Choose the correct answer from the given four options:
If $ \triangle \mathrm{ABC} \sim \Delta \mathrm{QRP}, \frac{\operatorname{ar}(\mathrm{ABC})}{\operatorname{ar}(\mathrm{PQR})}=\frac{9}{4}, \mathrm{AB}=18 \mathrm{~cm} $ and $ \mathrm{BC}=15 \mathrm{~cm} $, then $ \mathrm{PR} $ is equal to
(A) $ 10 \mathrm{~cm} $
(B) $ 12 \mathrm{~cm} $
(C) $ \frac{20}{3} \mathrm{~cm} $
(D) $ 8 \mathrm{~cm} $
Given:
\( \triangle \mathrm{ABC} \sim \Delta \mathrm{QRP}, \frac{\operatorname{ar}(\mathrm{ABC})}{\operatorname{ar}(\mathrm{PQR})}=\frac{9}{4}, \mathrm{AB}=18 \mathrm{~cm} \) and \( \mathrm{BC}=15 \mathrm{~cm} \).
To do:
We have to find the measure of \( \mathrm{PR} \).
Solution:
We know that,
The ratio of area of two similar triangles is equal to the ratio of square of their corresponding sides.
Therefore,
$\frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\Delta Q R P)}=\frac{(B C)^{2}}{(R P)^{2}}$
$\frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle P Q R)}=\frac{9}{4}$ (Given)
This implies,
$\frac{(15)^{2}}{(R P)^{2}}=\frac{9}{4}$
$(R P)^{2}=\frac{225 \times 4}{9}$
$(PR)^2=100$
$PR=10 \mathrm{~cm}$
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