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SOLUSI SOAL OLIMPIADE MATEMATIKA Persiapan Olimpiade Sains Provinsi dan Nasional Tingkat SMA
Departemen Matematika - Wardaya College
MMXVIII-VII
1. 111 1
2x
2 111 1x 2 1 111x ( Kuadratkan kedua ruas)
2
2
2
2
(2 1) 111
4 4 1 111
4 4 110 0
2 2 55 0...(1)
x
x x
x x
x x
Kalikan (1) dengan 3x ……. 5 4 32 2 55 0x x x …(2)
Kalikan (1) dengan x ……… 3 22 2 55 0x x x …(3)
Kalikan (1) dengan 1 …….. 22 2 55 0x x …(4) Jumlahkan (2)(3)(4), maka diperoleh:
5 4 32 2 53 57 55 0x x x x 5 4 3 2004(2 2 53 57 54)x x x x = 5 4 3 2004(2 2 53 57 55 1)x x x x
2004
2004
(0 1)
( 1)
1
Jawaban : E
2. sin18 = 1 5
4
maka nilai 8 cba Jawaban : A
3. 39
9)(
x
x
xf
xx
x
xf93
3
39
9)1(
1
1
11)( xfxf
1996
998
1996
999
1996
997...
1996
1995
1996
1
1996
1995...
1996
2
1996
1ffffffff
2
1995
33
3997
2
1997.1
1996
1995...
1996
2
1996
1
ffff
Jawaban : A
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4. Misalkan : 2x a , 2y b , 2z c . Substitusikan ke soal semula, sehingga
diperoleh : 2
2
2
42.........(1)
6...........(2)
30.......(3)
a bc
b ac
c ab
(1) – (2) : 2
2
2 2
42.........(1)
6...........(2)
36
( )( ) ( ) 36
( )( ) 36
36( ) .....(4)
( )
a bc
b ac
a b bc ac
a b a b c a b
a b a b c
a b ca b
(1) - (3) : 2
2
2 2
42.........(1)
30.......(3)
72
( )( ) ( ) 72
( )( ) 72
72( ) .....(5)
( )
a bc
c ab
a c bc ab
a c a c b a c
a c a b c
a b ca c
Substitusikan (4) dengan (5) :
36 72
( ) ( )a b a c
36( ) 72( )
( ) 2( )
2 2
2
....(6)2
a c a b
a c a b
a c a b
b a c
a cb
Substitusikan (6) ke (2) : 2
2
2 2
2 2
2 2
2
6...........(2)
( ) 62
26
4
2 4 24
2 24
( ) 24
( ) 24.....(7)
b ac
a cac
a ac cac
a ac c ac
a ac c
a c
a c
Substitusikan (6) ke (1) : (7) + (8) :
( ) 2 6.....(7)
(2 ) 7 6.....(8)
3 9 6
3 6
a c
a c
a
a
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2
2
22
2 2
42.........(1)
( ) 422
422
2 84
(2 )( ) 84
(2 ). 24 84
(2 ) 7 6........(8)
a bc
a ca c
ac ca
a ac c
a c a c
a c
a c
( ) 24.....(7)
3 6 2 6
6
a c
c
c
2
2
42.........(1)
(3 6) 6 42
54 6 42
6 12
2 6
a bc
b
b
b
b
Sesuai dengan pemisalan awal : 2
2(3 6)
54
x a
2
2(2 6)
24
y b
2
2( 6)
6
z c
54
24
6
x
y
z
maka 84 zyx
Jawaban : C
5. (√5 + 2 + √3 + √6)( √5 - √3 + 2 - √6)
= 5 -√15 + 2√5 - √30 + 2√5 - 2√3 + 4 - 2√6 + √15 - 3 + 2√3 - √18 + √30 - √18 + 2√6 – 6
= 5 + 4√5 + 4 – 3 - 2√18 - 6
= 4√5 - 6√2
(4√5 - 6√2)( 7√2 + 2√5)
= 28√10 + 40 – 84 - 12√10
= 16√10 - 44 Jawaban : A
6. baxba
xa
ab
xa
ab
ab
0,tan.
sin.1
sin.
.1 2
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.cos
sin
cos
sin).(.
sin).(
sin
cos
sin.)sin1(.
sin).(
sin
cos
sin.cos..
sin).(
sin.
