Answer:
In \[\Delta \text{ }DOQ\text{ }and\text{ }\Delta \text{ }BOP,\]
\[\angle QDO\text{ }=\angle PBO\] \[\left[ As\text{ }AB\text{ }\left| \left| \text{ }DC\text{ }in\text{ }this\text{ }way,\text{ }PB\text{ } \right| \right|\text{ }DQ. \right]\]
In this way, \[\angle DOQ\text{ }=\angle BOP\]\[\left[ Vertically\text{ }inverse\text{ }angles \right]\]
Thus, \[\vartriangle DOQ\text{ }\sim\text{ }\vartriangle BOP\]by \[AA\]rule for similitude
Since, relating sides of comparative triangles are corresponding we have
\[DO/BO\text{ }=\text{ }DQ/BP\]
\[DO/6\text{ }=\text{ }8/BP\]
\[BP\text{ }x\text{ }DO\text{ }=\text{ }48\text{ }c{{m}^{2}}\]