Perkalian bilangan asli dengan pecahan campuran Perkalian Pecahan Biasa dengan Pecahan Biasa

Kelas VII SMPMTs 134 1. Hitung 3 × 4 3 = …. ? Perhatikan gambar berikut × × × × × × × + + × × 4 3 4 3 3 4 1 4 2 2 1 4 Berdasarkan gambar di atas, 3 × 3 4 = 3 4 + 3 4 + 3 4 = 4 3 3 3 + + = 4 9 = 2 4 1 2. Hitunglah 5 3 × 4 = …. ? Jawab 1 2 3 4 Perhatikan gambar disamping banyak potongan daerah yang diarsir adalah 3 × 4 3 5 × 4 = 3 4 5 × = 12 5 = 2 2 5 Coba kamu ciptakan cara anda sendiri 1 2 3 4 5 1 5 2 5 3 1 5 4 × × × × × × × + + × ×

b. Perkalian bilangan asli dengan pecahan campuran

1 2 × 1 1 4 = … 1 2 1 × × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + × = = + = = × × × = = × × × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × + × = = = + = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × = × × × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + × = = = + = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × = × × × = = = × 4 1 1 4 1 2 4 1 1 1 1 4 1 1 4 2 1 4 1 2 3 4 Contoh 2.21 Contoh 2.21 Matematika 135 2 × 1 1 4 = 2 ×       + 4 1 1       × +       × =       + × = 4 1 2 4 4 2 4 1 4 4 2 8 2 10 1 2 4 4 4 4 = + = = Cara lain : 2 × 1 1 4 = 2 × 5 4 = 2 5 10 1 2 4 4 4 × = = 2 2 3 4 × 3 = …? × × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + × = = + = = × × × = = × × × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × + × = = = + = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × = × × × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + × = = = + = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × = × × × = = 3 3 4 1 4 3 2 3 = × = × 2 3 4 3 2 3 4 3 2 3 3 4 3 × = +      × = × + ×       = ×       + ×       = + = = 2 12 4 3 4 3 24 4 9 4 33 4 8 1 4 3 2 × 2 1 3 = … 1 2 5 4

I. 2 ×

2 1 3 = 2 ×       + 3 2 1       + × = 3 2 3 3 2 = 2 3 3 2 2 3 6 3 4 3 10 3 3 1 3 ×       + ×       = + = =

II. 2 ×

2 1 3 = 2 × 5 3 = 2 5 10 1 3 3 3 3 × = = 3 1 2 5 × 2 = 2 2 1 5 ×       +       × + × = 2 2 1 2 5 10 1 11 = + = 1 3 2 3 4 3 5 3 Kelas VII SMPMTs 136 × × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × + × = = + = = × = × × = × 2. Hitung ... 4 1 3 2 = × Penyelesaian: = × 12 2 4 3 1 2 4 1 3 2 = × × = × 1 3 1 4 1 1 3 2 × = × × = × × × × 4 1 2 3 1 4 2 1 3 4 × = × × = 2 12 Misalkan a, b, dan c bilangan asli, maka berlaku 1 c b a c b a × = × 2 c a b a c b × = × 3 1 × b a = b a × 1 = b a Sifat-2.15

c. Perkalian Pecahan Biasa dengan Pecahan Biasa

1 Hitung ... 4 3 5 3 = × × × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × + × = = + = = × Berdasarkan gambar di atas 9 3 3 3 3 = × × = × = × = × = × × = × × = × × = × × × 4 3 5 3 × 4 3 Berdasarkan gambar di atas 3 5 3 4 3 3 5 4 9 20 × = × × =

2 Hitung

... 4 1 3 2 = × 2 3 1 4 2 12 × = Penyelesaian Matematika 137 3 Hitung ... 3 2 4 3 = × × × × × × = × = × = × = × = × ଵ ଷ ଷ ସ = × × = × 4 3 = × × = × 1 3 1 4 3 1 3 2 3 4 2 3 3 2 4 3 × = × × = 6 12 3 4 2 3 6 12 × = 4 Hitung ... 4 1 3 1 = × × × × × × = × = × = × = × nyelesaian: = × ଵ ଷ ଷ ସ 3 1 = × × = × 1 3 1 4 1 1 = × × = × 1 3 4 1 1 3 4 × = × × 1 4 1 12 = 1 3 1 4 1 12 × = Misalkan a, b, c, dan d adalah bilangan asli, dan adalah pecahan biasa, maka berlaku: pecahan × pecahan = pembilang × pembilang penyebut × penyebut yaitu × = Sifat-2.16 Kelas VII SMPMTs 138

d. Perkalian pecahan biasa dengan pecahan campuran