Question

If $∆ABC~∆QRP$, $\frac{ar(∆ABC)}{ar(∆PQR)}=\frac{9}{4}$, $AB=18cm$ and $BC=15cm$, then find the value of $PR$.

Answer 1

Applying a property of areas of two similar triangles:

Given that $\frac{\mathrm{ar}(∆ABC)}{\mathrm{ar}(∆PQR)}=\frac{9}{4}$, $BC=15cm$

$∆ABC~∆QRP$, hence the ratio of areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$$\begin{gathered} \frac{\mathrm{ar}(△ABC)}{\mathrm{ar}(△PQR)}=\frac{B{C}^2}{P{R}^2} \\ \Rightarrow \frac{9}{4}=\frac{{\left(15\right)}^2}{P{R}^2} \\ \Rightarrow P{R}^2=\frac{{\left(15\right)}^2\times 4}{9} \\ \Rightarrow =\frac{900}{9} \\ \Rightarrow P{R}^2=100 \\ \Rightarrow PR=10cm \end{gathered}$$

Therefore, the value of $PR=10cm$.