Question

Question 11
If $\Delta ABC\sim \Delta QRP,\frac{ar\left(\Delta ABC\right)}{ar\left(\Delta PQR\right)}=\frac{9}{4}$, AB = 18 cm and BC = 15cm, then PR is equal to

(A) 10 cm
(B) 12 cm
(C) $\frac{20}{3}cm$
(D)​​​​​​​ 8cm

Answer 1

(A) 10 cm

Given, in $\Delta ABC\sim \Delta QRP,AB=18cmandBC=15cm$


We know that, the ratio of area of two similar triangles is equal to the ratio of square of their corresponding sides.
$\therefore \frac{ar\left(\Delta ABC\right)}{ar\left(\Delta QRP\right)}=\frac{(BC{)}^2}{(RP{)}^2}$
But given, $\frac{ar\left(\Delta ABC\right)}{ar\left(\Delta PQR\right)}=\frac{9}{4}$ [given]
$\Rightarrow \frac{(15{)}^2}{(RP{)}^2}=\frac{9}{4}$ [$\because$ BC=15cm, given]
$\Rightarrow (RP{)}^2=\frac{225\times 4}{9}=100$
$\therefore RP=10cm$