Question

Show that vector area of a quadrilateral $ABCD$ is $\frac{1}{2}(\overline{AC}\times \overline{BD})$, where $AC$ and $BD$ are its diagonals.

Answer 1

Let the diagonals intersect at ${}^{\prime }{0}^{\prime }$

$\overline{AC}=\overline{AO}+\overline{OC}$

$BD=\overline{BO}+\overline{OD}$

$\therefore \frac{1}{2}(\overline{AC}\times \overline{BD})=\frac{1}{2}(\overline{AO}+\overline{OC})(\overline{BO}+\overline{OD})$

$\Rightarrow \frac{1}{2}\overline{BO}+\frac{1}{2}\overline{AO}\times \overline{OD}+\frac{1}{2}\overline{OC}\overline{BO}+\frac{1}{2}\overline{OC}\overline{OD}$

$\Rightarrow$ area of $\Delta$ AOB$+$ar of $\Delta$AOD $+$ar of $\Delta$OBC $+$ ar of $\Delta$ DOC

$\Rightarrow$ ar of quadrilateral $=$ area of $(\Delta AOB+\Delta AOD+\Delta OBC+\Delta DOC)$

$\therefore$ Area of Quadrilateral $=\frac{1}{2}(\overline{AC}\times \overline{BD})$

Hence proved.