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How do you find the area of a parallelogram with the following vertices; $A(4,2)$, $B(8,4)$, $C(9,6)$ and $D(13,8)$.

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closed as off-topic by user21820, Xander Henderson, José Carlos Santos, YuiTo Cheng, RRL May 18 at 2:36

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  • $\begingroup$ Use the determinant. $\endgroup$ – Steven Gubkin Oct 13 '16 at 13:05
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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 $$A=\frac 12|13\cdot 6+9\cdot 2+4\cdot 4+8\cdot 8-9\cdot 8-4\cdot 6-8\cdot 2-13\cdot 4|\\=\frac 12\cdot 12=6$$

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  • $\begingroup$ You answered with another person's answer? xD $\endgroup$ – Billy Rubina Oct 13 '16 at 8:15
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    $\begingroup$ This is hitting a nail with a sledge hammer. There aren't many polygons that are as simple to handle as parallelograms. Or to be more specific, it is very probably the computation for parallelograms (and derived from that, triangles) that serves as basis for deriving the shoelace formula. $\endgroup$ – Marc van Leeuwen Oct 13 '16 at 9:14
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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

$$A_\text{parallelogram}= (b_1 - a_1)(d_2-a_2)-(b_2-a_2)(d_1-a_1)$$ $$= (8 - 4)(8-2)-(4-2)(13-4)=-24-(-18)=6$$

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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.

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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.

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For any quadrilateral the area is one-half the magnitude of the cross product of the two diagonal vectors.

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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

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  • $\begingroup$ I don't understand so much. What is 6sq unit? $\endgroup$ – El borito Jan 25 at 4:47
  • $\begingroup$ square unit as in square meter or any unit of length measurement...mm, cm, m..etc $\endgroup$ – Daniel Wasty Jan 25 at 4:49

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