Question

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

Answer 1

Given, $\Delta ABC\sim \Delta EDF$, so the corresponding sides of $\Delta ABC$ and $\Delta EDF$ are in the same ratio.
i.e., $\frac{AB}{ED}=\frac{AC}{EF}=\frac{BC}{DF}$ …..(i)


Also, AB = 5cm, AC = 7 cm
DF = 15 cm and DE = 12 cm
On putting these values in Eq. (i), we get
$\frac{5}{12}=\frac{7}{EF}=\frac{BC}{15}$
On taking first and second terms, we get
$\frac{5}{12}=\frac{7}{EF}$
$\Rightarrow EF=\frac{7\times 12}{5}=16.8cm$
On taking first and third terms, we get
$\frac{5}{12}=\frac{BC}{15}$
$\Rightarrow BC=\frac{5\times 15}{12}=625cm$
Hence, lengths of the remaining sides of the triangles are EF = 16.8 cm and BC = 625cm.