.1
2
2
2
22
2
2
22
2
x
x
x
xaba
xaba
x
x
xbxa
xaba
x
x
xbxa
a
xaba
xa
ab
ab
Jawaban : E
7. Ingkaran dari semua adalah ada Jadi ingkaran dari kalimat “ Semua anak – anak suka bermain air” adalah “Ada anak – anak yang tidak suka bermain air” Jawaban : C
8. QT = (𝑥 −3
2𝑦 4)
PQT = R
(12 40 −11
) (𝑥 −3
2𝑦 4) = (
96 −2066 −44
)
(𝑥 −3
2𝑦 4) = -
1
132(
−11 −40 12
) (96 −2066 −44
)
= (
11
132
4
132
0 −12
132
) (96 −2066 −44
)
= (
1056
132+
264
132−
220
132−
176
132
0 − 792
1320 +
520
132
)
= (10 −3−6 4
)
x = 10 2y = -6 y = -3 2x + y = 2(10) – 3 = 17 Jawaban : E
9. 100+10+1+0,1+0,01+0,001+...=9
1111
Jawaban : E 10. 4x – 12.2x + 32 = 0
22x – 12.2x + 32 = 0 Misal : 2x = a a2 – 12a + 32 = 0 ( a – 8 ) (a – 4 ) = 0 a = 8 a = 4 2x = 23 2x = 22 x = 3 x = 2
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x1x2 = 3 . 2 = 6 Jawaban : B
11. 4 1 -11 30 -8 4 -28 8
1 7 2 0 ( x – 4 ) (x2 – 7x + 2)
Jawaban : D
12.
Jika
2
0
20072007
2007
cossin
sin
dxxx
x kita misalkan dengan I, maka
4
2
2cossin
cossin
2
0
2
0
20072007
20072007
I
I
Idxxx
x
x
x
Jadi, 14
4
cossin
sin4
2
0
20072007
2007
dxxx
x
.
Jawaban : D
13. Misal : 1
𝑥 = a ,
1
𝑦 = b ,
1
𝑧 = c
2a + 4b + 7c = 31 3a + 2b + 5c = 22 a + 3b + 4c = 19
2
020072007
20072
0
20072007
2007
)2
(cos)2
(sin
)2
(sin
cossin
sin
dx
xx
xdx
xx
x
2
0
20072007
2007
sincos
cos
dxxx
x
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2a + 4b + 7c = 31 . 3 6a + 12b + 21c = 93 3a + 2b + 5c = 22 . 2 6a + 4b + 10c = 44 - 8b + 11 c = 49 2a + 4b + 7c = 31 .1 2a + 4b + 7c = 31 a + 3b + 4c = 19 .2 2a + 6b + 8c = 38 –
-2b – c = -7
8b + 11c = 49 .1 8b + 11 c = 49
-2b – c = -7 .4 -8b – 4c = -28 +
7c = 21 → c = 3
-2b – 3 = -7
-2b = -4→ b = 2
a + 3(2) + 4(3) = 19
a + 6 + 12 = 19 → a = 1
1
𝑥 = a
1
𝑦 = b
1
𝑧 = c
1
𝑥 = 1
1
𝑦 = 2
1
𝑧 = 3
𝑥 = 1 y = 1
2 z =
1
3
x + y + z = 1 + 1
2 +
1
3 =
6+3+2
6 =
11
6
Jawaban : B
14. x + 1
𝑥 = √3
(x + 1
𝑥)3 = x3 + 3x2.
1
𝑥 +3x.(
1
𝑥)2 + (
1
𝑥)3
= x3 + 3x + 3
𝑥 + (
1
𝑥)3
x3 + (1
𝑥)3 = (x +
1
𝑥)3 – 3x -
3
𝑥
= (x + 1
𝑥)3 – 3(x +
1
𝑥)
= (√3)3 – 3(√3)
= 3√3 - 3√3 = 0 Jawaban : A
15. prinsip: ))(()( 22 bababa
000.000.000.000.000.554.555.555.555.555
)554.555.555.555.555)(000.000.000.000.000.1(
223.222.222.222.222777.777.777.777.777 22
Maka, jumlah semua angkanya adalah (5x14)+(4x1) = 74 Jawaban : C
G H
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16.
