If I have a polygon with the sides given, is it able to calculate the area of it? If yes, how many sides do I need to know at least?

A formula is appreciated.

I know how to get the area with a triangle.

$$ \sqrt[2]{(\frac{1}{2}(a+b+c)\frac{1}{2}(-a+b+c)\frac{1}{2}(a-b+c)\frac{1}{2}(a+b-c)} $$

And I have somehow figured out a way for a trapezoid by splitting the area into two parts - a triangle consisting of two right triangles and a rectangle (area = smaller parallel side times height).

$$ \frac{(b+d)}{4(b-d)}\sqrt[2]{(a+b+c-d)(-a+b+c-d)(a+b-c-d)(-a-b+c+d)} $$

Where b and d are parallel sides.

Maybe it can be achieved by splitting the polygon into (total number of sides - 2) triangles.

Is there a more efficient way to do this?

  • 1
    $\begingroup$ Have a look at this Wikipedia page. $\endgroup$ – Peter Phipps Oct 16 '12 at 14:41
  • $\begingroup$ That's interesting. I never thought there was such a way to figure it out with its perimeter. $\endgroup$ – hello all Oct 17 '12 at 11:14

If only the lengths of the sides are known, the answer is no for more than three sides. Eg a square can be squashed to make an increasingly thin rhombus with the same side length and vanishingly small area. Unless a figure is rigid, you need more than the lengths of the sides to determine the area.

  • $\begingroup$ Good example. By 'more than...' you mean angles, or other elements? $\endgroup$ – hello all Oct 17 '12 at 11:17

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