(Long Questions) – Question 1


Question 1:
Table below shows the price indices and percentage of usage of four items, P , Q , R and S , which are the main ingredients in the production of a type of toy.


(a)  Calculate
(i) the price of S in the year 2004 if its price in the year 2007 is RM43.20 ,
(ii) the price index of P in the year 2007 based on the year 2002 if its price index in the year 2004 based on the year 2002 is 110.

(b) The composite index of the cost of toy production for the year 2007 based on the year 2004 is 125.
Calculate
(i) the value of  x ,
(ii) the price of the toy in the year 2004 if the corresponding price in the year 2007 is RM75.

Solution:
(a)(i)
135 = 43.2 P 2004 × 100 P 2004 = 43.2 × 100 135 = 32 The price of  S  in the year 2004 is RM32

(a)(ii)
Given  P 2004 P 2002 × 100 = 110   P 2004 P 2002 = 110 100 and  P 2007 P 2004 × 100 = 125   P 2007 P 2004 = 125 100 Price index of  P  in the year 2007 based on the year 2002, I = P 2007 P 2002 × 100 I = P 2007 P 2004 × P 2004 P 2002 × 100 I = 125 100 × 110 100 × 100 I = 137.50

(b)(i)
Given composite index  I ¯ = 125 I ¯ = I W W 125 = ( 125 ) ( 30 ) + ( x ) ( 20 ) + ( 115 ) ( 10 ) + ( 135 ) ( 40 ) 30 + 20 + 10 + 40 125 = 10300 + 20 x 100 12500 = 10300 + 20 x x = 110

(b)(ii)
Lets price of the toy in the year 2004 is  P 2004 P 2004 × 125 100 = 75 P 2004 = R M 60

11.2 Composite Index


11.2 Composite Index




Example 1

Noodle
Price index
Weightage
Fried noodle
112
4
Tomyam noodle
104
3
Seafood noodle
109
2
Wantan noodle
111
1







The above table shows the price indices of a few types of noodles sold at a shop for the year 2000 based on the year 1996 and their respective weightages. Calculate the composite index for the year 2000 based on the year 1996.

Solution:
Composite index for the year 2000 based on the year 1996
I ¯ = I W W I ¯ = ( 112 ) ( 4 ) + ( 104 ) ( 3 ) + ( 109 ) ( 2 ) + ( 111 ) ( 1 ) 4 + 3 + 2 + 1 I ¯ = 448 + 312 + 218 + 111 10 I ¯ = 1089 10 = 108.9



Example 2

Item
Price index
Weightage
P
110
4
Q
x
2
R
120
1
S
115
3







The above table shows the price indices of a few types of items, P, Q, R and S, for the year 2002 based on the year 1997 and their respective weightages. If the composite index for the year 2002 based on the year 1997 is 116.5, find the value of x.

Solution:
I ¯ = I W W 116.5 = ( 110 ) ( 4 ) + ( x ) ( 2 ) + ( 120 ) ( 1 ) + ( 115 ) ( 3 ) 4 + 2 + 1 + 3 116.5 = 440 + 2 x + 120 + 345 10 116.5 = 905 + 2 x 10 1165 = 905 + 2 x 2 x = 260 x = 130

11.1 Index Number

11.1 Index Number




Example 1
The price of the house Sulaiman bought in the year 1998 was RM 110000. In the year 2000, the price of the house went up to RM 130000. Calculate the price index of the house for the year 2000 based on the year 1998.

Solution:
P 0 = R M 110000 P 1 = R M 130000 I = P 1 P 0 × 100 I = 130000 110000 × 100 I = 118.18


Example 2
The price of a bottle of cooking oil in the year 1999 was RM15.00. Given that the price index for the year 2000 based on the year 1999 is 105, calculate the price of the cooking oil in the year 2000.

Solution:
P 0 = R M 15 P 1 = x  (Price 2000 ) I = P 1 P 0 × 100 105 = x 15 × 100 1575 = 100 x x = R M 15.75


Example 3
Based on the year 1995, the index number of an oven for the years 1998 and 2001 are 110 and 130 respectively. Find the index number for the year 2001 based on the year 1998.

Solution:
Given,  P 1998 P 1995 × 100 = 110  and  P 2001 P 1995 × 100 = 130 P 2001 P 1998 × 100 = P 2001 P 1995 × P 1995 P 1998 × 100                             = 130 100 × 100 110 × 100                             =118 .18

(Long Questions) – Question 6


Question 6:
Diagram below shows trapezium ABCD.
(a) Calculate
(i) ∠BAC.
(ii) the length, in cm, of AD.
(b) The straight line AB is extended to B’ such that BC = B’C.
(i) Sketch the trapezium AB’CD.
(ii) Calculate the area, in cm2, of ∆BB’C.  


