Module & More Add Math Tg 4

  Matematik Tambahan  Tingkatan 4  Bab 7 Geometri Koordinat SPM 7

PRAKTIS

Kertas 1 2. Rajah menunjukkan tiga garis lurus, dengan

1. Diberi bahawa A(6,−1) dan B(−2,5) terletak pada SPM keadaan k, p, q dan r ialah pemalar.
SPM suatu satah cartes. 2019
Diagram shows three straight lines, such that k, p, q and
2019 It is given that A(6, −1) and A(−2, 5) lie on a cartesian
plane. r are constants.

(a) Nyatakan jarak AB. y

State the distance AB. ky = 4x + 8

(b) Garis lurus AB dipanjangkan ke titik C dengan x
keadaan jaraknya dari titik B adalah dua kali 0
jarak AB. Cari koordinat C. –r –px + –qy = 1

The straight line AB is extended to point C such that
its distance from point B is twice the distance AB.
Find the coordinates of C.

(a) AB = √(6 + 2)2 + (–1 – 5)2
= √64 + 36 
= 10 unit

(b) Ungkapkan / Express
(a) k dalam sebutan q.
C(h, k) k in terms of q.
(b) r dalam sebutan k dan p.
2 r in terms of k and p.

BAB 7 B(–2, 5) 1

(a) Pintasan-y  / y-intercept = 8
k

A(6, –1) q = 8
k

(–2, 5) = 1 2(6) + 1(h) , 2(–1) + 1(k) 2 k = 8q
3 3
4
123+ h = –2 –2 + k = 5 (b) Kecerunan  / Gradient = k
12 + h = –6 3
h = –18 –r – 0 4
–2 + k = 15 0–p = k

k = 17 r 4
p k
\  C = (–18, 17) =

r = 4p
k

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  Matematik Tambahan  Tingkatan 4  Bab 7 Geometri Koordinat 

3. y (km) Kertas 2
O
SPM 1. Rajah menunjukkan kedudukan kilang K dan
2015

Rumah Ali SPM kilang L yang dilukis pada satah Cartes.
Ali’s House 2017
Diagram shows the location of factory K and factory L
x (km)
drawn on a Cartesian plane.

yT

Rumah Siti K(–4, 1)

Siti’s House x
0
Rajah di atas menunjukkan kedudukan rumah
Ali dan rumah Siti. Koordinat bagi rumah Ali S L(2, –1)
dan rumah Siti masing-masing ialah (14, 12) dan
(−10, −8). Ali dan Siti berbasikal dari rumah ke ST ialah jalan raya lurus dengan keadaan jarak
arah satu sama lain pada sebatang jalan raya yang dari kilang K dan kilang L ke mana-mana titik
lurus dengan halaju berbeza. Diberi halaju Ali pada jalan raya adalah sentiasa sama.
ialah tiga kali halaju Siti. Cari jarak antara rumah
Ali dengan tempat mereka bertemu. ST is a straight road such that the distance from factory

The diagram above shows the positions of Ali’s house K and factory L to any point on the road is always
and Siti’s house. The coordinates of Ali’s house and Siti’s
house are (14, 12) and (−10, −8) respectively. Ali and Siti equal.
cycle from their house towards each other on a straight
road with different velocity. Given that the velocity of Ali (a) Cari persamaan bagi ST.
is three times velocity of Siti. Find the distance between
Ali’s house and the place where they meet. Find the equation of ST.

(b) Satu lagi jalan raya lurus, MN dengan

persamaan y = 8 – 2x akan dibina.

Kedudukan mereka bertemu Another straight road, MN with an equation

The position where they meet y = 8 – 2x is to be built.
1
3

BAB 7
(i) Lampu isyarat akan dipasang di

= 1 3(–10) + 1(14)  ,  3(–8) + 1(12) 2 persimpangan kedua-dua jalan raya itu.
3+1 3 + 1
Cari koordinat bagi lampu isyarat itu.

= 1 –16  ,  –12 2 A traffic light is to be installed at the cross roads
4 4
A(14, 12) of the two roads. Find the coordinates of the

= (−4, –3) traffic light.

(ii) Antara dua jalan raya itu, yang manakah

Jarak / Distance melalui kilang R1– 4 , –12?
= [14 – (–4)]2 + [12 – (–3)]2 3
= 182 + 152 Which of the two roads passes through factory
= 549
= 23.43 unit (x, y) –1 2R4, –1 ?
B(–10, –8) 3

(a) Katakan P(x, y) ialah titik pada ST

Let P(x, y) is a point at ST.

PK = PL

  √[x – (–4)]2 + (y – 1)2 = √(x – 2)2 + [y – (–1)]2

(x + 4)4 + (y – 1)2 = (x – 2)2 + (y + 1)2

x2 + 8x + 16 + y2 – 2y + 1 = x2 – 4x + 4 + y2 + 2y + 1

12x – 4y + 12 = 0

3x – y + 3 = 0

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  Matematik Tambahan  Tingkatan 4  Bab 7 Geometri Koordinat

(b) (i) y = 3x + 3 …… 1 (a) persamaan bagi laluan titik P.

y = 8 – 2x …… 2 the equation of the path of point P.

Gantikan/Replace 1 ke dalam / into 2: (b) nilai-nilai yang mungkin bagi q.

3x + 3 = 8 – 2x the possible values of q.

5x = 5 (c) luas ∆ABC.

x = 1 the area of ∆ABC.

Apabila / When x = 1, y = 3(1) + 3 = 6

\  koordinat bagi lampu isyarat = (1, 6) (a) Jejari / Radius, AB = √(1 – 2)2 + (2 + 1)2

coordinates of the traffic light = √1 + 9  = √10 

(ii) Gantikan koordinat kilang R1– 4 , –12 ke PA = √10 
3
dalam kedua-dua persamaan jalan raya, √(x – 2)2 + (y – (–1)2 = √10 

ST dan MN. (x − 2)2 + (y + 1)2 − 10 = 0

1 2 4 , –1 x 2 − 4x + 4 + y2 + 2y + 1 − 10 = 0
Replace the coordinates of factory R – 3
x2 + y2 − 4x + 2y − 5 = 0

into both equations of the road, ST and MN.

⇒  ST : y = 3x + 3 (b) R(3, q): x = 3, y = q
4
–1 = 31– 3 2 + 3 9 + q2 – 12 + 2q – 5 = 0

–1 = –4 + 3 q2 + 2q – 8 = 0

–1 = –1 q = 2, q = –4

⇒ MN : y = 8 – 2x 4 (c) mAB × mBC = –1
3
–1 = 8 – 21– 2 1 –21––12 2 × mBC = –1

–1 = 8+ 8 –3 × mBC = –1
3
–1 ≠ 332
mBC = 1
3
\  hanya jalan raya ST melalui kilang R.
  only the road ST passes through factory R. Persamaan BC / Equation of BC

BAB 7 2. Rajah menunjukkan laluan titik bergerak P(x, y). y – 2 = 1 (x – 1)
3
SPM P sentiasa bergerak dengan jarak tetap dari titik A.
2019 3y – 6 = x – 1
Diagram shows the path of a moving point P(x, y). P is
3y = x + 5
always moving at a constant distance from point A.

y Pada C / At C, y = 0.
x + 5 = 0
B(1, 2) x = –5
\  C(–5, 0)

0 x Luas ∆ABC / Area of ∆ABC
A(2, –1)
u u=1 –5
2 21 0 2
–1 2 –1

P(x, y) = 1  z(4 + 0 + 5) – (–1 – 10 + 0)z
2

= 1  z9 – (–11)z
2

B(1, 2) dan R(3, q) terletak pada laluan titik P. = 1  z20z
2
Garis lurus BC ialah tangen kepada laluan itu dan Praktis
= 10 unit2 SPM
bersilang dengan paksi-x pada titik C. Cari Ekstra

B(1, 2) and R(3, q) lie on the path of point P. The straight
line BC is a tangent to the path and intersects the x-axis
at point C. Find

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  Matematik Tambahan  Tingkatan 4  Bab 7 Geometri Koordinat 

Sudut KBAT KBAT

Ekstra

Dalam rajah, SRT adalah selari kepada paksi-y. QR adalah berserenjang dengan SRT. Titik P(x, y) bergerak dengan
keadaan jaraknya dari Q(6, 5) adalah dua kali jaraknya dari garis lurus SRT.

In the diagram, SRT is parallel to the y-axis. QR is perpendicular to SRT. The point P(x, y) moves in such a way that its distance
from Q(6, 5) is twice its distance from the straight line SRT.

y

S

R Q(6, 5)

–4 0 x

T

(a) Cari persamaan lokus bagi titik P.

Find the equation of the locus of point P.

(b) Tentukan sama ada lokus itu menyilang paksi-x atau tidak.

Determine whether the locus intersect the x-axis.

(a) Katakan P(x, y) PQ = 2 × jarak SRT dengan P
(x – 6)2 + (y – 5)2 = 2[x – (–4)]2



(x – 6)2 + (y – 5)2 = 4(x + 4)2

x2 – 12x + 36 + y2 – 10y + 25 = 4(x2 + 8x + 16) BAB 7

x2 + y2 – 12x – 10y + 61 – 4x2 – 32x – 64 = 0

–3x2 + y2 – 44x – 10y – 3 = 0

3x2 – y2 + 44x + 10y + 3 = 0

(b) Pada paksi-x, y = 0

At x-axis, y = 0

3x2 – (0)2 + 44x + 10(0) + 3 = 0
3x2 + 44x + 3 = 0

a = 3, b = 44, c = 3
b2 – 4ac = 442 – 4(3)(3)
= 1 900 > 0

b2 – 4ac > 0
Lokus itu menyilang paksi-x pada dua titik.

The locus intersects the x-axis at two points.

Kuiz 7

143

BAB Vektor

8 Vectors

8.1 Vektor

Vectors

NOTA IMBASAN

NOTA IMBASAN

B
B

A
A

~a 2~a
–2~a

144

  Matematik Tambahan  Tingkatan 4  Bab 8 Vektor 

1. Nyatakan tembereng garis yang diberi dalam bentuk vektor. TP 2

State the line segments given in the vector form.

CONTOH

T m~ S Penyelesaian:
~n P→Q = ~q
U ~s Q→R = ~r
S→R = ~s
~p R S→T = –T→S = – m~
~r T→U = –U→T = –~n
P U→P = ~p
~q
Q



(a) B O→A = ~a (b) ~p K→L = ~q
~b O→B = –B→O = –~b K→N = – N→K = –~p
A ~d O→D = ~d K N M→N = ~r
O→C = –C→O = –~c ~r L→M = –M→L = –~s
~a D ~q
L
~c O
C ~s
M

2. Binakan vektor yang berikut. TP 2 8BAB BAB8

Construct the following vectors.

CONTOH

Diberi P→Q = ~p , bina Penyelesaian:

Given that P→Q = p~ , construct Q
~p
(i) A→B = 2~p
P B E
2~p
(ii) C→D = – 1 ~p – –23~p
2

(iii) E→F = – 3 ~p C
2 F

A D – –12~p



145

  Matematik Tambahan  Tingkatan 4  Bab 8 Vektor

(a) Diberi O→P = ~a , bina P C
~a
Given that O→P = a~ , construct E H –23~a
–2~a – –53~a G
(i) A→B = 1 ~a (ii) C→D = –2~a Q
3 D
B
(iii) E→F = – 5 ~a (iv) G→H = 2 ~a –31~a A
3 3
F

3. Tentukan magnitud vektor-vektor berikut. TP 3

Determine the magnitude of the following vectors.

CONTOH

r 3 unit / Penyelesaian: 12 unit / t | ~t | = 92 + 122
units |~r | = 32 + 42 units ~ = 15 unit
= 5 unit/ units
4 unit / units 9 unit / units

4. Nyatakan pasangan vektor-vektor yang sama. TP 2

State the pairs of vectors that are equal.

CONTOH

Penyelesaian:

BAB 8 ~p ~s ~u ~v ~p = m~ Tip
m~
~t ~u = ~x Vektor yang sama
(a) w~ ~x ~s = ~v mempunyai magnitud
dan arah yang sama.
~a Vectors that are equal
~s have equal magnitude
and same direction.

~b ~v ~c ~s = w~
~t = ~v
~t ~r ~a = ~c
w~ ~b = ~r

146

  Matematik Tambahan  Tingkatan 4  Bab 8 Vektor 

5. Ungkapkan vektor berikut dalam sebutan ~a atau ~b . TP 3

Express the following vectors in terms of a~ or b~ .

CONTOH

Q D Penyelesaian:
BC F
~a A→B = 1 ~a
P E 2

A C→D = –2~a

E→F = 1 1 ~a
2

A→B C→D
(a)

P 3~b (i) = – 1 ~a (ii) = 1 1 ~a
Q 2 2
S
~a V R (iii) Q→S = 7~b (iv) R→Q = –5~b
BC
D

U (v) S→R = –2~b
A

6. Tentukan vektor-vektor yang selari dan nyatakan hubungan mereka. TP 3

Determine the vectors that are parallel and state their relationships.

