MHB Find Area of Polygon: Simpler Method?

[Blandongstein]

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.

 

[sbhatnagar]

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

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

 

[Sudharaka]

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]

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 . . .

      |
      |           B
      |           o
      |          *: *
      |         * :   *
      |        *  :     *
      |       *   :   E   *
      |      *    :   o     *
      |     *     *   :*      *
      |    *  *   :   : *       o C
      | A o       :   :  *   *  :
      |   :       :   :   o     :
      |   :       :   :   D     :
      |   :       :   :   :     :
  - - + - + - - - + - + - + - - + - - -
      |   F       G   H   I     J

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

 

[Blandongstein]

Thank You!