1. SMP NEGERI 3 KALIBAGOR
PAKET 4
Alfa Kristanti
1

2. SMP NEGERI 3 KALIBAGOR
Alfa Kristanti
2

3. SMP NEGERI 3 KALIBAGOR
Pelangi Matematika Viandre
Al4-kristanty.blogspot.com
2−1 + 3−1 =
7 7 ×
1
1
1 1
3 2
5
+ 1= + = + =
1
2
3
2 3
6 6
6
14 = 7 7 × 14 = 7 7 × 7 × 2 = 7 49 × 2
= 7 7 2 = 49 2
1
2
1
1 12 6
1 12
5
3 + 2 ∶1 = 3 +
∶ = 3 +
×
2
5
5
2
5 5
2
5
6
1
= 32 +
5−3
Selisih = 5 + 3 × 24 =
2
1
2
8
=
7
2
4
+2 =
11
2
× 24 = 6
9
Bunga 9 bln = 12 × 12% × 𝑇 = 9% 𝑇
Tabungan setelah 9 bln = 100% T + 9% T = 109% T
109% T = 3.815.000  T =
3.815.000
109%
=
3.815.000
= 3.815.000 ×
109
100
100
109
= 3.500.000
Alfa Kristanti
3

4. SMP NEGERI 3 KALIBAGOR
b=
b=
𝑈7 − 𝑈3
7− 3
12
4
=
22 −10
4
=3
𝑛
Sn = 2 ( 2a + (n – 1)b )
U3 = a + 2b
10 = a + 2(3)
10 = a + 6  a = 10 – 6
a=4
S30 =
30
2
( 2(4) + (30 – 1) 3) = 15 (8 + (29) 3) = 15 (8 + 87)
= 15 (95) = 1.425
a = U1 = 3
Un = a + (n – 1)b
b = U2 – U1 = 7 – 3 = 4
U45 = 3 + (45 – 1) 4 = 3 + (44) 4
= 3 + 176 = 179
r=
𝑈2
𝑈1
=
32
64
=
1
2
Un = a × r n – 1 = 64 ×
1 𝑛−1
=
2
Un = 26 × 2−𝑛 + 1 = 26 −
a = U1 = 64
𝑛 +1
26 × 2
𝑛−1
= 27−𝑛
x – 3 ≤ 5 – 3x
x + 3x ≤ 5 + 3
4x ≤ 8
x≤2

×
×

m=
𝑦2 − 𝑦1
𝑥2 − 𝑥1
I. 2x2 + x – 3 = (2x + 3)(x – 1)
II. x2 + x – 6 = (x + 3)(x – 2)
III. 4x2 – 9 = (2x)2 – 32 = (2x + 3)(2x – 3)
IV. 6x2 – 15x = 3x(2x – 5)
=
2 − −5
3− 1
y – y1 = m (x – x1)
7
y – (5) = 2 (x – 1)
7
2
y + 5 = (x – 1)
2(y + 5) = 7(x – 1)
=
2 +5
2
=
7
2
2y + 10 = 7x – 7
2y – 7x = – 7 – 10
– 7x + 2y = – 17
7x – 2y = 17
Alfa Kristanti
4

5. SMP NEGERI 3 KALIBAGOR
ax + by + c = 0
a=6;b=3
m=
− 𝑎
𝑏
=
− −6
3
=
6
3
=2
Jumlah bilangan terbesar dan terkecil
2
= 3 × jml ketiga bilangan
2
= 3 × 84 = 56
M = { 1, 3, 5, 7 }
N = { 5, 7, 11 }
M N = { 1, 3, 5, 7, 11 }
f(5) = 15
5a + b = 15
f(5) =  5
5a + b =  5
2p + 6q = 23.000
2(4000) + 6q = 23.000
8.000 + 6q = 23.000
6q = 15.000
q = 2.500
3p + 2q = 3(4000)+2(2500)
= 12.000 + 5.000
= 17.000
D
A
Misal :
Harga pulpen = p
Harga pensil = q
F(1) =  2 (1) + 5
=2+5
=3
5p + 4q = 30.000 × 3 15p + 12q = 90.000
2p + 6q = 23.000 × 2 4p + 12q = 46.000 
11p
= 44.000
p = 4.000
S=
26
26
5a + b =  5
5( 2) + b =  5
 10 + b =  5
b =  5 + 10
b =5
5a + b = 15
5a + b =  5 
 10a = 20
a=2
𝐾
4
=
104
4
= 26
DE = 𝐴𝐷2 − 𝐴𝐸 2 = 262 − 102 = 676 − 100 = 576 = 24
BD = 2 × DE = 2 × 24 = 48
1
1
1
Lbelahketupat ABCD = 2 × d1 × d2 = 2 × AC × BD = = 2 × 20 × 48 = 480
10 10
C
E
26
26
B
M
A
80
80o
o
C
38
62
K
o
38o
o
62o
L
Pada gambar di samping:
ABC  KLM maka
AB = LM, BC = KL,
AC = KM
B
Alfa Kristanti
5

