jawaban praktis 1,2,3,4,5,6

Jawapan

Praktis 1 (ii) T20 = 7 × 20 – 5 (ii) 2n + 2 = 70
= 135 n = 34
8 (a) a, b, c…
Praktis Formatif 17 (a) Tn = 13n – 1 b = a + 7
(b) (i) 194 (iii) 467 c = b + 7
1 (a) (i) (b) (i) (c) (ii) (d) (i) (ii) 298 (iv) 649 a + b + c = 27
2 (a) Menambah 6 kepada nombor a + (a + 7) + (a + 14) = 27
sebelumnya 18 3a = 6
TTT231 = 17 = 4 × 1 + 13 a = 2
(b) Mendarab 2 kepada nombor = 21 = 4 × 2 + 13
sebelumnya = 25 = 4 × 3 + 13 (b) 2, 9, 16, 23, 30…
(c) Menolak 3 daripada nombor (c) Tnn ==17,n2,–35,,4…
sebelumnya (d) 7n – 5 = 205
(d) Membahagi nombor sebelumnya Sebutan tertentu bagi jujukan n = 30
dengan 2 17, 21, 25, 29…

3 (a) (i) Nombor ganjil Tn = 4 × n + 13
(ii) Nombor genap
(iii) Nombor Fibonacci n = 1, 2, 3, 4, …
19 (a) 161 (b) 24n + 41
(b) (i) Nombor yang tidak boleh (c) (i) 425 Praktis 2
dibahagi tepat dengan 2
(ii) Nombor yang boleh dibahagi (ii) 24n + 41 = 1313 Praktis Formatif
24n = 1272 1 (a) 2(3x + 4)
tepat dengan 2 n = 53
(iii) Nombor yang diperoleh = 2 × 3x + 2 × 4
dengan menambah dua 20 (a) 4, 10, 16, 22, … = 6x + 8
6n – 2 (b) 4(2x – 5)
nombor sebelumnya (b) (i) 6 × 15 – 2 = 88 = 4 × 2x – 4 × 5
4 (a) Menambah 9 kepada nombor (ii) 160 ⩽ 6n – 2 < 180 = 8x – 20
sebelumnya
(b) (i) 14, 32, 50, 68, … 162 ⩽ 6n < 182
2 (a) 2x2 3x
Menambah 18 kepada 27 ⩽ n < 3 0 13 10x 15
nombor sebelumnya
(ii) 5, 23, 41, 59, …
Menambah 18 kepada n = 27, 28, 29, 30, … (b) 2x2 + 13x + 15
3 (a) (i) x2 (ii) x (iii) 2x (iv) 2
nombor sebelumnya Praktis Sumatif
5 1 1 (b) (x + 2)(x + 1) = x2 + x + 2x + 2
1 21 1 (a) Menolak 2.3 daripada nombor = x2 + 3x + 2
sebelumnya 4 (a) ✗ (b) ✓ (c) ✗ (d) ✓
13 31 5 2h × 5k – 2h × 4 + 3 × 5k – 3 × 4
14 6 41 1
1 5 10 10 5 1 (b) Mendarab 4 kepada nombor

1 6 15 20 15 6 1 sebelumnya = 10hk – 8 h + 15k – 12
(c) Menambah 6 titik kepada corak
6 p = 2 t = 13 sebelumnya 6 (a) 4x2 + 20x + 21 (b) 4x2 – 31x – 8
q = 3 u = 21 (c) 6x2 – 29x + 9 (d) 6x2 + 7x – 10
r = 5 v = 34 (d) Membahagi –2 kepada nombor 7 (a) 2k2 + k – 10 (b) 8r2 – 18r + 9
sebelumnya (c) 9m2 + 12m +4 (d) 4p2 – 4p + 1
s = 8 2 (a) A: Nombor ganjil diperoleh
7 (a) Jujukan kerana terdapat pola.
(b) Bukan jujukan kerana tiada pola. dengan menambah 4 kepada 8 ( 8p )2 – ( 3r )2 – 9r × p + 9r2
nombor sebelumnya
8 (a) Bukan jujukan kerana tiada pola. B: Nombor genap diperoleh = 64 p2 – 9 r2 – 9 pr + 9 r2