CONTOH BAB 8

Penyelesaian:
(i) A→B dan R→S adalah selari.
A PQ A→B and R→S are parallel.
B R A→B R→S
= 1
G H 2

DS (ii) G→H dan C→D adalah selari.
F
G→H and C→D are parallel.
C G→H = −2C→D
E

(iii) EE→→FF dan P→Q adalah selari.
E→F
and P→Q are parallel.
P→Q
= −1 1
2


147

  Matematik Tambahan  Tingkatan 4  Bab 8 Vektor

(a) N (i) M→N dan / and K→L adalah selari / are parallel.
P
M Q M→N = 1 K→L
3
C (ii) C→D dan / and G→H adalah selari / are parallel.
C→D = −2G→H
L
D (iii) S→T dan / and P→Q adalah selari / are parallel.
KG T S→T = 3P→Q
S

H

7. Tentukan sama ada titik P, Q dan R adalah segaris. TP 4

Determine whether points P, Q and R are collinear.

CONTOH (a) Diberi / Given P→Q = 10~a dan / and P→R = 4~a .

Diberi / Given P→Q = 6~a dan / and Q→R = 2~a . P→Q = 5 (4 ~a )
Penyelesaian: 2
P→Q = 3(2~a )
P→Q = 3Q→R P→Q = 5 P→R
\ P→Q adalah selari dengan Q→R, dengan Q ialah titik 2
sepunya. Titik P, Q dan R adalah segaris.
\ P→Q adalah selari dengan P→R , dengan P ialah
\ P→Q is parallel to Q→R, with Q is common point.
Points P, Q and R are collinear. titik sepunya.
\ P→Q is parallel to P→R , with P is common point.

Titik P, Q dan R adalah segaris.
Points P, Q and R are collinear.

8.2 Penambahan dan Penolakan Vektor

Addition and Subtraction of Vectors

BAB 8 NOTA IMBASAN

1. Hasil tambah dua vveecktotrosr ~a~a adnadn~b~b, i,sdkinkoewnanliasserbesaugltaainvtevketcotorrpaandduisawnrdittaenn ditulis sebagai ~a + ~b .
The addition of two as ~a + ~b .

2. Apabila dua vektor selari ditambah, vektor paduan mempunyai magnitud tertentu dan selari dengan dua vektor tersebut.

When two parallel vectors are added, the resultant vector has a certain magnitude and parallel with the two vectors.

| ~a + ~b | = |~a | + |~b |

3. Hasil tambah vektor-vektor tidak selari / Addition of non-parallel vectors

(a) Hukum Segi Tiga (b) Hukum Segi Empat Selari (c) Hukum Poligon
Triangle Law Parallelogram Law Poligon Law

C DC B
~u + ~v
~u + ~v ~v A C
~v

A ~u B A ~u B OD

A→C = A→B + B→C A→C = A→B + B→C = A→D + D→C O→D = O→A + A→B + B→C + C→D

148

NOTA IMBASAN

  Matematik Tambahan  Tingkatan 4  Bab 8 Vektor 

4. Penolakan bagi suatu vektor, a~add–it~bio,niaolfavhecptoern~aamanbdahaannegvaetikvteovrec~atodr e~bn,gthaant vektor negatif ~b , iaitu
Subtraction of vector, ~a – ~b, is an is

(a) SPuebntoralactkioannovfepkatroarll-evlevkectotorrss,e~alaarin, d~a~bd.an ~b . ~a – ~b = ~a + (–~b )


~a
~a

–~b
~b ~a – b~

(b) Penolakan vektor-vektor tidak ~asealnadri,~b~a. dan ~b .
Subtraction of nonparallel vectors,

A C –~b A

~a ~a – b~ ~a

O ~b B B –~b O NOTA

8. Tentukan vektor paduan bagi vektor-vektor yang berikut. TP 3

Determine the resultant vector of the following vectors.

CONTOH 1 CONTOH 2

3~b + 2~a – 3 ~b + 4~a Hanya vektor yang selari boleh 5~x – 5 ~y – 7~x + 8~y
2 ditambahkan. 2
Only parallel vector can be added.
Penyelesaian: Penyelesaian:

1 23 – 3 ~b + (2 + 4)~a = 3 ~b + 6~a 1 2(5 – 7)~x + – 5 +8 ~y = –2~x + 11 ~y
2 2 2 2
BAB 8
(a) 6m~ – 3~n – 5m~ – 2~n (b) 8~p – 3~q + ~p – 2~q
= (6 – 5)m~ + (–3 – 2)~n = (8 + 1)~p + (–3 – 2)~q
= m~ – 5~n = 9~p – 5~q

(c) 2~s + 3 ~t – 1 ~s + 5~t (d) 3~x – 4~y – 7~x – 5~y
4 2 = (3 – 7)~x + (–4 – 5)~y
= –4~x – 9~y
1 2 1 2 2– 1 ~s + 3 +5 ~t
2 4 149

= 3 ~s + 23 ~t
2 4

= 1 1 ~s + 5 3 ~t
2 4

  Matematik Tambahan  Tingkatan 4  Bab 8 Vektor

9. Tentukan vektor paduan bagi vektor-vektor tidak selari yang berikut. TP 3

Determine the resultant vectors of the following non-parallel vectors.

CONTOH 1 CONTOH 2

U N M
PT J L

Q S K
R

Penyelesaian: Penyelesaian:
(i) P→Q + Q→R = P→R (i) J→N – M→N = J→N – (–N→M) = J→M
(ii) Q→S + S→U = Q→U (ii) J→M – K→M = J→M – (–M→K) = J→K
(iii) P→T + T→S = P→S (iii) K→N – L→N = K→N – (–N→L) = K→L
(iv) P→S + S→Q = P→Q

(a) PQRS ialah segi empat selari. (b) ~y L
K ~z
PQR is a parallelogram Q M

P

~x

S R J N

BAB 8 (i) P→Q + R→S = P→R (i) ~x + ~y = J→L
(ii) S→P + S→R = S→Q (ii) ~y + ~z = K→M
(iii) Q→P + Q→R = Q→S (iii) J→K + K→M = J→M
(iv) S→R + R→Q = S→Q (iv) K→N + N→J = K→J
(v) L→N + N→K = L→K

(c) K L (d) ~a Q
P ~b
JM
R ~c
PN
S
(i) J→K – M→K = J→K – (–K→M) = J→M T
(ii) K→N – P→N = K→N – (–N→P) = K→P
(iii) J→L – N→L – J→N = J→L + L→N + N→J = ~0 (i) Q→R – S→R = ~b – ~c / Q→S
(iv) K→M – P→M – J→P = K→M + M→P + P→J = K→J (ii) S→P = S→R + R→Q + Q→P = ~c – ~b – ~a
(iii) P→R = P→Q + Q→R = ~a + ~b
(iv) Q→S = Q→R + R→S = ~b – ~c

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  Matematik Tambahan  Tingkatan 4  Bab 8 Vektor 

1 0. Selesaikan masalah yang melibatkan penambahan dan penolakan vektor. TP 4

Solve the problems involving addition and subtraction of vectors.

CONTOH

Rajah menunjukkan sebuah segi tiga PQR. Q Penyelesaian:
Diberi bahawa (i) R→S = R→Q + Q→S

dTPP→→haTSen= =d/i3aaT→gt~xQar,au.mQ→C~yRas.hr=iowd6a~ysl,aamQ→trSiasen=bg6lue~ytPaQ–nR4~x.~xGdivaenn T = −6~y + (6~y − 4~x ) = − 4~x
R (ii) P→Q = P→R + R→Q
P S

(FP→ii)Snd=iR→n3S~xt e,rQm→ Rs (o=ifi6)~xy~aP→, nQQ→d S/= or6(yy~~ii.i–)4 that = [3~x – (– 4~x )] − 6~y = 7~x − 6~y
(iii) T→S = T→P + P→S
~x and P→T = T→Q .

T→S =– 1  (7~x − 6~y ) + 3~x
2

=– 1  ~x + 3~y
2

(b) Rajah di bawah menunjukkan sebuah sisi empat
(a) Rajah di bawah menunjukkan sebuah sisi PQRS.
empat OACD.
The diagram below shows a quadrilateral PQRS.
The diagram below shows a quadrilateral OACD.
R
C

O ~b D U
2~a B S

A ~x T Q
~y
Diberi bahawa O→A = 2~a , O→B = ~b , C→D = ~a − 3~b
dan 2A→B = B→C . Cari vektor-vektor berikut dalam P

sebutan ~a dan / atau ~b . Diberi P→S = ~x , P→P→TT ==~yT→,UR→.UC=ari21v(e~xkt–or3-~vy e),ktor
Given that O→A = 2 a~ , O→B = b~ , C→D = a~ − 3 b~ and R→U = 1 R→Q
2A→B = B→C . Find the following vectors in terms of 3 dan BAB 8

and / or b~ . ~a bGR→eiUvrei=knut13ht aRd→tQalPa→aSmnd=seP→~xbT,uP=→tTaT→nU= ~x.y~Fd,inaR→ndU/tah=tea21fuo(ll~~xyow.– i3nyg~

(i) A→B  (ii) O→C  (iii) D→A ),
vectors

(i) A→B = A→O + O→B = –A→O + O→B in terms of ~x and / or y~ . P→Q

= −2~a + ~b (i)  S→T        (ii) 

(ii) O→C = O→A + A→C = O→A + 3A→B (i) S→T = S→P + P→T = −~x + ~y

= −24~a~a++3(3−~b2~a + ~b )
=
P→Q = P→U + U→Q = 2~y 2 R→Q
(iii) D→A = D→C + C→A = –C→D + (–A→C) (ii) + 3

= –C→D + (–3A→B) = 2~y + 2 (3R→U)
3
= 5−(~a~a − 3~b ) − 3(−2~a + ~b ) 1 2 1
= = 2~y + 2 2 (~x – 3~y )

= ~x − ~y

151

  Matematik Tambahan  Tingkatan 4  Bab 8 Vektor

8.3 Vektor dalam Satah Cartes

Vectors in a Cartesian Plane

NOTA IMBASAN

1. Vektor unit ialah vektor dengan magnitud 1 unit. O→P = x ~i + y ~j = 1 x 2
A unit vector is a vector with a magnitude of 1 unit. y

2. Semua vektor dalam satah Cartes boleh diungkapkan Magnitud bagi O→P / Magnitude of O→P
| O→P | = x2 + y2
duAdlnaal llviaatNemmcdOtaoalrsTasreamAibnhuaIaptMrCaoaansBhrittAep~iifsoiSpasdnaAiatkpinNfslaip-n~xjae,k|cds~aiii-n|ym=b, e|1a~jenux|ap=nrie~t1isdsueiaadnnliiatn~.hjteivramelakshtoofvr~ei ukatnnoidtr
5. Vektor unit pada arah ~a ditandakan saesb~a^agwahier~a^e dengan /
Unit vector in the direction of ~a is denoted

|~j=w1huenreit ~i is a unit vector in the positive direction of the x-axis, | ~i ~a^ = |~~aa|  = (x~i + y~j )  
and is a unit vector in the positive direction of the y-axis, x2 + y2

| ~j | unit.

3. Vektor unit ~i dinyatakan dalam bentuk vektor lajur, 6. Penambahan dan penolakan dua atau lebih vektor:
1 2~i =1
0 dan vektor unit ~j dinyatakan dalam bentuk Addition of two or more vectors:
Jika/ if  ~a = x1~i + y1~j = xy11
1 2 1 2vektor lajur, ~j = 0 .
1

1 2 1 2 1 yx22
Unit vector ~i can be written as a column vector, ~i = 0 and dan/ and ~b = x2~i + y2~j =
maka, / then,
1 2unit vector ~j can be written as a column vector, ~j = 0 .
1

1 2 1 2 1 2 1 2x~i + y~j atau yx11 yx22 yx11 ± yx22
4. Vektor dalam satah cartes boleh diwakili dalam bentuk ~a + ~b = ± = ±

x . 7. Pendaraban vektor dengan skala
y

Any vector on a Cartesian plane can be represented in the form Multiplication of vectors with a scalar
x
1 2x ~i + y ~j or y . Jika / if ~a = x1~i + y1~j = yx11
1 2y

P(x, y) Maka / Then 1 x 2 1kkxy2
y
BAB 8 k ~a = k =

y~j k ialah pemalar.
k is a constant.

