MCQ
$\triangle\text{ABC}\sim\triangle\text{PQR}$ such that $\text{ar}(\triangle\text{ABC})=4\ \text{ar}(\triangle\text{PQR}).$ If $\text{BC} = 12\ cm,$ then $\text{QR} =$
- A$9\ cm.$
- B$10\ cm.$
- ✓$6\ cm.$
- D$8\ cm.$
✓
Answer
Correct option: C.
$6\ cm.$
Given: $\triangle\text{ABC}\sim\triangle\text{PQR}$ and $\text{BC} = 12\ cm$
$\text{ar}(\triangle\text{ABC})=4\ \text{ar}(\triangle\text{PQR})$
Now, $\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{PQR})}=\Big(\frac{\text{BC}}{\text{QR}}\Big)^2$
$\Rightarrow\frac{4\text{ar}(\triangle\text{PQR})}{\text{ar}(\triangle\text{PQR})}=\frac{(12)^2}{\text{QR}^2}$
$\Rightarrow\text{QR}^2=\frac{144}{4}$
$\Rightarrow\text{QR}=\frac{12}{2}$
$\Rightarrow\text{QR}=6\text{cm}.$
$\text{ar}(\triangle\text{ABC})=4\ \text{ar}(\triangle\text{PQR})$
Now, $\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{PQR})}=\Big(\frac{\text{BC}}{\text{QR}}\Big)^2$
$\Rightarrow\frac{4\text{ar}(\triangle\text{PQR})}{\text{ar}(\triangle\text{PQR})}=\frac{(12)^2}{\text{QR}^2}$
$\Rightarrow\text{QR}^2=\frac{144}{4}$
$\Rightarrow\text{QR}=\frac{12}{2}$
$\Rightarrow\text{QR}=6\text{cm}.$
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