Let $A(0,0), B(7,2), C(5,9), D(12,11)$
$\therefore $ Area of $\Delta A B C=\frac{1}{2}\begin{vmatrix}0 & 0 & 1 \\ 7 & 2 & 1 \\ 5 & 9 & 1\end{vmatrix}$
$=\frac{1}{2} \cdot 1(63-10)=\frac{53}{2}$ sq unit
$\therefore $ Area of $\Delta A C D=\frac{1}{2}\begin{vmatrix}0 & 0 & 1 \\ 5 & 9 & 1 \\ 12 & 11 & 1\end{vmatrix}$
$=\frac{1}{2} \cdot 1(55-108)$
$=\frac{53}{2}$ sq unit
$\therefore $ Area of parallelogram $A B C D$
$=$ Area of $\triangle A B C+$ Area of $\Delta A C D$
$=\frac{53}{2}+\frac{53}{2}$
$=53$ sq unit