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The shoelace formula found here or here tells you how to calculate the area of any polygon given its coordinates.

The second link I mentioned gives a proof of it, but it is a bit beyond my level of comprehension. Could anyone try to simplify the proof (or provide their own) to a level up to and including single variable calculus?

1110101001
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    Take a look at this article, which provides a very accessible proof. You don't need much background other than a familiarity with basic vector geometry and the basic notion of a determinant (just for $2\times 2$ matrices). – EuYu Nov 29 '13 at 03:23
  • @EuYu One part I didn't understand was why you need to go in a counter-clockwise order. Isn't the area of the large polygon the sum of the areas of the triangles? Why would it matter which points you choose to take the area? – 1110101001 Nov 29 '13 at 20:01
  • Each triangle's area breaks up into three terms, one for each edge. The terms are positive or negative depending on the orientations of the triangles. Since the diagonals of the triangulation are not a part of the original polygon you need them to cancel. The only way to do that is for all the triangles to be oriented the same way. By convention we choose counterclockwise as positive, so all the triangles are oriented that way. – EuYu Nov 29 '13 at 22:29
  • Another way to think about it is in terms of signed areas. The area is positive if oriented counterclockwise and negative if oriented clockwise. The triangles need to be oriented the same way otherwise the areas have different sign and cancel each other. – EuYu Nov 29 '13 at 22:32
  • Updated Link: http://steiner.math.nthu.edu.tw/disk5/js/cardioid/12.pdf – 1110101001 Sep 15 '16 at 05:09

2 Answers2

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A simple induction proof can be based on this fact [*]:

Every simple polygon has a diagonal, that is, a segment joining two non-consecutive vertices that lies totally inside the polygon.

Take the polygon and split it along a diagonal. The area of the polygon is the sum of the areas of the two parts. By induction, these areas are given by the formula. When you combine them, the terms corresponding to the diagonal cancel.

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lhf
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Here's a really good response I found on YahooAnswers a while ago:

The proof in the link is sheer madness.

For this clockwise order to make sense, you need a point $O$ inside the polygon so that the angles form $(OA_i, OA_{i+1})$ and $(OA_n,OA_1)$ be all positive.

Then the formula is just adding up the areas of the triangles $OA_iA_{i+1}$ and $OA_nA_1$.

So all you need is area of $ OA_1A_2 = \dfrac {b_2a_1-b_1a_2}{2} $, which is elementary.

So here's my summary of the above answer:

  1. I don't quite agree that the AoPS proof is sheer madness $-$ it is pretty rigorous and I like it, in the long run.
  2. Pick a point $O$ inside the polygon so that we have positive angles.
  3. You want the sum of the areas $\left[\triangle OA_iA_{i+1}\right]+\left[\triangle OA_nA_1\right]$ over all points $A_i$.
  4. Using the area of a triangle, we have $ \left[\triangle OA_1 A_2\right] = \dfrac {b_2a_1 - b_1 a_2}{2} $. Summing over all indices, you get the desired result.

$ \blacksquare $

Ahaan S. Rungta
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