(b) Jujukan kerana terdapat pola. dengan menambah 8 kepada = 64p2 – 9pr
9 (a) 80, 84, 88, 92
(b) 133, 123, 113, 103 nombor sebelumnya 9 A
(b) 35, 47, 59
(c) 20, 100, 500, 2500 Menambah 12 kepada nombor 10 (4x + 5)(3x + 8) – 2(x + 3)
= 12x2 + 32x + 15x + 40 – 2x – 6
(d) 64, 32, 16, 8 sebelumnya = 12x2 + 45x +34
1
10 (a) 11, 35 (b) 4 , 4 (c) 52 , 49 3 11 (a) y = 5 + x

11 (a) 34, 22 (b) 2, 2 12 1 5 10 10 5 1 (b) z = x + (5 + x)
1 6 15 20 15 6 1 = 5 + 2x
1 7 21 35 35 21 7 1 (c) z2 – y = (5 + 2x)2 – (5 + x)
(c) 1.5, 0.75 = 4x2 + 19x + 20
12 (a) 6, 6, 6 (b) 7n – 6
4 (a) (b)
13 (a) n + 6, n = 1, 2, 3, 4, …
(b) 3n, n = 1, 2, 3, 4, … 12 (a) 3(2k + 1) 3, 2k + 1

(c) 10n – 5 , n = 1, 2, 3, 4, … (b) (y + 3)(y – 3) y + 3, y – 3
14 (a) 1 , 2 , 3 , 4
(b) n2 + 2 , n = 1, 2, 3, 4, … (c) (a – 2c)2 a – 2c
15 T2 = 7 T7 = 17 5 (a) 22, 29, 36 (b) 20, 17, 8
(c) 320, 1280 (d) 450, 0.72 (d) (3x – 2)(x – 4) 3x – 2, x – 4
T3 = 9 T8 = 19 6 (a) 13, 13, 13, 13
(b) (i) Menambah 13 kepada
T4 = 11 T9 = 21 13 5p2 – 20qr3 = 5(p r2 – 4 qr3)
nombor sebelumnya
T5 = 13 T10 = 23 (ii) 13n – 11, n = 1, 2, 3, 4… = 5 r2 (p – 4 q r )
7 (a) 1, 2, 3, 4…
T6 = 15 14 (a) 7 (r – 4) (b) 2m(2n + m)
(b) Tn = 2n + 2, n = 1, 2, 3, 4…
16 (a) 7 , 7 , 7 (c) (i) 52 (c) 92 – w2
(b) (i) T10 = 7 × 10 – 5 = 65
= (9 + w)(9 – w)

© Oxford Fajar Sdn. Bhd. (008974-T) 2018 J1 Drill in KSSM Matematik Tingkatan 2 (Jawapan)

(d) 3(9k2 – 4) 15 (a) ✓ (b) ✗ (c) ✗ (d) ✓ 12 (a) 2 + 3f – 3
= 3(3k + 2)(3k – 2) 4ac – 9a2 + 8bc – 18ab f(f + 3) f+3
16 2 + 3f – 3f
= f(f + 3)

Keadah 1 Keadah 2 = 2
f(f + 3)
1 + 4p 1 – 3p
(4ac – 9a2) + (8bc – 18ab) (4ac – 8bc) + (9a2 + 18ab) (b) 3(1 + 3p) + 6(1 – 4p)
= 8 ( 4 c – 9a) + 2b ( 4 c – 9 a) = 4c (a + 2b ) – 5a (a + 2b )
2(1 + 4p)(1 – 4p) +
= ( 4 c – 9 a)( 8 + 2b ) = (a – 2b )( 4c – 8a ) = (1 – 3p)(1 + 3p)
6(1 + 3p)(1 – 4p)

17 2y + 3b, 3x + 5a = 1 – 3x = 2(1 + 16p2) + 1 – 9p2
6 2(3x + 1) 6(1 + 3p)(1 – 4p)
18 (a) 16(4p2 + 1)
(b) 16(2p + 1)(2p – 1) 3 – 41p2
19 (a) (x + 6)(x + 1) (b) (3x – 2)(x + 4) = 3x + 1 – 9x = 6(1 + 3p)(1 – 4p)
6(3x + 1)
20 (a) x2, 4x, 3x, 12
(b) x = 1 – 6x 13 (a) h + 3k ÷ 9h + 3k
4 6(3x + 1) 10hk 25k2

x Praktis Sumatif = h + 3k × 25k2
1 (a) 6x – 3x2y (b) 4p2 + pr – 18r2 10hk 3(3h + k)