O x~i x

1 2 11. Ungkapkan vektor-vektor berikut dalam bentuk x dan x~i + y~j . TP 3
y

1 2Express the following vectors in the form x and x ~i + y ~j .
y

CONTOH Penyelesaian:

y Dari koordinat (x, y), tukar ke A(6, 3)
4
A 1 2bentuk vektor lajur x . 1 2O→A = 6
2 y 3

From the coordinates (x, y), O→A = 6~i + 3~j

O x 1 2change to column vector x .
246 y



152

  Matematik Tambahan  Tingkatan 4  Bab 8 Vektor 

y B(5, 5) C(−3, 4)

5 B
C4
1 2O→B = 5 1 2O→C =–3
3 5 4

2 O→B = 5i + 5 j O→C = −3i + 4 j

1 x D(−6, −2) E(3, −3)
–7 –6 –5 –4 –3 –2 ––11O 12345

D –2 1 2O→D = –6 1 2O→E =3
–3 E –2 –3
–4
O→D = −6i − 2 j O→E = 3i − 3 j

12. Ungkapkan vektor-vektor berikut dalam bentuk:

Express the following vectors in the form:

1 2(i) x~i + y~j (ii) yx

Seterusnya, cari magnitud. TP 4

Hence, find the magnitude.

CONTOH

y Tip Penyelesaian:
B4
C Guna titik tamat (i) A(−4, 1), B(−2, 4)
menolak titik
2 mula. A→B = (−2 − (−4))~i + (4 − 1)~j = 2~i + 3~j
–2 O Use ending point
subtracting C(−1, 3), D(4, −2)
starting point.
A x C→D = (4 − (−1))~i + (−2 − 3)~j = 5~i − 5~j
–6 –4 24

1 2 1 2(ii) A→B =2 5
–2 D 3 , C→D = –5

A→B = 22 + 32 = 13 , C→D = 52 + (–5)2­ = 50 BAB 8



(a) (i) ~a = 3~i + 2~j ,  ~b = −3~i − 6~j ,  ~c = 4~i − 3~j

y

4 1 2 1 2 1 2(ii) ~a =3 –3 4
~a 2 ~b 2 , ~b = –6 , ~c = –3
x
–6 –4 –2 O ~a  = 32 + 22 = 13
24

–2 ~b  = (–3)2 + (–6)2 = 45
~c
~c  = 42 + (–3)2­ = 5
–4

153

  Matematik Tambahan  Tingkatan 4  Bab 8 Vektor

13. Berdasarkan vektor-vektor g~aivdena,nfin~bd, yang diberikan, cari vektor yang berikut. TP 4
Based on the vectors ~a and ~b
the following vectors.

CONTOH (a) ~a = 3~i – 4~j , ~b = –2~i + 5~j
(i) ~a + 3~b
~a = 2~i – 3~j , ~b = 5~i + 2~j (ii) 2~a – 5~b
(i) ~a + ~b (iii) 4~a + 2~b
(ii) 3~a – 2~b
(iii) 2~a – 5~b 1 2 1 2(i) ~a + 3~b =3 +3 –2
–4 5
Penyelesaian:

1 2 1 2(i) ~a + ~b = 2 + 5 1 2 1 2 = 3 + –6
–3 2 –4 15

1 2 = 7 = 7~i – ~j 1 2 = –3 = –3~i + 11~j
–1 11

1 2 1 2(ii) 3~a – 2~b = 3 2 –2 5 1 2 1 2(ii) 2~a – 5~b = 23 –5 –2
–3 2 –4 5

1 2 1 2 = 6 – 10 1 2 1 2 = 6 – –10
–9 4 –8 25

1 2 = –4 = –4~i – 13~j 1 2 = 16 = 16~i – 33~j
–13 –33

1 2 1 2(iii) 2~a – 5~b = 2 2 –5 5 1 2 1 2(iii) 4~a + 2~b = 43 +2 –2
–3 2 –4 5

1 2 1 2 = 4 – 25 1 2 1 2 = 12 + –4
–6 10 –16 10

1 2= –21 = –21~i – 16~j 1 2= 8 = 8~i – 6~j
–16 –6

BAB 8 14. Tentukan vektor unit pada arah vektor yang diberi. TP 4

Determine the unit vector in the direction of given vector.

CONTOH 1 CONTOH 2

A→B = –5~i + 12~j 1 2P→Q = 5
–8
Penyelesaian:
A→B = (–5)2 + 122 = 13 Penyelesaian:

Vektor unit pada arah A→B P→Q = 52 + (–8)2 = 89
Vektor unit pada arah P→Q
Unit vector in the direction
of A→B Unit vector in the direction of P→Q

= 113 (–5~i + 12~j ) Cari magnitud vektor dan 5
gantikan ke rumus.
Find the magnitude of the 1 2= 1 5 = 89
vector and substitute to the 89 –8 –8
153 ~i 12 formula.
= – + 13 ~j 89

Vektor unit / Unit vector
(x~i + y~j )
= |~~aa| = x2 + y2

154

  Matematik Tambahan  Tingkatan 4  Bab 8 Vektor 

(a) C→D = 6~i – 8~j (b) E→F = 5~i + 6~j 1 2(c) G→H = 8
C→D = 62 + (–8)2­ = 10 –15

 E→F  = 52 + 62 = 61 G→H = 8­2 + (–15)2 = 17
Vektor unit pada arah G→H
Vektor unit / Unit vector Vektor unit pada arah E→F

= 1 (6~i – 8~j ) Unit vector in the direction of E→F Unit vector in the direction of G→H
10
8
= 3 ~i – 4 ~j = 1 (5~i + 6~j )
5 5 61 1 2=1 8 = 17
17 –15
5 6 – 15
= 61 ~i + 61 ~j 17

= 8 ~i – 15 ~j
17 17

1 5. Selesaikan masalah-masalah berikut melibatkan vektor. TP 5

Solve the following problems involving vectors.

CONTOH 1 CONTOH 2

Diberi bahawa A→B = 2~i − 5~j dan Rajah di sebelah menunjukkan
C→D = nilai m jika A→B vektor-vektor O→S dan O→T.
selari ~di e+ng(a4n−C→mD).~j . Cari adalah
The diagram shows vectors
O→S and O→T .

Given that A→B = 2 ~i − A5→B~j and C→D = ~i + (4 − m) ~j . Ungkapkan / Express
Find the value of m if is parallel to C→D .
1 2(i)
S→T dalam bentuk x ,
y
1 2 x
Penyelesaian: S→T in the form y ,
A→B = λC→D
(ii) vektor unit pada arah S→T .
2~i − 5~j = λ(~i + (4 − m)~j ) BAB 8
= λ~i + (4 − m)λ~j unit vector in the direction of S→T .
2 = λ , −5 = (4 − m)λ
Penyelesaian:

(i) S→T = S→O + O→T = − O→S + O→T

= (4 − m)(2) 1. Jika selari, maka 1 2 1 2 1 2 =−–4 + 6 = 10
−13 = −2m If parallel, then 7 5 –2
A→B = λC→D
m = 6 1 2. Bandingkan nilai- (ii)  S→T  = 102 + (–2)2 = 104
2
Cniolami p~iadreanth~ej .
Vektor unit pada arah S→T / Unit vector S→T .
values of ~i and ~j .
1 2 = 1 10
104 –2

= 10 ~i – 2 ~j
104 104

155

  Matematik Tambahan  Tingkatan 4  Bab 8 Vektor

CONTOH 3

Dalam rajah, diberi QR

In the diagram, given that

P→Q = 2~x , P→S = 5~y dan/and R→S = 3~x . T Penyelesaian:
S (a) P→R = P→S + S→R = P→S + (– R→S )
(a) Ungkapkan dalam sebutan ~x dan / atau ~y , P
P→R dan S→Q .
Express in terms of ~x and / or y~ , P→R and S→Q . = S→P + = (5–~yP→−S 3~x P→Q
S→Q P→Q = )+

= −5~y + 2~x

(b) Diberi P→T = h P→R dan Q→T = k Q→S , nyatakan P→T (b) (i) P→T = h P→R

Given P→T = h P→R and Q→T = k Q→S , state P→T = h(5~y − 3~x )
= 5h~y − 3h~x
(i) dalam sebutan h, ~x dan ~y ,

in terms of h, ~x and y~ , (ii) P→T = P→Q + Q→T

(ii) dalam sebutan k, ~x dan ~y . = 2~x + k Q→S
in terms of k, ~x and y~ . = (22~x −−2kk()−~x5 ~+y + 2~x )
= 5k~y



(a) sDK→eiLlbaer=rii(d2bepanh+gaaw1n)a~Ki→GL→−H. 6=~j .~iC+ar3i ~nj idlaainp jika G→H adalah (b) Diberi bahawa P→Q = 12 ~i h−ji2k0a~jP→dQanadalah
Cari nilai
dR→eSn=ga3n~i R→+S h~j . selari
.

Given that G→H = ~i + 3G→~Hj and K→L = (2p + 1)~i − 6 ~j . Given that P→Q = 12~i P−→Q20is~jpaanradlleR→lSto=R→3S~i. + h~j .
Find the value of p if is parallel to K→L . Find the value of h if

BAB 8    G→H = λK→L    P→Q = λ R→S

~i + 3~j = λ((2p + 1) ~i − 6~j ) 12~i − 20~j = λ(3~i + h~j )
= (2p + 1)λ ~i − 6λ~j
= 3λ~i + hλ~j

3 = − 6λ , 1 = (2p + 1)λ 12 = 3λ , −20 = hλ

λ=− 1 1 21 = (2p + 1) − 1  λ = 4 −20 = 4h
2 2
−2 = 2p + 1 h = −5

p = − 3
2

156

  Matematik Tambahan  Tingkatan 4  Bab 8 Vektor 

(c) Rajah menunjukkan P (–8, 6) y (d) Rajah menunjukkan y
vektor-vektor O→P dan Q→O. vektor-vektor O→R dan O→S . R (–3, 5)
Q (4, 2)
The diagram shows Ox The diagram shows O x
vectors O→P and Q→O. vectors O→R and O→S . S (–8, –7)

Ungkapkan / Express Ungkapkan / Express

(i)

1 21 2
1 2(i) R→O dalam bentuk x ,
Q→O dalam bentuk x , y
y R→O in the form x
Q→O in the form x y ,
1 2 y ,

(ii) vektor unit pada arah Q→P. (ii) vektor unit pada arah R→S .
unit vector in the direction of Q→P .
unit vector in the direction of R→S .

1 2 1 2(i) 4 –4
Q→O = −O→Q = − 2 = –2 1 2 1 2(i) –3 3
R→O = −O→R = − 5 = –5

(ii) Q→P = Q→O + O→P (ii) R→S = R→O + O→S

1 2 1 2 1 2 =–4+ –8 = –12 1 2 1 2 1 2 =3+ –8 = –5
–2 6 4 –5 –7 –12

Q→P = √(–12)2 + 42 = √160  R→S  = (–5)2 + (–12)2 = 13
Vektor unit pada arah Q→P Vektor unit pada arah R→S

Unit vector in the direction of Q→P . Unit vector in the direction of R→S .

1 21 2 =1 –12 – 12 1 21 2 = – 5
√160 4 √160 13
= 1 –5
4 13 –12 =
12
√160 – 13

(e) Diberi bahawa ~a = 2~i + 7~j dan ~b = ~i − 3~j . (f) Diberi bahawa O(0, 0), C(8, −3), D(5, 4), cari
Given that a~ = 2~i + 7~j and b~ = ~i − 3~j .
Cari / Find dalam sebutan ~i dan −~j3b),aDgi(5, 4), find in terms of BAB 8

(i) ~a + 5~b , Given that O(0, 0), C(8,
(ii) vektor unit pada arah ~a + 5~b .
(~ii)a ndC→~jDf,o r
unit vector in the direction of a~ + 5b~. (ii) vektor unit pada arah C→D.

unit vector in the direction of C→D .

(i) ~a + 5~b = 2~i + 7~j + 5(~i − 3~j ) (i) C→D = C→O + O→D = −O→C + O→D
= 7~i − 8~j
= −(8~i − 3~j ) + (5~i + 4~j )
(ii) 7~i − 8~j  = 7­2 + (–8)2 = 113 = −3~i + 7~j

Vektor unit pada arah 7~i 7−~i8–~j8~j . (ii) C→D = (–3)2 + 72 = 58
Vektor unit pada arah C→D
unit vector in the direction of
Unit vector in the direction of C→D .
= 1 (7~i − 8~j )
113 1
= 58 (–3~i + 7~j )

157

  Matematik Tambahan  Tingkatan 4  Bab 8 Vektor

(g) Rajah menunjukkan sebuah segi tiga ABC. Garis lurus AD bersilang dengan garis B
D
lurus BE pada titik F. Diberi bahawa
F
The diagram shows a triangle ABC. The straight line AD intersects the straight line BE at
point F. Given that AE

A→E = 1 A→C , 2B→D = D→C , A→E = 2~x dan / and A→B = 6~y C
2

(a) UE(ixn)p grekB→saEsp,i knatnermdasloafm~x saenbduy~ta: n ~x dan ~y b(iai)g i:A→D.


(b) Diberi B→F = hB→E dan F→D = kA→D,
Given B→F = h B→E and F→D = kA→D,

(i) ungkapkan B→F dalam sebutan h, ~x dan ~y ,
k, ~x dan ~y .
express B→F in terms of h, ~x and y~,
(ii)
ungkapkan F→D dalam sebutan

express F→D in terms of k, ~x and y~.

