Question

If ΔABC ~ ΔQRP, $\frac{ar\left(\mathrm{\Delta }ABC\right)}{ar\left(\mathrm{\Delta }PQR\right)}=\frac{9}{4},AB$= 18 cm and BC = 15 cm, then PR = _________.

Answer 1

ΔABC ~ ΔQRP (Given)

$\therefore \frac{\mathrm{ar}\left(\mathrm{\Delta ABC}\right)}{\mathrm{ar}\left(\mathrm{\Delta PQR}\right)}={\left(\frac{\mathrm{BC}}{\mathrm{RP}}\right)}^2$ (The ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides)

$$\begin{gathered} \Rightarrow {\left(\frac{15\mathrm{cm}}{\mathrm{RP}}\right)}^2=\frac{9}{4} \\ \Rightarrow \frac{15\mathrm{cm}}{\mathrm{RP}}=\frac{3}{2} \\ \Rightarrow \mathrm{RP}=\frac{15\times 2}{3}=10\mathrm{cm} \end{gathered}$$

If ΔABC ~ ΔQRP, $\frac{\mathrm{ar}\left(\mathrm{\Delta ABC}\right)}{\mathrm{ar}\left(\mathrm{\Delta PQR}\right)}=\frac{9}{4},$ AB = 18 cm and BC = 15 cm, then PR = ____10 cm____.