MCQ
$\triangle\text{ABC}\sim\triangle\text{PQR}$ such that $\text{ar}(\triangle\text{ABC})=4\ \text{ar}(\triangle\text{PQR}).$ If BC = 12cm, then QR =
- A9cm.
- B10cm.
- ✓6cm.
- D8cm.
✓
Answer
Correct option: C.
6cm.
Given: $\triangle\text{ABC}\sim\triangle\text{PQR}$ and BC = 12cm
$\text{ar}(\triangle\text{ABC})=4\ \text{ar}(\triangle\text{PQR})$
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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