If ∆ABC and ∆DEF are two triangles such that ABDE=BCEF=CAFD=34,

GIVEN: $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DEF}$ are two triangles such that $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{CA}}{\mathrm{FD}}=\frac{3}{4}$.

TO FIND: Area (ABC):Area(DEF)

We know that two triangles are similar if their corresponding sides are proportional.

Here, $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DEF}$ are similar triangles because their corresponding sides are given proportional, i.e. $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{CA}}{\mathrm{FD}}=\frac{3}{4}$

Since the ratio of the areas of two similar triangle is equal to the ratio of the squares of their corresponding sides.

$\Rightarrow \frac{\text { Area }(\mathrm{ABC})}{\text { Area }(\mathrm{DEF})}=\frac{3^{2}}{4^{2}}$

$\frac{\text { Area }(\mathrm{ABC})}{\text { Area }(\mathrm{DEF})}=\frac{9}{16}$