Luas Persegi pada sisi miring
Teorema Pythagoras
A. Teorema Pythagoras
Gambar Luas Persegi pada sisi Luas Persegi pada sisi Luas Persegi pada sisi siku-siku ke-1 siku-siku ke-1 miring
2
2
2
(i) 3 ⋅ 3 = 9 = 3 4 ⋅ 4 = 16 = 4 5 ⋅ 5 = 25 = 5
2
2
2
(ii) 6 ⋅ 6 = 36 = 6 8 ⋅ 8 = 64 = 8 10 ⋅ 10 = 100 = 10
5
3
4 9 + 16 = 25
2
2
2
2
2
2
3 + 4 = 5 ⇔ 5 = 3 + 4
2
2
2
= + Contoh: Jika diketahui
= 6 cm dan = 8 cm, maka tentukan panjang ! Jawab:
6
8
2
2
2
- =
2
= √
- 2
2
2
= √6 + 8 = √36 + 64 = √100 = 10 cm Jadi, panjang
= 10 cm
B. Jarak antara Dua Titik
2
2
2
) ) = ( − + ( −
2
1
2
1
jika dan hanya jika
2
2
) ) = √( − + ( −
2
1
2
1 Contoh: Jika
(−4,5), (−6, −3), maka tentukan panjang ! Jawab: (−4,5), maka = 1 dan = −2
2
2
(−6, −3), maka = −2 dan = −6
1
1
2
2
= √( − ) + ( − )
2
1
2
1
2
2
= √(1 − (−2)) + (−6 − (−2))
2
2
= √(1 + 2) + (−6 + 2)
2
2
= √(3) + (−4) = √9 + 16 = √25 = 5 Jadi, panjang
= 5
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.
⇔ −
−
2
2
2
= − ⋯ (1)
2
2
2
= − ( − ) ⋯ (2) Dari
(1) dan (2)
2
2
2
2
2
2
dan , maka = − = − ( − )
2
2
2
2
− ( − ) = −
2
2
2
− ( − )( − ) = −
2
2
2
2
2
− ( − 2 + ) = −
2
2
2
2
2
− + 2 − = −
2
2
2
2
2
2 = + − − +
2
2
2
2 = + −
- 2
- 2
- 2
- 2
- 2
- 2
- 2
- 2
- 2
- 2
- 2
−
−
2
2 )
2
= (
2 2 +
2
2
⋅ 2 2 + −
2 ) (
2 2 + −
2
−
2
2 )
2
2
2 ) (
2
−
2 )]
2
= ( +
2
−
2
2 ) ( +
2
2
−
2
2 )
2
= ( ⋅ 2 2 +
2
−
= ( 2 +
−
2
2
2
2 ) (
−(
2
− 2 +
2
) +
2 )
2
2
= [ ( + )
2
−
2
2 ] [
−( − )
) −
2
2
2
2 ) ( 2 −
2
−
2
2 )
2
= (
2
= ( (
−
2
2 ) (
−
2
2
2 )
2
2
2
−
2
2
−
2
dan
2
= (
2 −
2
2
)
2
, maka
2
2
2
−
=
(1) dan (3)
−
2
2 2 =
2
−
2
2 =
2
−
2
⋯ (3) Dari
2
= (
2
−
2
2 )
2
2
2
2
−
= [ + (
2
−
2
2 )] [ + (
−1 1 ) (
2
2
2
2 )]
2
= [ + (
2
−
2
2 )] [ + (
2
2 )]
2
−
= [ + (
2
−
2
2 )] [ − (
2
−
2
2 )]
2
= [ + (
2
−
2
2 )] [ + (−1) (
2
- 2
- 2
- 2
- 2
2
- 2
- 2
- 2
- 2
- 2
- 2 +
- 2 −
- 2
- 2 +
- 2
2 ]
2 − )]
2 = [(
2 ⋅ ( + + − 2 )
2 ⋅ ( + + − 2 )
2 ⋅ ( + + )
2 )
2 = [(
2 ⋅ 2 ⋅ 2 ⋅ 2 ]
2 ⋅ ( + + )( + + − 2 )( + + − 2 )( + + − 2 )
4 2 )
16 ]
2 ]
2 ⋅ ( + + )( + + − 2 )( + + − 2 )( + + − 2 )
2 (2 )
4
2 = [
1 16 ⋅ {( + + )( + + − 2 )( + + − 2 )( + + − 2 )}]
2 ⋅
16 (2 )
2 = [
2 ⋅ ( + + − 2 )
2 = [(
2 ⋅
2 − 2
2 ) ⋅ (
2 − ) ⋅ (
2 − ) ⋅ (
2 ) ⋅ (
2 ⋅ (
2 )
2 = [(
2 − 2 2 )]
2 − 2 2 ) ⋅ (
2 ) ⋅ (
2 )
2 ⋅ (
2 )
2 = [(
2 )]
2 ) ⋅ (
2 ) ⋅ (
2 ) ⋅ (
2 ⋅ (
16 16 ⋅ {( + + )( + + − 2 )( + + − 2 )( + + − 2 )}]
