Question

ABCD is a trapezium in which ABIIDC and P, Q are points on AD and BC respectively, such that PQIIDC, if PD = 18 cm, BQ = 35cm and QC = 15cm, find AD.

Answer 1

Given, a trapeszium ABCD in which ABIIDC. P and Q are points on AD and BC, respectively such that PQII DC. Thus, ABIIPQIIDC.


Join BD.
In $\Delta ABD$ PO || AB $[\because PQIIAB]$
By basic proportionally theorem, $\frac{DP}{AP}=\frac{DO}{OB}$ ….(i)
In $\Delta BDC$ OQ II DC $[\because PQIIDC]$
By basic proportionally theorem,
$\frac{BQ}{QC}=\frac{OB}{OD}$
$\Rightarrow \frac{QC}{BQ}=\frac{OD}{OB}$ ….(ii)
From Eqs .. (i) and (ii), $\frac{DP}{AP}=\frac{QC}{BQ}$
$\Rightarrow \frac{18}{AP}=\frac{15}{35}$
$\Rightarrow AP\frac{18\times 35}{15}=42$
$\therefore$ $AD=AP+DP=42+18=60cm$