# 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) $8 \mathrm{~cm}$

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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,

$\frac{(15)^{2}}{(R P)^{2}}=\frac{9}{4}$

$(R P)^{2}=\frac{225 \times 4}{9}$

$(PR)^2=100$

$PR=10 \mathrm{~cm}$

Updated on 10-Oct-2022 13:27:54