KERTAS 1

ii 1. amxan=am+n Iii ^^ 7. an =\a' 2. a^+a'=Y- =gtn-n'ii an -- 8. (a* X anY = a*p X Anp

,: 3. (a^)" = a*nt: 4. (ab) = a*b g. grY = o^otJ \an ) anP1: s. ao=1,:rm-/\z' 6. a-t - ! 10. an - \la^ - lryr;\"'

- \Y"/j:i A

SPM m Question 231 Simplify (*2)o + ms.

Ringkaskan (m2)4 + m5.

Am Bm3c mtr D mr3

SPM ffi Question 24

2 Given that 3r = +. find the value of x.3zx'

Diberi 3. = +, carikan nilai x.

SPM m7 Question 23

4 +can be written as3P

L Uorr, ditulis sebagai3rr

A trr'C 2lrs

B!;D 2lxs

A2)C+J

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/ I \3

3Simprify(t#)/ I \rI g2x7t l-Ringkaska"\ *_1.

A 2ax7-3B 26x7-2C 212 * 7-sD 216 y 7-r

SPM Xn7 Question 24

rr1s Simplify lAf^)3 x ,-2f3.

/ 'lRingkaska" lAfu)' x e-2f3.

A e1f B ,rf-,C ef D ef-3

SPM m Question 23

6 Given that *t = 3-2, find the value of m and

of n.

Diberibahaw I ^' 'o *, = 3-', cari nilai m dan nilai n.

A m=2,n---3 B m=2,n=3C m=3,,n=-2 D m=3,n=2

BID+

28

SPM m Question 24

7 Simplify:Ringkaskan:

A m-2n7

B m2n4C m-6n-rD m6n

J

Diberi 184nilai n.

A m=4,nB m=3,nC m=4,n

D m=3,n

SPM mg Question 24

(mzn\-r__l 1m 'n"

= *tEF, nyatakan nilai m dan

SPM m Question 233

8 Given 18T = ^,EBn , state the value of m andof n.

-3-4

1

J

1

4

3

(+)'3

(*r2

(+)'2

(+r

(+l'

SPM nlo Question 23

tO 3Gr =

2

A (+), B

2

C55 D

SPM NlO Question 24

11 Simplify:Ringkaskan:

I(sr')5 1 .t xs-r

A s3/

B 4s3t

C s3fD 4s3f

SPM 2oll Question 2312 Simplify:

Ringkaskan:

('*^

A 6mnB 6m2n

C 36m2nD 36msn

SPM 2o1l Questlon 24I

13 t5=A_1

I

5l

c-+5r

3

/r\zt_t\5/3n

5'

l.o

" z"')m3

B1I

5:

D+5r

29

KERTAS 1

l1ry.,'ry.,,--ffi . (q*), :6nni, .Ofr+Off:Om-n

(*r)o + ms

=x:lr*,=m3Answer: B

".Tji'ffi . Express 9 in index form with base i.

'^me j{_: nn1 -t1

on

' . Equate the indices to find the value of x.

oJ-3u

^a'1,x _ "J-3x

3x - 32-2xAnswer: C

Answer: C

,fl,+1..ffi,.t] . I :g-,,,t O

7 7 -s-Y'3f - 3*

Answer: A

i}..-Tr1fry . (o* x orY : emp x onpt,.orxen:om+n

r rllrrfu)t x e-2f3 - rtf-, x e-zf3

= s3+(a\7-2+3-ef

Answer: C

x=2-Zx3x=2

21-4

J

x$*avffi

r.: ll ii

ii:

. Express each term in index form as a term withbase 2 or 7.

/ t\3 / -!-\3(s'' 7r \ (e\', z5\-\--w-) = le x7V1

u I t3

- (r' "l' I_ \2, x7z I218 x7

- 26 x76

=)18-6*71-6= )r2 x 7-s

16 m^ = m-'= 3-l

.'. m=3,n=2Answer: D

n (m2n+1-r - m-2n4,-____________' m4n3 m4n3

= m-2 * (4)n4 - 3

= pr2n-7

Answer: B

m+-ru

"Joo : oT

3n184=^W=18il.'. m=4rn=3Answer: A

Answer: A

t2l0 3fi,

= (52)r = 5rAnswer: C

sffi=(+l+=(+)+

W . Express 8 as a term with base 2.

lt(8s3)T 1 ., (23s3)T 1 1

27 xs-r=- 2- xs-r

=AxszP2t

=sl*2P'r= s3r

Answer: A

/ t\z\l*oxznz) 9mBx4n_-7-=-_a-

=9ms x4n= 36msn

Answer: D

--+ 1

"l

5T

Answer: B

t2

30

13

FORM 3

KERTAS 1

> is greater thanlebih daripada

< is less thankurang daripada

> is greater than or equal tolebih daripada atau sama dengan

S is less than or equal tokurang daripada atau sama dertgan

5. o is used for > and <digunaknn untuk > dan <

6. o is used for ) and <digunakan untuk) dan I

-1 0r23x<33 is included in the inequality.3 termasuk dalam ketaksamaan itu.

x = 3, 2, 1, 0, -1, ...