Solution:
(a)(i)
5 2 = 4 2 + 7 2 2( 4 )( 7 )cosBAC 25=16+4956cosBAC 56cosBAC=40 cosBAC= 40 56  BAC= cos 1 40 56    = 44 o 25'


(a)(ii)
AD sinDCA = 7 sin 115 o AD sin 44 o 25' = 7 sin 115 o ( DCA=BAC )   AD= 7 sin 115 o ×sin 44 o 25'   AD=5.406 cm


(b)(i)




(b)(ii)
sinABC 7 = sin 44 o 25' 5 sinABC= sin 44 o 25' 5 ×7    = 78 o 28' ABC= 180 o 78 o 28' ABC= 101 o 32'( obtuse angle ) CBB'= 180 o 101 o 32'= 78 o 28' BCB'= 180 o 78 o 28' 78 o 28'= 23 o 4' Area of BB'C= 1 2 ×5×5× 23 o 4'  =4.898  cm 2

(Long Questions) – Question 5


Question 5:
The diagram below shows a triangle ABC.

(a) Calculate the length, in cm, of AC.
(b) A quadrilateral ABCD is now formed so that AC is a diagonal, ACD = 45° and AD = 14 cm. Calculate the two possible values of ADC.
(c) By using the acute ADC from (b), calculate
 (i) the length, in cm, of CD,
 (ii) the area, in cm2, of the quadrilateral ABCD


Solution:
(a)
Using cosine rule,
AC2 = AB2 + BC2 – 2 (AB)(BC) ABC
AC2 = 162 + 122 – 2 (16)(12) cos 70o
AC2 = 400 – 131.33
AC2 = 268.67
AC = 16.39 cm


(b)

Using sine rule, sin A D C 16.39 = sin 45 14 sin A D C = 16.39 × sin 45 14 sin A D C = 0.8278 A D C = 55.87  or  ( 180 55.87 ) A D C = 55.87  or 124 .13


(c)(i)
Acute angle of  A D C = 55.87 C A D = 180 45 55.87 = 79.13 C D sin 79.13 = 14 sin 45 C D = 14 × sin 79.13 sin 45 = 19.44  cm


(c)(ii)
Area of quadrilateral  A B C D = Area of  Δ   A B C + Area of  Δ   A C D = 1 2 ( 16 ) ( 12 ) sin 70 + 1 2 ( 16.39 ) ( 14 ) sin 79.13 = 90.21 + 112.67 = 202.88  cm 2

(Long Questions) – Question 4


Question 4:
Diagram below shows a quadrilateral PQRS.


(a) Find
(i) the length, in cm, of QS.
(ii) ∠QRS.
(iii) the area, in cm2, of the quadrilateral PQRS.
(b)(i) Sketch a triangle S’Q’R’ which has a different shape from triangle SQR such that S’R’ = SR, S’Q’ = SQ and ∠S’Q’R’ = ∠SQR.
(ii) Hence, state ∠S’R’Q’.

Solution:
(a)(i)
P=1807634=70 QS sin70 = 8 sin34 QS= 8×sin70 sin34  =13.44 cm

(a)(ii)
13.44 2 = 6 2 + 9 2 2( 6 )( 9 )cosQRS 108cosQRS= 6 2 + 9 2 13.44 2 cosQRS= 6 2 + 9 2 13.44 2 108  QRS= cos 1 ( 0.5892 )    = 126 o 6'

(a)(iii)
Area of PQRS =Area of PQS+Area of QRS =( 1 2 ×8×13.44×sin76 )+( 1 2 ×6×9×sin 126 o 6' ) =52.16+21.82 =73.98  cm 2


(b)(i)



(b)(ii)
S'R'Q'=S'RR'    =180 126 o 6'    = 53 o 54'

(Long Questions) – Question 3


Question 3:


The diagram shows a trapezium PQRS. PS is parallel to QR and QRS is obtuse. Find
(a) the length, in cm, of QS,
(b) the length, in cm, of RS,
(c) QRS,
(d) the area, in cm2, of triangle QRS.


Solution:
(a)
Q S sin P = P S sin Q Q S sin 85 = 13.1 sin 28 Q S = 13.1 × sin 85 sin 28 Q S = 27.8  cm

(b)
 RQS = 180o – 85o – 28o
 RQS = 67o
Using cosine rule,
RS2 = QR2 + QS2 – 2 (QR)(QS) RQS
RS2 = 6.42 + 27.82 – 2 (6.4)(27.8) cos 67o
RS2 = 813.8 – 139.04
RS2 = 674.76
RS = 25.98 cm

(c)
Using cosine rule, Q S 2 = Q R 2 + R S 2 2 ( Q R ) ( R S ) cos Q R S 27.8 2 = 6.4 2 + 25.98 2 2 ( 6.4 ) ( 25.98 ) cos Q R S 772.84 = 715.92 332.54 cos Q R S cos Q R S = 715.92 772.84 332.54 cos Q R S = 0.1712 Q R S = 99.86

(d)
Area of triangle QRS
= ½ (QR)(RS) sin R
= ½ (6.4) (25.98) sin 99.86o
= 81.91 cm2

(Long Questions) – Question 2


Question 2:
In the diagram below, ABC is a triangle. AGJB, AHC and BKC are straight lines. The straight line JK is perpendicular to BC.


