$ A $ and $ B $ are respectively the points on the sides $ P Q $ and $ P R $ of a triangle $ P Q R $ such that $ \mathrm{PQ}=12.5 \mathrm{~cm}, \mathrm{PA}=5 \mathrm{~cm}, \mathrm{BR}=6 \mathrm{~cm} $ and $ \mathrm{PB}=4 \mathrm{~cm} . $ Is $ \mathrm{AB} \| \mathrm{QR} $ ? Give reasons for your answer.

Given:

\( A \) and \( B \) are respectively the points on the sides \( P Q \) and \( P R \) of a triangle \( P Q R \) such that \( \mathrm{PQ}=12.5 \mathrm{~cm}, \mathrm{PA}=5 \mathrm{~cm}, \mathrm{BR}=6 \mathrm{~cm} \) and \( \mathrm{PB}=4 \mathrm{~cm} . \)

To do:

We have to find whether \( \mathrm{AB} \| \mathrm{QR} \).

Solution:

$Q A=Q P-P A$

$=12.5-5$

$=7.5 \mathrm{~cm}$

$\frac{P A}{A Q}=\frac{5}{7.5}$

$=\frac{50}{75}$

$=\frac{2}{3}$..............(i)

$\frac{P B}{B R}=\frac{4}{6}$

$=\frac{2}{3}$..........(ii)

From(i) and (ii), we get,

$\frac{P A}{A Q}=\frac{P B}{B R}$

By converse of basic proportionality theorem,

$AB \| Q R$

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Updated on: 10-Oct-2022

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