ABCD is a trapezium in which AB || DC and P and Q are points on AD and BC, respectively such that PQ || DC. If PD = 18cm, BQ = 35cm and QC = 15cm, find AD.
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Given, a trapezium ABCD in which AB || DC. P and Q are points on AD and BC, respectively such that PQ || DC. Thus, AB || PQ || DC.
Join BD
$\text{In }\triangle\text{ABD},$
$\text{PO }\parallel\text{ AB}\ [\because\text{PO}\parallel\text{AB}]$
By basic proportionality theorem,
$\frac{\text{DP}}{\text{AP}}=\frac{\text{DO}}{\text{OB}}\ .....(\text{i})$
$\text{In }\triangle\text{BDC},$
$\text{OQ }\parallel\text{ DC}\ [\because\text{PQ}\parallel\text{DC}]$
By basic proportionality theorem,
$\frac{\text{BQ}}{\text{QC}}=\frac{\text{OB}}{\text{OD}}$
$\Rightarrow\frac{\text{QC}}{\text{BQ}}=\frac{\text{OB}}{\text{OD}}\ .....(\text{ii})$
From Eq. (i) and (ii),
$\frac{\text{DP}}{\text{AP}}=\frac{\text{QC}}{\text{BQ}}$
$\Rightarrow\frac{\text{18}}{\text{AP}}=\frac{\text{15}}{\text{35}}$
$\Rightarrow\text{AP}=\frac{18\times35}{15}=42$
$\therefore$ $\text{AD}=\text{AP}+\text{DP}$
$\text{AD}=42+18$
$\text{AD}=60\text{cm}$
Join BD
$\text{In }\triangle\text{ABD},$
$\text{PO }\parallel\text{ AB}\ [\because\text{PO}\parallel\text{AB}]$
By basic proportionality theorem,
$\frac{\text{DP}}{\text{AP}}=\frac{\text{DO}}{\text{OB}}\ .....(\text{i})$
$\text{In }\triangle\text{BDC},$
$\text{OQ }\parallel\text{ DC}\ [\because\text{PQ}\parallel\text{DC}]$
By basic proportionality theorem,
$\frac{\text{BQ}}{\text{QC}}=\frac{\text{OB}}{\text{OD}}$
$\Rightarrow\frac{\text{QC}}{\text{BQ}}=\frac{\text{OB}}{\text{OD}}\ .....(\text{ii})$
From Eq. (i) and (ii),
$\frac{\text{DP}}{\text{AP}}=\frac{\text{QC}}{\text{BQ}}$
$\Rightarrow\frac{\text{18}}{\text{AP}}=\frac{\text{15}}{\text{35}}$
$\Rightarrow\text{AP}=\frac{18\times35}{15}=42$
$\therefore$ $\text{AD}=\text{AP}+\text{DP}$
$\text{AD}=42+18$
$\text{AD}=60\text{cm}$
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