2 a2 – 4a + 4 + 4a2 + 16a + 12 = 5k(h + 3k)
6h(3h + k)
3 = 5a2 + 12a + 16

(x + 4)(x + 3) 3 1 (5x + 8)(4x – 6) (b) 4n – m – 3m – 2n
2 2m – 5n 4m – 10n
21 (a) 3x 2
= (5x + 8)(2x – 3) = 4n – m – 3m – 2n
x 3x2 2x = 10x2 + x – 24 2m – 5n 2(2m – 5n)
4 (a) 6a(2b + c) (b) (3p + 5)(3p – 5)
2 6 x 4 5 (i) 5k – 10 = 2(4n – m) – 3m + 2n
(ii) 5(k – 2) 2(2m – 5n)

(b) (3x + 2)(x + 2) (iii) x2 – 9 = 10n – m
22 (4x – y)(x + 5y), 295.5 (iv) (x + 3)(x –3) 2(2m – 5n)
23 (a) (4x – 3) + (3x + 2)2 + 2(4x – 3)
(v) m2 + 8m + 16 = 5(2n – m)
= 9x2 + 24x – 5 (vi) (m + 4)2 2(2m – 5n)
(b) 5(4x – 3) = 270
x = 14.25 (vii) 3ac + ad – 6bc – 2bd
24 (a) 2x2 + 15x – 6x – 18 (viii) (a – 2b)(3c + d)
= 2x2 + 9x – 18 6 (a) n(100m2 – 9) 14 (a) 9 × (x + 1)(x – 1)
x+1 12
= n(10m + 3)(10m – 3)
= (2x – 3)(x + 6) (b) 39x2 – 4xy + 4y2) = 3(x – 1)
(b) 25k2 – 20k + 4 – 9 = 3(x – 2y)2 4
= 25k2 – 20k – 5 7 (a) (2p – 5)(3r – p) 4t2
= (5(5k2 – 4k – 1) (b) (4x – 3y)(z + 2w) (b) (2t – 3)2 × 3(2t – 3)
= 5(5k + 1)(k – 1) 2t(4t + 3)
8 (a) 5r(2t + 3) (b) (2n + 1)(n – 7) 6t
25 (a) Benar (b) Palsu 9 3(y2 – 8y + 16) + 10y – 53 =
(c) Palsu (d) Benar
= 3y2 – 14y – 5 (2t – 3)(4t + 3)
3p – 8
26 1 – p2 + 2p – 3 = (3y + 1)(y – 5) 15 (a) y+2 +y
p+3 a = 3, b = 1, c = –5 y2 – 1 2y + 2

= 1 – 3p – 8 10 (a) PQ = 7x – 4 y+2 +y
p+3 (p + 3)(p – 1) QR = 2x + 8
1 = (y + 1)(y – 1) 2(y + 1)
2
= p – 1 – 3p + 8 [7x – 4) + (3x + 5)] = 2(y + 2) + y(y – 1)
(p + 3)(p – 1) 2(y + 1)(y – 1)
(2x + 8) – (x + 1)2
= (10x2 + 1)(x + 4) – (x + 1)2
7 – 2p = 10x + 41x + 4 – x2 – 2x – 1 = 2y + 4 + y2 – y
= 2(y + 1)(y – 1)
(p + 3)(p – 1) = 9x2 + 39x + 3
= 3(3x2 + 13x + 1) = y2 + y + 4
27 (a) ✓ (b) ✓ (c) ✗ (d) ✓ (b) (x + 1)2 = 16 2(y + 1)(y – 1)
28 (a) Benar (b) Palsu
x = 3
29 k Luas = 3(27 + 39 + 1)
3(k + 2) = 201 cm2 18m2 + 42mn
(b) (42m + 98n) ÷ n2
30 (a) ✓ (b) ✗ 11 (a) x + 2y n2
31 A 2y x = 14(3m + 7n)
6 4r s × 6m(3m + 7n)
32 (a) rs + 3 × 2r2 = x2 + 4y2
2xy
= 6 + 2s = 7n2
rs 3r 5k – 1 2k – 6 3m
18 + 2s2 (b) 9 – 3k
= 3rs
= k(5k – 1) – 3(2k – 6) Praktis 3
2(9 + s2) 9k
= 3rs 5k2 – k – 6k + 18 Praktis Formatif
= 1 (a) m = p2
(b) 1 – x ÷ 3x + 1 9k (c) z = y – w (b) k = 3(r + x)
6 4x – 4 6x – 6 = 5k2 – 7k + 18 (d) s = a
9k
= 1 – x × 6(x + 1) x
6 4x – 4 3x + 1

