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$ABCD$ is a trapezium in which $AB||DC$ and $P,\ Q$ are points on $AD$ and $BC$ respectively such that $PQ || DC$. If $PD=18\ cm,\ BQ = 35\ cm$ and $QC = 15\ cm$, find $AD$.
Given: $ABCD$ is a trapezium in which $AB||DC$ and $P,\ Q$ are points on $AD$ and $BC$ respectively such that $PQ||DC$. If $PD = 18\ cm,\ BQ = 35\ cm$ and $QC=15\ cm$.
To do: To find $AD$.
Solution:
Construction: Join $BD$
In $\vartriangle ABD,\ PO∥AB$ [ $\therefore PQ || AB$ ]
$\Rightarrow \frac{DP}{AP}=\frac{DO}{OB}\ ---- ( i)$ [ By basic proportionality theorem ]
$\Rightarrow In \vartriangle BDC,\ OQ||DC$ [ $\therefore PQ||DC$ ]
$\frac{BQ}{QC}=\frac{OB}{OD}$ [ By basic proportionality theorem ]
$\Rightarrow \frac{QC}{BQ}=\frac{OD}{OB}\ \ ---- ( ii)$
$\Rightarrow \frac{DP}{AP}=\frac{QC}{BQ}$ [ From $( i)$ and $( ii)$ ]
$\Rightarrow \frac{18}{AP}=\frac{15}{35}$
$\therefore AP=\frac{18\times 35}{15}=42$
$\therefore AD=AP+DP=42+18=60\ cm$