If ΔABC∼ΔFDE such that AB = 9 cm, AC = 11 cm, DF = 16 cm and DE = 12 cm, then the length of BC is:

$6\frac{3}{4}$ cm

AB = 9 cm,

AC = 11 cm,

DF = 16 cm and DE = 12 cm

So, \(\frac{ AB }{ DF }\) = \(\frac{ AC }{ FE }\)  = \(\frac{ BC }{ DE }\) 

So,

\(\frac{ AB }{ DF }\) = \(\frac{ BC }{ DC }\) 

= \(\frac{ 9 }{ 16 }\) = \(\frac{ BC }{ 12 }\) 

= BC = 27/4 = $6\frac{3}{4}$ cm