# $A$ 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.

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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,

$AB \| Q R$

Updated on 10-Oct-2022 13:27:56