(a) (i) B→E = B→A + A→E (b) (i) B→F = hB→E
= −6~y + 2~x
= −h(6−h6~y~y++22h~x~x)
A→D = A→B + B→D = A→B + 1 B→C =
3
(ii) (ii) F→D = kA→D

= 6~y + 1 (B→A + A→C ) 1 2 4
3 3
=k 4~y + ~x
1
= 6~y + 3 (−6~y + 4~x ) 4
3
= 4~y + 4 ~x = 4k~y + k~x
3

BAB 8

158

PRAKTIS   Matematik Tambahan  Tingkatan 4  Bab 8 Vektor 

SPM 8

Kertas 1 2. Rajah di bawah menunjukkan tiga vektor, O→P ,
O→Q dan O→R, dilukis pada grid segi empat sama
1. Rajah menunjukkan vektor-vektor O→A, O→B dan yang sama besar bersisi 1 unit.
O→D
SPM yang dilukis pada grid segi empat sama bersisi The diagram below shows three vectors, O→P , O→Q and
2017 O→R, drawn on a grid of equal squares with sides of
1 unit. 1 unit.
Diagram shows vectors O→A, O→B and O→D drawn on a
square grid with sides 1 unit. P

B R
Q
~x ~y

OA O



Tentukan / Determine
(a) O→P,
D

(a) Cari /  Find |–B→O| (b) O→P dalam sebutan ~x dan ~y .
( b) sDGeiivbbeuentraiOn→O→AA~a ==daa~~anadn~badnO→O→BB==b~~b,
, ungkapkan dalam O→P in terms of ~x and y~.

express in terms of ~a 1 2(a) O→P = –2
4

and b~ O→P = (–2)2 + 42
(i) A→B
(ii) O→D = 4.472 unit

1 2 1 2(b) ~x =1 3
(a) |–B→O| = |O→B| 2 , ~y = 3 BAB 8

= 32 + 42 Biar O→P = h~x + k~y,
= 5 unit 1 –2 2 = h1 1 2 k1 3 2
4 2 3
(b) (i) A→B = –O→A + O→B +
= –~a + ~b
= ~b – ~a Maka, −2 = h + 3k..............a
4 = 2h + 3k ...........b
(ii) O→D = O→B + B→D
= O→B + 2B→A a − b ; −2 − 4 = −h
= O→B + 2(–A→B)
= ~b + 2(~a – ~b ) h = 6
= 2~a – ~b
Gantikan dalam a,

−2 = 6 + 3k

k = – 8
3
O→P 8
\ = 6~x – 3 ~y

159

  Matematik Tambahan  Tingkatan 4  Bab 8 Vektor P→Q = 3~v + kw~
O→Q – O→P = 3 3 8
3. Rajah menunjukkan sebuah segi empat selari 1 2 1 2 –2 +k 4
SPM ABCD.

2019 Diagram shows a parallelogram ABCD.

10 9 + 8k
h –6 + 4k
DE C 1 2 1 2 1 2 – m =
4

10 – m 9 + 8k
h–4 –6 + 4k
T 1 2 1 2 =
A
B

10 – m = 9 + 8k h – 4 = –6 + 4k

Titik E terletak pada DC dengan keadaan DE : EC 8k = 1 – m h = –2 + 4k

= 1 : 2. Diberi bahawa D→E = 6~u, A→D = 5~v dan T k = 1 – m 1 2h = –2 + 4 1–m
8 8
ialah titik tengah AC. Ungkapkan dalam sebutan ~u
dan / atau ~v , 1 ­m2
h = –2 + 2 –

Point E lies on DC such that DE : EC = 1 : 2. It is given that h = –m – 3
D→E = 6u~, A→D = 5v~ and T is the midpoint of AC. Express 2

in terms of u~ and / or ~v,

(a) E→C
(b) E→T
5. Rajah menunjukkan vektor-vektor O→C, O→D dan
O→N dilukis pada grid segi empat
(a) DE→→CE = 1 SPM sama.
2 2018 Diagram shows the vectors O→C, O→D
and O→N
drawn on a

E→C = 2D→E square grid.

= 2(6~u) = 12~u N

(b) E→T = E→C + C→T

= 12~u + 1 C→A D
2

= 12~u + 1 ( C→D + D→A) ~y C
2 M ~x

BAB 8 = 12~u + 1 (18~u + (–5~v )) O
2
5
= 12~u + 9~u – 2 ~v (a) Ungkapkan O→N dalam
keadaan h dan k ialah
= 21~u – 5 ~v bpeenmtualkarh.~x + k~y , dengan
2
Express O→N in
are constants. the form of h~x + ky~, where h and k

(b) Pada rajah, tanda dan label titik M dengan
keadaan N→M + 2O→C = O→D.
On the diagram, mark and label the point M such
4. mDw~i,b=eℎr8id~i ab+nah4ka~jwidaaalanPh(Pm→pQ,em=4)a3, l~vaQr(.+1U0k ,wn~gℎ,k)d,aep~nvkga=ann3ℎk~ieda−adla2aam~jn, that N→M + 2O→C = O→D.

SPM (a) O→N = 2O→B + (–O→C) = 2~y – ~x
2019
(b) N→M + 2O→C = O→D
sebutan m. − 2~j , w~ = 8~i + N→M = O→D – 2O→C
It is given that P(m, 4), Q(10, ℎ), = 3~i k are constants. = ~y – 2~x
P→Q = 3~v + v~ and
E4x~jparensds ℎ in terms kw~ , where m, ℎ
of m.

160

  Matematik Tambahan  Tingkatan 4  Bab 8 Vektor 

Kertas 2 (b) B→F = n(~a – 4~b )
B→A + A→F =
1. Rajah menunjukkan segi tiga ABC. Garis lurus AE –A→B + mA→D = n(~a – 4~b )
SPM bersilang dengan garis lurus BC di titik D. Titik F nnnn((~~aa~~aa––––4444nn~~bb~~bb))
2017 terletak pada garis lurus AE. ––344m~~bb ~a++ +3mm((23~am~a++–242m)~b~~bb) ===

Diagram shows a triangle ABC. The straight line AE Bandingkan / Compare:
intersects with the straight line BC at point D. Point F lies
on the straight line AE.

E

3m = n ,  2m – 4 = –4n

B Gantikan n = 3m  → 2m – 4 = –4n
D Replace 2m – 4 = –4(3m)

14m = 4

CF m = 4 = 2
A 14 7

1 2n = 3 2 = 6
7 7
Diberi bahawa C→D 1 C→B , A→C A→B
= 2 = 6~a dan = 4~b . (c) A→F = k~a + 4 ~b = mA→D
7
It is given that C→D = 1 C→B , A→C = 6~a and A→B = 4b~. 4 2
2 k~a + 7 ~b = 7 (3~a + 2~b )
(a) Ungkapkan dalam sebutan ~a dan/ atau ~b :
Express in terms of a~ 4 6 4
(i) B→C , and / or b~: k~a + 7 ~b = 7 ~a + 7 ~b

(ii) A→D. Bandingkan / Compare:
6
(b) Diberi bahawa A→F = mA→D dan B→F = n(~a – 4~b ) , k = 7
dengan keadaan m dan n ialah pemalar. Cari
2. Rajah menunjukkan segi tiga OBQ dan OPA
nilai m dan n. SPM dengan keadaan titik A berada pada OQ dan titik B
2018 berada pada OP. Garis lurus BQ dan garis lurus AP
It is given A→F = mA→D and B→F = n(a~ – 4b~), where m
and n are constants. Find the value of m and of n. bersilang pada titik C.

(c) Diberi A→F = k~a + 4 ~b , dengan keadaan k ialah Diagram shows triangles OBQ and OPA where point A BAB 8
7 lies on OQ and point B lies on OP. The straight lines BQ
pemalar, cari nilai k. and AP intersect at point C.
Given A→F = ka~ +
the value of k. 4 b~, such that k is a constants, find Q
7

(a) (i) B→C = B→A + A→C A
C
= –A→B + A→C
= −4~b + 6~a

(ii) A→D = A→C + C→D PB O

= 6~a + 1 C→B ODBibe: rBiPba=h3aw: a1,O→A→QC==12h~Ax→,PO→dPan= 8B→~yC, OA : AQ = 2 : 1,
2 = kB→Q, dengan

= 6~a – 1 B→C keadaan h dan k ialah pemalar.
2 that O→Q = =12h~xA→, PO→aPn=d8B→y~C,
It is given 3 : 1, A→C OA : AQ = 2 : 1,
= 6~a – 1 (–4~b + 6~a ) OB : BP = = kB→Q, where h
2
and k are constants.
= 3~a + 2~b

161

  Matematik Tambahan  Tingkatan 4  Bab 8 Vektor

(a) Ungkapkan O→C dalam sebutan 3. Rajah di bawah menunjukkan sebuah trapezium
SPM ABCD dan titik E terletak pada AC.
Express O→C in terms of
2014 The diagram below shows a trapezium ABCD and point
(i) h, ~x dan / and ~y , E lies on AC.
(ii) k, ~x dan / and ~y .
(b) Seterusnya, cari nilai h dan nilai k. BC

Hence, find the value of h and of k. E

(c) Diberi u~x u = 1 unit, u~y u = 2 unit dan OQ
berserenjang kepada OP, hitung u A→C u.

Given u~xu = 1 unit, uy~u = 2 units and OQ is AD
perpendicular to OP, calculate u A→C u.
Diberi bahawa A→B 8=~a8, A~a→D, A→=D12=b~1a2n~bd dan A→D = B→C
(a) (i) O→C = O→A + A→C 2 .
It is given that A→B = A→D = 2 B→C .
2 O→Q h A→P
= 3 + (a) Ungkapkan dalam sebutan ~a dan ~b ,

= 2 (12~x ) + h( O→P – O→A) Express in terms of ~a and b~,
3
= (888~~xx –++88hhh()8~~xy~y+––888hh~x~~xy) (i) A→C,
=
= (ii) B→D.

(b) Diberi bahawa A→E = kA→C, dengan keadaan k

(ii) O→C = O→B + B→C ialah pemalar. Cari nilai k jika titik-titik B, E

= 3 O→P + kB→Q dan D adalah segaris.
4
k(O→Q O→B) It is given that A→E = kA→C, where k is a constant.
= 3 (8~y ) + – Find the value of k if the points B, E and D are
4
collinear.

= 6~y + k(12~x – 6~y ) (a) (i) A→C = A→B + B→C
= 12k~x + (6 – 6k)~y
= A→B + 1 A→D
(b) Bandingkan / Compare: 2

8 – 8h = 12k ; 8h = 6 – 6k = 8~a + 6~b

2 – 2h = 3k … 1 3 – 4h = 3k … 2 (ii) B→D = B→A + A→D

BAB 8 2 – 1: 1 – 2h = 0 = –A→B + A→D
1 = –8~a + 12~b
h = 2

Gantikan / Replace h = 1 ke dalam / into 1;
2

1 2 1 (b) A→E = kA→C
2–2 2 = 3k

1 = 3k A→E = 8k(k8~a~a++66k~b~b)
=
k = 1
3
Katakan / Let B→E = lB→D,
(c) uO→Qu = u12~x u = u12(1)u = 12 Maka / Hence B→E = l(–8~a + 12~b )
u O→P u = u8~y u = u8(2)u = 16 = –8l~a + 12l~b
A→E = A→B + B→E
u u u u O→Q
uO→Au = 2 = 2 (12) =8
3 3
uA→Pu = √82 + 162 = √320 = 8√5
8k~a + 6k~b = 8~a + (–8l~a + 12l~b )
\  uA→Cu = 1 u A→P u = 1 (8√5) = 4√5 8k~a + 6k~b = (8 – 8l)~a + 12l~b
2 2

162

  Matematik Tambahan  Tingkatan 4  Bab 8 Vektor 

8k = 8 – 8l dan / and 6k = 12l (a) (i) Q→A = Q→P + P→A
1
l = 2 k = –~a + 2~b

RGeapnlatickea n l = 21 k kinetodalam 8k = 8 – 8l , (ii) Q→B = 2 Q→P
5

1 2 8k = 8 – 8 1 k = 2 (–a)
2 5

8k = 8 – 4k =– 2 ~a
5
1 2k = 8
(b) Q→A = –~a + 2~b
k = 2 = –3~i + 2(~i – 3~j )
3 = –~i – 6~j
Q→A = (–1)2 + (–6)2

= 37
= 6.08 unit

4. Rajah di bawah menunjukkan sebuah segi tiga (c) Q→C = mQ→A = m(–~a + 2~b ) = –m~a + 2m~b
SPM PQR.
C→B = n( R→P + P→B ) = n(–10~b + 3 ~a )
2015 The diagram below shows a triangle PQR. 5
3
P = 5 n~a – 10n~b
A

R Q→C = Q→B + B→C = – 2 ~a – 3 n~a + 10n~b
CB 5 5

= 1 (–2 – 3n)~a + 10n~b
5

Q Oleh itu, 1 (–2 – 3n) = –m..................... a
5
4, P→Q = ~a 10n = 2m
Diberi PB : BQ = 3 : 2, PA : AR = 1: dan
P→A = 2~b . : 4, P→Q = a~ n = 1 m................... b BAB 8
= 3 : 2, PA : AR =1 and 5
It is given PB : BQ
P→A = 2b~. Gantikan b ke dalam a,
(a) Ungkapkan dalam sebutan ~a dan/atau ~b :
Replace b into a,

Express in terms of ~a and/or b~: 15 1–2 – 3 m2 = –m
(i) Q→A, 5

(ii) Q→B. 22 m = 2
25 5

(b) Diberi ~a = 3~i dan ~b = ~i – –33~j~j, ,cfainridQ→Q→AA.. m = 5
11
Given that ~a = 3~i and b~ = ~i
n = 1
(c) Diberi Q→C = mQ→A dan C→B = nR→B, dengan 11

keadaan m dan n ialah pemalar, cari nilai m

dan nilai n. Praktis
SPM
Given that Q→C = mQ→A and C→B = n R→B , where m and Ekstra

n are constants, find the value of m and of n.

