Question

∆ABC ∼ ∆PQR and ar(∆ABC) = 4 ar(∆PQR). If BC = 12 cm, find QR.

Answer 1

^{$$\begin{gathered} \mathrm{Given}:\mathrm{ar}(△ABC)=4\mathrm{ar}(∆PQR) \\ \frac{\mathrm{ar}(△ABC)}{\mathrm{ar}(∆PQR)}=\frac{4}{1} \\ \because ∆ABC~∆PQR \\ \therefore \frac{\mathrm{ar}(△ABC)}{\mathrm{ar}(∆PQR)}=\frac{B{C}^2}{Q{R}^2} \\ \therefore \frac{B{C}^2}{Q{R}^2}=\frac{4}{1} \end{gathered}$$}
$$\begin{gathered} \Rightarrow Q{R}^2=\frac{{12}^2}{4} \\ \Rightarrow Q{R}^2=36 \\ \Rightarrow QR=6\mathrm{cm} \end{gathered}$$
Hence, QR = 6 cm