Lihat segitiga OCG
Maka OC = 1
2𝑎√2 , CG = a
tan α = 𝑂𝐶
𝐶𝐺 =
1
2𝑎√2
𝑎 =
1
2√2
Jawaban : A
17. a + b = -c b + c = -a c + a = -b
maka (𝑎+𝑏)(𝑏+𝑐)(𝑐+𝑎)
𝑎𝑏𝑐 =
−𝑐.−𝑎.−𝑏
𝑎𝑏𝑐 =
− 𝑎𝑏𝑐
𝑎𝑏𝑐 = -1
Jawaban : B
18. 𝑎
𝑏 +
𝑎+10𝑏
𝑏+10 𝑎 = 2
x b (b+10a)
a(b+10a) + b(a+10b) = 2b(b+10a)
ab +10a2 + ab + 10b2 = 2b2 + 20ab
10a2 -18ab + 8b2 = 0 : b2
10(𝑎
𝑏)2 – 18
𝑎
𝑏 + 8 = 0
: 2
5(𝑎
𝑏)2 – 9
𝑎
𝑏 + 4 = 0
(5(𝑎
𝑏) - 4)( (
𝑎
𝑏) - 1)= 0
𝑎
𝑏 =
4
5 atau
𝑎
𝑏 = 1
Karena dikatakan a dan b adalah bilangan berbeda maka 𝑎
𝑏 =
4
5
Jawaban : B
19. ( 1 - 1
4 ) ( 1 -
1
5 ) ( 1 -
1
6 ) ... ( 1 -
1
100 )
= 3
4 .
4
5 .
5
6 .....
99
100
= 3
100
Jawaban : C
20. 𝑎𝑏
𝑐 .
𝑐𝑎
𝑏 = a2 = 1 . 3 = 3
A B
C D
E F
O
α
O C
G
α
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𝑎𝑏
𝑐 .
𝑏𝑐
𝑎 = b2 = 1 . 2 = 2
𝑏𝑐
𝑎 .
𝑐𝑎
𝑏 = c2 = 2 . 3 = 6
Maka a2 + b2 + c2 = 3 + 2 + 6 = 11 Jawaban : C
21. 1! = 1 2! = 2 . 1 = 2 3! = 3 . 2 . 1 = 6 4! = 4 . 3 . 2 . 1 = 24 5! = 5 . 4 . 3 . 2 . 1 = 120 Jumlah satuan = 1 + 2 + 6 + 4 + 0 = 13 Jadi angka satuannya = 3 Jawaban : D
22. 𝐶3
4 − 𝐶2 6
𝐶5 10 =
4 .15
252 =
60
252 =
5
21
Jawaban : A
23.
a √𝑎2 + 1
α
1
(sinα + cosα)2 = sin2α + cos2α + 2sinαcosα
= 1 + 2 𝑎
√𝑎2+ 1
1
√𝑎2+ 1
= 1 + 2𝑎
𝑎2 + 1
= 𝑎2+ 2𝑎+1
𝑎2+1
Jawaban : B
24.
A a C O 11
11
7
7
D E
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DE = DC + CE = 22 + a
OD = OE = 1
2 (22 + 𝑎) = 11 +
1
2𝑎
AO = OD – DA = (11 + 1
2𝑎) - 11 =
1
2𝑎
Sehingga OC = 11 – 1
2𝑎
Perhatikan ∆ ACB DAN OCB yang mempunyai sisi CB yang sama
(18)2 – (11)2 = (4 + 1
2𝑎)2 – (11 -
1
2𝑎)2
324 – 121 = 16 + 4a + 1
4𝑎2 – 121 + 11a -
1
4𝑎2
203 = 15a – 105 308 = 15a
a = 208
15
Jawaban : A
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25.
