Area of triangle with vertex $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by :

$$\frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 & y_3 & 1 \end{vmatrix}$$

In this determinant if we take all the coordinates as rational numbers, we will never get an irrational number as an answer.

Does that mean a triangle's area can't ever be irrational if its coordinates are rational? (because I don't think so) Also,is there anything even remotely similar in 3d to the shoelace formula?


Yes indeed, a direct consequence of the shoelace formula is that a triangle (convex polygon) having all its vertices with rational coordinates has a rational area. It is a nice technique for showing that there is no equilateral triangle in $\mathbb{Z}\times\mathbb{Z}$, too.

  • $\begingroup$ Beautifully simple formula :). I was always splitting polygons in triangles to compute their area, but this seems to be much more efficient. $\endgroup$ – Surb Oct 30 '15 at 16:30
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    $\begingroup$ is there anything even remotely similar in 3d or higher to the shoelace formula? $\endgroup$ – hyportnex Oct 30 '15 at 17:06
  • $\begingroup$ @hyportnex: yes, the (hyper-)volume of a symplex can still be computed through a determinant. The Cayley-Menger determinant (mathworld.wolfram.com/Cayley-MengerDeterminant.html) gives an analogue of Heron's formula in dimension $3$. $\endgroup$ – Jack D'Aurizio Oct 30 '15 at 17:33

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