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,
$\frac{\operatorname{ar}(\triangle P R Q)}{\operatorname{ar}(\triangle B C A)}=\frac{(Q R)^{2}}{(B C)^{2}}$
$=(\frac{Q R}{B C})^{2}$
$=(\frac{3}{1})^{2}$
$=\frac{9}{1}$
$=9$
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