1
A. Teorema Pythagoras
Gambar Luas Persegi pada sisi siku-siku ke-1
Luas Persegi pada sisi siku-siku ke-1
Luas Persegi pada sisi miring
(i)
(ii)
3 ⋅ 3 = 9 = 32
6 ⋅ 6 = 36 = 62
4 ⋅ 4 = 16 = 42
8 ⋅ 8 = 64 = 82
5 ⋅ 5 = 25 = 52
10 ⋅ 10 = 100 = 102
9 + 16 = 25
32+ 42 = 52 ⇔ 52 = 32+ 42
𝐴𝐶2 = 𝐴𝐵2+ 𝐵𝐶2
Contoh: Jika diketahui 𝐴𝐵 = 6 cm dan 𝐵𝐶 = 8 cm, maka tentukan panjang 𝐴𝐶! 𝐴
𝐵 𝐶
3
2 Jawab:
𝐴𝐶2 = 𝐴𝐵2+ 𝐵𝐶2
𝐴𝐶 = √𝐴𝐵2+ 𝐵𝐶2
𝐴𝐶 = √62+ 82
𝐴𝐶 = √36 + 64 𝐴𝐶 = √100 𝐴𝐶 = 10 cm
Jadi, panjang 𝐴𝐶 = 10 cm
B. Jarak antara Dua Titik
𝐴
𝐵 𝐶
6
3 𝑟2 = (𝑥
2− 𝑥1)2+ (𝑦2− 𝑦1)2
jika dan hanya jika
𝑟 = √(𝑥2− 𝑥1)2+ (𝑦2− 𝑦1)2
C. Garis Tinggi pada Segitiga
Definisi
Garis tinggi adalah garis yang melalui salah satu titik sudut dan tegak lurus terhadap sisi dihadapan titik sudut tersebut.
6
jika dan hanya jika
7 𝐴𝐷 =17 ⋅ √2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 3 ⋅ 7 ⋅ 7
𝐴𝐷 =17 ⋅√24⋅ 32⋅ 72
𝐴𝐷 =17 ⋅√24⋅ √32⋅ √72
𝐴𝐷 =17 ⋅ 242⋅ 322⋅ 722
𝐴𝐷 =17 ⋅ 22⋅ 3 ⋅ 7
𝐴𝐷 = 22⋅ 3
𝐴𝐷 = 4 ⋅ 3 𝐴𝐷 = 12 cm
Jadi, panjang 𝐴𝐷 = 12 cm
D. Luas Segitiga
𝑡 =𝐵𝐶 ⋅2 √(𝑠𝑘)(𝑠𝑘 − 𝐴𝐵)(𝑠𝑘 − 𝐴𝐶)(𝑠𝑘 − 𝐵𝐶)
𝐿 = 12 ⋅ 𝑎 ⋅ 𝑡
𝐿 = 12 ⋅ 𝐵𝐶 ⋅𝐵𝐶 ⋅2 √(𝑠𝑘)(𝑠𝑘 − 𝐴𝐵)(𝑠𝑘 − 𝐴𝐶)(𝑠𝑘 − 𝐵𝐶)
𝐿 = √(𝑠𝑘)(𝑠𝑘 − 𝐴𝐵)(𝑠𝑘 − 𝐴𝐶)(𝑠𝑘 − 𝐵𝐶)
Contoh: Jika diketahui 𝐴𝐵 = 13, 𝐴𝐶 = 15 cm dan 𝐵𝐶 = 14 cm, maka luas 𝐴𝐵𝐶! 𝐶
𝐴
𝐵
𝑡
𝐷
𝐶 𝐴
𝐵 𝐷
13 15
8 Jawab:
Cara 1
𝑠𝑘 = 𝐴𝐵+𝐴𝐶+𝐵𝐶2 𝑠𝑘 = 13+15+142 𝑠𝑘 = 422 𝑠𝑘 = 21
𝐴𝐷 =𝐵𝐶 ⋅2 √(𝑠𝑘)(𝑠𝑘 − 𝐴𝐵)(𝑠𝑘 − 𝐴𝐶)(𝑠𝑘 − 𝐵𝐶)
𝐴𝐷 =14 ⋅2 √(21)(21 − 13)(21 − 15)(21 − 14)
𝐴𝐷 =17 ⋅√(21)(8)(6)(7)
𝐴𝐷 =17 ⋅√(3 ⋅ 7) ⋅ (2 ⋅ 2 ⋅ 2) ⋅ (2 ⋅ 3) ⋅ (7)
𝐴𝐷 =17 ⋅ √2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 3 ⋅ 7 ⋅ 7
𝐴𝐷 =17 ⋅√24⋅ 32⋅ 72
𝐴𝐷 =17 ⋅√24⋅ √32⋅ √72
𝐴𝐷 =17 ⋅ 242⋅ 322⋅ 722
𝐴𝐷 =17 ⋅ 22⋅ 3 ⋅ 7
𝐴𝐷 = 22⋅ 3
𝐴𝐷 = 4 ⋅ 3 𝐴𝐷 = 12
𝐿 = 12 ⋅ 𝑎 ⋅ 𝑡
𝐿 = 12 ⋅ 𝐵𝐶 ⋅ 𝐴𝐷
𝐿 = 12 ⋅ 14 ⋅ 12 𝐿 = 7 ⋅ 12 𝐿 = 84 cm2
atau
Cara 2
𝐿 = √(𝑠𝑘)(𝑠𝑘 − 𝐴𝐵)(𝑠𝑘 − 𝐴𝐶)(𝑠𝑘 − 𝐵𝐶) 𝐿 = √(21)(21 − 13)(21 − 15)(21 − 14) 𝐿 = √(21)(8)(6)(7)
9 𝐿 = √24⋅ 32⋅ 72
𝐿 = √24⋅ √32⋅ √72
𝐿 = 242⋅ 322⋅ 722
𝐿 = 22⋅ 3 ⋅ 7
𝐿 = 4 ⋅ 3 ⋅ 7 𝐿 = 12 ⋅ 7 𝐿 = 84 cm2
