Question 1143813
.


            The method to which Alan refers in his post is very robust,  but Alan made a number of errors on the way,
            so I came to fix it and to get the correct answer/solution.



Given the points A(1,5), B(5,-2) C(-5,-5) and D(-15,1).
1.Find the area of the quadrilateral ABCD.
Here's a better way to do these:
<pre>
     A    B    C    D    A
x    1    5   -5   -15   1
Y    5   -2   -5    1    5       <----
</pre>
---
Add the diagonal products starting at the upper left
1*(-2) + 5*(-5) + (-5)*1 + (-15)*5 = -2 -25 -5 -75 = -107    &nbsp;&nbsp;&nbsp;&nbsp;<----
----
Add the diagonal products starting at the lower left
5*5 + (-2)*(-5) + (-5)*(-15) + 1*1 = 25 + 10 + 75 + 1 = 111 &nbsp;&nbsp;&nbsp;&nbsp;<----
------
The difference between the 2 sums is 11 - (-107) = 111 + 107 = 218.                     &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<----
The area is 1/2 that:   Area = 218/2 = 109 sq units                   &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<----



<U>Answer for the area</U> :  &nbsp;&nbsp;109 square units.


By the arrows  &nbsp;" <---- "  &nbsp;I showed the lines where I edited Alan's calculations.


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For this method, &nbsp;see the site

https://www.mathopenref.com/coordpolygonarea.html


It works for non-convex polygons, &nbsp;too.


It works, &nbsp;practically, &nbsp;for all and any polygons.


The major and the only restriction is &nbsp;<U>THIS</U> : &nbsp;&nbsp;a polygon should not have self-intersections.