Sebuah belah ketupat panjang sisinya (3x+8) cm.jika kelilingnya 68 cm dan panjang salah satu diagonalnya 16CM tentukan. a.nilai x b.panjang diagonal lainnya c.luasnya
HilmaRosyidah9
A.) 68 = 4(3x+8) , 68 = 12x + 32 , 68 - 32 = 12x , 36 = 12x , 3 = x . b.) 16/2 = 8 , (3x3)+8 = 17 , 8, ..., 17 = 15 , p : 15x2 = 30 . c.) 1/2 x 16 x 30 = 24 .
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whongaliema) keliling = 4 x s 68 = 4 (3x + 8) 3x + 8 = 68 : 4 3x + 8 = 17 3x = 17 - 8 3x = 9 x = 9 : 3 x = 3 b) s = 3x + 8 = 17 cm s² = + 17² = 8² + 289 = 64 + = 289 - 64 = 225 = √225 = 15 d2 = 15 x 2 d2 = 30 cm c) L = x d1 x d2 = x 16 x 30 = 8 x 30 = 240 cm²
68 = 4 (3x + 8)
3x + 8 = 68 : 4
3x + 8 = 17
3x = 17 - 8
3x = 9
x = 9 : 3
x = 3
b) s = 3x + 8 = 17 cm
s² = +
17² = 8² +
289 = 64 +
= 289 - 64
= 225
= √225
= 15
d2 = 15 x 2
d2 = 30 cm
c) L = x d1 x d2
= x 16 x 30
= 8 x 30
= 240 cm²