Question

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

Answer 1

Given vertices are $(4,3),(-5,6),(0,-7),(3,-6)$ and $(-7,-2)$


By plotting the vertices we get order of vertices as $(4,3),(-5,6),(-7,-2)(0,-7)$ and $(3,-6)$
So the required area is,
$\Delta =\frac{1}{2}\left|\begin{array}{cccccc}{x}_1& x_2& x_3& x_4& x_5& x_1\\ y_1& y_2& y_3& y_4& y_5& y_1\end{array}\right|=\frac{1}{2}\left|\begin{array}{cccccc}4& -5& -7& 0& 3& 4\\ 3& 6& -2& -7& -6& 3\end{array}\right|=\frac{1}{2}|39+52+49+21+33|=97\text{sq. units}$