© Oxford Fajar Sdn. Bhd. (008974-T) 2018 J2 Drill in KSSM Matematik Tingkatan 2 (Jawapan)

2 s = 500 + 2h (c) 3p(5) = 1 (40) 8 (a) y = R (b) y = 8
3 (a) p = 2(m + n + 6) 4 m
1 2
(b) L= 2 (xy – 16) p = 3 9 m = 4
3
4 (a) 10 = 3 × 1 + 7 18 (a) 4r – 3rt = t
13 = 3 × 2 + 7 4r = t + 3rt 10 3yz – 6x = 2xy + 8z
3yz – 8z = 2xy + 6x
16 = 3 × 3 + 7 4t = t(1 + 3r) z(3y – 8) = 2x(y + 3)
19 = 3 × 4 + 7 t = 4r z = 2x(y + 3)
(b) y = 3x + 7
5 (a) k (b) y (c) m (d) w 5 1 + 3r
4 3y – 8
(b) t=
= 8(6 + 3)
6 (a) 12 a + 5b = 60 19 (a) (i) w1 = q+p 18 – 8
pq
(b) 12 a = 5 ( 12 – b) pq = 7.2
p+q
(c) 5 ( 12 – b) w = 11 (a) S = 2(ab + bc + ac)
12 (b) 112 = 2(3a + 6 + 2a)
1 1 1 56 = 5a + 6
(ii) p = w – q a = 10

7 (a) Palsu (b) Benar 1 qw 12 2ac = (9b – 4c)d
8 (a) ✓ (b) ✓ p q–w
9 (a) ✗ (b) ✓ (c) ✓ = (9b – 4c)d
2c ✓
a =
10 3k + 4 = 4 m3 3
(b) (i) w = 2 (ii) p = 10 2ac = 9bd – 4cd

3k = 4 (m3 – 1 ) 20 (a) s = 24(3) – 4.9(3)2 9bd = 2ac + 4cd
9bd = 2c(a + 2d)
k = 4 (m3 – 1 ) = 27.9 m b = 2c(a + 2d)
(b) 52.5 = 5u – 4.9(5)2 9d
5u = 175
3 u = 35
11 D 2ac + 4cd = 9bd

12 9v2 – p2 = 8t2 Praktis Sumatif 2c(a + 2d) = 9bd
2   1 (a) Ya c = 9bd

9v2 – p2 = 16t2 (b) 5pr + 3r = 10p 2(a + 2d)
3r = 5p(2 – r)
(a) p2 = 9v2 – 16t2 2ac = d(9b – 4c)

p = 9v2 – 16t2 p = 3r d = 2ac ✓
(b) v2 = 16t2 + p2 5(2 – r) 9b – 4c

9 2 2y – 6yk = 1 Praktis 4
2y(1 – 3k) = 1
y = 1
v = 16t2 + p2 2(1 – 3k) Praktis Formatif

3 3 2x + 16 = 4xy – y 1 (a) ✓ (b) ✓ (c) ✓
y + 16 = 4xy – 2x 2 (a) (i) berbeza (ii) sama
13 (a) 1 – x = 1 w3 y + 16 = 2x(2y – 1) (iii) poligon tak sekata
4y 8 (b) (i) sama (ii) sama
x
4y = 1 – w3 x = y + 16 (iii) poligon sekata
2(2y – 1) 3 (a) 3, =, poligon sekata
x = 4y(1 – 1 w3) (b) 1, ≠, bukan poligon sekata
8 4 2f = 1 – 3h
2 4 (a) ✓ (c) ✓
= 4y(8 – w3) 5 (a)
8 2f = 1 – 6h
y(8 – w3) C
2
= (1 2 A
– 6h)2
x 8 – w3 2f =
(b) 4y = 8 4
f = (1 – 6h)2
y = 2 8
x 8 – w3
B
y = 2x 5 10ac – 15ab = 9b + c
8 – w3 10ac – c = 9b + 15ab (b)
x(10a – 1) = 3b(3 + 5a) S R
14 (a) x = 99 (b) s = –3 (c) t = 5
15 (a) 2m = m – 4 c = 3b(3 + 5a)
(10a – 1)
m = –4
1 3
(b) 1 = 1 k – 4 6 16 g2 = 2– y
2
3 1 g2 P Q
k = 10 y = 2– 10 D C
16 V = 10t2y – 165 6
3 32 – g2 B
315 = 10t2(3) – 165 y =
16 E
30t2 = 480
48
t2 = 16 y = 32 – g2