0,21
0,14 0,14
0,21
X . y = 2
Bilangan 1 : x → x – o,28
Bilangan 2 : 2
𝑥 → (
2
𝑥 - 0,42)
Luas = (x – o,28) (2
𝑥 - 0,42)
Luas = 2 – 0,42x – 0,56
𝑥 + 0,1176
Luas = – 0,42x – 0,56
𝑥 + 2,1176
Luas’ = o = -0,42 + 0,56
𝑥2
0,56
𝑥2 = 0 , 42
𝑥2 = 0,56
0,42 =
9
3
x = 2
3 √3 , y = √3
Jawaban : A
26. ∫𝑥
𝑥+1 dx = ∫
𝑥+1−1
𝑥+1 dx
= ∫𝑥+1
𝑥+1 dx - ∫
1
𝑥+1 dx
= ∫ 𝑑𝑥 - ∫ 1
𝑥+1 dx
= x – ln |x + 1| + c
Jawaban : C
27. | 2x-5 | < | x+4 | ( 2x-5 )2 < ( x + 4 )2 4𝑥2 - 20x + 25 < 𝑥2 + 8x + 16 3𝑥2 – 28x + 9 < 0 ( 3x -1 ) ( x – 9 ) < 0 3x – 1 = 0 x – 9 = 0
x = 1
3 x = 9
Hp = {x | 1
3 < x < 9 }
Jawaban : E
9 1
3
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28. √2,5+ √0,4
√10 − √2,5+ √0,4 .
√10
√10 =
√25+ √4
√100 − √25+ √4
= 5+2
10−5+2 =
7
7 = 1
Jawaban : A
29. x + √𝑥 = 1 → √𝑥 = 1 – x
(√𝑥 = 1 – x )2 x = 1 – 2x + 𝑥2 : x
1 = 1
𝑥 - 2 + x
x + 1
𝑥 = 3
Jawaban : D
30. Kenaikan bakso = 16
100 . 5000 = 800
Harga bakso menjadi = Rp. 800,- + Rp. 5.000,- = Rp. 5.800,-
Kenaikan jus = 4
100. 5000 = 200
Harga jus menjadi = Rp. 200,- + Rp. 5.000,- = Rp. 5.200,- Harga bakso + jus = Rp. 11.000,- Selisih kenaikan = Rp. 11.000,- - Rp. 10.000,- = Rp. 1.000,-
Persen kenaikan = 1000
10000 . 100% = 10%
Jawaban : B
31. (𝑠𝑖𝑛4 750 - 𝑐𝑜𝑠4 750) + (𝑠𝑖𝑛4 750 + 𝑐𝑜𝑠4 750) (𝑠𝑖𝑛2 750 - 𝑐𝑜𝑠2 750) (𝑠𝑖𝑛2 750 + 𝑐𝑜𝑠2 750) ((𝑠𝑖𝑛2 750 + 𝑐𝑜𝑠2 750)2 - 2𝑠𝑖𝑛2 75 𝑐𝑜𝑠2 750)
(𝑠𝑖𝑛2 750 - 𝑐𝑜𝑠2 750) (𝑠𝑖𝑛2 750 + 𝑐𝑜𝑠2 750) ((𝑠𝑖𝑛2 750 + 𝑐𝑜𝑠2 750)2 - 1
2 .4𝑠𝑖𝑛2 75 𝑐𝑜𝑠2 750)
(𝑠𝑖𝑛2 750 - 𝑐𝑜𝑠2 750) (𝑠𝑖𝑛2 750 + 𝑐𝑜𝑠2 750) ((𝑠𝑖𝑛2 750 + 𝑐𝑜𝑠2 750)2 -1
2 (2𝑠𝑖𝑛750 𝑐𝑜𝑠 750)2)
− 𝑐𝑜𝑠 1500 . 1 ( 1 - 1
2(sin 1500)2)
1
2√3 ( 1-
1
8 ) =
1
2√3 .
1
8 =
7
16 √3
Jawaban : A
32. Perhatikan ∆BCD
Sin 300 = 1
𝐵𝐷
1
2 =
1
𝐵𝐷
BD = 2
BC = √22 − 12 = √3
BR = 1
2BA =
1
2√3
B
A C
D
T 30o Ѳ
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CR = √(√3)2 − (1
2√3)
2
= 3
2
Jadi , tan Ѳ = 𝐶𝐷
𝐶𝑇 =
13
2
= 2
3
Jawaban : D
33.
