0
$\begingroup$

is possible to calculate the overlapping polygons area of two polygons in cartesian plan

coordinate:

polygon 1: $(1,1) - (2,2) - (3,3) - (4,2)$

polygon 2: $(1,0) - (2,3) - (3,2) - (4,1)$

percentual area of overlapping polygons = ?

Thanks so much!! ;-)

$\endgroup$

closed as off-topic by Michael Galuza, user91500, Did, Claude Leibovici, N. F. Taussig Aug 23 '15 at 8:36

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Michael Galuza, user91500, Did, Claude Leibovici, N. F. Taussig
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Did you try plotting these points on a graph? What did you find? $\endgroup$ – David K Aug 23 '15 at 2:58
0
$\begingroup$

Let $$A(1,1),B(3,3),C(4,2),D(1,0),E(2,3),F(4,1).$$

Then, you want to know the overlapping area of $\triangle{ABC}$ and $\triangle{DEF}$.

  • The intersection point $G$ of $AB$ with $ED$ is $(3/2,3/2)$.

  • The intersection point $H$ of $AB$ with $EF$ is $(1/2,1/2)$.

  • The intersection point $I$ of $AC$ with $ED$ is $(11/8,9/8)$.

  • The intersection point $J$ of $AC$ with $EF$ is $(1/4,3/4)$.

Now since the overlapping area is the sum of the areas of $\triangle{GHI}$ and $\triangle{HIJ}$, we have $$\small\frac 12\left|\left(\frac 32-\frac 12\right)\left(\frac 98-\frac 12\right)-\left(\frac 32-\frac 12\right)\left(\frac{11}{8}-\frac{1}{2}\right)\right|+\frac 12\left|\left(\frac{11}{8}-\frac{1}{2}\right)\left(\frac 34-\frac 12\right)-\left(\frac 98-\frac 12\right)\left(\frac 14-\frac 12\right)\right|$$$$=\color{red}{\frac{5}{16}}.$$

$\endgroup$
  • $\begingroup$ excellent .... work with polygons too? $\endgroup$ – Stefano Vet Aug 23 '15 at 8:45
  • $\begingroup$ @StefanoVet: Yes. (but it must need a lot of calculations) $\endgroup$ – mathlove Aug 23 '15 at 10:07

Not the answer you're looking for? Browse other questions tagged or ask your own question.