Prove the following


If $\triangle A B C \sim \triangle P Q R$ with $\frac{B C}{Q R}=\frac{1}{3}$, then $\frac{\operatorname{ar}(\Delta P R Q)}{\operatorname{ar}(\Delta B C A)}$ is equal to

(a) 9

(b) 3

(c) $\frac{1}{3}$

(d) $\frac{1}{9}$



(a) Given, $\triangle A B C \sim \Delta P Q R$ and $\frac{B C}{Q R}=\frac{1}{3}$

We know that, the ratio of the areas of two similar triangles is equal to square of the ratio of their corresponding sides.

$\therefore$ $\frac{\operatorname{ar}(\Delta P R Q)}{\operatorname{ar}(\Delta B C A)}=\frac{(Q R)^{2}}{(B C)^{2}}=\left(\frac{Q R}{B C}\right)^{2}=\left(\frac{3}{1}\right)^{2}=\frac{9}{1}=9$

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