-4-3-2-10t2-3 <x<2

x = -2, -1,0, 1,2

SPM 2006 Question 242 List all the integers -r which satisfy both the

inequalitiesf - I <xand tO*4)>x.Senarail<an semuo integer xyang memuasl<nn kedua-

dua lcetaksanutan i - t 1 x dan tU * 4) > x.

A -1, 0, 1,2,3B 0, 1,2,3c -1,0, 1

D 0, I

Ia:g

,II

8.1.

2.

3.

4.

9.

7.

12345x>22 is not included in the inequality.2 tidak termasuk dalam ketaksamaan itu.

x=3,4,5,,.,

10.< I I I I I I F>-1 012345

-l(x<5x = -1,0, 1,2,3, 4

i

i

SPM n6 Question 25I List all the integers x that

inequalities 3x - 8 < x 3 5 + 4x.Senaraikan semua integer x langketaksamaan 3x* 8 < x < 5 + 4x.

A 0, 1,2,3B -1, 0, 1,2, 3

c -1, 0, 1,2,3,4D -2, -1,0, 1, 2,3, 4

satisfy the

memuaskan

31

SPM m7 Question 25

3 The solution for 4x + 5 <

Penyelesaian bagi 4x + 5 <

,7

A *.-TB *.-+

IC *.-T

)D *.-iSPM m Question 254 Which number line represents the solution

of the simultaneous linear inequalities2x- 3>5 and4-x>4?Garis nombor manakah lan9 mewakilipenyelesaian bagi ketaksamaan linear serentak2x-3> 5 dan4-x>-6?A

,2xr- 3 rs

r-+iahh

-54-3-2-1 0 | 2 3 4 5 6 7 8 910

B

-5 4-3-2-t O t 2 3 4 5 6 1 8 9 10

C +

-54-3-2-10 | 2 3 4 5 6 7 8 9 10

D<---.-<

-5 4-3-2-1 0 I 2 3 4 5 6 7 8 9 10

SPM mg Question 255 Find the solution for 5 - 3x > 6 + x.

Cari penyelesaian bagi 5 - 3x > 6 + x.

A r>-14

B *<-l-4I

C x2-fZID xS-z

SPM m Question 26

6 List all the integer values of x that satisfyboth the simultaneous linear inequalities3 -2x < 1 and 4x- 3 < 13.Senaraikan semua nilai integer x yangmemuaskan kedua-dua ketaksamaan linearserentak 3 - 2x < I dan 4x - 3 < 13.

A 1,2B 3,4c 1,2,3D 2,3,4

SPM nlO Question 25

7 It is given that 2 < r< 9 and 2 - s < 4, wherer and.r are integers.Find the maximum value of r - s.

Diberi bahawa 2 < r < 9 dan 2 - s 1 4, dengankeadaan r dan s ialah integer.Cari nilai terbesar bagi r - s.

A8c10

B9D11

SPM nlo Question 268 List all the integers x which satisfy both the

simultaneous linear inequaliti.r {, < 3 andZ

8-3x<2.Senaraikan semua integer x yang memuaskan

kedua-dua ketaksamaan linear serentak 5 . fZ

danS-3x<2.A 2,3,4,5,6B 2,3,4,5c 3,4,5D 4,5

SPM 2oll Question 259 List all the integers y that satisfy both the

simultaneous linear inequalit i"t tr; > 9 and

y+1<10.Senaraikan semua integer y yang memuaskan

kedua-dua ketaksamaan linear serentak It a,dany + I < 10.

A 7,8B 6,7,8c 7,8,9D 6,7,8,9

32

REVIEW AND

KERTAS 1

''.i.i,,,. . Solve the inequalities 3x- 8 <xand x<5 + 4x.i 'x < 4 means 4 is not incuded'

,i.r.,.n :r

3x-8<x and x<5+4x2x<8 -5<3xx<4 _5 a"

{3-i<x<4 '

5

.'.x=-1,0, 1,2,3Answer: B

lr?i.t . First integer that is greater than -a: -l-iue2

L-l<x and3x-3<3x

-3 32x_1 <*

2?

-;=x<4.'. r = -1, 0, 1,2,3Answer: A

3 4x+5<t-U3

lZx + 15 <3 -Zxl4x < -12

*. -9t4*'-L

7

Answer: B

iry . o is used for > and <.I . o is used for > and <.

' . lnequality sign is reversed when both sides are. divided by a negative number.

I*(x+4)>xI

x+4>2x4>xx<4

2x-3>52x28

x> 44<x

:.4<x<10

and 4-x>4-x > -10-x - -10----1 -1r< 10

}-+

-54-3-2-t O I 2 3 4 5 6 7 8 9 10

Answer: A

33

5 5-3x26+x-4x21

,<- I4

Answer: B

f'ft,ij,, .-ffi o I < x means I is not included.:i . x ( 4 means 4 is included.

3-Zx<L and2<2xl<x

l<xS4:. x=2,3,4Answer: D

4x-3<134x<16x34

'f'*ry .Ic get the maximum of r - s, r must be the

' maximum value and s must be the minimumi value.