t = 4 7 p2 = 9r – 4q FA
32 – g2
17 (a) 3(1)(2) = 1 y G
4
2p2q = 9r – 4q 7 (a) (i) 3 1 180°
y = 24 2p22q + 4q = 9r 360°
540°
(b) 3(4)r = 1 (8) 2q(p2 + 2) = 9r (ii) 4 2 720°
4 q = 9r
1 (iii) 5 3
6 2(p2 +
r = 2) (iv) 6 4

© Oxford Fajar Sdn. Bhd. (008974-T) 2018 J3 Drill in KSSM Matematik Tingkatan 2 (Jawapan)

(b) (i) (n – 2) 9 x = 115° Luas bulatan = x × j
(ii) (n – 2) × 180° y = 360° – 115° – 100° – 90°
=π×jxj
8 (a) a = 100°, b = 110°, c = 75°, = 55° = π × j2
d = 56°, e = 19° z + 10° = 25° 14 (a) 2π × jejari (b) π × jejari2
(b) 360° z = 15°
9 (a) 360° (b) 720° 1 15 (a) 8.80 cm (b) 13.20 cm
2 (c) 9.42 cm (d) 125.60 cm
10 Bilangan sisi = 5 10 x = × 120°
22
Jumlah sudut pedalaman = 60° 16 2 × 7 ×r= 132

= ( 5 – 2) × = 180° = 540° y = 60° – 45° r = 132 × 1 × 7
= 15° 2 22
x + 70° + 100° + 130° + 140° = 540° z = x + 45°
x = 100° = 105°
Praktis 5 = 21

11 (a) 540° 104° Praktis Formatif 17 (a) 44 cm, 154 cm2
(b) 35 mm, 3 850 mm2
(b) 720° 137° 1 (a) Pusat (b) Diameter (c) 2.1 m, 13.2 m

12 n = 6 (c) Jejari (d) Sektor 18 (a) 2π cm (b) 8π cm (c) 7π cm
x = (6 – 2) × 180° = 120° 19 (a) 11 cm (b) 9 cm (c) 120°
6 2 (a) Garis lurus yang menyambungkan 20 72°
dua titik pada bulatan.
(b) Sebahagian daripada lilitan. 200° 22
13 y = 135° (ii) 74 (c) Perimeter sebuah bulatan. 21 360° ×2× 7 × j = 440
14 (a) (i) 360°
(b) (i) 360° (ii) 113 (d) Rantau yang dibatasi oleh j = 126 m
22 (a) 22 cm2 (b) 594 cm2
15 360°, 70° lengkok dan perentas.
16 x = 360° = 72° 3 (a) (b) P π 22
23 360° × 7 × 8.42 = 36.96
5
17 30° 140 22 π = 60°
360° 7
18 n = 360° = 18 O × × j2 = 99
20° Q
19 (a) 30° j = 9 cm
(b) 12 24 (a) PR = 15 cm (b) 31.4 cm
20 108° (c) 91.4 cm
21 ∠RST = 108° 4 K N
22
x = 36° 25 142 – 7 × 72 = 42 cm2

y + 63° = 108° H 26 π × 62 – π × 32 = 27π cm2 (C)
y = 45°
x + y = 36° + 45° = 81° 27 (a) 60°, 150°
(b) Pajang lengkok PQ
22 C
23 C M 150° 22
5 B
Praktis Sumatif = 360° × 2 × 7 × 42
125°
1 (a) 9 (b) 140° O A = 110 cm
(c) A B

IC (c) Perimeter bagi rantau berlorek

= 110 + 21 + 21 + 21

HD 6 (a) Palsu (b) Benar + 42

E (c) Palsu (d) Benar = 215 cm
7 (a) ✓ (b) ✓ (c) ✗
G 8 (a) Benar (b) Benar 28 120° × π(OR2 – 92) = 48π
F 360° OR2 – 81 = 144