163

  Matematik Tambahan  Tingkatan 4  Bab 8 Vektor KBAT KBAT

Sudut Ekstra

1. Vektor-vektor 1 a 2 dan 1–132 adalah selari. Diberi bahawa 1 a 2 mempunyai magnitud √250 dan a . 0, cari
b b
nilai a dan nilai b.

1 2 1 2 1 2The vectorsa 1 a has a magnitude of √250 and a . 0, find the value of a and of b.
b and –3 are parallel. Given that b

1 a 2 = k 1 –132 Apabila / When a = 5, b = –3(5)
b = –15
Apabila / When a = –5, b = –3(–5)
k = a, –3k = b = 15
b = –3a Memandangkan a . 0,

Magnitud = √a2 + b2 It is seen that a . 0,

Magnitude Maka / Thus, a = 5 dan b = –15

√250 = √a2 + (–3a)2
√10a2 = √250
10a2 = 250

a2 = 25

 a = ±5

BAB 8

Kuiz 8

164

BAB Penyelesaian Segi Tiga

9 Solution of Triangles

9.1 Petua Sinus

Sine Rule

NOTA IMBASAN Petua Sinus / Sine rule

C
ba
A cB

NOTA IMBASAN c a
A
C
ba c a ca
AB A
A
C aуc
ba
A B1 c a ca
A A
C
ba
A B2

165

  Matematik Tambahan  Tingkatan 4  Bab 9 Penyelesaian Segi Tiga

1. Hitung panjang x bagi setiap segi tiga yang berikut. TP 4

Calculate the length of x in each of the following triangles.

CONTOH (a)

x cm

53°

9.6 cm x cm 8.7 cm
118°
82°

23°

Penyelesaian: x = 8.7
sin 82° sin 53°
x 9.6 Sudut Kalkulator
sin 118° = sin 23° x = 8.7 × sin 82°
Tekan 9   .   6   × sin 53°
Press
x = 9.6 × sin 118 = 10.79
sin 23° sin   1   1   8   ÷
sin   2   3   =
= 21.69

(b) x cm (c) 16.7 cm
123° 39°
42° 109°
15.9 cm x cm

Q = 180° − 123° − 39° = 18° Q = 180° − 109° − 42° = 29°

x = 15.9 x = 16.7
sin 18° sin 123° sin 29° sin 109°

x = 15.9 × sin 18° x = 16.7 × sin 29°
sin 123° sin 109°

= 5.859 = 8.563

2. Hitung nilai y bagi setiap segi tiga yang berikut. TP 4

Calculate the value of y in each of the following triangles.

CONTOH (a)

12.1 cm y

BAB 9 65° x
9.1 cm

y 8.5 cm 72°

x 10.3 cm

Penyelesaian: sin y = sin 72°
8.5 12.1

sin x = sin 65° sin y = 8.5 × sin 72°
9.1 10.3 12.1

sin x = 9.1 × sin 65° = 53°12’ y = 41°55’
10.3
x = 180° – 72° – 41°55’

y = 180° − 65° − 53°12’ x = 66°59
y = 61°48’

166

  Matematik Tambahan  Tingkatan 4  Bab 9 Penyelesaian Segi Tiga 

(b) 15.3 cm (c) x 11.5 cm
xy y

110° 10.8 cm 41° 16.3 cm

sin x = sin 110° sin x = sin 41°
10.8 15.3 16.3 11.5

sin x = 10.8 × sin 110° sin x = 16.3 × sin 41°
15.3 11.5

x = 41°33’ x = 180° – 68°25’

y = 180° – 110° – 41°33’ x = 111°35’

y = 28°27’ y = 180° – 41° – 111°359

y = 27°259

3. Selesaikan setiap yang berikut. TP 5

Solve each of the following.

CONTOH 1

Cari dua nilai yang mungkin bagi ∠C dalam segi tiga 8.2 cm A
ABC. Seterusnya, lakar segi tiga ABC’. Diberi C’ terletak 34° 5.7 cm
pada BC dengan keadaan AC = AC’. B
53°33ʹ
Find two possible values of ∠C in triangle ABC. Hence, 8.2 cm C
34°126°27ʹ
sketch triangle ABC’, given C’ lies on BC such that B Cʹ A
5.7 cm
AC = AC’.
53°33ʹ
Penyelesaian: C

sin C = sin 34°
8.2 5.7

sin C = 8.2 × sin 34°
5.7

sin C = 0.8045

∠C = 53°33’ atau 180° – 53°33’

= 126°27’

(a) Cari dua nilai yang mungkin bagi ∠C dalam segi tiga C B BAB 9
ABC. Seterusnya, lakar segi tiga ABC9, diberikan C9 8 cm 42°
terletak pada BC dengan keadaan AC = AC9. 11 cm

Find two possible values of ∠C in triangle ABC. Hence, sketch A
C
triangle ABC9, given C9 lies on BC such that AC = AC9.
66°56’
sin C = sin 42° 8 cm Cʹ B
11 8 113°4’ 42°

sin C = 11 × sin 42° 11 cm
8

sin C = 0.9201

∠C = 66°56’ atau 180 – 66°56’ A

= 113°4’

167

  Matematik Tambahan  Tingkatan 4  Bab 9 Penyelesaian Segi Tiga

CONTOH 2

Dalam rajah di sebelah, BCD ialah garis lurus. Hitung A

In the diagram, BCD is a straight line. Calculate

(i) ∠ABC, 73° 9 cm
(ii) panjang AD. 8 cm

the length of AD. B C 28° D

Penyelesaian: 10.5 cm

(i) Dalam/ In ∆ABC, (ii) Dalam/ In ∆ABD,

sin ∠ABC = sin 73° AD = 8
9 10.5 sin 55°3’ sin 28°

sin ∠ABC = 9 × sin 73° AD = 8 × sin 55°3’
10.5 sin 28°

= 0.8197 = 13.97 cm

∠ABC = 55°3’

(b) Dalam rajah di sebelah, ABC ialah garis lurus. A 11.2 cm B C
Hitung 43° 13 cm

In the diagram, ABC is a straight line. Calculate 85°
D
(i) panjang BD,

the length of BD,

(ii) ∠BCD.

(i) Dalam ∆ABD, (ii) Dalam ∆BCD,
∠DBC = 43° + 85°
BD = 11.2 = 128°
sin 43° sin 85°

BD = 11.2 × sin 43° sin ∠BCD = sin 128°
sin 85° 7.668 13

= 7.668 cm sin ∠BCD = 7.668 × sin 128°
13

= 0.4648

∠BCD = 27°42’

(c) Rajah di sebelah menunjukkan sebuah sisi empat C
ABCD. Hitung
BAB 9 22 cm 9° B
The diagram shows a quadrilateral ABCD. Calculate D 6 cm
138° A
(i) panjang AC, 55°

the length of AC,

(ii) ∠ADC.



(i) Dalam ∆ABC, (ii) Dalam ∆ACD,

AC = 6 sin ∠ADC = sin 55°
sin 138° sin 9° 25.66 22

AC = 6 × sin 138° sin ∠ADC = 25.66 × sin 55°
sin 9° 22

= 25.66 cm = 0.9554

∠ADC = 72°509

168

9.2 Petua Kosinus   Matematik Tambahan  Tingkatan 4  Bab 9 Penyelesaian Segi Tiga 

Cosine Rule Petua kosinus / Cosine rule

NOTA IMBASAN a
Cb
NOTA IMBASAN

A b
c

C
Ba

ab
c

4. Hitung nilai x bagi setiap segi tiga yang berikut. TP 4

Calculate the value of x in each of the following triangles.

CONTOH (a) BAB 9

x cm

8.3 cm x cm

7.5 cm 119° 9.3 cm 81°
Gunakan petua kosinus 11.6 cm
Penyelesaian: Use cosine rule
x2 = 7.52 + 9.32 − a2 = b2 + c2 – 2bc kos A x2 = 8.32 + 11.62 – 2(8.3)(11.6) kos /cos 81°
2(7.5)(9.3) kos 119° a2 = b2 + c2 – 2bc cos A x = 13.17 cm

2(7.5)(9.3) cos 119°

x = 14.50 cm

169

  Matematik Tambahan  Tingkatan 4  Bab 9 Penyelesaian Segi Tiga

(b) (c) x cm

9.5 cm 108° 12.1 cm

10.5 cm 123° 16.3 cm

x cm x2 = 10.52 + 16.32 – 2(10.5)(16.3) kos/cos 123°
x = 23.71 cm
x2 = 9.52 + 12.12 – 2(9.5)(12.1) kos/cos 108°
x = 17.54 cm

5. Selesaikan setiap yang berikut. TP 4

Solve each of the following.

CONTOH (a) R

P 9 cm 8 cm

17.6 cm

8.9 cm Q 10 cm P

Q 11.4 cm R Hitung ∠PRQ dalam segi tiga PQR.

Hitung sudut terbesar dalam segi tiga PQR. Calculate ∠PRQ in the triangle PQR.

Calculate the largest angle in the triangle PQR.

Penyelesaian: 102 = 82 + 92 – 2(8)(9) kos/cos ∠PRQ

∠Q ialah sudut terbesar kerana ∠Q bertentangan 100 = 145 – 144 kos/cos ∠PRQ
dengan sisi terpanjang.
kos ∠PRQ = 145 – 100
∠Q is the biggest angle because ∠Q is opposite to the 144
longest side. cos ∠PRQ
= 0.3125

17.62 = 8.92 + 11.42 – 2(8.9)(11.4) kos/cos Q ∠PRQ = 71°479
309.76 = 209.17 – 202.92 kos/cos Q

k os/cos Q = 209.17 – 309.76
202.92

= –0.4957

∠Q = 119°43’

(b) Q (c) P 11.5 cm Q

8.2 cm 10 cm

P R 6.5 cm 9.8 cm

15 cm

BAB 9 Hitung sudut terbesar dalam segi tiga PQR. R

Calculate the largest angle in the triangle PQR. Hitung sudut terkecil dalam segi tiga PQR.

∠Q ialah sudut terbesar kerana ∠Q bertentangan Calculate the smallest angle in the triangle PQR.

dengan sisi terpanjang ∠Q ialah sudut terkecil kerana ∠Q bertentangan

∠Q is the biggest angle because ∠Q is opposite to the dengan sisi terpendek
longest side
∠Q is the smallest angle because ∠Q is opposite to the
152 = 8.22 + 102 – 2(8.2)(10) kos/cos Q shortest side

225 = 167.24 – 164 kos/cos Q 6.52 = 9.82 + 11.52 – 2(9.8)(11.5) kos/cos Q

kos Q = 167.24 – 225 42.25 = 228.29 – 225.4 kos/cos Q
164
cos Q kos Q = 228.29 – 42.25
= –0.3522 225.4
cos Q
= 0.8254
∠Q = 110°379
∠Q = 34°22’

170

  Matematik Tambahan  Tingkatan 4  Bab 9 Penyelesaian Segi Tiga 

6. Selesaikan setiap yang berikut. TP 5

Solve each of the following.

CONTOH

A Penyelesaian:

8 cm 14 cm (i) Dalam/ In ∆ABC,
B 7 cm
6 cm 82 = 62 + 72 – 2(6)(7) kos/cos ∠ACB
C
D 64 = 85 – 84 kos/cos ∠ACB

Dalam rajah di atas, BCD ialah garis lurus. kos ∠ACB = 85 – 64
cos ∠ACB = 84
Hitung
0.25
In the diagram, BCD is a straight line. Calculate
∠ACB = 75°319
(i) ∠ACB,
(ii) ∠ADC. (ii) Dalam/ In ∆ACD,

∠ACD = 180° – 75°319

= 104°299

sin ∠ADC = sin 104°299
6 14

sin ∠ADC = 6 × sin 104°299
14

∠ADC = 24°319

(a) S (b) D C

7 cm 15 cm 130°
12 cm 8 cm

P 8 cm Q 10 cm R

A 5 cm B

Dalam rajah di atas, PQR ialah garis lurus. Hitung Dalam rajah di atas, ABCD ialah sebuah trapezium.
In the diagram, PQR is a straight line. Calculate

(i) ∠RQS, Hitung

(ii) panjang PS. In the diagram, ABCD is a trapezium. Calculate

the length of PS. (i) ∠BAC,

(i) Dalam ∆QRS, (ii) panjang AD.

the length of AD.

