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 $PQ$. "
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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