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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 $D(13,8)$.

For this, we plan to use the Shoelace formula.

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**Shoelace Formula:** Given the coordinates of vertices of a polygon, its area is found by $$A=frac 12left|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

ight|$$ Or, in other words, we have $$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 $i=1,2,3dots$ 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 $$A=frac 12|13cdot 6+9cdot 2+4cdot 4+8cdot 8-9cdot 8-4cdot 6-8cdot 2-13cdot 4|\=frac 12cdot 12=6$$

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edited Oct 13 “16 at 12:24

answered Oct 13 “16 at 0:01

FrankFrank

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The absolute value of the cross product of two vectors $vec{a}, vec{b} in occupychristmas.orgbb{R}^3$ spanning the parallelogram is its area:

$$A_ ext{parallelogram}= left|vec{a} imesvec{b}

ight|$$

So in your case we have to write the points in $occupychristmas.orgbb{R}^2$ as vectors in $occupychristmas.orgbb{R}^3$ and apply the formula:

$vec{AB} = egin{pmatrix}8\4\0end{pmatrix} -egin{pmatrix}4\2\0end{pmatrix} =egin{pmatrix}4\2\0end{pmatrix}$

$vec{AD} = egin{pmatrix}13\8\0end{pmatrix} -egin{pmatrix}4\2\0end{pmatrix} =egin{pmatrix}9\6\0end{pmatrix}$

$A_ ext{parallelogram}= left|vec{AB} imesvec{AD}

ight| = left| egin{pmatrix}4\2\0end{pmatrix} imes egin{pmatrix}9\6\0end{pmatrix}

ight| = left|egin{pmatrix}0\0\6end{pmatrix}

ight| = 6$

You might have noticed that this simplifies to

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

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edited Oct 13 “16 at 14:40

answered Oct 12 “16 at 23:52

adjanadjan

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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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answered Oct 12 “16 at 23:51

suomynonAsuomynonA

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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

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

Or you can use distance formula

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

and then heron”s formula

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

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answered Oct 13 “16 at 8:10

user377111user377111

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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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answered Oct 13 “16 at 16:15

chowdahchowdah

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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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answered Jan 25 “19 at 4:28

Daniel WastyDaniel Wasty

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