It is given that $\triangle ABC \sim \triangle EDF$ such that $AB = 5 \, cm$,$AC = 7 \, cm$,$DF = 15 \, cm$,and $DE = 12 \, cm$. Find the lengths of the remaining sides of the triangles.

(N/A) Given,$\triangle ABC \sim \triangle EDF$. Since the triangles are similar,their corresponding sides are in the same ratio.
i.e.,$\frac{AB}{ED} = \frac{AC}{EF} = \frac{BC}{DF}$ ...... $(i)$
Given values are $AB = 5 \, cm$,$AC = 7 \, cm$,$DF = 15 \, cm$,and $DE = 12 \, cm$.
Substituting these values in Eq. $(i)$,we get:
$\frac{5}{12} = \frac{7}{EF} = \frac{BC}{15}$
Taking the first and second terms:
$\frac{5}{12} = \frac{7}{EF}$
$\Rightarrow EF = \frac{7 \times 12}{5} = \frac{84}{5} = 16.8 \, cm$
Taking the first and third terms:
$\frac{5}{12} = \frac{BC}{15}$
$\Rightarrow BC = \frac{5 \times 15}{12} = \frac{75}{12} = 6.25 \, cm$
Hence,the lengths of the remaining sides are $EF = 16.8 \, cm$ and $BC = 6.25 \, cm$.