How do you find the area of a parallelogram with the vertices? How do you find the area of a parallelogram with the following vertices; $A(4,2)$, $B(8,4)$, $C(9,6)$ and $D(13,8)$.
 A: There are plenty of ways, such as the Shoelace Theorem and Pick's Theorem.
If you have a graph, you can also simply draw a rectangle around the shape and subtract the parts you don't want.
A: The absolute value of the cross product of two vectors $\vec{a}, \vec{b} \in \mathbb{R}^3$ spanning the parallelogram is its area:
$$A_\text{parallelogram}= \left|\vec{a}\times\vec{b}\right|$$

So in your case we have to write the points in $\mathbb{R}^2$ as vectors in $\mathbb{R}^3$ and apply the formula:
$\vec{AB} = \begin{pmatrix}8\\4\\0\end{pmatrix}  -\begin{pmatrix}4\\2\\0\end{pmatrix} =\begin{pmatrix}4\\2\\0\end{pmatrix}$
$\vec{AD} = \begin{pmatrix}13\\8\\0\end{pmatrix}  -\begin{pmatrix}4\\2\\0\end{pmatrix} =\begin{pmatrix}9\\6\\0\end{pmatrix}$
$A_\text{parallelogram}= \left|\vec{AB}\times\vec{AD}\right| = \left| \begin{pmatrix}4\\2\\0\end{pmatrix} \times \begin{pmatrix}9\\6\\0\end{pmatrix} \right| = \left|\begin{pmatrix}0\\0\\6\end{pmatrix}  \right| = 6$

You might have noticed that this simplifies to 
$$A_\text{parallelogram}= (b_1 - a_1)(d_2-a_2)-(b_2-a_2)(d_1-a_1)$$
$$= (8 - 4)(8-2)-(4-2)(13-4)=-24-(-18)=6$$
A: For this, we plan to use the Shoelace formula.

Shoelace Formula: Given the coordinates of vertices of a polygon, its area is found by $$A=\frac 12\left|\sum_{i=1}^{n-1}x_iy_{i+1}+x_ny_1-\sum_{i=1}^{n-1}x_{i+1}y_i-x_1y_n\right|$$
  Or, in other words, we have $$A=\frac 12|x_1y_2+x_2y_3+\ldots x_{n-1}y_n+x_ny_1-x_2y_1-x_3y_2-\ldots -x_ny_{n-1}-x_1y_n|$$
  Where $A$ is the area of the polygon, and $(x_i,y_i)$ with $i=1,2,3\dots$ are the vertices of the polyon

So with your case, the vertices are $A(4,2), B(8,4), C(9,6)$ and $D(13,8)$. We let $x_1=13,y_1=8,x_2=9,y_2=6,x_3=4,y_3=2,x_4=8,y_4=4$ and the area is given by $$A=\frac 12|13\cdot 6+9\cdot 2+4\cdot 4+8\cdot 8-9\cdot 8-4\cdot 6-8\cdot 2-13\cdot 4|\\=\frac 12\cdot 12=6$$
A: I think this is a special case of shoelace theorem. 
A quad is made up of two triangle and area of a triangle is 

$${1\over 2}{|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|}$$

Or you can use distance formula 

$$distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$$

and then heron's formula

$$A = {1\over 2}\sqrt{s(s-a)(s-b)(s-c))}$$
  Where s is the semi-perimeter of the triangle and, a,b,c are the length of its sides.

A: For any quadrilateral the area is one-half the magnitude of the cross product of the two diagonal vectors.
A: I am just providing you with the simplest shortcut to doing this
Pick the first three points A(4,2), B(8, 4) and C(9, 6)
Negate point A to get (-4, -2) and add to the other two points B and C. Add x's and y's so you have a new point
(4, 2)
(5, 4)   Now use determinant to find the area. 
16-10 = 6sq unit
Neglect any negative sign that arises
