Question
$\triangle\text{ABC}\sim\triangle\text{PQR}$ and $\text{ar}(\triangle\text{ABC})=4\text{ar}(\triangle\text{PQR}).$ If $BC\ 12\ cm,$ find $QR.$
✓
Answer
Given: $\text{ar}(\triangle\text{ABC})=4\text{ar}(\triangle\text{PQR})$
$\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{PQR})}=\frac{\text{4}}{\text{1}}$
$\therefore\triangle\text{ABC}\sim\triangle\text{PQR}$
$\therefore\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{PQR})}=\frac{\text{BC}^2}{\text{QR}^2}$
$\therefore\frac{\text{BC}^2}{\text{QR}^2}=\frac{\text{4}}{\text{1}}$
$\Rightarrow\text{QR}^2=\frac{12^2}{4}$
$\Rightarrow\text{QR}^2=36$
$\Rightarrow\text{QR}=6$
Hence, $QR = 6\ cm$
$\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{PQR})}=\frac{\text{4}}{\text{1}}$
$\therefore\triangle\text{ABC}\sim\triangle\text{PQR}$
$\therefore\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{PQR})}=\frac{\text{BC}^2}{\text{QR}^2}$
$\therefore\frac{\text{BC}^2}{\text{QR}^2}=\frac{\text{4}}{\text{1}}$
$\Rightarrow\text{QR}^2=\frac{12^2}{4}$
$\Rightarrow\text{QR}^2=36$
$\Rightarrow\text{QR}=6$
Hence, $QR = 6\ cm$
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