Question

ABCD is a trapezium in which AB $\parallel$ DC. P and Q are points on sides AD and BC such that PQ $\parallel$ AB. If PD = 18cm, BQ = 35cm and QC = 15cm, find AD.

Answer 1

Construction: Join $BD$

In $△ABD$, $PO\parallel AB$ [$\because$ $PQ\parallel AB$ ]

$\Rightarrow$ $\frac{DP}{AP}=\frac{DO}{OB}$ [ By basic proportionality theorem ] ------ ( 1 )

$\Rightarrow$ In $△BDC$, $OQ\parallel DC$ [

$\because$ $PQ\parallel DC$ ]

$\frac{BQ}{QC}=\frac{OB}{OD}$ [ By basic proportionality theorem ]

$\Rightarrow$ $\frac{QC}{BQ}=\frac{OD}{OB}$ ----- ( 2 )

$\Rightarrow$ $\frac{DP}{AP}=\frac{QC}{BQ}$ [ From ( 1 ) and ( 2 ) ]

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

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

$\therefore$ $AD=AP+DP=42+18=60cm$