It is given that ΔABC ~ ΔEDF 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,
Given, ΔABC ~ ΔEDF, so the corresponding sides of ΔASC and ΔEDF are in the same ratio.
i.e., $\frac{A B}{E D}=\frac{A C}{E F}=\frac{B C}{D F}$ $\cdots$ (i)
Also, $\quad A B=5 \mathrm{~cm}, A C=7 \mathrm{~cm}$
$D F=15 \mathrm{~cm}$ and $D E=12 \mathrm{~cm}$
On putting these values in Eq. (i), we get
$\frac{5}{12}=\frac{7}{E F}=\frac{B C}{15}$
On taking first and second terms, we get
$\frac{5}{12}=\frac{7}{E F}$
$\Rightarrow$ $E F=\frac{7 \times 12}{5}=16.8 \mathrm{~cm}$
On taking first and third terms, we get
$\frac{5}{12}=\frac{B C}{15}$
$\Rightarrow$ $B C=\frac{5 \times 15}{12}=625 \mathrm{~cm}$
Hence, lengths of the remaining sides of the triangles are EF = 16.8 cm and SC = 625 cm.