$ABCD$ 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 $( ii)$ ]

$\Rightarrow \frac{18}{AP}=\frac{15}{35}$

$\therefore  AP=\frac{18\times 35}{15}=42$

$\therefore  AD=AP+DP=42+18=60\ cm$

Tutorialspoint
Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

24 Views

Kickstart Your Career

Get certified by completing the course

Get Started
Advertisements