152 = 72 + 102 – 2(7)(10) kos/cos ∠RQS (i) Dalam / In  ∆ABC,

225 = 149 – 140 kos/cos ∠RQS 82 = 52 + 122 – 2(5)(12) kos/cos ∠BAC

kos ∠RQS = 149 – 225 64 = 169 – 120 kos/cos ∠BAC BAB 9
140
cos ∠RQS 169 – 64
= –0.5429 120
kos ∠BAC =
∠RQS = 122°539
cos ∠BAC
= 0.875

(ii) Dalam / In ∆PQS, ∠BAC = 28°579

∠PQS = 180° – 122°539 (ii) Dalam / In ∆ACD,
= 57°7’
∠ACD = 28°579
PS2 = 72 + 82 – 2(7)(8) kos/cos 57°79
= 52.19 sin AD = 12
28°57’ sin 130°
PS = 7.224 cm

AD = 12 × sin 28°57’
sin 130°

AD = 7.583 cm

171

  Matematik Tambahan  Tingkatan 4  Bab 9 Penyelesaian Segi Tiga

9.3 Luas Segi Tiga

Area of a Triangle

NOTA IMBASAN

C B
ba

Ac

NOTA

7. Hitung luas bagi setiap segi tiga yang berikut. TP 4 (a) K
9.5 cm
Calculate the area of each of the following triangles.

CONTOH 1

B

BAB 9 9 cm C 63° L

116° 14 cm 11.6 cm
A J

Penyelesaian: Luas ∆JKL / Area of ∆JKL

Luas ∆ABC / Area of ∆ABC 1
2
= 1 (9)(14) sin 116° = (11.6)(9.5) sin 63°
2
= 49.09 cm2
= 56.62 cm2

172

  Matematik Tambahan  Tingkatan 4  Bab 9 Penyelesaian Segi Tiga 

(b) D 12.6 cm E (c) Q

42° 121° 18.7 cm

12.1 cm

R

13.5 cm P

F Luas ∆PQR / Area of ∆PQR

Luas ∆DEF / Area of ∆DEF = 1 (12.1)(18.7) sin 121°
2
= 1 (13.5)(12.6) sin 42°
2 = 96.98 cm2

= 56.91 cm2

CONTOH 2

R a+b+c
2
9 cm 8 cm s = Luas/ Area ∆PQR

Q P s = 9 + 8+ 10 = √s(s – a)(s – b)(s – c) 
2
10 cm = √13.5(13.5 – 8)(13.5 – 9)(13.5 – 10) 

= 13.5 = √13.5(5.5)(4.5)(3.5) 

Tip = √1 169.44 
= 34.20 cm2
Guna Hukum Heron untuk
mencari luas segi tiga.
Use Heron law to find the
area of triangle.

(d) 12 cm B s = 5.7 + 7.5 + 12 Luas / Area ∆ABC
C 2
A = √12.6(12.6 – 5.7)(12.6 – 7.5)(12.6 – 12) 
7.5 cm = 12.6
5.7 cm = √12.6(6.9)(5.1)(0.6) 

= √266.04 
= 16.31 cm2

(e) s = 4.5 + 6 + 10.1 Luas / Area ∆ABC BAB 9
2
P 10.1 cm = √10.3(10.3 – 4.5)(10.3 – 6)(10.3 – 10.1) 
= 10.3
6 cm = √10.3(5.8)(4.3)(0.2) 

Q 4.5 cm R = √51.376 
= 7.168 cm2

173

  Matematik Tambahan  Tingkatan 4  Bab 9 Penyelesaian Segi Tiga

9.4 Aplikasi Petua Sinus, Petua Kosinus dan Luas Segi Tiga

Application of Sine Rule, Cosine Rule and Area of a Triangle

8. Selesaikan setiap yang berikut. TP 5

Solve each of the following.

CONTOH (a) U

T W

U V W V
5 cm S 12 cm
R T
6 cm
8 cm

P 10 cm Q R 8 cm S

Rajah di atas menunjukkan sebuah kuboid. Cari Rajah di atas menunjukkan sebuah prisma tegak.

The diagram shows a cuboid. Find Cari

(i) ∠TQU, The diagram shows a right prism. Find
(ii) luas ∆TQU.
(i) ∠WSU,
the area of ∆TQU. (ii) luas segi tiga WSU.

Penyelesaian: the area of triangle WSU.
(i) QU = 52 + 102
= 11.18 cm Teoem Pythagoras (i) SU = 122 + 62
Pythagorean Theorem = 13.42 cm
QS = 102 + 82
= 12.81 cm a2 = b2 + c2 WS = 122 + 82
Petua kosinus = 14.42 cm
QT = 52 + 12.812 Cosine rule
= 13.75 cm WU = 82 + 62
= 10 cm

 82 = 11.182 + 13.752 – 2(11.18)(13.75) kos/cos ∠TQU 102 = 13.422 + 14.422 – 2(13.42)(14.42) kos/cos ∠WSU

kos ∠TQU = 314.0549 – 64 kos ∠WSU = 388.0328 – 100
307.45 387.0328
cos ∠TQU cos ∠WSU
= 0.8133
= 0.7442
∠TQU = 35°359
∠WSU = 41°55’

(ii) Luas/ Area ∆TQU = 1 (11.18)(13.75) sin 35°359 (ii) Luas ∆WSU = 1 (13.42)(14.42) sin 41°55’
2 2
Area ∆WSU
= 44.73 cm2 = 64.64 cm2

BAB 9 (b) PQR ialah sebuah segi tiga dengan PQ = 7.3 cm, (i) s= 7.3 + 9.6 + 14.7 = 15.8
QR = 9.6 cm dan PR = 14.7 cm. Hitung 2
PQR is a triangle where PQ = 7.3 cm, QR = 9.6 cm and
PR = 14.7 cm. Calculate Luas / Area ∆PQR

(i) luas ∆PQR, = √15.8(15.8 – 7.3)(15.8 – 9.6)(15.8 – 14.7) 
the area of ∆PQR,
(ii) tinggi P dari QR. = √15.8(8.5)(6.2)(1.1) 

the height of P from QR. = √915.926
= 30.264 cm2

(ii) Katakan h = tinggi P dari QR
Let h = height of P from QR

Luas / Area ∆PQR = 21 × QR × h = 30.264
1
2 × (9.60) × h = 30.264
h = 6.305 cm

174

  Matematik Tambahan  Tingkatan 4  Bab 9 Penyelesaian Segi Tiga 

(c) Rajah menunjukkan sebuah prisma tegak. (i) Luas / Area ∆ABC = 62 cm2

Segi tiga ABC ialah keratan rentas prisma itu 1 (8)(20) sin ∠ABC = 62
2
dengan keadaan ∠ABC ialah sudut cakah. 62 × 2
sin ∠ABC = 8 × 20
The diagram shows a right prism. Triangle ABC is the
uniform cross-section of the prism where ∠ABC is an = 0.775
obtuse angle.

∠ABC = 180° – 50°489 (sudut cakah)

A C = 129°129 (obtuse angle)

F B 5 cm (ii) AC2 = 82 + 202 – 2(8)(20) kos/cos 129°129
8 cm E D = 666.25
AC = 25.81 cm
20 cm

Diberi luas segi tiga ABC ialah 62 cm2, hitung

Given the area of triangle ABC is 62 cm2, calculate

(i) ∠ABC,
(ii) panjang AC.

the length of AC.

PRAKTIS SPM 9

Kertas 2 P R (b) (i) Lakar sebuah ∆P’Q’R’ yang bentuknya
115° berbeza daripada ∆PQR dengan keadaan
1. P’Q’ = PQ, P’R’ = PR dan ∠P’Q’R’ = ∠PQR.

SPM Sketch a ∆P 9Q9R9 which is different in shape
2014 to ∆PQR such that P 9Q9 = PQ, P 9R9 = PR and
∠P 9Q9R9 = ∠PQR.
T
(ii) Seterusnya, nyatakan saiz ∠Q’P’R’.
20° 50°
Hence, state the size of ∠Q9P 9R9.
S
Q (a) (i) QR = 3.3
sin 115° sin 20°
Rajah di atas menunjukkan dua buah segi tiga
PQR dan RST. Diberi PR = 3.3 cm, RS = 5.5 cm dan QR = 8.745 cm
ST = 4.5 cm.
(ii) RT 2 = 5.52 + 4.52 – 2(5.5)(4.5) kos/cos 50° BAB 9
The diagram shows two triangles PQR and RST. It is = 18.68
given that PR = 3.3 cm, RS = 5.5 cm and ST = 4.5 cm. RT = 4.322 cm
QT = QR – RT
(a) Hitung = 8.745 – 4.322
= 4.423 cm
Calculate
(iii) ∠PRQ = 180° – 115° – 20°
(i) panjang, dalam cm, bagi QR,
= 45° 1
the length, in cm, of QR, 2
Luas ∆PQR = (3.3)(8.745) sin 45°
(ii) panjang, dalam cm, bagi QT, Area ∆PQR =
10.20 cm2
the length, in cm, of QT,

(iii) luas, dalam cm2, bagi ∆PQR.

the area, in cm2, of ∆PQR.

175

  Matematik Tambahan  Tingkatan 4  Bab 9 Penyelesaian Segi Tiga

(b) (i) PЈ R (a) (i) ∠ADC = 180° – 105° = 75°

AC2 = 112 + 102 – 2(11)(10) kos 75°
AC = 12.809 cm

(ii) sin ∠BAC = sin 105°
5 12.809

RЈ ∠BAC = 22°9’
20°
∠ACB = 180° – 105° – 22°9’
QЈ
= 52°51’

(b) (i) Luas / Area DACD
(ii) ∠Q’P’R’ + 20° = 45° 1
∠Q’P’R’ = 45° – 20° = 2 × 11 × 10 × sin 75°
= 25°
= 53.126 cm2

(ii) D

t

A 12.809 cm C

Luas / Area DACD = 53.126
12 × 12.809 × t = 53.126
2. Rajah menunjukkan sisi empat kitaran ABCD. 53.126 × 2
t = 12.809
2S0P1 M6 Diagram shows a cyclic quadrilateral ABCD.
= 8.295 cm
A

10 cm

B 105° D 3. Rajah menunjukkan sebuah segi empat ABCD di
SPM mana AC dan BD adalah garis lurus.
5 cm C 11 cm
2019 Diagram shows a quadrilateral ABCD such that AC and
BAB 9 (a) Hitung BD are straight lines.

Calculate B
5 cm
(i) panjang, dalam cm, AC, C
the length, in cm, of AC,
(ii) ∠ACB. 6 cm

(b) Cari AD

Find Diberi bahawa luas ∆BCD = 12.58 cm2 dan ∠BCD
adalah sudut cakah.
(i) luas, dalam cm2, bagi ∆ACD,
the area, in cm2, of ∆ACD, It is given that the area of ∆BCD = 12.58 cm2 and ∠BCD
(ii) jarak terdekat, dalam cm, dari titik D ke is obtuse angle.

AC. (a) Cari / Find
the shortest distance, in cm, from point D to (i) ∠BCD,
(ii) panjang, dalam cm, bagi BD,
AC.
the length, in cm, of BD,

(iii) ∠CBD.

176

  Matematik Tambahan  Tingkatan 4  Bab 9 Penyelesaian Segi Tiga 

(b) Diberi AC = 11.2 cm dan ∠ADC = 90°, hitung Cari / Find

luas, dalam cm2, bagi ∆ABD. (a) ∠QSR,
(b) panjang, dalam m, bagi QS,
Given AC = 11.2 cm and ∠ADC = 90°, calculate the
area, in cm2, of ∆ABD. the length, in m, of QS,

(a) (i) Luas / Area of ∆BCD = 12.58 cm2 (c) luas, dalam m2, bagi satah condong QVS,

the area, in m2, of inclined plane QVS.

