$\triangle\text{ABC}\sim\triangle\text{DEF}$ such that $\text{ar}(\triangle\text{ABC})=64\text{cm}^2$ and $\text{ar}(\triangle\text{DEF})=169\text{cm}^2.$ If BC = 4cm, find EF.
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$\therefore\triangle\text{ABC}\sim\triangle\text{DEF}$
$\therefore\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\frac{\text{BC}^2}{\text{EF}^2}$
$\Rightarrow\frac{64}{169}=\frac{4^2}{\text{EF}^2}$
$\Rightarrow\text{EF}^2=\frac{16\times169}{\text{64}}$
$\Rightarrow\text{EF}=\frac{4\times13}{\text{8}}=6.5\text{cm}$
$\therefore\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\frac{\text{BC}^2}{\text{EF}^2}$
$\Rightarrow\frac{64}{169}=\frac{4^2}{\text{EF}^2}$
$\Rightarrow\text{EF}^2=\frac{16\times169}{\text{64}}$
$\Rightarrow\text{EF}=\frac{4\times13}{\text{8}}=6.5\text{cm}$
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