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} $

AcademicMathematicsNCERTClass 10

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,

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

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

$(PR)^2=100$

$PR=10 \mathrm{~cm}$

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

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