23r<9 and 2-s<4-s <2s>-2

Maximum value of r - s

= maximum value of r - minimum value of s=8-(-1)-oAnswer: B

ry.ltrtr .2 1x means 2 is included."' .x < 6 means 6 is not included.

1r.: and2x<6

23x<6.'. x = 2,3, 4, 5

Answer: B

?9 *v>9z

)v>9x+J

y>-6

63y<9.'. y = 6,'7,8Answer: B

8-3x<26<3x2<x

and y+l<10y <9

KERTAS 1

SPM M Question 221 Given that 10 - 3(2 - w) = 9w + 2, calculate

the value of w.Diberi l0 - 3(2 - w)nilai w.

.1Ai B

ICT D

SPM mG Question 225

2 Given that * - 4k = -2(3 - k), then k =

,,Diberi * - 4k = -2(3 - k), maka k =/.

= 9w + 2, hitungkan

IT2-=)

A#CY

f)

SPM m Question 24 Given that p + 2

B#D+

- NA?A, calculate the

SPM m7 Question 22

3 Given *" iL+ - 2x *1, find the value of x.

Diberi+ = 2x - l, cari nilai x.

)/,Aq B1l5)C; D;

value of p.

Diberi bahawa p + 2 - 3(I+d, hitung nilai p.

34

FORMS 2 &3

1-z1

m

B-+)D -l-6

SPM mg Question 25 Given :'* - I - x, calculate the value

of x.\*-)DiberiT -,

?A;5c6

SPM nlo Question 26 Given that 3 - 2(x : l) - +, find the value32

of x.

Diberi bahawa, -2(x - l) = L, cari nilai x.32

A+ B

c+ D

SPM frll Question 2,

7 Given Y;6 -!;4 = l,calculatethe32value of y.

Diberi+-+A-lc*6

- x, hitung nilai x.

B+D*

22T225

- l, hitung nilai y.

B-3D-8

KERTAS 2rr trl I S ISI l) :l i : I :Irl r Il I5E].I rI I f"'t :J rvl I Y{ il I ! tl

r#(!it. ii ffi iffi ffi ,B' ffi ,

LinearEquationsl&ll 1 1 1 1 1 1 1

B"ehos!?nffiSPM M Question 2I Calculate the value of p and of q that satisfy the following simultaneous linear equations:

Hitungkan nilai p dan nilai q yang memuaskan persamaan linear serentak berikut:

2p-3q=134p+e=5

Answer/,Iawapani

SPM mO Question 42 Calculate the value of x and of y that satisfy the following simultaneous linear equations:

Hitungkan nilai x dan nilai y yang memuaskan persamaan linear serentak berikut:

x+2y-63

Zx-!=-r

[4 marks]14 markah)

[4 marksl14 markah)

14 marksl14 markah)

Answer/,Iawapan:

SPM m7 Question 23 Calculate the value of g and of h that satisfy the following simultaneous linear equations:

Hitung nilai g dan nilai h yang rnemuaskan persamaan linear serentak berikut:

g+2h=l49-3h=*18

Answer//awapan:

35

SPM m Question 24 Calculate the value of x and of y that satisfy the following simultaneous linear equations:

Hitung nilai x dan nilai y yong memuaskan prrio*oan linear ierentak berikut:

*+|t--34x-!=16

Answer/Jawapan:

SPM mg Question 35 Calculate the value of .r and of y that satisfy the following simultaneous linear equations:

Hitung nilai x dan nilai y yang memuaskan persamaan linear ierentak berikut:

4x+!=l2x+3y=8

14 marks)[4 markah]

14 marksl14 markahl

14 marksl[4 markah]

Answer/"Iawapan:

SPM nlO Question 46 Calculate the value of x and of y that satisfy the following simultaneous linear equations:

Hitung nilai x dan nilai y yang memuaskan persamaan linear serentak berikut:

2x-!=4x+3y--5

Answer/,/awapan:

SPM Nl l Question 27 Calculate the value of m and of n that satisfy the following simultaneous linear equations:

Hitung nilai m dan nilai n yang memuaskan persamaan linear serentak berikut:

m+3n-12)-;l|L-n=/,J

Answer/,Iawapan:

36

14 marks)14 markah)

REVIEW AND

KERTAS 1

ffi . Expand the brackets first.. Croup the unknowns on the left of the equation

and the numbers on the right.i

l0-3(2-w)=9w+210-6 +3w--9w+2

3w-9w=2-4-6w = -2

*=*3

Answer: C

lilrlfixr$ . Multiply both sides of the equation by 2 first.. Then, expand the brackets.

r,

5+-4k=-2(3-k\,2

5-8ft=-4(3-k)5-8t=-12+4k-l2k = -17

,L7,\--t2

Answer: A

, x+l =2.r-l)x+l=10x-5

-9x = -62

L- ^J

Answer: D

4 p+2=3(l-2P)4

4p+8=3-6pl0p - -5

1P=- 2

Answer: A

- 5x-2_rr

-=

[ --t3

5x-2=3-3x8x=5

5

8

Answer: D

37

6r_2(x-t)-x329-2x+2 _ x

32ll-2x=x

322(11 -2x)=3a

22-4x=3x-7x = -22

__22

Answer: B

L'lffi . IVlultip y both s des of the equation by 6

y-6 _!-4 =t3220 - 6) - 3Cy - 4) = 62y-12-3y+12=6

-!=6!=-6

Answer: C

KERTAS 2+:11 . To solve trruo simultaneous linear equations is to

i find a pair of values for the variables which satisfy, both the equations.

