Question:
$\triangle A B C \sim \triangle D E F$ and their areas are respectively $64 \mathrm{~cm}^{2}$ and $121 \mathrm{~cm}^{2}$. If $E F=15.4 \mathrm{~cm}$, find $B C$.
Solution:
It is given that $\triangle A B C \sim \triangle D E F$.
Therefore, ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
$\frac{a r(\triangle A B C)}{a r(\triangle D E F)}=\frac{B C^{2}}{E F^{2}}$
Let $B C$ be $x \mathrm{~cm}$.
$\Rightarrow \frac{64}{121}=\frac{x^{2}}{(15.4)^{2}}$
$\Rightarrow x^{2}=\frac{64 \times 15.4 \times 15.4}{121}$
$\Rightarrow x=\sqrt{\frac{64 \times 15.4 \times 15.4}{121}}$
$=\frac{8 \times 15.4}{11}$
$=11.2$
Hence, BC = 11.2 cm