2 2x + 5x + 52° + 60° + 73° = 360° 9 1.7 cm
10 HM = 5 cm
x = 35° OR2 = 225
3 x + (360° – x) + 3x + 90° + 143° + 73°
= 720° KO = OM OR = 15 cm
GH HM PR = 6 cm
x = 18°
4 (a) x = 120 (b) 6 KO = 1 29 10° 40°
5 x = 180°, y = 24° 42 360° × π × j2 = 360° × π × 32

6 (a) (n – 2) × 180° = 1260° KO = 2 cm J = 6 cm
n = 9 3 0 32600°°
x = 140° KL = 2 + 2 12 × π × x2 = 80° × π × 52
(b) x : y = 140° : 40° 360°
= 4 21
= 7 : 2 x = 10 (B)
31 21 × 9 –
7 ∠BCD = (8 – 2) × 180° 11 OH2 = 512 – 452 30° × π × 182
8 = 576 360°
= 135° = 189 – 27π (C)
x = 45° OH = 576
∠BAE = 67.5° = 24 cm
y + 67.5° + 135° + 100° = 360° Tinggi air = 51 + 24 Praktis Sumatif
= 75 cm 1 (a) Jejari
y = 57.5° 12 (a) 3.14, 3.14, 3.14, 3.14, 3.14
(b) Tembereng minor
∠FEH = 1 × (360° – 135° – 135°) (b) 3.14, 3.14 (c) Sektor major
2 (d) Perentas
= 45° 13 12x = 1 × lilitan 2
z = 22.5° 2 × 2π × j
8 x = 108° = 1
y = 60 2 C 140° A
= π × j BO
z = 108° – 30° y = j
= 78°



© Oxford Fajar Sdn. Bhd. (008974-T) 2018 J4 Drill in KSSM Matematik Tingkatan 2 (Jawapan)

3 C 4 (a) ✓ (b) ✓ (c) ✗ (d) ✓ 17 C
B 5 I, III 18 2 × π × 6 2 + 2π × 6 × y = 216π
6 C
D   7 P Silinder 72 + 12 y = 216
O
Q Piramid 12 y = 144

A R Kuboid y = 12
19 D
4 (a) 16.97 cm (b) 13.42 cm S Prisma
5 (a) 34.2 cm (b) 5.63 cm
6 (a) x = 5 cm 20 (a) 2π × j × 8 = 80π
8 (a) c cm b cm c cm j = 5 cm
QM2 = 132 – 102 Q (b) π × 52 + 80π + π × 5 × 8
QM = 8.31 cm = 455.7 cm2
QR = 16.6 cm a cm
21 (a) ABCD = 160 cm2
P 13 cm ABFE = 120 cm2
c cm BCGF = 192 cm2
x
12 cm x R BFKJ = 28 cm2

a cm FGLK = 80 cm2

JKLM = 64 cm2
(b) 106 400 mm2
12 cm c cm
1
S (b) ac + ab + bc +b camc + ab + bc 22 (a) 2 × isi padu kuboid

(b) 1 × (24 + 16.6) × 15 = 2ac + 2ab + 2bc = 1 ×a×b×c
2 = 2(ac + ab + bc) cm2 2
= 305 cm2 9 (a)
7 (a) x = 12.12 cm j cm = luas tapak × tinggi
(b) luas bulatan × tinggi
BC = 26.1 cm = πj2t

x 14 23 (a) 3 × isi padu piramid
1
(b) 3 × luas tapak × tinggi

j cm 24 (a) 35 cm3 (b) 48 cm3
(c) 120 cm3 (d) 60 cm3
7 7 25 96 cm3
(b) 2 × (28 + 26.1) + 3× 2 × 22 × 7 (b) Panjang segi empat tepat = lilitan 26 I, II dan IV
7 bulatan 27 (a) 28 cm3 (b) 12 cm3 (c) 8 cm
= 240.2 cm 28 (a) 179.7 cm3 (b) 493 cm3
= 2πj cm (c) 5 352.4 cm3
22 Lebar segi empat tepat = tinggi
8 (a) 7 × 142 silinder

