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

Given, △ABC ~ △EDF

$\therefore \dfrac{AB}{DE} = \dfrac{AC}{EF} = \dfrac{BC}{DF}. \\[1em] \text{Consider, } \dfrac{AB}{DE} = \dfrac{AC}{EF} \\[1em] \Rightarrow \dfrac{5}{12} = \dfrac{7}{EF} \\[1em] \Rightarrow EF = \dfrac{84}{5} \\[1em] \Rightarrow EF = 16.8 \text{ cm.}$

Now consider,

$\dfrac{AB}{DE} = \dfrac{BC}{DF} \\[1em] \Rightarrow \dfrac{5}{12} = \dfrac{BC}{15} \\[1em] \Rightarrow BC = \dfrac{5 \times 15}{12} \\[1em] \Rightarrow BC = \dfrac{75}{12} \\[1em] \Rightarrow BC = 6.25.$

Hence, EF = 16.8 cm and BC = 6.25 cm.