It is given that $\overrightarrow{AB}=\begin{pmatrix}-12\\a\end{pmatrix}$, $a>0$ and $|\overrightarrow{BA}|=37$ units.

1. Form an equation using the magnitude of the vector

Given $\overrightarrow{AB}=\begin{pmatrix}-12\\a\end{pmatrix}$ and $|\overrightarrow{BA}|=37$.

Since reversing a vector does not change its magnitude,
$|\overrightarrow{AB}|=|\overrightarrow{BA}|=37.$

Using the magnitude formula,
$\sqrt{(-12)^2+a^2}=37.$

2. Solve the equation for the positive value of $a$

$(-12)^2+a^2=37^2$
$144+a^2=1369$
$a^2=1225$
$a=35 \quad \text{since } a>0.$

1. Substitute the value of $a$ into the vector $\overrightarrow{AB}$

From part (a), $a=35$.

So
$\overrightarrow{AB}=\begin{pmatrix}-12\\35\end{pmatrix}.$

Since $\overrightarrow{AB}=\overrightarrow{DC}$,
$\overrightarrow{DC}=\begin{pmatrix}-12\\35\end{pmatrix}.$

2. Find the coordinates of $D$ using $\overrightarrow{DC}=C-D$

Let $D=(x,y)$ and $C=(0,10)$.

$\overrightarrow{DC}=\begin{pmatrix}0-x\\10-y\end{pmatrix}=\begin{pmatrix}-12\\35\end{pmatrix}$

Equating components,
$0-x=-12 \Rightarrow x=12,$
$10-y=35 \Rightarrow y=-25.$

Therefore, $D=(12,-25)$.

1. Classify the quadrilateral using the equal vector relationship

Since $\overrightarrow{AB}=\overrightarrow{DC}$, sides $AB$ and $DC$ are equal in length and parallel in the same direction.

Therefore, $ABCD$ is a parallelogram.