1 × 5 ×6 × sin ∠BCD = 12.58 (a) sin ∠QSR = sin 70°
2 20 22

sin ∠BCD = 12.58 sin ∠QSR = 20 × sin 70°
15 22

∠BCD = 180° − 57° = 0.8543

= 123° ∠QSR = 58°419

(ii) BD2 = 52 + 62 − 2(5)(6) kos / cos 123° (b) ∠QRS = 180° – 58°41’ – 70°
BD = 9.679 cm
= 51°19’

(iii) sin ∠CBD = sin 123° QS2 = 202 + 222 – 2(20)(22) kos 51°19’
6 9.679 = 333.986
QS = 18.28 m
sin ∠CBD = 6 × sin 123°
9.679

∠CBD = 31°19’ (c) VQ = √82 + 52  = 9.434 m
VS = √122 + 52  = 13 m
(b) ∠BDC = 180° − 123° − 31°19’ = 25°41’
∠ADB = 90° − 25°41’ = 64°19’ V
AD2 = 11.22 – 62
9.434 m 13 m
AD = 9.457 cm
Q 18.28 m S

Luas / Area of ∆ABD 9.434 + 18.28 + 13
2
= 1 × 9.457 × 9.679 × sin 64°19’ s =
2
= 20.36
= 41.25 cm2

4. Rajah menunjukkan sisi empat PQRS pada suatu Luas / Area ∆QVS
= √20.36(20.36 – 9.434)(20.36 – 18.28)(20.36 – 13)
SPM satah mengufuk. = √20.36(10.926)(2.08)(7.36)
= √3 405.49
2017 Diagram shows a quadrilateral PQRS on a horizontal = 58.36 m2
plane.
BAB 9
V S
12 m 22 m

P
Q 70°

20 m

R Praktis
SPM
VQSP ialah sebuah piramid dengan keadaan Ekstra
PQ = 8 m dan V adalah 5 m tegak di atas P.

VQSP is a pyramid such that PQ = 8 m and V is 5 m
vertically above P.

177

  Matematik Tambahan  Tingkatan 4  Bab 9 Penyelesaian Segi Tiga KBAT KBAT

Sudut Ekstra

Rajah menunjukkan sebuah kuboid. T U
W
Diagram shows a cuboid. V
Hitung / Calculate 50 cm

(a) luas, dalam cm2, bagi satah condong PRT. SR
20 cm
the area, in cm2, of the inclined plane PRT.
P 30 cm Q
(b) jarak serenjang, dalam cm, dari T ke PR.

the perpendicular distance, in cm, from T to PR.

(c) sudut di antara satah condong PRT dan tapak PQRS.

the angle between the inclined plane PRT and the base PQRS.

(a) T PR2 = 302 + 202
PR = √1300
2900 3400
TR2 = 302 + 502
TR = √3400

TP2 = 202 + 502
TP = √2900

PK R
1300

(√3400)2 = (√2900)2 + (√1300)2 – 2(√2900)(√1300) kos/cos P
2(√2900)(√1300) kos/cos P = 2900 + 1300 − 3400

kos/cos P = 800
2 (√2900)(√1300)

∠P = 78°7’

Luas / Area of ∆PRT = 1 (√2900)(√1300) sin 78°7’
2

= 950.02 cm2

(b) Luas / Area of ∆PRT = 950.02 cm2

1 × √1300 × t = 950.02, di mana / where t = Titik tengah / Midpoint PR
2
950.02 × 2
t= √1300 = 52.70 cm

BAB 9 (c) L sin ∠TKS = 50
52.70

∠TKS = 71°35’

50 cm 52.70

SK Kuiz 9

178

BAB

10 Nombor Indeks
NOTA IMBASIAnNdex Numbers

10.1 Nombor Indeks

Index Numbers

NOTA IMBASAN

Nombor Indeks
Index Numbers

1. Nombor indeks / Index number, 2. Indeks harga / Price index,

I= Q1 × 100  dengan keadaan / where I= P1 × 100   dengan keadaan / where
Q0 P0

Q1 = Kuantiti pada masa tertentu P1 = Harga barang pada masa tertentu
Quantity at a specific time Price of item at a specific time

Q0 = Kuantiti pada masa asas P0 = Harga barang pada masa asas
Quantity at base time Price of item at base time

NOTA

1. Selesaikan. TP 3

Solve.

CONTOH

Kilang P menghasilkan 250 0  00 dan 287 5  00 buah telefon bimbit masing-masing pada tahun 2017 dan
2019. Hitung nombor indeks untuk menunjukkan perubahan pengeluaran pada tahun 2019 menggunakan
tahun 2017 sebagai tahun asas.

Factory P produced 250 000 and 287 500 mobile telephones in the year 2017 and 2019 respectively. Calculate the index
number to show the change in the production in the year 2019 using 2017 as the base year.

Penyelesaian:

Katakan / Let

Q 17 = Bilangan telefon bimbit pada tahun 2017

Number of mobile telephone in the year 2017

Q19 = Bilangan telefon bimbit pada tahun 2019

Number of mobile telephone in the year 2019

Nombor indeks/ Index number  I = Q19 × 100
Q17

= 287 500 × 100
250 000

= 115

179

  Matematik Tambahan  Tingkatan 4  Bab 10 Nombor Indeks

(a) Bilangan buku latihan yang dijual di sebuah koperasi sekolah pada tahun 2016 dan tahun 2018 masing-

masing adalah 25000 dan 30500 buah. Hitung indeks jualan pada tahun 2018 berasaskan tahun 2016.

The number of exercise books sold by a school cooperative store in the year 2016 and 2018 were 25000 and 30500
respectively. Calculate the index number in the year 2018 based on the year 2016.

Q16 = 25000 buku / books

Q18 = 30500 buku / books Q18
Q16
Nombor indeks / Index number, I = × 100

= 30 500 × 100
25 000

= 122

2. Hitung indeks harga bagi setiap yang berikut. TP 3

Calculate the price index for each of the following.

CONTOH

Harga sebotol sos tomato pada tahun 2007 dan tahun 2011 masing-masing ialah RM2.50 dan RM3.25. Cari

indeks harga bagi sos tomato itu pada tahun 2011 dengan menggunakan tahun 2007 sebagai tahun asas.

The prices of a bottle of tomato sauce in the year 2007 and the year 2011 were RM2.50 and RM3.25 respectively. Find the
price index of the tomato sauce in the year 2011 by using the year 2007 as the base year.

Penyelesaian:
Katakan P07 = Harga sebotol sos tomato pada tahun 2007/ Price of a bottle of tomato sauce in the year 2007
Let P11 = Harga sebotol sos tomato pada tahun 2011/ Price of a bottle of tomato sauce in the year 2011

Indeks harga, I = P11 × 100
P07
Price index

= 3.25 × 100
2.50

= 130

(a) Harga sebuah kereta pada tahun 2011 dan (b) Barangan Harga / Price (RM)
tahun 2013 masing-masing ialah RM80 000 dan Item 2001 2003
RM60 0 00. Cari indeks harga bagi kereta itu
pada tahun 2013 berasaskan tahun 2011. A 3.50 4.90

The prices of a car in the year 2011 and the year 2013 B 6.00 8.10
were RM80 000 and RM60 000 respectively. Find the
price index of the car in the year 2013 based on the Jadual di atas menunjukkan harga bagi dua
year 2011.
barangan A dan B pada tahun 2001 dan 2003.
Indeks harga, I = P13 × 100 Hitung indeks harga bagi A dan B pada tahun
P11 2003 berasaskan tahun 2001.
Price index
60 000 The table shows the prices of two items A and B in the
BAB 10 = 80 000 × 100 years 2001 and 2003. Calculate the price indices of A
and B in the year 2003 based on the year 2001.

= 75 Indeks harga barangan A, IA = P03 × 100
P01 × 100
Price index item A 4.90

= 3.50

= 140

Indeks harga barangan B, IB = P03 × 100
P01 × 100
Price index item B
8.10
=
6.00

= 135

180

  Matematik Tambahan  Tingkatan 4  Bab 10 Nombor Indeks 

3. Selesaikan setiap yang berikut.

Solve each of the following.

CONTOH 1

Indeks harga sejenis jam tangan pada tahun 2009 dengan menggunakan tahun 2006 sebagai tahun asas

ialah 130. Jika harga jam tangan itu pada tahun 2006 ialah RM420, cari harganya pada tahun 2009.

The price index of a watch in the year 2009 by using the year 2006 as the base year was 130. If the price of the watch in
the year 2006 was RM420, find its price in the year 2009. TP 4

Penyelesaian:
Katakan P06 = Harga jam tangan pada tahun 2006/ Price of the watch in the year 2006
Let P09 = Harga jam tangan pada tahun 2009/ Price of the watch in the year 2009

Indeks harga = 130

Price index
P09
P06 × 100 = 130

P09 × 100 = 130
420
130 × 420
P09 = 100

= RM546

(a) Indeks harga bagi sejenis mentol LED pada tahun 2003 berasaskan tahun 2001 ialah 85. Jika harga bagi

mentol itu pada tahun 2001 ialah RM22, cari harganya pada tahun 2003.

The price index of a LED bulb in the year 2003 based on the year 2001 was 85. If the price of the bulb in the year 2001
was RM22, find its price in the year 2003.

Indeks harga = 85

Price index

PP0013 × 100 = 85

P2023 × 100 = 85
85 × 22
P03 = 100

= RM18.70

(b) Indeks harga bagi sebuah rumah pada tahun 2012 berasaskan pada tahun 2009 ialah 150. Jika harga

rumah itu ialah RM240 000 pada tahun 2012, cari harganya pada tahun 2009.

The price index of a house in the year 2012 based on the year 2009 was 150. If the price of the house was RM240 000
in the year 2012, find its price in the year 2009.

Indeks harga = 150

Price index BAB 10

P12 × 100 = 150
P09

240P009 00 × 100 = 150
P06 = 240
000 × 100
150

= RM160 000

181

  Matematik Tambahan  Tingkatan 4  Bab 10 Nombor Indeks

(c) Jadual menunjukkan harga satu kilogram barangan A, B dan C pada tahun 2005 dan 2006.

Table shows the price per kilogram of items A, B and C in the year 2005 and 2006.

Barangan Harga / Price (RM) Indeks harga pada tahun 2006
berasaskan tahun 2005
Item 2005 2006
Price index in the year 2006
A RM1.20 RM1.60 based on the year 2005

B x RM2.30 z

C RM0.60 y 110
Cari nilai z, x dan y.
102
Find the value of z, x and y.

z = P06 × 100 Indeks harga B = 110 Indeks harga C = 102
P05
  Price index B   Price index C

= 1.60 × 100 P06 × 100 = 110 P06 × 100 = 102
1.20 P05 P05

= 133.33 2.30 × 100 = 110 y × 100 = 102
x 0.60
2.30 × 100 102 × 0.60
x = 110 y = 100

= RM2.09 = RM0.61

(d) Harga / Price (RM) Indeks harga pada tahun 2012
Barangan
Item berasaskan tahun 2010
Price index in the year 2012
2010 2012 based on the year 2010

R x 360 125

S 125 y 140

Jadual di atas menunjukkan harga bagi barangan R dan S pada tahun 2010 dan 2012 dan indeks harga

pada tahun 2012 berasaskan tahun 2010. Cari nilai x dan y.

The table shows the prices of items R and S in the years 2010 and 2012 and the price indices for both items in the
year 2012 based on the year 2010. Find the values of x and y.

Indeks harga barangan R = 125 Indeks harga barangan S = 140

Price index of item R Price index of item S

P12 × 100 = 125 P12 × 100 = 140
P10 P10
BAB 10
360 × 100 = 125 y × 100 = 140
x 125
360 × 100 140 × 125
x = 125 y = 100

= 288 = 175

182

  Matematik Tambahan  Tingkatan 4  Bab 10 Nombor Indeks 

CONTOH 2

Barangan Indeks harga

Item Price index

M 2010 2012 2012
N
(2005 = 100) (2005 = 100) (2010 = 100)

120 150 x

110 y 140

Jadual di atas menunjukkan indeks harga bagi dua barangan M dan N. Cari nilai x dan y. TP 5

The table shows the price indices of two items M and N. Find the values of x and y.  

Penyelesaian:

Bagi barangan M x = P12 × 100
P10
For item M

P10 × 100 = 120  ⇒  P10 = 120 = P12 × P05 × 100
P05 P05 100 P05 P10

P12 × 100 = 150  ⇒   P12 = 150 = 150 × 100 × 100
P05 P05 100 100 120

= 150 × 100
120

x = 125

Bagi barangan N y = P12 × 100
P05
For item N

P10 × 100 = 110  ⇒  P10 = 110 = P12 × P10 × 100
P05 P05 100 P10 P05

P12 × 100 = 140  ⇒   P12 = 140 = 140 × 110 × 100
P10 P10 100 100 100

y = 154

(e) Indeks harga /Price index
Barangan
Item 2002 2008 2008

(2001 = 100) (2001 = 100) (2002 = 100)

J 125 175 x

K 120 y 110 BAB 10

L z 130 150


Jadual di atas menunjukkan indeks harga bagi tiga barangan J, K dan L. Cari nilai x, y dan z.

The table shows the price indices of three items J, K and L. Find the values of x, y and z.

