Choose the correct answer from the given four options:
It is given that $ \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR} $, with $ \frac{\mathrm{BC}}{\mathrm{QR}}=\frac{1}{3} $. Then, $ \frac{\text { ar (PRQ) }}{\operatorname{ar}(\mathrm{BCA})} $ is equal to
(A) 9
(B) 3
(C) $ \frac{1}{3} $
(D) $ \frac{1}{9} $

Given:

\( \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR} \), with \( \frac{\mathrm{BC}}{\mathrm{QR}}=\frac{1}{3} \)

To do:

We have to find \( \frac{\text { ar (PRQ) }}{\operatorname{ar}(\mathrm{BCA})} \).

Solution:

$\triangle A B C \sim \triangle P Q R$

$\frac{B C}{Q R}=\frac{1}{3}$

We know that,

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Therefore,

$\frac{\operatorname{ar}(\triangle P R Q)}{\operatorname{ar}(\triangle B C A)}=\frac{(Q R)^{2}}{(B C)^{2}}$

$=(\frac{Q R}{B C})^{2}$

$=(\frac{3}{1})^{2}$

$=\frac{9}{1}$

$=9$