It is given that BG= 40cm, GA = 33 cm, AH = 30 cm, GAH = 85o and JBK= 45o.
(a) Calculate the length, in cm of
  i.      GH
    ii.   HC
(b) The area of triangle GAH is twice the area of triangle JBK. Calculate the length, in cm, 
  of BK.
(c) Sketch triangle  which has a different shape from triangle ABC such that, A’B’ = AB
 A’C’ = AC and A’B’C’ = ABC.


Solution:
(a)(i)
Using cosine rule,
GH2 = AG2 + AH2 – 2 (AG)(AH) GAH
GH= 332+ 302 – 2 (33)(30) cos 85o
GH2 = 1089 + 900 – 172.57
GH2 = 1816.43
GH = 42.62 cm

(a)(ii)
A C D = 180 45 85 = 50 Using sine rule, A C sin 45 = 73 sin 50 A C = 73 × sin 45 sin 50 A C = 67.38  cm H C = 67.38 30 = 37.38  cm


(b)
Area of  Δ   G A H = 1 2 ( 33 ) ( 30 ) sin 85 = 493.12  cm 2 Let length of  B K = J K = x ×  Area of  Δ   J B K  = Area of  Δ   G A H 2 × [ 1 2 ( x ) ( x ) ] = 493.12 x 2 = 493.12 x = 22.21  cm B K = 22.21  cm


(c)

(Long Questions) – Question 1


Question 1:

The diagram shows a quadrilateral ABCD. The area of triangle BCD is 12 cm2 and BCD is acute. Calculate
(a) ∠BCD,
(b) the length, in cm, of BD,
(c) ABD,
(d) the area, in cm2, quadrilateral ABCD.


Solution:
(a) Given area of triangle BCD = 12 cm2
½ (BC)(CD) sin C = 12
½ (7) (4) sin C = 12
14 sin C = 12
sin C = 12/14 = 0.8571
C = 59o
BCD = 59o
 
(b)
Using cosine rule,
BD2 = BC2 + CD2 – 2 (7)(4) cos 59o
BD2 = 72+ 42 – 2 (7)(4) cos 59o
BD2 = 65 – 28.84
BD2 = 36.16
BD = 36.16
BD = 6.013 cm

(c)
Using sine rule,
A B sin 35 = 6.013 sin A 10 sin 35 = 6.013 sin A sin A = 6.013 × sin 35 10 sin A = 0.3449 A = 20.18 A B D = 180 35 20.18 A B D = 124.82

(d)

Area of quadrilateral ABCD
= Area of triangle ABD + Area of triangle BCD
= ½ (AB)(BD) sin B + 12 cm
= ½ (10) (6.013) sin 124.82 + 12
= 24.68 + 12
= 36.68 cm²

10.3 Area of Triangle


10.3 Area of Triangle

Example:
Calculate the area of the triangle above.

Solution:
A r e a = 1 2 a b sin C A r e a = 1 2 ( 7 ) ( 4 ) sin 40 o A r e a = 9  cm 2



Example:

Diagram above shows a triangle ABC, where AC = 8 cm and ∠C = 32o. Point D lies on straight line BC where BD = 10 cm and ∠ADB = 70o . Calculate
(a) the length, in cm, of CD,
(b) the area, in cm2 of ∆ ADC,
(c) the area, in cm2 of ∆ ABC,
(d) the length, in cm, of AB.

Solution:
(a)
ADC+ 70 o = 180 o ADC= 110 o CAD= 180 o 110 o 32 o CAD= 38 o Using Sine Rule, 8 sin 110 o = CD sin 38 o CDsin 110 o =8sin 38 o CD( 0.940 )=8( 0.616 ) CD=5.243

(b)
Area of ADC = 1 2 ( 8 )( 5.243 )sin 32 o =20.972( 0.530 ) =11.12  cm 2

(c)
Area of ABC = 1 2 ( 8 )( 10+5.243 )sin 32 o =60.972( 0.530 ) =32.315  cm 2

(d)
Use Cosine Rule for ABC, A B 2 = 15.243 2 + 8 2 2( 15.243 )( 8 )cos 32 o A B 2 =232.35+64206.83 A B 2 =89.52 AB=9.462 cm