SEGITIGA SOHCAHTOA

Problem 1 HOME EXIT
Problem 2
Problem 3
Problem 4
Problem 5
BC EXAM
TEXAS A&M HIGH SCHOOL MATH CONTEST
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Problem 7
Problem 8
NOVEMBER 12, 2011
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Problem 10 Editing by - Muhammad Yusuf
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Problem 12 GO
Muhammad Yusuf, S.Pd.
Problem 13 SMP NEGERI 1 BOLO

Problem 1 HOME EXIT
Problem 2
Problem 3 Orang yang gagal selalu
mencari jalan untuk
Problem 4
Problem 5
Problem 6 menghindari kesulitan,
Problem 7
Problem 8
sementara orang yang sukses
Problem 9 selalu menerjang kesulitan
Problem 10
untuk menggapai kesuksesan.
Problem 11
Problem 12

Problem 1 Problem 1 Solusi HOME EXIT
Problem 2
Problem 3
Jika
Problem 4
Problem 5
Problem 6
Problem 7
Maka x = ...
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12

Problem 2
Problem 3
Problem 4
Problem 5 Berapakah jumlah semua pembagi dari 36,
Problem 6
termasuk 1 dan dirinya sendiri
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12

Problem 2
Problem 3
Keliling sebuah persegi panjang adalah 28 m.
Problem 4
Sebuah persegi panjang yang lain dengan
Problem 5 panjang 3 kali panjang persegi panjang
Problem 6 pertama dan lebar 2 kali lebar persegi panjang
Problem 7 pertama memiliki luas 72 m. Berapakah luas
Problem 8 persegi panjang pertama?
Problem 9
Problem 10
Problem 11
Problem 12

Problem 2
Problem 3
Problem 4 Sebuah operasi “ * “ didefinisikan dengan :
Problem 5 a * b = a2 + 3b
Problem 6
Tentukan nilai dari (2 * 0) * (0 * 1)
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12

Problem 2
Problem 3
Problem 4
Segitiga ABC dengan panjang AB = 25
Problem 5
meter, AC = 24 meter, dan BC = 23 meter.
Problem 6
Suatu titik D yang merupakan titik pada
Problem 7
AC sehingga BD tegak lurus dengan AC.
Problem 8
Berapakah AD − DC?
Problem 9
Problem 10
Problem 11
Problem 12

Problem 2
Problem 3
Jika
Problem 4
Problem 5
Problem 6
Problem 7
Maka x = ...
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12

Problem 2
Problem 3
Problem 4
Tentukan penyelesaian dari :
Problem 5
Problem 6
Problem 7
Untuk 0 ≤ x ≤ 2
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12

Problem 2
Problem 3
Problem 4
What is the last digit of the sum
Problem 5 1! + 2! + 3! + · · · + 2010! + 2011! ?
Problem 6
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12

Problem 2
Problem 3
Problem 4
Suppose that f(x) is a function such that
Problem 5 for every real number x :
Problem 6
Problem 7
i) f(x) + f(1 − x) = 11
Problem 8 ii) f(1 + x) = 3 + f(x)
Problem 9
Then f(x) + f(−x) =
Problem 10
Problem 11
Problem 12

Problem 2
Problem 3
Problem 4 Let a, b and c be the three roots of
Problem 5
x3 − 64x− 14 . What is the value of
Problem 6
Problem 7 a3 + b3 + c3 ?
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12

Problem 2
Problem 3
Problem 4
Let ABC be an isosceles right triangle
Problem 5
with right angle at C. Let P be a point
Problem 6
inside the triangle such that AP = 3, BP =
Problem 7
5, and CP = 2√2. What is the area of the
Problem 8
triangle ABC?
Problem 9
Problem 10
Problem 11
Problem 12

Problem 2
Problem 3
Problem 4 In how many distinct ways can one write
Problem 5
1,000,000 as the product of three positive
Problem 6
Problem 7 integers? Treat all orderings of the same
Problem 8
set of factors as one way
Problem 9
Problem 10
Problem 11
Problem 12

Problem 2
Problem 3
Problem 4 A cube is inscribed in a ball. What is the
Problem 5
ratio of the volume of the cube to the
Problem 6
Problem 7 volume of the ball?
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12

Problem 1 HOME EXIT
Problem 2
Problem 3
Problem 4
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12

Problem 1 HOME EXIT
Problem 2
Problem 3
Problem 4
Pembagi dari 36 :
Problem 5 1, 2, 3, 4, 6, 9, 12, 18, 36
Problem 6
Problem 7 Jumlahnya :
Problem 8 1 + 2 + 3 + 4 + 6 + 9 + 12 + 18 + 36
Problem 9
Problem 10
= 91
Problem 11
Problem 12

Problem 1 HOME EXIT
Problem 2
Persegi panjang A, dengan panjang = p dan lebar = l
Problem 3
Problem 4 Kelilingnya = 28, sehingga 2p + 2l = 28
Problem 5 Persegi panjang B, dengan panjang = 3p
Problem 6
dan lebar = 2l
Problem 7 Kelilingnya = 72, sehingga 2(3p + 2l) = 72
Problem 8
Disederhanakan menjadi 3p + 2l = 36
Persamaan 2 3p + 2l = 36
Problem 9 p = 8 m dan l = 6 m
Problem 10 Persamaan 1 2p + 2l = 28
Problem 11 Sehingga luas segitiga pertama = 8 × 6 = 48 m2
Problem 12

