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 = 15 cm, then PR is equal to :

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

Answer 1

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


We know that the ratio of areas 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 it is given that
$\frac{ar\left(\Delta ABC\right)}{ar\left(\Delta PQR\right)}=\frac{9}{4}...\left(i\right)$
$\Rightarrow \frac{(15{)}^2}{(RP{)}^2}=\frac{9}{4}[\because BC=15cm[given\left]\right]...\left(ii\right)$
$\Rightarrow (RP{)}^2=\frac{225\times 4}{9}=100(from\left(i\right)and\left(ii\right))$
$\therefore RP=10cm$