The mid-point of the coordinates $(x_1,y_1),(x_2,y_2)$ is given by $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
$\therefore$ The mid-points of AB, BC and CA are $(\frac{-4+2}{2}$,  $\frac{3+3}{2})$, $(\frac{4+2}{2}$,  $\frac{3+5}{2})$ and $(\frac{-4+4}{2}$,  $\frac{5+3}{2})$ respectively. Let us denote these by D, E and F.
i.e.,  D(-1, 3), E(3, 4) and F(0, 4).

Area of the triangle formed by $(x_1,y_1),(x_2,y_2)(x_3,y_3)$ is given by
$\text{Area}=\frac{1}{2}$|$x_1$($y_2$-$y_3$)+$x_2$($y_3$-$y_1$)+$x_3$($y_1$-$y_2$)|
Now, let $x_1$ = -1,  $y_1$ = 3, $x_2$ = 3,  $y_2$ = 4 $x_3$ = 0 and  $y_3$ = 4

$\Rightarrow \text{Required area}=\frac{1}{2}$| -1(4 - 4) + 3(4 - 3) + 0(3 - 4)|  = 1.5 square units

$\therefore$ Area of the triangle = 1.5 square units