Question

Find the area of quadrilateral ABCD, the co ordinates of whose vertices are $A(-3,2),B(5,4),C(7,-6),D(-5,-4)$

Answer 1

Using the shoelace formula

$A\left(\square ABCD\right)=\frac{1}{2}\left|\begin{array}{cc}-3& 2\\ 5& 4\\ 7& -6\\ -5& -4\end{array}\right|$

$=\frac{1}{2}\left|[-3\times 4+5\times -6+7\times -4+2\times -5-5\times 2-(-3)\times (-4)-7\times 4-(-5)\times -6]\right|$

$=\frac{1}{2}\left|[-12-30-28-10-10-12-28-30]\right|$

$=\frac{1}{2}\times 160$

$A\left(\square ABCD\right)=80$ sq. units.