Question

The area (in sq. units) of a pentagon whose vertices are $(4,3),(-5,6),(-7,-2),(0,-7)$ and $(3,-6)$ taken in order, is ?

Answer 1

Given vertices are $(4,3),(-5,6),(-7,-2),(0,-7)$ and $(3,-6)$, so the area is,
$\Delta =\frac{1}{2}|\left|\begin{array}{cc}{x}_1& y_1\\ x_2& y_2\end{array}\right|+\left|\begin{array}{cc}{x}_2& y_2\\ x_3& y_3\end{array}\right|+\left|\begin{array}{cc}{x}_3& y_3\\ x_4& y_4\end{array}\right|+\left|\begin{array}{cc}{x}_4& y_4\\ x_5& y_5\end{array}\right|+\left|\begin{array}{cc}{x}_5& y_5\\ x_1& y_1\end{array}\right||$

$=\frac{1}{2}|\left|\begin{array}{cc}4& 3\\ -5& 6\end{array}\right|+\left|\begin{array}{cc}-5& 6\\ -7& -2\end{array}\right|+\left|\begin{array}{cc}-7& -2\\ 0& -7\end{array}\right|+\left|\begin{array}{cc}0& -7\\ 3& -6\end{array}\right|+\left|\begin{array}{cc}3& -6\\ 4& 3\end{array}\right||$

$=\frac{1}{2}\left|\right[24+15]+[10+42]+[49-0]+[0+21]+[9+24\left]\right|$

$=\frac{1}{2}|39+52+49+21+33|$

$=\frac{1}{2}\left|194\right|$

$=97\text{sq. units}$