Misalkan AB = 2 , maka BP = √5 dan BG = 2√2 Aturan kosinus (PG)2 = (PB)2 + (BG)2 – 2.PB.BG. cosѲ
1 = 5 + 8 - 2√5 . 2√2 cos Ѳ
1 = 13 - 4√10 cos Ѳ
Cos Ѳ = 3
√10
√10 1
3
Jadi , tan Ѳ = 1
3
Jawaban : C 34. Persamaan garis melalui titik (7,1) adalah
y – 1 = m ( x – 7 ) y = mx + 1 – 7m , maka c2 = R2 ( 1 + m2 ) ( 1 – 7m )2 = 25 ( 1 + m2 ) 1 – 14m + 49m2 = 25 + 25m2 24m2 – 14m – 24 = 0 ( 4m + 3 ) ( 3m – 4m )= 0
m1 = - 3
4 , m2 =
4
3
Untuk m1 = - 3
4 Untuk m2 =
4
3
y = - 3
4x + 1 – 7 (-
3
4) y =
4
3x + 1 – 7 (
4
3 )
4y = -3x + 4 +21 3y = 4x + 3 - 28 3x + 4y = 25 4x – 3y = 25 Jawaban : B
35. Karena akar – akar membentuk deret aritmatika dengan beda 2 , maka : Misal : x1 = p , x2 = p + 2 , x3 = p + 4 , x4 = p + 6
x1 + x2 + x3 + x4 = - 𝑏
𝑎
p + ( p + 2 ) + ( p + 3 ) + ( p + 4 ) = 8 4p + 12 = 8
A B
C D
E
G H
F
Ѳ
P Q
Ѳ
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Maka p = - 1 Dengan demikian , maka x1 = -1 , x2 = 1 , x3 = 3 , x4 = 5
x1x2 + x1x3 + x1x4 + x2x3 + x2x4 + x3x4 = 𝑐
𝑎
(-1)(1) + (-1)(3) + (-1)(5) + (1)(3) + (1)(5) + (3)(5) = 𝑎
1 , → a = 14
x1x2x3 + x1x2x4 + x1x3x4 + x2x3x4 = - 𝑑
𝑎
(-1)(1)(3) + (-1)(1)(5) + (-1)(3)(5) + (1)(3)(5) = − (−𝑏)
1 , → b = - 8
x1x2x3x4 = 𝑐
𝑎
(-1)(1)(3)(5) = 𝑐
1 , → c = - 15
Jawaban : E
36. Misalkan x + y = a , x – y = b , maka :
a + √𝑎 = 30
a + √𝑎 - 30 = 0
(√𝑎 + b ) (√𝑎 - 5 ) = 0
√𝑎 = - 6 (TM) , √𝑎 = 5
b + √𝑏 = 12
b + √𝑏 - 12 = 0
(√𝑏 + 4 )( √𝑏 - 3 ) = 0
√𝑏 = - 4 (TM) , √𝑏 = 3
Maka nilai dari √𝑥2 − 𝑦2 = √(𝑥 + 𝑦)(𝑥 − 𝑦)
= √𝑥 + 𝑦 . √𝑥 − 𝑦
= √𝑎 . √𝑏
= 5 . 3 = 15
Jawaban : D
37. Misalkan : x! =a maka : 𝑎!
𝑎 = 120
𝑎 .(𝑎−1)!
𝑎 = 120
a ( a – 1 )! = 120a ( a – 1 )! = 5! a – 1 = 5 a = 6 x! = a x! = 6 x = 3
sin𝜋𝑥
2 = sin
3𝜋
2 = -1
Jawaban : A
38. 𝑏
𝑎 = 2005
𝑐
𝑏 = 2005
b = 2005a c = 2005b b = 2005a
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c = 2005b + b + c = 2005a + 2005b b + c = 2005 ( a + b ) 𝑏+𝑐
𝑎+𝑏 = 2005
Jawaban : A 39. Misalkan : a = 𝑥2 - 10x – 29
Maka : 𝑥2 - 10x – 45 = a – 16 𝑥2 - 10x – 69 = a – 40 1
𝑎 +
1
𝑎−16 -
2
𝑎−40 = 0
(a – 16) (a – 40) + a (a – 40) – 2a(a – 16) = 0 (a2 – 56a + 640) + (a2 – 40a) – 2a2 + 32a = 0 -56a + 640 – 40a + 32a = 0 64a = 640 a = 10 10 = x2 – 10x – 29 x2 – 10x – 39 = 0
x + y = - 𝑏
𝑎 = 10
