or

Area of ∆BDE =

= 13.3007 cm2 Or

Area of ∆CEF =

= 9.7581 Area of ∆DEF = 39.4671 – 6.5778 – 13.3007 – 9.7581 = 9.8305 = 9.831 cm2

1

1

1

1

1

1

1


12

(ii) Using the Cosine Rule in ∆CEF, EF2 = 3.52 + 7.82592 – 2(3.5)(7.8259)Cos45.44o EF = 5.9209 cm Let the perpendicular distance from D to EF be h cm. Area of ∆DEF = 9.8305 ½ (5.9209)(h) = 9.8305 h = 3.3206 = 3.321 cm

1

1

1

10


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