183

  Matematik Tambahan  Tingkatan 4  Bab 10 Nombor Indeks

Bagi barangan J, Bagi barangan K, Bagi barangan L,

For item J, For item K, For item L,

PP0012 × 100 = 125 PP0012 × 100 = 120 PP0081 × 100 = 130
P02 P02 P08
⇒  P01 = 125 ⇒  P01 = 120 ⇒  P01 = 130
100 100 100

PP0018 × 100 = 175 PP0028 × 100 = 110 PP0082 × 100 = 150
P08 P08 P08
⇒  P01 = 175 ⇒  P02 = 110 ⇒  P02 = 150
100 100 100

P02
P01
P08 y = P08 × 100 z = × 100
P02 P01
x = × 100 P02 P08
P08 P01
P08 P01 y = P08 × P02 × 100 z = × × 100
P01 P02 P02 P01
x = × × 100 100 130
150 100
175 100 = 110 × 120 × 100 z = × × 100
100 125 100 100
x = × × 100 130
150
175 = 132 z = × 100
125
x = × 100 z = 86.67

= 140

(f) Indeks harga bagi dua barangan untuk tahun-tahun tertentu diberi dalam jadual berikut.

Price indices of two items for certain years are given in the following table.

Barangan Indeks harga /Price index

Item 1995 1999 1999

P (1990 = 100) (1990 = 100) (1995 = 100)
Q
125 140 x

135 y 120

Hitungkan nilai bagi x dan nilai bagi y.

Calculate the value of x and of y.

Bagi barangan P: Bagi barangan Q:

For item P For item Q

PP9950 × 100 = 125  ⇒  P95 = 125 PP9950 × 100 = 135  ⇒  P95 = 135
P90 100 P90 100

PP9909 × 100 = 140  ⇒  P99 = 140 PP9995 × 100 = 120  ⇒  P99 = 120
P90 100 P95 100
BAB 10
x = P99 × 100 y = P99 × 100
P95 P90

x = P99 × P90 × 100 y = P99 × P95 × 100
P90 P95 P95 P90

x = 140 × 100 × 100 y = 120 × 135 × 100
100 125 100 100

x = 112 y = 162

184

  Matematik Tambahan  Tingkatan 4  Bab 10 Nombor Indeks 

(g) Indeks harga bagi Perodua Kancil bagi tahun 1996 berasaskan tahun 1990 ialah 140 dan indeks harga
bagi Perodua Kancil bagi tahun 2000 berasaskan tahun 1996 ialah 105. Cari indeks harga bagi Perodua
Kancil bagi tahun 2000 berasaskan tahun 1990.

The price index for Perodua Kancil in the year 1996 based on the year 1990 is 140 and the price index for Perodua
Kancil in the year 2000 based on the year 1996 is 105. Find the price index of Perodua Kancil in the year 2000 based
on the year 1990.

PP9960 × 100 = 140  ⇒  P96 = 140 P00 × 100 = P00 × P96 × 100
P90 100 P90 P96 P90

PP9006 × 100 = 105  ⇒  P00 = 105 = 105 × 140 × 100
P96 100 100 100

= 147

(h) Indeks harga bagi suatu barangan elektrik pada tahun 2000 dan 2002 berasaskan tahun 1998 masing-
masing ialah 116 dan 125. Hitungkan indeks harga bagi tahun 1998 dan 2000 jika tahun 2002 digunakan
sebagai tahun asas.

Price indices for a certain electric product in the year 2000 and 2002 based on the year 1998 are 116 and 125
respectively. Calculate the price index in the year 1998 and 2000 if the year 2002 is used as based year.

PP9008 × 100 = 116  ⇒  P00 = 116 P00 × 100 = P00 × P98 × 100
P98 100 P02 P98 P02

PP9082 × 100 = 125  ⇒  P02 = 125 = 116 × 100 × 100
P98 100 100 125

P98 × 100 = 100 × 100 = 92.8
P02 125

= 80

NOTA IMBASAN

10.2 Indeks Gubahan

Composite Index

NOTA IMBASAN

Indeks Gubahan
Composite Index

1. Indeks gubahan / Composite index 2. Pemberat ialah nilai atau kuantiti yang
diberikan kepada setiap barangan untuk
I = I1w1 + I2w2 + …… + Inwn menunjukkan kepentingan relatif setiap BAB 10
= Iiwiw1 + w2 + …… + wn barangan itu.
Weightage is the value or quantity assigned to each
wi item to show the relative importance of each item.

dengan keadaan / where

I = Nombor indeks / Index number

w = Pemberat / Weightage

185

  Matematik Tambahan  Tingkatan 4  Bab 10 Nombor Indeks

4. Hitung indeks gubahan bagi setiap yang berikut. TP 3

Calculate the composite index for each of the following.

CONTOH (a) Bahan Indeks harga Pemberat

Komponen Indeks harga Pemberat Ingredient Price index Weightage

Component Price index Weightage J 125 4

P 120 5 K 135 2
Q 115 6
R 130 7 L 130 1
S 145 2
M 140 5


Indeks gubahan,/ Composite index,

Penyelesaian: I = Ii wi
wi
Indeks gubahan,/ Composite index,
= 125(4) + 135(2) + 130(1) + 140(5)
Ii wi 4+2+1+5
I = wi 1 600
= 12

= 120(5) + 115(6) + 130(7) + 145(2) = 133.3
5+6+7+2
2 490
= 20 = 124.5

(b) Makanan Indeks harga Pemberat (c) Barangan Indeks harga Pemberat

Food Price index Weightage Item Price index Weightage

A 115 2 E 123 30
B 120 3 F 145 20
C 140 5 G 151 10
D 105 6 H 115 40

Indeks gubahan,/ Composite index, Indeks gubahan,/ Composite index,

I = Ii wi I = Ii wi
wi wi

= 115(2) + 120(3) + 140(5) + 105(6) = 123(30) + 145(20) + 151(10) + 115(40)
2+3+5+6 30 + 20 + 10 + 40
1 920
BAB 10 = 16 = 12 700
100
= 120
= 127

186

  Matematik Tambahan  Tingkatan 4  Bab 10 Nombor Indeks 

5. Selesaikan setiap yang berikut.

Solve each of the following.

CONTOH 1

Barangan Jumlah perbelanjaan sebulan (RM) Pemberat

Item Total expenditure per month (RM) Weightage

Elektrik 2007 2009 4

Electricity 200 270

Pendidikan 350 420 1

Education 220 242 3

Pengangkutan 340 493 2

Transportation

Makanan

Food

Jadual di atas menunjukkan perbelanjaan bulanan sebuah keluarga pada tahun 2007 dan tahun 2009 dan
pemberatnya masing-masing. Hitung indeks gubahan pada tahun 2009 dengan menggunakan tahun 2007
sebagai tahun asas. TP 4

The table shows the monthly expenditure of a family in the years 2007 and 2009 and their respective weightages.
Calculate the composite index in the year 2009 using year 2007 as the base year.

Penyelesaian: 270 493
200 340
Indeks harga bagi elektrik = 135 × 100 Indeks harga bagi makanan = 145 × 100
Price index of electricity = Price index of food =

Indeks harga bagi pendidikan = 420 × 100 Indeks gubahan/ Composite index

Price index of education = 350 = 135 × 4 + 120 × 1 + 110 × 3 + 145 × 2
120 4+1+3+2

Indeks harga bagi pengangkutan = 242 × 100 = 1 280
220 10
Price index of transportation 110
= = 128

(a) Harga / Price Pemberat 37.50
Barangan IA = 30.00 × 100 = 125
(RM) Weightage
Item
2010 2013
IB = 49.50 × 100 = 110
A 30.00 37.50 2 45.00

B 45.00 49.50 3 IC = 70.00 × 100 = 140
50.00
C 50.00 70.00 4
32.50
D 25.00 32.50 3 ID = 25.00 × 100 = 130

Jadual di atas menunjukkan harga bagi empat Indeks gubahan  BAB 10
barangan pada tahun 2010 dan tahun 2013 dan Composite index
pemberatnya masing-masing. Hitung indeks
gubahan pada tahun 2013 berasaskan tahun = 125 × 2 + 110 × 3 + 140 × 4 + 130 × 3
2010. 2 + 3 + 4+3

The table shows the prices of four items in the years = 1 530
2010 and 2013 and their respective weightages. 12
Calculate the composite index in the year 2013 based
= 127.5

on the year 2010.

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  Matematik Tambahan  Tingkatan 4  Bab 10 Nombor Indeks

(b) Harga / Price 120
(RM) 80
Barangan Pemberat IE = × 100 = 150

Item 2012 2014 Weightage

E 80 120 3 IF = 46 × 100 = 115
6 40
7
F 40 46 IG = 66 × 100 = 120
55
G 55 66

H 95 133 4 IH = 133 × 100 = 140
Jadual di atas menunjukkan harga bagi empat 95

komponen pada tahun 2012 dan tahun 2014 Indeks gubahan / Composite index 140
150 × 3 + 115 × 6 + 120 × 7 + × 4
dan pemberatnya masing-masing. Hitung indeks I = 3 + 6 + 7+4
gubahan pada tahun 2014 berasaskan tahun
2012. = 2 540
20
The table shows the prices of four components in the
years 2012 and 2014 and their respective weightages. = 127
Calculate the composite index in the year 2014 based
on the year 2012.

CONTOH 2 (c) Barangan Indeks harga Pemberat
Item Price index Weightage
Barangan Indeks harga Pemberat
P 125 4
Item Price index Weightage
Qx 3
A 120 1
B 114 y R 120 y

Cx5 S 150 3

D 130 2 Jadual di atas menunjukkan indeks harga bagi
empat barangan pada tahun 2010 berasaskan
Jadual di atas menunjukkan indeks harga bagi tahun 2007 dan pemberatnya masing-masing.
empat barangan pada tahun 2008 berasaskan Harga bagi barangan Q pada tahun 2007 dan
tahun 2004 dan pemberatnya masing-masing. tahun 2010 masing-masing ialah RM8.00 dan
Harga bagi barangan C pada tahun 2004 dan tahun RM10.40. Diberi indeks gubahan pada tahun
2008 masing-masing ialah RM30 dan RM42. Diberi 2010 berasaskan tahun 2007 ialah 128.75, cari
indeks gubahan bagi tahun 2008 berasaskan tahun nilai x dan nilai y.
2004 ialah 128, cari nilai x dan nilai y. TP 5
The table shows the price indices of four items in the
The table shows the price indices of four items in the year 2010 based on the year 2007 and their respective
year 2008 based on the year 2004 and their respective weightages. The prices of item Q in the years 2007 and
weightages. The prices of item C in the years 2004 and 2010 were RM8.00 and RM10.40 respectively.
2008 were RM30 and RM42 respectively. Given the Given the composite index in the year 2010 based on
composite index in the year 2008 based on the year 2004 the year 2007 is 128.75, find the value of x and of y.
is 128, find the value of x and of y.  

Penyelesaian: x = P10 × 100 = 10.40 × 100 = 130
P08 42 P07 8.00
BAB 10 x = P04 × 100 = 30 × 100 = 140

Indeks gubahan/ Composite index = 128 Indeks gubahan / Composite index = 128.75

120 × 1 + 114 × y + 140 × 5 + 130 × 2 = 128 125 × 4 + 130 × 3 + 120 × y + 150 × 3 = 128.75
1+y+5+2 4 + 3 + y+3

114y + 1 080 = 128 120y + 1 340 = 128.75
y+8 y + 10

114y + 1 080 = 128y + 1 024 120y + 1 340 = 128.75y + 1 287.5
8.75y = 52.5
14y = 56 y = 6

y = 4

188

  Matematik Tambahan  Tingkatan 4  Bab 10 Nombor Indeks 

CONTOH 3

Rajah menunjukkan carta palang yang mewakili kos bagi tiga komponen A, B, C dan indeks harga bagi

komponen-komponen bagi tahun 2003 berasaskan tahun 2001 diberi dalam jadual di bawah. Diberi indeks

gubahan ialah 107. Cari nilai m. TP 5

Diagram shows a bar chart which represents the cost for three components A, B and C and price indices for the
components in the year 2003 based on the year 2001 are given in table below. Given the composite index is 107. Find the
value of m.  

Kos / Cost Komponen Indeks harga

m Component Price index

6 A 122
4 B 80

ABC Komponen C 110
Component

Penyelesaian: 122m + 920 = 1 070 + 107m
15m = 150
Indeks gubahan/ Composite index = 107 m = 10

122 × m + 80 × 6 + 110 × 4 = 107
m+6+4

122m + 920 = 107
10 + m

(d) Jadual menunjukkan indeks harga dan peratus penggunaan empat bahan P, Q, R dan S, yang digunakan

dalam pengeluaran sejenis biskut.

Table shows price indices and percentage of usage for four items P, Q, R and S, used in a production of a biscuit.

Bahan Indeks harga pada tahun 1995 berasaskan Peratus penggunaan
tahun 1993
Item Percentage of usage
Price index in the year 1995 based on the year 1993
P (%)
Q 135 40
R 30
S x 10
105 20
130

Diberi indeks gubahan pada tahun 1995 berasaskan tahun 1993 ialah 128. Cari nilai x.

Given the composite index in the year 1995 based on the year 1993 is 128. Find the value of x.

Indeks gubahan / Composite index = 128

135 × 40 + x × 30 + 105 × 10 + 130 × 20 = 128
40 + 30 + 10 + 20
BAB 10
9 050 + 30x = 128
100

9 050 + 30x = 12 800
30x = 3 750
x = 125

189