Write the two vectors as the columns of a matrix,
$$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, \qquad A_1 = \begin{bmatrix} a \\ c \end{bmatrix}, \quad A_2 = \begin{bmatrix} b \\ d \end{bmatrix}.$$
The green region is the parallelogram with sides formed by $A_1$ and $A_2$, and its vertices are $0$, $A_1$, $A_2$, and $A_1 + A_2 = (a+b,\, c+d)$. The claim is that its area is exactly $|ad - bc|$, the absolute value of the determinant of $A$. The proof just relies on basic geometry (no fancy trig or linear algebra needed). We make some mild assumptions for ease of proof and omission of cases.
\begin{proposition}
If $a, b, c, d > 0$ and $A_1$ is clockwise of $A_2$, then the parallelogram with sides formed by $A_1$ and $A_2$ has area $ad - bc$.
\end{proposition}
\begin{proof}
Consider the outlined rectangle (in black), stretching from $0$ to $(a+b, c+d)$. This shape has area $(a+b)(c+d)$. Everything in the rectangle that is not green comes in pairs:
\begin{itemize}
\item two purple $b \times c$ rectangles in the top-left and bottom-right corners, each with area $bc$;
\item two red right triangles with legs $a$ and $c$, each with area $\tfrac12 ac$;
\item two blue right triangles with legs $b$ and $d$, each with area $\tfrac12 bd$.
\end{itemize}
Removing all six pieces from the rectangle leaves the parallelogram, and subtracting all the areas from the total $(a+b)(c+d)$ resulting in the parallelogram's area. Algebraically, this looks like
\begin{align*}
\text{Area of parallelogram} &= (a+b)(c+d) - 2bc - 2\cdot\tfrac12 ac - 2\cdot\tfrac12 bd \\
&= ac + ad + bc + bd - 2bc - ac - bd \\
&= ad - bc.
\end{align*}
\end{proof}
\begin{remark}
Notably, we can make a stronger statement, that the area (which is always nonnegative) is the absolute value of the determinant. This requires considering more cases and being careful about orientation, so we leave this generalization as an exercise. So the determinant is really a \textbf{signed} area. Moreover, this is why swapping two columns flips the sign of the determinant, and why $\det A = 0$ exactly when $A_1$ and $A_2$ are parallel (the parallelogram collapses to a segment of zero area). The same story holds in $\R^3$, where $|\det A|$ is the volume of the parallelepiped formed by the three columns, with an analogous proof.
\end{remark}