. There are two methods of solving simultaneoust linear equations in two variables:

f O Elimination method(ii) Substitution method

i: -, . , .

', *th ,fi'Eli ,-

2P - 3q = 13 .'... @4P+q=5 ""'@

@x3: l2p+3q=t5 .....@@+@: l4p=)g

P=2Substitute p=2 into@:

aQ)+Q=58+q=5

Q=-3]. P=2,Q=-3

,fi{$ &bs*itu.t"ig*,,

2P-3q=tf ""'O4P+q=5 ""'@

From@: q=5-4p.....@

Substitute @ into @:

2p-3(5-4p)-t32p-15+l2p-1t

l4P = )gP=2

Substitute p = Z,irl P_ o,r,

=5-8I

j. P=2,q=-3

ilIethod 1r Elimination method

x+2y-6 .....O3

7-! =-7 ""' @

@x2: 3x-2y=-14 .....@o+@: ^--=-i

Substitute x = -2 into O:

-2+2Y=$2v -B!=4

... x=-2,y=4hlethod 2: Substitution method

x+2Y=$3ix-t*=-7/.

From@: x=6-2ySubstitute @ into @:

?+(6-2Y)-!=-7

^!.

9-3Y-!=-7-4y - -16

!=4Substitute ! = 4 into @:

x=6_2(4)=6-8

,t

.'. r= -2,y=4g+2h=I

49-3h=-18Ox4: 49+8h=4@ - @: -llh = -22

h=2Substitute h=2 into @:

g+2(2)=1g+4=l

o--7:.9--3,h=2

38

I

I

2'"

, -r. 1, = -3 ..... O

2"4x-Y=16 .....@

Ox4: 4x+6y=-12.....@@-@: -7y-28

!=4Substitute ! = -4 into @:

4x-(4)=164x+4=16

4x=12x=3

|. x-3,Y=1

4x+y-l2x+3Y=$

@x2: 4x+6y=l$@-@: 5y= 15

!=3Substitutelu=3intoO:

4x+3=l4x=-2

2x-Y=4x+3y--5

@x3: 6x-3y=l)@+@: 7x=7

x=lSubstitute x=l into@:

1+3Y=-J3Y=-6Y=-2

... r= l,!=-2

m+3n=12 .....O2--:-lll-n=2 .....@3

@x3: 2m-3n=6 .....@@+@: 3m=18

m=6

Substitute m=6 into@:6+3n=12

3n=6n=2

.'. m=6,n=2

FORM 2

KERTAS 2.l I7,! l'&'1l.ill : I T,JTIIIsE},III I I*.T I II I 3:I i] I IT.

flos ,, sos *ffiz 2008 ,'ie(l00:.' s10,. r r },,..:i',?0{.1.,,',.i

,$,,'r ,!::*.., i ,,:'*, I .,,,81 .,fi", ,,i,$: . B *t.,r,

.E.".,,,ii,.f_t

Circles I 1 1 1 1 1 1 -11

:;:

:rj

ii f. Circumference of a circle = ltd = Znr!,; Lilitan bulatan = nd = 21tj

o:1 2. Area of a circle = TCr2l1t Luas bulatan = tlj2:iiri

iit,: t.

liii

; u.t igngth- - angle subtended at centre{-circumference of circle 360"panjang lengkok _ sudut pusat

lilitan bulatan 360'

a area of sector angle subtended at centre+.=-area of circle 360"luas sektor

_ - sudut pusat

tr^ brlrtr" - 360"

SPM ffi Question 7

I Diagram 4 shows two sectors, ORST and OUV, with the same centre O. RWO is

a semicircle with diameter RO pnd RO = zOV. ROV and OUT are straight lines.

Rajah 4 menunjukkan dua sektor bulatan ORST dan OUV, kedua-duanya berpusat O. RWO ialah

semibulatan dengan RO sebagai diameter dan RO = zOV. ROV dan OUT ialah Saris lurus.

OV = 7 cm and ZUOV = 60o.

OV - 7 cm dan IUOV = 60o.

T)2' Using fi = T, calculate))

Dengan menggunakan n = 1, hitungkan

(a) the perimeter, in cm, of the whole diagram,perimeter, dalam cm, seluruh rajah itu,

(b) the area,, in cm2, of the shaded region.luas, dalam cmz, kawasan yang berlorek.

RO

Diagram 4Rajah 4

Answer/,Iawapan;(a)

[6 marks)16 markahT

(b)

39

SPM ffi Questrbn 82 In Diagraffi 3, OMRN is a quadrant of a circle with centre O and PQ is an arc of another circle

with centre O. OMP and ORQ are straight lines.Dalam Rajah 3, OMRN ialah sukuan bulatan berpusat O dan PQ ialah lengkok suatu bulatan lain,juga berpusat O. OMP dan ORQ ialah garis lurus.