Jawaban : B 40. 432 = 264
(44)10 = 280 1618 = 272 (83)8 = 272 Yang paling besar adalah 281 Jawaban : A
41. Bila keterangan – keterangan dinyatakan dalam diagram venn , maka diagramnya adalah Tring
Trang
Trung
Jadi , yang paling benar adalah Y dan Z
Jawaban : E
42. 5 ekor ~ 5 lapangan bola dalam 5 hari 1 ekor ~ 1 lapangan bola dalam 5 hari Jadi , 3 ekor kambing dapat menghabiskan rumput seluas 3 kali lapangan bola dalam 5 hari. Jawaban : D
43. (2x + 𝑦
2 )-1 . [ (2x)-1 + (
𝑦
2)-1 ] =
1
2𝑥+ 𝑦
2
. (1
2𝑥 +
1𝑦
2
)
= 1
2𝑥+ 𝑦
2
(1
2𝑥 +
2
𝑦)
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= 1
2𝑥+ 𝑦
2
(𝑦+4𝑥
2𝑥𝑦)
= 1
2𝑥+ 𝑦
2
(𝑦
2+ 2𝑥
𝑥𝑦)
= 1
𝑥𝑦 = (xy)-1
Jawaban : D
44. 1
𝑚 +
1
𝑛 =
4
7 →
1
𝑚 +
1
𝑛 =
8
14
1
𝑚 +
1
𝑛 =
7
14 +
1
14
1
𝑚 +
1
𝑛 =
1
2 +
1
14
Didapat m = 2 , n = 14 Jadi m2 + n2 = 22 + 142 = 4 + 196 = 200
Jawaban : D 45. C
750 E
A 550 D B
Sudut B = 1800 – (550 + 750) = 500
Karena BD = BE , maka sudut D = sudut E
Sudut D + sudut E + 500 = 1800
Sudut E = 650
Jadi Sudut BED = 650
Jawaban : C
46. C 3x
X 2x
A B
x0 + 2x0 + 3x0 = 1800
x0 = 300
𝐴𝐵
sin 3𝑥 =
𝐵𝐶
sin 𝑥
𝐴𝐵
𝐵𝐶 =
sin 3𝑥
sin 𝑥 =
sin 90
sin 30 =
112
= 2
Maka AB : BC = 2 : 1
Jawaban : C
47. y = 𝑥−1
2𝑥+3
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2xy + 3y = x – 1 2xy – x = - 3y -1 (2y – 1)x = -3y -1
x = −3𝑦−1
2𝑦−1 =
1+3𝑦
1−2𝑦
Jawaban : A
48. Misalkan keliling = 12x 4x 4x 4x
L = 1
2 . 4x . 4x sin 600
= 1
2 . 4x . 4x .
1
2√3
= 4x2√3 3x 3x 3x 3x
L = 3x.3x = 9x2
Keliling = 2𝜋R = 12x
= R = 6𝑥
𝜋
Luas = 𝜋R2
= 𝜋.36𝑥2
𝜋2
= 11,1x2
Jawaban : C
49. x2 ≥ 4 |x – 1| ≤ 2 x2 – 4 ≥ 0 ( x – 1 )2 ≤ 4 ( x+2 )(x – 2) ≥ 0 x2 – 2x – 3 ≤ 0 x1 = 2 , x2 = -2 ( x – 3 ) ( x + 1 ) ≤ 0 x1 = 3 , x2 = -1
+ + + + - -
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Yang memenuhi kedua – duanya adalah 2 ≤ x ≤ 3 Jawaban : B
50.
OD = jari – jari lingkaran dalam ∆ABC
= √𝑠(𝑠−𝑎)(𝑠 − 𝑏)(𝑠−𝑐 )
𝑠
= √27(27−12)(27 − 24)(27−18 )
27
= 27
27√15 = √15
AB2 = BC2 + AC2 – 2BC.AC cos C 144 = 576 + 324 – 2.24.18 cos C 2.24.18 cos C = 756
cos C = 7
8
sin C = √1 − 𝑐𝑜𝑠2𝐶
= √1 − 49
64
= √15
8
Perhatikan ∆NPC
sin C = 𝑁𝑃
𝑁𝐶 =
√15
8
√15
𝑁𝐶 =
√15
8
Maka NC = 8 AN = 18 – 8 = 10
Keliling ∆AMN = 𝐴𝑁
𝐴𝐶. Keliling ∆ABC
= 10
18 (54)
= 30 Jawaban : C
-2 2 1 3
C P
O
D B
A
N M