= 616 cm2 = t cm 29 2 21 m
(b) AB = x cm Luas permukaan silinder
x2 + x2 = 282 = luas segi empat tepat + 2 30 (a) 1 ×8× 16 × 11 = 704 cm3
2
x = 19.8 cm × luas bulatan
j = 9.9 cm = 2πj × t + 2 × πj2 (b) 22 × j2 × 14 = 704 cm3
22 10 (a) 7 j2 = 16
19.82 – 7 × 9.92 = 8.4 cm2 6 cm 4.8 c8mcm
9 (a) 45° × 2× 6 cm
360° 22 × 49 8 cm j = 4 cm
7 1
10 cm 31 2 × (9 + 7) × 4 × p = 83

= 38.5 cm 12 cm 12 cm p = 16 cm

45° × 22 × 492 4 10 cm 32 t= 3 j
360° 7 2 8 cm 4
(b) =
135° 22 × 24.52 6 cm 8 cm 1 πj2t = 2000π
360° × 7 6 cm 3

4:3 (b) 12 × 6 + 12 × 10 + 12 × 8 + 2 31 πj2 × 3 j = 2000π
10 (a) 79.1 cm (b) 135.975 cm2 4
22 j3 = 8000
× 7 × 10 × 4.8 j = 20
Praktis 6
= 336 cm2 2
Praktis Formatif 11 (a) 216 cm2 (b) 100 cm2 33 π× 32× 4 + 3 π × 33

1 (a) : Kuboid (b) : Silinder (c) 204 cm2 (d) 84 cm2 = 36 π + 18 π = 54 π cm2
12 Luas permukaan
(c) : Kon (d) : Piramid 34 (a) 15 × 3 × x + 10 × 8 × x = 1500
2 (a) Bentuk I: silinder tegak
Bentuk II: silinder serong = 2 × π × 4 2 + 2π × 4 × 10 125x = 1500

(b) A = 32 π + 80 π x = 12
(b) 13 × 12 + 15 × 12 + 15 × 3 × 2 +
= 112 π cm2 3 × 12 + 10 × 8 × 2 + 8 × 12 + 10
C Bentuk I Bentuk II D × 12 + 12 × 7 = 922 cm2
13 (a) 8 cm (b) 144 cm2
(c) 48 cm2 (d) 336 cm2
3 B 14 (a) 314 cm2 (b) 942 cm2 35 (a) 1 × (8 + 15) × 13 × 6 +
CB (c) 1256 cm2 3

15 2, 4, 2, 16 1 × 22 × 32 × 7 = 963 cm3
16 616 cm2 3 7

Piramid A Kon (b) 43 × 22 × j3 = 963
7
j = 613 m
ED 2464 cm2

© Oxford Fajar Sdn. Bhd. (008974-T) 2018 J5 Drill in KSSM Matematik Tingkatan 2 (Jawapan)

Praktis Sumatif 5 x = 13 (b) 22 × 52 + 440 + 22 × 5 × 13
1 (a) Keratan rentas seragam adalah 7 6 cm2 7
7
bentuk trapezium. Dua muka x cm = 722
bertentangan yang berbentuk
trapezium adalah kongruen dan 12 cm 7 (a) 2 × 22 × 3.5 × k = 2 × 22 × 14 × 8
selari. Muka-muka lain adalah 7 7
berbentuk segi empat tepat.
(b) A 5 cm k = 32 cm
2 (a) y = 12.65 cm
(b) 2 × 22 × 142 + 2 × 2 × 22
7 7
× 14 × 8 = 2640 cm2

y
(b) (i) θ ×2× 22 × 25 = 44 12 cm 8 196 cm3
360° 7

θ = 100.8° 9 22 × 4.22 × 25.2 – 3 × 4 × 22 × 4.23
22 4 cm 7 3 7
(ii) 2× 7 × j = 44 = 465.696 cm3
10 × 8 + 2 × 1
j = 7 2 × 8 × 13 + 2 × –1 22
102 0 × 20 × 15 3 7
36θ0° × 22 × 252 + 22 × 72 1 × 10 × 12.65 = 310.5 cm2 154 t = × × 72 × t = 5384
7 7 2 616
= 704 cm2 22
3 1440 cm2 6 (a) 2 × 7 × j × 14 = 440 3 t = 12 cm

4 (a) 96 cm2 (b) 1 : 9 j = 5 cm 11 (a) 4 cm (b) 552 cm3

© Oxford Fajar Sdn. Bhd. (008974-T) 2018 J6 Drill in KSSM Matematik Tingkatan 2 (Jawapan)