OM = MP = 7 cm and Z.POQ = 60".OM = MP = 7 cm dan IPOQ = 60".

'))Using n = 7, calculate

Dengan menggunakan n = +, hitungkan

(a) the perimeter, in cm, of the whole diagram,perimeter, dalam cm, seluruh rajah itu,

(b) the area, in cm2, of the shaded region.luas, dalam cmz, kawasanyang berlorek.

[6 marks]16 markahf

(b)

SPM m7 Question 63 Diagram 3 shows quadrant OST and semicircle PQR, both with centre O.

Rajah 3 menunjukkan sukuan bulatan OST dan semibulatan PQR, yang kedua-duanya berpusat O.

OS = 2l cm and OP = 14 cm.OS = 2l cm dan OP = 14 cm.t?21lUselGuna Tc = !L 7)CalculateHitung

(a) the area, in cm2, of the shaded region,luas, dalam cmz, kawasan yang berlorek,

(b) the perimeter, in cm, of the whole diagram.perimeter, dalam cm, seluruh rajah itu.

M

Diagram 3Rajah 3

Answer/,Iawapan:(a)

Diagram 3

Rajah 3

Answer/,/awapani(a)

t6 marksT

16 markahl

(b)

40

SPM m Question 7

4 In Diagram 7, PQ and RS are arcs of two different circles which have the same centre O.

OPR is a straight line.Dalam Rajah 7, PQ dan RS ialah lengkok bagi dua bulatan yang berlainan tetapi mempunyai pusat yang

sama O. OPR ialah garis lurus.

s It is given that ZPOQ = 36" and ZROS = 60o.Diberi bahawa ZPOQ = 36" dan IROS = 60".

,r r)

Using n = 7, calculate

Menggunakan n = +, hitung

(a) the perimeter, in cm, of the sector ORS,perimeter, dalam cm, sektor ORS,

(b) the area, in cm2, of the coloured region.luas, dalam cmz, kawasan yang berwarna.

R P<-L4 cm ----+

Diagram 7Rajah 7

Answer/"Iawapan:(a)

[6 marks)16 markahl

(b)

SPM 20@ Question 1O

5 In Diagram lO, POQ is a sector of a circle with centre O and CDEF is a semicircle withcentre C. OFCDQ is a straighP line.Dalam Rajah 10, POQ ialah sektor kepada bulatan berpusat O dan CDEF ialah semibulatan berpusat C.

OFCDQ ialah garis lurus.

P It is given that OQ = 12 cm, CD - 3.5 cm and ZPOQ = 70o.))

Use n - 7 and give the answer correct to two decimal places.

CalculateDiberi bahawa OQ = 12 cm, CD = 3.5 cm dan IPOQ = 70o.

'))Guna n = 7 dan beri jawapan betul kepada dua tempat perpuluhan.

Hitung

(a) the area, in cm2, of the coloured region,luas, dalam cmz, kawasan yang berwarna,

(b) the perimeter, in cm, of the coloured region.perimeter, dalam cm, kawasanyang berwarna.

16 marksl16 marknhT

Answer/,Iawapan:(a) (b)

CDFCLDiagram l0

Rajah l0

4t

SPM r-10 Question 96 Diagram 9 shows quadrant oPO and sector oRS, with common centre o. oeRis a straight line.

Raiah 9 menuniukkan sukuan bulatan oPQ dan sektor oRS, dengan pusat ,rpuryi O. OeR ialah garislurus.

Diagram 9Rajah 9

use n = +and give the answers coffect to two decimalplaces.Calculate

S Guna n = ? dan beri jawapan betul kepada dua tempat

prrpulrhon.THitung

(a) the perimeter, in cm, of the whole diagram,perimeter, dalam cm, seluruh rajah itu,

(b) the area, in cm2, of the whole diagram.luas, dalam cmz, seluruh rajah itu.

[6 marks][6 markah]

It is given that MP = 14 cm.t)')

Use n = *, calculateI

Diberi bahawa MP = 14 cm.

Gunan=4,hitung7

(a) the perimeter, in cm, of the whole diagram,perimeter, dalam cm, seluruh rajah itu,

(b) the area, in cm2, of the shaded region.luas, dalam cmz, kawasan yang berlorek.

16 marksl16 markah)

(b)

Answer/,/awapan:(a)

SPM 2fll Question 97 In Diagram 9, PMQL is a sector of a circle with centre P and OPRQ is a semicircle with

centre O.Dalam Rajah 9, PMQL ialah satu sektor bulatan dengan pusat P dan OPRQ ialah semibulatan denganpusat O.

(b)

Diagram 9Rajah 9

Answer/Jawapan;(a)

P 7cm O

42

REVIEWAND

KERTAS 2

ry (a) ' tRor= ]33. _:!9'

: 120"

i . RO:2x7 Cfl't:14Cm

sum of the lengths of arc RSI, line TU, arc

UV and line ROtl

. Length of an arc : S x Znr

(b) . Note that only the area of the shaded region

is required. Do not find the area of thewhole diagram.

