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How quickly we forget - basic trig. Calculate the area of a polygon
Calculate area of a figure based on vertices

I saw a formula in a book, $$\mathrm{area}=\frac{1}{2}\left|\sum_{i}(x_iy_{i-1}-x_{i-1}y_i)\right|.$$ Where $x_iy_i$ are the vertices of the polygon. Since it was an exercise in the book (no proof) I would very much like to see a proof, or maybe an outline? I can't come up with anything. This is no homework, I just got very curious and wanted to know.

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marked as duplicate by joriki, t.b., Américo Tavares, Rudy the Reindeer, J. M. Aug 4 '12 at 19:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Its very hard to figure out the answer, without knowing whether you are looking at a regular or irregular polygon . But I think that typing " derive area of polygon " in Google may fetch you lots of links. Prefer Google before moving your question to Math.SE. Thank you. – Iyengar Aug 4 '12 at 18:06
@t.b. : You are exactly right , But OP could have added a reference , about the book in which he has seen the formula, in order to facilitate others . – Iyengar Aug 4 '12 at 18:08
@t.b.: Please post the link as an answer, so I can accept! Thanks! – Satyas Aug 4 '12 at 18:09
up vote 5 down vote accepted

On OP's request:

This is called the Shoelace Formula or the Shoelace method and it works as long as your polygon is simple (non-selfintersecting). There is a nice explanation on the Wikipedia page and it refers (among other things) to the article

Bart Braden, The Surveyor’s Area Formula, The College Mathematics Journal, September 1986, Volume 17, Number 4, pp. 326–337

which looks pretty nice at first glance.

Some related questions on this site:

  1. How quickly we forget - basic trig. Calculate the area of a polygon
  2. How to calculate the area of a polygon?
  3. Calculate area of a figure based on vertices
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I salute your sincerity in keeping Community wiki, which shows that you are not interested in reputation . – Iyengar Aug 4 '12 at 18:25

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