∆ABC ∼ ∆DEF, ar(∆ABC) = 9 cm2, ar(∆DEF) = 16 cm2.

Given: $\operatorname{Ar}(\triangle \mathrm{ABC})=9 \mathrm{~cm}^{2}, \operatorname{Ar}(\triangle \mathrm{DEF})=16 \mathrm{~cm}^{2}$ and $\mathrm{BC}=2.1 \mathrm{~cm}$

To find: measure of EF

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

$\frac{\operatorname{Ar}(\triangle \mathrm{ABC})}{\operatorname{Ar}(\Delta \mathrm{DEF})}=\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}}$

$\frac{9}{16}=\frac{2.1^{2}}{E F^{2}}$

$\frac{3}{4}=\frac{2.1}{\mathrm{EF}}$

$\mathrm{EF}=2.8 \mathrm{~cm}$

Hence the correct answer is