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3.4.4 SPM Praktis, Penjelmaan (Soalan Panjang)

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Soalan 4:
Rajah menunjukkan titik J(1, 2) dan sisi empat ABCD dan sisi empat EFGH, dilukis pada suatu satah Cartes.



(a) Penjelmaan U ialah satu putaran 90o, ikut arah  jam pada pusat O.
Penjelmaan T ialah satu translasi ( 2 3 )  
Penjelmaan R ialah satu pantulan pada garis x = 3.

Nyatakan koordinat imej bagi titik J di bawah setiap penjelmaan berikut:
(i) RU,
(ii) TR.

(b) EFGH ialah imej bagi ABCD di bawah gabungan penjelmaan MN.
Huraikan selengkapnya penjelmaan:
(i) N,   
(ii) M.

(c) Diberi bahawa sisi empat ABCD mewakili suatu kawasan yang mempunyai luas 18 m2.
Hitungkan luas, dalam m2, kawasan yang diwakili oleh rantau berlorek.


Penyelesaian:



(a)
(i) J (1, 2) → U → (2, –1 ) → R → (4, –1)
(ii) J (1, 2) → R → (5, 2) → T → (7, 5)

(b)(i)
N: Satu pantulan pada garis lurus x = 6.

(b)(ii)
M: Satu pembesaran pada pusat (8, 7) dengan faktor skala 3.

(c)
Luas EFGH = (faktor skala)2 x Luas objek ABCD
  = 32 x 18
 = 162 m2

Oleh itu,
Luas rantau berlorek
= Luas EFGH – luas ABCD
= 162 – 18
= 144 m2


3.4.3 SPM Praktis, Penjelmaan (Soalan Panjang)

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Soalan 3:
(a) Rajah di bawah menunjukkan dua titik, M dan N, pada suatu satah Cartesan.



Penjelmaan T ialah satu translasi (  3 −1 ) dan penjelmaan R ialah satu putaran 90o lawan arah jam pada pusat (0, 2).
(i) Nyatakan koordinat imej bagi titik M di bawah penjelmaan R.
(ii) Nyatakan koordinat imej bagi titik N di bawah penjelmaan berikut:
(a) T2,
(b) TR,

(b) Rajah di bawah menunjukkan tiga pentagon, A, B dan C, dilukis pada suatu satah Cartesan.


(i) C ialah imej bagi A di bawah gabungan penjelmaan WV.
Huraikan selengkapnya penjelmaan:
(a) V    (b) W
(ii) Diberi bahawa A mewakili suatu kawasan yang mempunyai luas 12 m2, hitung luas, dalam m2, kawasan yang diwakili oleh C.



Penyelesaian:
(a)




(b)


(b)(i)(a)
V: Satu pantulan pada garis x  = 8

(b)(i)(b)
W: Satu pembesaran pada pusat (14, 0) dengan faktor skala.

(b)(ii)
Luas B = luas A = 12 m2
Luas C = (faktor skala)2 x Luas objek
 = 22 x luas B
  = 22 x 12
 = 48 m2


6.8.4 Statistik, SPM Praktis (Kertas 2)

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Soalan 7:
Data dalam rajah di bawah menunjukkan markah diperoleh sekumpulan 30 orang murid dalam suatu ujian sejarah.

(a) Berdasarkan data dalam rajah di atas, lengkapkan Jadual di ruang jawapan.

(b) Berdasarkan Jadual di (a)
(i) cari find the modal class,
(ii) hitung min markah bagi seorang murid.

Untuk ceraian soalan ini, gunakan kertas graf.
(c) Dengan menggunakan skala 2cm kepada 10 markah pada paksi mengufuk dan 2cm kepada 1 murid pada paksi mencancang, lukiskan satu poligon kekerapan bagi data tersebut.

Jawapan:





Penyelesaian:
(a)


(b)(i)
Kelas modal = 50 – 54

(b)(ii)

Min markah =  1615 30  = 53.83
(c)



10.3 SPM Practice (Long Questions)

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Question 4:

The diagram above shows a solid consisting of a right prism and a half-cylinder which are joined at the plane HICB. The base ABCDEF is on a horizontal plane. The rectangle LKJG is an inclined plane. The vertical plane JDEK is the uniform cross-section of the prism. AB = CD = 2 cm. BC = 4 cm. CM = 12 cm.
Draw to full scale
(a)  The plan of the solid
(b) The elevation of the solid on a vertical plane parallel to ABCD as viewed from X.
(c)  The elevation of the solid on a vertical plane parallel to DE as viewed from Y.
 
Solution:
(a)


(b)



(c)


10.3 SPM Practice (Long Questions)

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Question 1:



The diagram above shows a solid consisting of a right prism and a half-cylinder which are joined at the plane EFGH. EF is the diameter of the semi-circle and is equal to 3 cm.

