# find area of polygon

#### DigitalComputer

##### New member
Find the area of the polygon formed by the points (3,5), (5,11), (14,7), (8,3), and (6,6).

I can find the area of the polygon by dividing it into 3 triangles and then finding area of each triangle separately. I want to know if there is any simpler way of doing this.

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#### sbhatnagar

##### Active member
Hello DigitalComputer! Such problems are solved using the shoelace formula. If $A_r (x_r , y_r); \ r = 1,2,3, \cdots , n$ be the vertices of a polygon, taken in order then the area of the polygon $A_1 A_2 A_3 \cdots A_n$ is given by

$\text{area}= \Bigg|\frac{1}{2}\left( \sum_{r=1}^{n-1} \begin{vmatrix}x_i & y_i \\ x_{i+1} & y_{i+1}\end{vmatrix}+\begin{vmatrix}x_n & y_n \\ x_{1} & y_{1}\end{vmatrix}\right) \Bigg|$

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#### Sudharaka

##### Well-known member
MHB Math Helper
Find the area of the polygon formed by the points (3,5), (5,11), (14,7), (8,3), and (6,6).

I can find the area of the polygon by dividing it into 3 triangles and then finding area of each triangle separately. I want to know if there is any simpler way of doing this.
Hi DigitalComputer,

To elaborate more on the Shoelace method, suppose you have a set of points, $$(x_i,\,y_i)\mbox{ where }i=1,\,2,\,\cdots,\,n$$, which are vertices of a polygon. Then the Shoelace formula is,

$A={1 \over 2}|x_1y_2 + x_2y_3 + \cdots + x_{n-1}y_n + x_ny_1 - x_2y_1 - x_3y_2 - \cdots - x_ny_{n-1} - x_1y_n|$

where $$A$$ is the area of the polygon.

Note that in the Shoelace formula, the positive terms are obtained by the following manner;

The first $$x$$ coordinate is multiplied by the second $$y$$ coordinate, the second $$x$$ coordinate is multiplied by the third $$y$$ coordinate and so on. Finally the nth, $$x$$ coordinate is multiplied by the first $$y$$ coordinate.

And the negative terms are obtained by,

The second $$x$$ coordinate is multiplied by the first $$y$$ coordinate, the third $$x$$ coordinate is multiplied by the second $$y$$ coordinate and so on. Finally the first $$x$$ coordinate is multiplied by the nth, $$y$$ coordinate.

In your case, you have the points, $$(3,5),\,(5,11),\, (14,7),\, (8,3)\mbox{ and }(6,6)$$. Therefore by the Shoelace formula,

$A=\frac{1}{2}|(3\times 11)+(5\times 7)+(14\times 3)+(8\times 6)+(6\times 5)-(5\times 5)-(14\times 11)-(8\times 7)-(6\times 3)-(3\times 6)|=41.5$

Kind Regards,
Sudharaka.

#### soroban

##### Well-known member
Hello, DigitalComputer!

Find the area of the polygon formed by the points;
. . A (3,5), B(5,11), C(14,7), D(8,3), and E(6,6).

I use trapezoids . . .

Code:
[SIZE=3]
|
|           B
|           o
|          *: *
|         * :   *
|        *  :     *
|       *   :   E   *
|      *    :   o     *
|     *     *   :*      *
|    *  *   :   : *       o C
| A o       :   :  *   *  :
|   :       :   :   o     :
|   :       :   :   D     :
|   :       :   :   :     :
- - + - + - - - + - + - + - - + - - -
|   F       G   H   I     J[/SIZE]

First, I find the total area under the tent-shaped figure:
. . trapezoids $$ABGF + BCJG.$$

Then I subtract the areas of the three lower trapezoids:
. ..$$AEHF + EDIH + DCJI$$

Thank You!