1 (2 )
2
2 = [
( + − )( − + )
2 ] [
( + + )( + − )
2 = [
2 ]
{ +( − )}{ −( − )}
2
] [
{( + )+ }{( + )− }
2 ]
2 = [
2
− ( − )
2
2 ] [
2
−
2
= [ ( + )
2 ]
( + + )( + − ) ⋅
2 = [
⋅ ( + + )( + − )( + − )( + − )]
2 ⋅ 1 ⋅ {( + + )( + + − 2 )( + + − 2 )( + + − 2 )}]
1 (2 )
2 = [
1 (2 ) 2 ⋅ ( + + )( + − + − )( + − + − )( + − + − )] 2 = [ 1 (2 ) 2 ⋅ ( + + )( + + − − )( + + − − )( + + − − )]
= [
2 ⋅ ( + + )( + − + 0)( + − + 0)( + − + 0)] 2
1 (2 )
2 = [
2
( + − )( + − ) ]
4
1
= [
2
2 ]
4
( + + )( + − )( + − )( + − )
2 = [
- − 2
- − 2
- − 2
1
misal: = keliling
2
- , maka
=
2
2 2 + + + + + + + +
2
= [( ⋅ ( ) ⋅ ( − ) ⋅ ( − ) ⋅ ( − )]
)2
2
2
2
2
2
2
= [( ⋅ ( ) ⋅ ( − ) ⋅ ( − ) ⋅ ( − )] )
2
2
2
= ( ⋅ ( ) ⋅ ( − ) ⋅ ( − ) ⋅ ( − ) ) jika dan hanya jika
2
2 = √( ⋅ ( ) ⋅ ( − ) ⋅ ( − ) ⋅ ( − )
)
2
2 = √( ⋅ √( )( − )( − )( − )
)
2 = √( )( − )( − )( − )
⋅
2 = √( )( − )( − )( − )
⋅ Contoh: Jika diketahui
= 13, = 15 cm dan = 14 cm, maka tentukan panjang !
13
15
14 Jawab:
=
2 13+15+14
=
2
42
=
2
= 21
2 =
√( )( − )( − )( − ) ⋅
2 =
√(21)(21 − 13)(21 − 15)(21 − 14) 14 ⋅
1 =
√(21)(8)(6)(7) 7 ⋅
1 =
√(3 ⋅ 7) ⋅ (2 ⋅ 2 ⋅ 2) ⋅ (2 ⋅ 3) ⋅ (7) 7 ⋅
1 = 7 ⋅ √2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 3 ⋅ 7 ⋅ 7
1
4
2
2
= √2 ⋅ 3 ⋅ 7 7 ⋅
1
4
2
2
= √2 ⋅ √3 ⋅ √7 7 ⋅
4
2
2
1
2
2
2
= ⋅ 3 ⋅ 7 7 ⋅ 2
1
2
= ⋅ 3 ⋅ 7 7 ⋅ 2
2
= 2 ⋅ 3 = 4 ⋅ 3 = 12 cm Jadi, panjang
= 12 cm
D. Luas Segitiga
2 = √( )( − )( − )( − )
⋅
1 = 2 ⋅ ⋅
1
2 =
√( )( − )( − )( − ) 2 ⋅ ⋅ ⋅ = √( )( − )( − )( − ) Contoh: Jika diketahui
= 13, = 15 cm dan = 14 cm, maka luas !
13
15
14 Jawab:
Cara 1
=
2 13+15+14
=
2
42
=
2
= 21
2 = √( )( − )( − )( − )
⋅
2 = √(21)(21 − 13)(21 − 15)(21 − 14) 14 ⋅
1 = √(21)(8)(6)(7) 7 ⋅
1 = √(3 ⋅ 7) ⋅ (2 ⋅ 2 ⋅ 2) ⋅ (2 ⋅ 3) ⋅ (7) 7 ⋅
1 = 7 ⋅ √2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 3 ⋅ 7 ⋅ 7
1
4
2
2
= √2 ⋅ 3 ⋅ 7 7 ⋅
1
4
2
2
= √2 ⋅ √3 ⋅ √7 7 ⋅
4
2
2
1
2
2
2
= ⋅ 3 ⋅ 7 7 ⋅ 2
1
2
= ⋅ 3 ⋅ 7 7 ⋅ 2
2
= 2 ⋅ 3 = 4 ⋅ 3 = 12
1 = 2 ⋅ ⋅
1 = 2 ⋅ ⋅
1 = 2 ⋅ 14 ⋅ 12 = 7 ⋅ 12
2
= 84 cm atau
Cara 2
= √( )( − )( − )( − ) = √(21)(21 − 13)(21 − 15)(21 − 14) = √(21)(8)(6)(7) = √(3 ⋅ 7) ⋅ (2 ⋅ 2 ⋅ 2) ⋅ (2 ⋅ 3) ⋅ (7) = √2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 3 ⋅ 7 ⋅ 7
4
2
2
= √2 ⋅ 3 ⋅ 7
4
2
2
= √2 ⋅ √3 ⋅ √7
4
2
2
2
2
2
= 2 ⋅ 3 ⋅ 7
2
= 2 ⋅ 3 ⋅ 7 = 4 ⋅ 3 ⋅ 7 = 12 ⋅ 7
2
= 84 cm
2 Jadi, luas
= 84 cm