6. SMP NEGERI 3 KALIBAGOR
P
12
Q
B
10
16
R
𝐴𝐶
𝑄𝑅
5
6
8
C
KL =
=
=
8
16
1
=2 ,
𝐴𝐵
𝑃𝑅
=
5
10
1
=2 ,
𝐵𝐶
𝑃𝑄
=
6
12
1
=2
A
3 × 12 + 2 × 27
3+2
36 + 54
5
=
90
5
= 18
L daerah tdk diarsir
= LEFGH + LABCD – 2 × L daerah diarsir
= 42 + 12 × 8 – 2 × 8
= 16 + 96 – 16
= 96
Banyak pohon cemara =
=
𝐾𝑒𝑙𝑖𝑙𝑖𝑛𝑔 𝑡𝑎𝑚𝑎𝑛
𝑗𝑎𝑟𝑎𝑘 𝑡𝑎𝑛𝑎𝑚
2(80)
4
=
2(48+ 32)
4
= 40
P
Q
(3x  6) + (4x + 25) = 180
7x + 19 = 180
7x = 180 – 19
7x = 161
x = 23
R
Pelurus AOC = COB = 4x + 25
= 4(23) + 25 = 92 + 25 = 117
Alfa Kristanti
6

7. SMP NEGERI 3 KALIBAGOR
AOE = 2 × ABE
= 2 × 32
= 64
Sudut keliling menghdp
busur yg sama maka
ABE = ACE =ADE
96
ABE = = 32
3
dalam = 172 − 5 + 3
=
289 − 64 =
 𝐴𝑂𝐷
 𝐵𝑂𝐶
LAOD =
=
40
100
2
=
289 − 82
225 = 15
× 𝐿 𝐵𝑂𝐶
× 50
= 20
Panjang kawat minimal = Panjang rusuk balok
= 4 (p + l + t) = 4 (40 + 40 + 35)
= 4 (115)
= 460
Garis pelukis
BC =
𝐾 𝐴𝐵𝐶𝐷
4
=
72
4
t = 152 − 92 =
t = 144 = 12
1
3
1
3
= 18
225 − 81
t
15
9
2
Vlimas = La × t = × 18 × 12 = 324 × 4 = 1.296
Alfa Kristanti
7

8. SMP NEGERI 3 KALIBAGOR
dbola = rusuk kubus
dbola = 24, maka :
24
r = 2 = 12
4
4
4
Vbola = π r3 = π × 123 = π × 1.728
3
3
3
= 4 π × 576 = 2.304 π
Ldinding = 2 (pt + lt) = 2(6 × 4 + 8 × 4)
= 2(24 + 32) = 2(56) = 112
Biaya = 112 × Rp 50.000,- = Rp 5.600.000,-
10
s
s
Lkubus = 6 s2
s2 + s2 = 102
2 s2 = 100
s2 = 50
s = 50 = 25 × 2 = 5 2
=6× 5 2
= 6 × 50
= 300
2
22
Ltabung = 2πr (r + t) = 2 × 7 × 35 (35 + 10)
= 220 (45) = 9.900
Jumlah siswa = 18 + 22 = 40 anak
Jml tinggi = 18 × 156 + 22 × 152 = 2.808 + 3.344 = 6.152
Rata-rata tinggi =
𝑗𝑚𝑙 𝑡𝑖𝑛𝑔𝑔𝑖
𝑏𝑎𝑛𝑦𝑎𝑘 𝑠𝑖𝑠𝑤𝑎
=
6.152
40
= 153,8
Data
5 6 7 8 9
frekuensi 3 2 3 4 2
Modus
Alfa Kristanti
8

9. SMP NEGERI 3 KALIBAGOR
32
30
Banyak siswa dapat nilai kurang dari 8 = 1 + 4 + 5 = 10
Alfa Kristanti
9

10. SMP NEGERI 3 KALIBAGOR
n(S) = 2 × 2 × 2 = 8
3 gambar = GGG
1
P (3 gambar) = 8
6
5
5
4
3
3
2
2
Banyak kelereng = 6 + 5 + 3 + 3 + 2 + 4 + 2 + 5 = 30
Kelereng merah = 6
6
P ( 1 merah) = 30 × 100 % = 20 %
Alfa Kristanti
10