The base ABCD is on a horizontal plane and AB = 6 cm, BC = 4 cm. The vertical plane ABFE is the uniform cross-section of the prism. 

Draw to full scale, the plan of the solid.
 
Solution:



Question 2:


Diagram above shows a solid right prism with rectangular base ABCD on a horizontal table. The vertical plane ABEHIL is the uniform cross-section of the prism. Rectangle LIJK, IHGJ and HEFG are inclined planes. AL, DK, BE and CF are vertical edges. 

Given BC = 4 cm, AB = 6cm. EB = FC = LA = KD = 4cm, The vertical height of  I and J from the rectangular base ABCD = 3cm, while the vertical height of H and G from the rectangular base ABCD = 5cm. 

Draw to full scale, the elevation of the solid on a vertical plane parallel to BC as viewed from X.

Solution:


10.2 Plans and Elevations

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10.2 Plans and Elevations
 
1.  Plan is the image formed when solid is viewed from the top. Its orthogonal projection lies on the horizontal plane.
2.  Elevation is the image formed when a solid is viewed from the front or from the side. Its orthogonal projection lies on the vertical plane.
3.  In drawing plans and elevations of solids,
    (a)  Visible edges should be drawn using solid lines(──),
    (b)  Hidden edges whose views are blocked should be drawn using dashed lines (- - -).
 
Example:









10.3 SPM Practice (Long Questions)

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Question 3:
The diagram above shows a right prism attach to a cuboid at one of its plane. ABCG and CDEF, IJNH and KLMJ are horizontal planes, ABIH, AGNH and GCJN are vertical planes while IBCJ is an inclined plane. HA = 3 cm, BC = 3 cm, CD = 1.5 cm, HI = NJ = JK = 3 cm. Draw to full scale

(a) The elevation of the solid on a vertical plane parallel to AB as viewed from X.

(b) A solid semi-cylinder is joined to the solid in A as show in the diagram above. JPON is a vertical plane and JP = 1.5 cm. Draw to full scale
i. The plan of the solid
ii.   The elevation of the solid on a vertical plane parallel to BCD as viewed from Y.


Solution:
(a)



(b)(i)



(b)(ii)

9.4.3 Shortest Distance between Two Points

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9.4.3 Shortest Distance between Two Points
 
1. The shortest distance between two points on the surface of the earth is the distance measured along a great circle.


  Shortest distance between points D and M
   = ( θ × 60 ) nautical miles

Example:


In the above diagram, calculate
(a) The distance from P to Q, measured along the parallel of latitude 48o S,
(b) The distance from P to Q, measured along the route PSQ, where S is the South Pole.
 State the shorter distance.

Solution:

(a)
Distance from Pto Q, measured along the parallel of latitude 48o S
= 180 × 60 × cos 48o← (angle PMQ = 180o)
= 7266.61 n.m.
 
(b)
Distance from Pto Q, measured along the route PSQ, where S is the South Pole
= 84 × 60 ← (angle POQ = 180o – 48o – 48o = 84o)
= 5040 n.m.
 
The distance from P to Q, measured along the route PSQ in (b), where S is the South Pole, is shorter than the distance measured along the parallel of latitude in (a). 
 
  The shortest distance in the above example is    
  the distance along the arc of a great circle,   
  which passes through the South (or North) Pole.
 

9.6 SPM Practice (Long Questions)

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Question 1:
Diagram below shows four points P, Q, R and M, on the surface of the earth. P lies on longitude of 70oW. QR is the diameter of the parallel of latitude of 40o N. M lies 5700 nautical miles due south of P.
(a) Find the position of R.

(b) Calculate the shortest distance, in nautical miles, from Q to R, measured along the surface of the earth.

(c) Find the latitude of M.

(d) An aeroplane took off from R and flew due west to P along the parallel of latitude with an average speed of 660 knots.
Calculate the time, in hours, taken for the flight.
 
Solution:
(a)
Latitude of R = latitude of Q = 40o N
Longitude of Q = (70o – 25o) W = 45o W
Longitude of R = (180o – 45o) E = 135o E
Therefore, position of R = (40o N, 135oE).
 
(b)
Shortest distance from Q to R
= (180 – 40 – 40) x 60
= 100 × 60
= 6000 nautical miles
 
(c)
∠ P O M = 5700 60 = 95 o ∴ Latitude of M = ( 95 o − 40 o ) S = 55 o S

(d)
Time taken = distance from R to P average speed = ( 180 − 25 ) × 60 × cos 40 o 660 = 155 × 60 × cos 40 o 660 = 10.79 hours