The document discusses the Pythagorean theorem and provides examples of using it to solve for unknown sides of right triangles. It defines the theorem as: for a right triangle with sides a, b, and hypotenuse c, c^2 = a^2 + b^2. Several practice problems are shown applying the theorem to find missing lengths. Solutions to challenging problems involving using the theorem to find perimeters and areas are also provided.

1. TEOREMA
PHYTAGORAS

2. Identifikasi Segitiga Siku-siku
Gambarlah segitiga-segitiga di atas kemudian identifikasi bagian-bagiannya.

3. RIGHT-ANGLED TRIANGLE
𝑨
𝑪
𝑩
Sisi siku-siku
hypotenusa
Sudut siku-siku

4. A. Teorema Phytagoras
Untuk sebuah segitiga siku-siku dengan panjang sisi
a, b dan sisi miring c, maka berlaku:
Teorema pythagoras
𝑐2
= 𝑎2
+ 𝑏2
𝒂
𝒃
𝒄

5. LATIHAN SOAL
Hitung sisi yang ditanyakan
𝑐2 = 𝑎2 + 𝑏2
𝑐2 = 62 + 82
𝑐2 = 36 + 64
𝑐 = 100
𝑐 = 10
6
8
?
𝒂
𝒃
𝒄

6. LATIHAN SOAL
Hitung sisi yang ditanyakan
𝑐2 = 𝑎2 + 𝑏2
𝑎2 = 𝑐2 − 𝑏2
𝑎2
= 102
− 82
𝑎2 = 100 − 64
𝑎2 = 36
𝑎 = 36
𝑎 = 6
?
8
10
𝒂
𝒃
𝒄

7. LATIHAN SOAL
Hitung sisi yang ditanyakan
𝑐2 = 𝑎2 + 𝑏2
𝑏2 = 𝑐2 − 𝑎2
𝑏2
= 102
− 62
𝑏2 = 100 − 36
𝑏2 = 64
𝑏 = 64
𝑏 = 8
6
?
10
𝒂
𝒃
𝒄

8. a. 6,8 dan 10
𝑐2
= 𝑎2
+ 𝑏2
102
= 62
+ 82
100 = 36 + 64
100 = 100
(Tripel Pyhtagoras)
B. Tripe Phytagoras

9. C. Penggunaan Teorema Phytagoras
Pada Bangun Datar
12
16

10. SOAL TANTANGAN
1. When each side of a triangle has a length which is a
prime factor of 2001, how many different such triangles
are there?
2. Given that in a right triangle the length of a leg of the right
angle is 11 and the length of the other two sides are both
positive integers. Find the perimeter of the triangle.
3. Pada ABC dengan siku-siku di A terdapat titik D dan E
sehingga AD = BE = EB, jika panjang CD = 221 cm dan
CE = 521 cm, hitung luas ABC!

11. SOAL TANTANGAN
4. Buktikan teorema pythagoras 𝑎2 + 𝑏2 = 𝑐2 pada gambar
di bawah.
𝒂
𝒂
𝒃
𝒃
𝒄
𝒄

12. SOLUSI 1
Solustion
Since 2001 = 3  23  29, the triangles with sides of
the following lengths exist.
{3, 3, 3}; {23, 23, 23}; {29, 29, 29}
{3, 23, 23}; {3, 29, 29}; {23, 29, 29};
{23, 23, 29}
There are 7 possible triangles in total.

13. SOLUSI 2
Solution. From the given conditions we have
n2 = m2 + 112
n2 – m2 = 112
(n – m)(n + m) = 121 = 1  121 = 11  11,
Therefore
n – m = 1, n + m = 121 or n – m = 11, n + m = 11,
n = 61, m = 60 n = 11, m = 0 (is not available)
Thus, the perimeter is 11 + 61 + 60 = 132

14. SOLUSI 3
A D B
C
AD = DE = EB
CD = 221
CE = 521
Diketahui:
Misal: AD = DE = EB  X

15. SOLUSI
ADC dan AEC siku-siku di “A”
Dengan teorema pythagoras, akan dicari panjang AC
ADC
AC2 = CD2 – AD2
AEC
AC2 = CE2 – AE2
CD2 – AD2 = CE2 – AE2

16. SOLUSI
CD2 – AD2 = CE2 – AE2
221 – X2 = 521 – (2X) 2
221 – X2 = 521 – 4X2
4X2 – X2 = 521 – 221
3X2 = 300
X2 = 100
X = 100
X = 10
AC2 = CD2  AD2
= ( 221)2  102
= 221  100
= 121
AC = 121
AC = 11

17. SOLUSI
X = 10
AD = DE = EB → AD = 10
AB = AD + DE + EB
= 10 + 10 + 10
AC = 11
AB = 30
A B
C
30 cm
11 cm
LABC =
1
2
x Alas x Tinggi
=
1
2
x 30 x 11
=
1
2
x 330
LABC = 165 cm2

18. TERIMA KASIH