If $\Delta A B C \sim \Delta Q R P, \frac{\operatorname{ar}(\Delta A B C)}{\operatorname{ar}(\Delta P Q R)}=\frac{9}{4}, A B=18 \mathrm{~cm}$ and $B C=15 \mathrm{~cm}$, then $P \mathrm{R}=?$
(a) 8 cm
(b) 10 cm
(c) 12 cm
(d) $\frac{20}{3} \mathrm{~cm}$
(b) 10 cm
$\because \triangle A B C \sim \triangle Q R P$
$\therefore \frac{A B}{Q R}=\frac{B C}{P R}$
Now,
$\frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle Q R P)}=\frac{9}{4}$
$\Rightarrow\left(\frac{A B}{Q R}\right)^{2}=\frac{9}{4}$
$\Rightarrow \frac{A B}{Q R}=\frac{3}{2}$
Therefore,
$\frac{A B}{Q R}=\frac{B C}{P R}=\frac{3}{2}$
Hence, $3 P R=2 B C=2 \times 15=30$
$P R=10 \mathrm{~cm}$
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