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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$\begingroup$ Use the determinant. $\endgroup$– Steven GubkinCommented Oct 13, 2016 at 13:05
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6 Answers
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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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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 $\endgroup$ Commented Oct 13, 2016 at 8:15
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4$\begingroup$ 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. $\endgroup$ Commented Oct 13, 2016 at 9:14
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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.
answered Oct 12, 2016 at 23:51
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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? $\endgroup$ Commented Jan 25, 2019 at 4:47
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$\begingroup$ square unit as in square meter or any unit of length measurement...mm, cm, m..etc $\endgroup$ Commented Jan 25, 2019 at 4:49