Question
If $\triangle\text{ABC}\sim\triangle\text{DEF}$ such that $AB = 5cm,$ area $(\triangle\text{ABC}) = 20cm^2 $ and area $(\triangle\text{DEF}) = 45cm^2, $determine $DE.$
✓
Answer
$\triangle\text{ABC}\sim\triangle\text{DEF}$ area $(\triangle\text{ABC}) = 20cm^2 $ area $(\triangle\text{DEF}) = 45cm^2$
$AB = 5cm$ Let $DE = x\ cm$ Now
$\because\triangle\text{ABC}\sim\triangle\text{DEF} $
$\therefore\frac{\text{area}(\triangle\text{ABC})}{\text{area}(\triangle\text{DEF})}=\frac{\text{AB}^2}{\text{DE}^2} $
$\Rightarrow\frac{20}{45}=\frac{(5)^2}{\text{x}^2}\Rightarrow\frac{20}{45}=\frac{25}{\text{x}^2} $
$\Rightarrow\text{x}^2=\frac{25\times45}{20}=\frac{225}{4}=\Big(\frac{15}{2}\Big)^2 $
$\therefore\text{x}=\frac{15}{2}=7.5$
$ \therefore\text{DE}=7.5\text{cm}$
$AB = 5cm$ Let $DE = x\ cm$ Now
$\because\triangle\text{ABC}\sim\triangle\text{DEF} $
$\therefore\frac{\text{area}(\triangle\text{ABC})}{\text{area}(\triangle\text{DEF})}=\frac{\text{AB}^2}{\text{DE}^2} $
$\Rightarrow\frac{20}{45}=\frac{(5)^2}{\text{x}^2}\Rightarrow\frac{20}{45}=\frac{25}{\text{x}^2} $
$\Rightarrow\text{x}^2=\frac{25\times45}{20}=\frac{225}{4}=\Big(\frac{15}{2}\Big)^2 $
$\therefore\text{x}=\frac{15}{2}=7.5$
$ \therefore\text{DE}=7.5\text{cm}$
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