# Luas Persegi pada sisi miring

## Teks penuh

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1

A. Teorema Pythagoras

Gambar Luas Persegi pada sisi siku-siku ke-1

Luas Persegi pada sisi siku-siku ke-1

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(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

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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)

3 𝑟2 = (𝑥

2− 𝑥1)2+ (𝑦2− 𝑦1)2

jika dan hanya jika

𝑟 = √(𝑥2− 𝑥1)2+ (𝑦2− 𝑦1)2

Definisi

Garis tinggi adalah garis yang melalui salah satu titik sudut dan tegak lurus terhadap sisi dihadapan titik sudut tersebut.

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6

jika dan hanya jika

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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

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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)

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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

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