Question : Let $\triangle ABC \sim \triangle RPQ$ and $\frac{{area}(\triangle {ABC})}{{area}(\triangle {PQR})}=\frac{4}{9}$. If AB = 3 cm, BC = 4 cm and AC = 5 cm, then RP (in cm) is equal to:
Option 1: 6
Option 2: 5
Option 3: 4.5
Option 4: 12
Correct Answer: 4.5
Solution :
When $Δ ABC ∼ Δ RPQ$
$\frac{AB}{RP}=\frac{BC}{PQ}=\frac{AC}{QR}=\frac{\sqrt{area(ABC)}}{\sqrt{area(RPQ)}}$
$\therefore \frac{AB}{RP} = \sqrt\frac{4}{9}$
⇒ $\frac{AB}{RP} = \frac{2}{3}$
⇒ $RP = \frac{3}{2} × AB=\frac{3}{2} × 3=4.5$ cm
Hence, the correct answer is 4.5.
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