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In a $Δ\ ABC$, $P$ and $Q$ are the points on sides $AB$ and $AC$ respectively, such that $PQ\ ∥\ BC$. If $AP\ =\ 2.4\ cm$, $AQ\ =\ 2\ cm$, $QC\ =\ 3\ cm$ and $BC\ =\ 6\ cm$. Find $AB$ and $PQ$.
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Given:


In a $Δ\ ABC$, $P$ and $Q$ are the points on sides $AB$ and $AC$ respectively, such that $PQ\ ∥\ BC$.


$AP\ =\ 2.4\ cm$, $AQ\ =\ 2\ cm$, $QC\ =\ 3\ cm$ and $BC\ =\ 6\ cm$.

To do:


We have to find $AB$ and $PQ$.


Solution:


$PQ\ ∥\ BC$  (given)


Therefore,


By Basic proportionality theorem,


$\frac{AP}{PB}=\frac{AQ}{QC}$


$\frac{2.4}{PB}=\frac{2}{3}$


$PB=\frac{2.4\times3}{2}$


$PB=\frac{7.2}{2}$


$PB=3.6 cm$


From the figure,


$AB=AP+PB$


$AB=(2.4+3.6) cm$


$AB=6 cm$


In $\vartriangle APQ$ and $\vartriangle ABC$,


$ \angle A=\angle A$ 


$\angle APQ=\angle ABC $    ($PQ\ ∥\ BC$, corresponding angles are equal)


$\vartriangle APQ\sim \vartriangle ABC$    (By AA criteria)


Therefore,


$\frac{AP}{AB} =\frac{PQ}{BC}$   (CPCT)


$PQ=\frac{2.4\times6}{6}$


$PQ=2.4 cm$.

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Updated on: 10-Oct-2022

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