∆ ABC ∼ ∆ DEF and their areas are respectively 64 cm2

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

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