Problem 1 HOME EXIT
Problem 2
Problem 3
Problem 4 a * b = a2 + 3 b
Problem 5
Problem 6 (2 * 0) = 22 + 30 = 4 + 1 = 5
Problem 7 (0 * 1) = 02 + 31 = 0 + 3 = 3
Problem 8 (2 * 0) * (0 * 1) = (5 * 3) = 52 + 33 = 25 + 27 = 52
Problem 9
Problem 10
Problem 11
Problem 12

Problem 1 HOME EXIT
Problem 2
Problem 3
C BD2 = BC2 – CD2
BD2 = 242 – CD2
Problem 4
D Disamping itu,
23 meter 24 meter
Problem 5
BD2 = AB2 – AD2
BD2 = 252 – AD2
Problem 6
BD2 = 625 – AD2
Problem 7 A 25 meter B Sehingga
Problem 8 576 – CD2 = 625 – AD2
Problem 9 Sehingga AD2 – CD2 = 49
Problem 10 49 (AD – CD)(AD + CD) = 49
AD – CD =
Problem 11 23 (AD – CD) × 23 = 49
Problem 12

Problem 1 HOME EXIT
Problem 2
Problem 3
Problem 4
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
Pangkat disamakan,
Problem 10
Problem 11
Problem 12

Problem 1 HOME EXIT
Problem 2
Problem 3
Problem 4
Kedua ruas dikuadratkan,
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11 Penyelesaian persamaan tersebut

Problem 12

Problem 1 HOME EXIT
Problem 2
Problem 3
Problem 4
Problem 5
Sehingga nilai x
Problem 6
adalah
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12

Problem 1 HOME EXIT
Problem 2
Problem 3 Mulai dari 5! Dan seterusnya, angka
Problem 4 satuanya adalah nol.
Problem 5
Problem 6 Sehingga yang perlu diperhatikan adalah
Problem 7 jumlah dari
Problem 8 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33
Problem 9
Problem 10 Sehingga angka satuannya adalah 3
Problem 11
Problem 12

Problem 1 HOME EXIT
Problem 2
Problem 3 f(x) + f(1 − x) = 11 f(-x) + f(1 + x) = 11
Problem 4
Problem 5
f(1 + x) = 3 + f(x) f(1 - x) = 3 + f(-x)
Problem 6 Sehingga :
Problem 7 f(x) + 3 + f(-x) = 11
f(x) + f(1 − x) = 11
Problem 8
atau f(x) + f(-x) = 8
Problem 9
Problem 10 f(-x) + f(1 + x) = 11 f(-x) + 3 + f(x) = 11
Problem 11 f(-x) + f(x) = 8
Problem 12

Problem 1 HOME EXIT
Problem 2
Problem 3 x3 − 64x− 14
Problem 4
= (x − a)(x − b)(x − c)
Problem 5
= x3 − (a + b + c)x2 + (ab + bc + ac)x − abc
Problem 6 Dengan (a + b + c) = 0, maka
Problem 7 a3 + b3 + c3 = (64a + 14) + (64b + 14) + (64c + 14)
Problem 8 = 64(a + b + c) + 42
Problem 9 = 64(0) + 42
Problem 10
= 42
Problem 11
Problem 12

Problem 1 HOME EXIT
Problem 2 A
Misalkan AC = CB = l
Problem 3
PD2 = 9 – (l – DC)2
Problem 4
P PD2 = 8 – CD2
Problem 5 D 9 – (l – DC)2 = 8 – DC2
Problem 6 l2 - 2DC = 1
Problem 7 C E B
Karena DC = CE , maka
Problem 8 PE2 = 25 – (l – CE)2 2l2 - 4DC = 18
Problem 9 PE2 = 8 – CE2 l2 - 2DC = 1
Problem 10 25 – (l – CE)2 = 8 – CE2
Problem 11
l2 - 2CE = 17
Problem 12

Problem 1 HOME EXIT
Problem 2 A
l2 - 2DC = 1 dan l2 - 2CE = 17
Problem 3
DC = ½ (l2 – 1) CE = ½ (l2 – 17)
Problem 4
Problem 5 D P Karena DC2 + CE2 = 8l , maka
Problem 6 ¼(l2 – 1)2 + ¼(l2 – 17)2 = 8l
Problem 7 C E B (l2 – 1)2 + (l2 – 17)2 = 32l2
Problem 8 l4 – 2l2 + 1+ l4 – 34l2 + 289 = 32l2
Problem 9 l4 – 34l2 + 145 = 0 l2 = 5 atau 29
Problem 10 Yang memenuhi hanya l2 = 29
Problem 11 Sehingga luasnya = 29/2
Problem 12

Problem 1 HOME EXIT
Problem 2
Let (x, y) be the coordinates of P . Then x2 + y2 = 8 , (s −
Problem 3
x)2 + y2 = 9 and (s − y)2 + x2 = 25 . We obtain s2 − 2sx = 1
Problem 4 and s2 − 2sy = 17 . Solve for x and y in terms of s , and
Problem 5 place in the first equation, obtaining (s2 − 1)2 + (s2 − 17)2
Problem 6 = 32s2 . This simplifes to s4 − 34s2 + 145 = 0 , hence s2 = 5
Problem 7 or 29 . Only s2 = 29 is consistent with the given data
(notice that if s = √5 , then √10 = √2s = AB > BP = 5, a
Problem 8
contradiction). Thus, the area is 29/2
Problem 9
Problem 10
Problem 11
Problem 12

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