. Area of a sector : * x Tcr2

. Area of a semicircle: lC

Perimeter of the whole diagram

= arc RSI+ TU + arc UV + RO + OV

= (#" zxlxr\+t* (#" 2"+"7)+14+7

-rn++7+r**14+l)

= 64i cm

Area of the shaded region

= area of sector ORST - area of semicircle

RWO + area of sector OUV

=(# "+xra2) -(+ "+x?)+( 60 *22 x z2)\360 7 t

= 205! -T + 25?3 "'-"3= 154 cmz

(a) . The perimeter of the whole diagram is

the sum of the len$hs of line OP, arc PQ,

line QR, arc RIV and line tVO.

. Length of an arc : S x z*

. OP:2x7 Cm: 14 cm

(b) . Note that onlythe area of the shaded region is

required. Do not find the area of the whole

diagram.

. Areaof asector:*xnr2

(a)

(b)

r"'*',=-:

(a) Perimeter of the whole diagram

=OP +arc PQ+QR+ arcRN+NO160 ?? \

= t4+ (m xTxf xlro-)+1 +

(# x2x+"t)+t

=t4+ruI+t+3t+t

(b) Area of the shaded region

= area of sector ORN + area of sector OPQ

- area of sector OMR

= (;* "+ "','). (# "+ x u2) -t 60 22 -,\[roo' t ' '')

= tZA + to2L - zsz-"6 3 '"35^

= 89? cm'

'r'3,'l (a) . Note that only the area of the shaded region

is required, that is the area of quadrant OSf' minus the area of sector OPQ.

(b) . Note that the perimeter of the wholediagram is the sum of the lengths of arc SI,arc QR, lines OS, Qf and OR.

' Lengh of an arc : * x 2nr

' TQ:OT-OQ, :21 -14, -7cm

(a) Area of the shaded region

= area of quadrant OST - area of sector OPQ

=(#"+xzr'?) -(#"+xr+'z)1)

= 346T - 1021

5

- 243i cm2

= +o! cm

43

Lwlffi'.I& (a) . The perimeter of sector ORS is the sum ofj the lengths of arc RS and radii OR and OS.

. Lengh of an arc : fi x Zru

(b) . The area of the coloured region is thearea of sector ORS minus the area of sectorOPQ,

. Areaof asector:*xnFr

Perimeter of sector OR,S

=arcRS+OR+OS

=(g x2x22 *zl)+ 2L+21\360 7 -l -=22+21+2I=64cmArea of the coloured region

= area of sector ORS - area of sector OPQ

=(#"+xzP) (#"+xw)?

- 231 - 6r;

= l69L cm25

(a) . The area of the coloured region is the areaof sector POQ minus the area of semicircleCDEF,

. Areaof asector:*xnp

. Area of a semicircle : Lf,(b) . The perimeter of the coloured region is

the sum of the lengths of arc PQ, radius Op,lines OF and DQ and arc FED.

. Length of an arc : S x Zv. Remember to give the answers correct to two

decimal places.

(a)

(b) Perimeter of the whole diagratn

= arc S7+ arc QR + OS + eT + OR

=(# x2x+"2t)+

(# x2x+"14) + 2t +7 + t4

=33+rr++21+7+t4

= loa* cm

(a) Area of the coloured region

= area of sector POQ - area of semicircleCDEF

= (9 x Zx r22) * (L * 22 x r.s2)\roo " 7 " " I' \7"7"u' )

= 88 - 19.25

= 69.75 cm212 d.p.;

(b) Perimeter of the coloured region

= arc PQ + OP + (OF + De) + arc FED

= (Jg-x|xs"rz) + D + (12 -2x 3.5)\roo 7 '-") '

*(Lx2x22 x:.s)\2 7 I)

-14++12+5+ll5)

= 4l-; ClllJ

= 42.67 cm (2 d.p.)

.ffi.,... (a) . The perimeter of the whole diagram is thei sum of the lengths of line Op, irc pe, line

i OR, arc RS and line SO.

. . Length of an arc : * * Zrui"360. o QR:14-7:7Cm

. (b) . Area of a sector : a# x nr2

, . Remember to give the answers correct to two

(a) Perimeter of the whole diagram

=OP +arc PQ+QR+ arcRS+SO

-i + (# xlx+"t) + t

*(9x2x?"la) +ta\360 7 'l=7 +ll+7 +14.67 +14= 53.67 cm (2 d.p.)

(b) Area of the whole diagram

= area of quadrant OPQ + area of sector ORS

-(eo "+"',,). (;* "+x ruz)\360

= 381 + rczZ23= AIL

6

= l4l.t7 cm2 (2 d.p.)

(b)

rI#:,.K,i-:

44

ffi (a).

a

(b) .

a

a

a

The perimeter of the whole diagram is

the sum of lenghs of line lP, line PM andarc MQL.

Reflex ZMPL : 360o - 150':21Oo

Lengh of an arc : * * Zm360

Notethatonlythearea of theshaded region isrequired. Do not find the area of the wholediagram.

