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Area of a (planar) quadrilateral using two opposite sides and four angles

Posted 1 month ago

I guessed the following area formula for a (planar) quadrilateral using two opposite sides and four angles:(When $a=AB$ and $c=CD$,)

$S = \frac{a^2}{2(\cot A + \cot B)} + \frac{c^2}{2(\cot C + \cot D)}$

enter image description here

This formula can be applied to the quadrilateral satisfying $A+B \ne \pi$ and $C+D \ne \pi$, including convex, concave, and even self-intersecting cases.

I'd like to know the applicable case , inapplicable case and reasons for the validity of those cases. Are there any known papers or more elegant ways to derive this formula?I guess there are many ways to explain those reasons. It greatly resembles $S = \frac{1}{2}(ab \sin B + cd \sin D)$. It seems that $b$ and $d$ can be determined using $a, c, A, B, C, D$.

POSTED BY: Ryo Takayama
2 Replies
Posted 1 month ago

This follows cleanly from the law of sines on each triangle formed by a diagonal. b and d are indeed determined (up to the constraint A+B+C+D=2π), and the formula breaks down when A+B=π or C+D=π because the diagonal length becomes indeterminate (parallel opposite sides, essentially a degenerate split).

POSTED BY: Connor Hayes

I just played around with your formula with this little program to check it on random quadrilaterals:

area[{a_, b_, c_, d_}] :=
  RegionMeasure[Polygon[{a, b, c, d}]];
formula[{a_, b_, c_, d_}] :=
  EuclideanDistance[a, b]^2/(
   2 (Cot[VectorAngle[b - a, d - a]] + 
      Cot[VectorAngle[c - b, a - b]])) +
   EuclideanDistance[d, c]^2/(
   2 (Cot[VectorAngle[b - c, d - c]] + Cot[VectorAngle[a - d, c - d]]));
Graphics[Polygon[rndmQuadrilateral =
   RandomInteger[{0, 10}, {4, 2}]], Frame -> True]
area[rndmQuadrilateral]
formula[rndmQuadrilateral]
POSTED BY: Gianluca Gorni
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