$BC = 2.1$ cm, not $BC = 2:1$. Using the formula $\frac{A_{ABC}}{A_{DEF}} = (\frac{BC}{EF})^2$, we have $$ \frac{9}{16} = \left( \frac{2.1}{EF} \right)^2 $$ Taking the square root of both sides, $$ \frac{3}{4} = \frac{2.1}{EF} $$ $$ EF = \frac{4 \times 2.1}{3} = \frac{8.4}{3} = 2.8 $$ So, $EF = 2.8$ cm.

In the following figure \(\angle\)MNP = 90\(^\circ\), seg NQ \(\perp\) seg MP, MQ = 9, QP = 4, find NQ. 
Solve the following sub-questions (any four): In \( \triangle ABC \), \( DE \parallel BC \). If \( DB = 5.4 \, \text{cm} \), \( AD = 1.8 \, \text{cm} \), \( EC = 7.2 \, \text{cm} \), then find \( AE \). 