Question : Let $\triangle {ABC} \sim \triangle {RPQ}$ and $\frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle P Q R)}=\frac{4}{9}$. If ${AB}=3 {~cm}, {BC}=4 {~cm}$ and ${AC}=5 {~cm}$, then ${PQ}$ (in ${cm}$ ) is equal to:
Option 1: 12
Option 2: 4.5
Option 3: 5
Option 4: 6
Correct Answer: 6
Solution :
$\triangle$ ABC ∼ $\triangle$ RPQ
$\frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle P Q R)}=\frac{4}{9}$
AB = 3 cm
BC = 4 cm
AC = 5 cm
Concept used:
$\triangle$ ABC ∼ $\triangle$ RPQ
$\frac{\text{AB}}{\text{RP}} = \frac{\text{BC}}{\text{PQ}} = \frac{\text{AC}}{\text{QR}} = \sqrt{\frac{ar(\text{ABC})}{ar(\text{RPQ})}}$
$\frac{\text{AB}}{\text{RP}} = \sqrt{\frac{4}{9}}$
Calculation:
$\frac{\text{AB}}{\text{RP}} = \frac{\text{BC}}{\text{PQ}}= \frac{2}{3}$
PQ $=\frac{3}{2}\times \text{BC}$
⇒ BC $=\frac{3}{2}\times 4 =6$
$\therefore$ The value of PQ is 6 cm.
Hence, the correct answer is 6.
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