Question

If $∆ABC~∆QRP,\frac{\mathrm{ar}(∆ABC)}{\mathrm{ar}(∆PQR)}=\frac{9}{4},$ AB = 18 cm and BC = 15 cm, then PR = ?
(a) 8 cm
(b) 10 cm
(c) 12 cm
(d) $\frac{20}{3}\mathrm{cm}$

Answer 1

(b) 10 cm

$$\begin{gathered} \because △ABC~△QRP \\ \therefore \frac{AB}{QR}=\frac{BC}{PR} \\ \mathrm{Now}, \\ \frac{ar(△ABC)}{ar(△QRP)}=\frac{9}{4} \\ \Rightarrow (\frac{AB}{QR}{)}^2=\frac{9}{4} \\ \Rightarrow \frac{AB}{QR}=\frac{3}{2} \\ \mathrm{Therefore}, \\ \frac{AB}{QR}=\frac{BC}{PR}=\frac{3}{2} \\ \mathrm{Hence},3PR=2BC=2\times 15=30 \\ PR=10\mathrm{cm} \end{gathered}$$