Area of a sector:Lxw2360

Area of a semicircl I "": 2*

OQ:14cm+2:7cm

Perimeter of the whole diagram

=LP+PM+arcMQL

=14+ru*(2lo x2x?xru\\360 7 t

=14+14+51!3

= 79L cm3

Area of the shaded region

= area of sector PMQL - area of semicircleoPRQ

_ (210 *22 x r+2) _ (L ,22 x z2)\360 7 l\2 7 t

=x9L -n3

= 282L cmz3

(a)

(b)

45

FORM 3

KERTAS 2

2005, tr i ,,"2007 n,rffi8,,- ,,,,,20O9- ,..201o. ,mlit',fi,. fi,.r{,_Bf. *,i a fi , fi: t,,El". ,fi B A,il B,

Solid Geometry lll 1

I

1iI

1 1 1 1 1

1. Area of a trapeziuml^

= ; * sum of parallel sides x height

Luas trapezium1

= ; * hasil tambah sisi selari x tinggi

Curved surface area of a cylinder = 2nrhLuas permukaan melengkung silinder = Zxjt

Surface area of a sphere = 4nPLuas permukaan sfera = 4nj2

Volume of a right prism

= cross sectional area x lengthIsi padu prisma tegak = luas keratan rentas x panjang

5. Volume of a cylinder = nPhIsi padu silinder = fij2t

7.

volume of a cone = lnfhIsi padu kon = *ni',

Volumeofaspher"=lnfIsi padu sfera = *"fVolume of a right pyramid

= f, base area x height

Isi padu piramid tegak = f " fuas tapak x tinggi

Bahagian ASPM ffi Question 61 Diagram 3 shows a solid cone with radius 9 cm and height 14 cm. A cylinder with radius 3 cm

and height 7 cm is taken out of the solid.Rajah 3 menunjukkan sebuah pepejal berbentuk kon berjejari 9 cm dan tinggi 14 cm. Sebuah silinder yang

berjejari 3 cm dan tinggi 7 cm dikeluarkan daripada pepejal itu.

Calculate the volume, in cffi3, of the remaining solid.Hitungkan isi padu, dalam cm3, pepejal yang tinggal.| _221lUselGunakan Tt 7l

t4marks)[4 markah]

Answer/"Iawapan:

Diagram 3Rajah 3

/%\

46

SPM Xm Question 52 Diagram 2 shows a combined solid consists of a right prism and a right pyramid which are joined

at the plane EFGH. V is vertically above the base EFGH. Trapezium ABGF is the uniform cross

section of the prism.Rajah 2 menunjukkan sebuah gabungan pepejal yang terdiri daripada sebuah prisma tegak dan sebuah

piramid tegak yang tercantum pada satah EFGH. V terletak tegak di atas tapak EFGH. Trapezium ABGFialah keratan rentas seragam prisma itu.

The height of the pyramid is 8 cm and FG = 14 cm.Tinggi piramid ialah 8 cm dan FG = 14 cm.

(a) Calculate the volume, in cffi3, of the right pyramid.Hitungkan isi padu, dalam cm3, piramid tegak itu.

(b) It is given that the volume of the combined solidis 584 cm3.

G Calculate the length, in cm, of AF.Diberi bahawa isi padu gabungan pepeial itu ialah 584 cm3.

Hitung panjang, dalam cm, AF.

14 marksl[4 markah]

Diagram 2Raiah 2

Answer/,Iawapan:(a)

SPM m7 Question 1l3 Diagram 6 shows a solid, formed by joining a cylinder to a right prism. Trapezium AFGB is the

uniform cross section of the prism.AB - BC = 9 cm. The height of the cylinder is 6 cm and its diameter is 7 cm.Rajah 6 menunjukkan suatu pepejal yang terdiri daripada cantuman sebuah silinder kepada sebuah prisma

tegak. Trapezium AFGB ialah keratan rentas seragam prisma itu.

AB = BC = 9 cm. Tinggi silinder ialah 6 cm dan diameternya ialah 7 cm.

Calculate the volume, in cffi3, of the solid.Hitung isi padu, dalam cm3, pepejal itu.| 221llJselGuna fi =:7l

t4marks)14 markahl

Answer/,Iawapan:

l}cm G

Diagram 6Raiah 6

I*r.

tqhI

(b)

.'+--

47

SPM ffi Question 44 Diagram 4 shows a composite solid formed by the combination of a right prism and a half circular

cylinder at the rectangular plane ABFE. Right-angled triangle DFE is the uniform cross sectionof the prism.Rajah 4 menunjukkan sebuah gabungan pepejal yang dibentuk daripada cantuman sebuah prisma tegakdan sebuah separuh silinder pada satah segi empat tepat ABFE. Segi tiga bersudut tegak DFE ialahkeratan rentas seragam prisma itu.

Diagram 4Raiah 4

Answer/,/awapan;(a)

The diameter of the half circular cylinder is 7 cm and the volume

of the composite solid is 451.5 cm3. Using n = ?,calculatecmDiameter separuh silinder itu ialah 7 cm dan isi padu gabungan

pepejal itu ialah451.5 cm3. Menggunakan n = ?, hitung

(a) the volume, in cm3, of the half circular cylinder,isi padu, dalam cm3, bagi separuh silinder itu,

(b) the length, in cm, of BC.panjang, dalam cm, BC.

