If ∆ABC and ∆DEF are two triangles such that ABDE=BCEF=CAFD=25, then Area (∆ABC) : Area (∆DEF) =

Given: $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DEF}$ are two triangles such that $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{CA}}{\mathrm{FD}}=\frac{2}{5}$.

To find: $\operatorname{Ar}(\triangle \mathrm{ABC}): \operatorname{Ar}(\triangle \mathrm{DEF})$

We know that if the sides of two triangles are proportional, then the two triangles are similar.

Since $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{CA}}{\mathrm{FD}}=\frac{2}{5}$, therefore, $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DEF}$ are similar.

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

$\frac{\operatorname{Ar}(\Delta \mathrm{ABC})}{\operatorname{Ar}(\Delta \mathrm{DEF})}=\frac{\mathrm{AB}^{2}}{\mathrm{DE}^{2}}$

$\frac{\operatorname{Ar}(\triangle \mathrm{ABC})}{\operatorname{Ar}(\triangle \mathrm{DEF})}=\frac{2^{2}}{5^{2}}$

$\frac{\operatorname{Ar}(\Delta \mathrm{ABC})}{\operatorname{Ar}(\Delta \mathrm{DEF})}=\frac{4}{25}$

Hence the correct answer is $(b)$.