Area

$\large A =\dfrac{1}{2} \begin{vmatrix} x_1 & x_2 & x_3 \cdots & x_{n} & x_1\\ y_1 & y_2 & y_3 \cdots & y_n & y_1 \\ \end{vmatrix}$

$\large A =\dfrac{1}{2} \begin{vmatrix} 4 & 2 & -12 & -16 & 4\\ 12 & 26 & 24 & 2 & 12 \\ \end{vmatrix}$

$\large A=\dfrac{1}{2}\left[4\cdot26+2\cdot24-12\cdot2-16\cdot12-(12\cdot2-26\cdot12-24\cdot16+2\cdot4)\right]$

$\large A=\dfrac{1}{2}[104+48-24-192-(24-312-384+8)]$

$\large A=\dfrac{1}{2}(-64+664)=\dfrac{1}{2}(600)=$ $\boxed{300}$

Let $A:(4,12), \: B:(2,26), \: C:(-12,24), \: D:(-16,2)$

Let $E_{1}$ be point on $AD$ such that $CE_{1} \perp AD$ and $E_{2}$ be point on $AC$ such that $BE_{2} \perp AC$

For $\bigtriangleup {ACD}:$

Slope $m_{AD} = \dfrac{1}{2} \implies m_{CE_{1}} = -2 \implies$

$2y - x = 20$ $y + 2x = 0$

$\implies y = 8 \implies x = -4$

$\therefore CE_{1} = 8\sqrt{5}$ and $AD = 10\sqrt{5} \implies Area_{\bigtriangleup {ACD}} = 200$

For $\bigtriangleup {ABC}:$

Slope $m_{AC} = -\dfrac{3}{4} \implies m_{BE_{2}} = \dfrac{4}{3} \implies$

$4y + 3x = 60$ $3y - 4x = 70$

$\implies y = 18 \implies x = -4$

$\therefore BE_{2} = 10$ and $AC = 20 \implies Area_{\bigtriangleup {ABC}} = 100$

$\therefore$ The area $A_{ABCD} = \boxed{300}.$