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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