[5 marks)15 markah)

SPM mg Question 85 Diagram 8 shows a solid hemisphere. A solid cone is taken out from the hemisphere.

Rajah 8 menunjukkan sebuah pepejal berbentuk hemisfera. Sebuah pepejal berbentuk kon dikeluarkandaripada hemisfera itu.

The radius of the hemisphere is 7 cm. The radius of the cone is4 cm and the height of the cone is 3.5 cm.

')')Using n = t, calculate the volume, in cffi3, of the remaining solid.

Give the answer correct to two decimal places.Jejari hemisfera itu ialah 7 cm. Jejari kon itu ialah 4 cm dan tinggi kon

itu ialah 3.5 cm. Menggunakan n = +,hitung isi padu, dalam cm3,

bagi pepejal yang tinggal itu. Beri jawapan betul kepada dua tempatperpuluhan.

14 marksl14 markahl

Answer/,/awapani

(b)

Diagram 8Rajah 8

F ---.

48

SPM n10 Question 56 Diagram 5 shows the remaining solid after a cylinder is removed from a cube. The diameter of

the cylinder is 7 cm and the height of the cylinder is l0 cm.Raiah 5 menunjukkan pepejal yang tinggal setelah sebuah silinder dikeluarkan daripada sebuah kubus.Diameter silinder itu ialah 7 cm dan tinggi silinder ialah l0 cm.

Using n = ?,calculate the volume, in cm3, of the remaining

solid. Give the answer correct to two decimal places.

Menggunakan n = 2, hitung isi padu, dalam cm3, pepejal yang7

l0 cm tinggal itu. Beri jawapan betul kepada dua tempat perpuluhan.

14 marksl14 markah)

Diagram 5Rajah 5

Answer/,/awapan:

SPM 2oll Question 7

7 Diagram 7 shows a solid formed by joining a cuboid and a half cylinder at the rectangular planeEFGH.Rajah 7 menunjukkan sebuah pepejal yang terbentuk daripada cantuman sebuah kuboid dan sebuahseparuh silinder pada satah segi empat tepat EFGH.

The volume of the solid is 483 cm3.

Using n = +,calculate the height, in cm, of the cuboid.

Isi padu pepejal itu ialah 483 cm3.

Menggunakan n = +, hitung tinggi, dalam cm, kuboid itu.

14 marksl14 markahl

Diagram 7Rajah 7

Answer/,Iawapan:

--ir- - - -lt..- - -)

---,

' El,4-

I

,!, --

49

REVIEW AND +NSWERS

KERTAS 2iIffi . Volume of the remaining solid

' : volume of the cone - volume of the cylinder

. Volume of a cone : L nrz1,

3. Volume o{ a cylinder : nr2h

,.,

Volume of the remaining solid

= volume of the cone - volume of the cylinder

=(!r22 xe2x 14) - (2"z2xt\\3 7 - I \7 l

=1188-198= 990 cm3

*lfirill (a) . volume of a right pyramidI::- X base area x height3

(b) . Volume of a right prismi : cross sectional area x length

. Area of a trapezium

_t: -= X sum of parallel sides x height2

(a) Volume of the pyramid = | * 6 x 14 x 8

= 224 cm3

(b) Volume of the combined solid = 584 cm3

I x t10 + 14) x AFx 6 + 224 = 5842 TzxAF= 360

AF = 5 cm

3 Volume of the solid

= volume of the cylinder + volume of the rightprism

=l+ " (+)' x o] + l+ "(e + rzl x 8 x e]

=231 + 756

= 987 cm3

. (b) . Area of the right-angled triangle ,4BC

, - LxABxBC'2.1

(a) Volume of the half circular cylinder

=1r 22 r3.52x627= 115.5 cm3

(b) Total volume = 45L5 cm3

lxZ xBCx6+ 115.5=451.52 zrxBC=336BC=16cm

+tffi . Volume of the remaining solid: volume of the hemisphere - volume of

, the cone

, . Volume of a hemisphere : Z*,,3. Volume of a cone : !ru2h5. Remember to give the answer correct to

' two decimal places.

Volume of the remaining solid

= volume of the hemisphere - volume of the cone

=(Z*22 xz3) -(t *22 x42x3.5)\3 7 I \3 7 -'-l)12-7r8i-58t

= 660.00 cm3 12 d.p.;

i'.t . Volume of a cube

. :length

lbreadth x height

Volume of the remaining solid

= volume of the cube - volume of the cylinder

= (10 x l0 x 10) - (4"3.52 x 10)

=1000-385 \7 I

= 615.00 cm3 12 d.p.;

ry . Volume of a half cylinder : ,{'O. Radius of the semicircle

:FG:Jcm 7-^*: -T: --:

5.) Cm

Let the height of the cuboid be h cm.Volume of the solid

= volume of the cuboid + volume of the halfcylinder

483 = (7 x t2x h) +(+. 4"3.52 x 12)' \2 7 I483=84h+23184h = 252

h=3.'. Height of the cuboid = 3 cm

50