Formulae for calculating the area of a quadrilateral non-cyclic

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Calculating the area of a non-cyclic quadrilateral requires six values: the lengths of the four edges and the lengths of the two diagonals. A formula that only uses the four edge lengths is insufficient, as demonstrated by examples where the same edge lengths can yield different areas based on the shape's configuration. Testing with software like GeoGebra confirms that the simplified formula does not produce accurate results. The discussion emphasizes that both diagonals are necessary to account for the quadrilateral's shape. Accurate area calculation must consider the arrangement of the vertices and the diagonals involved.
Bruno Tolentino
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I want knows if a formula for calculate the area of a quadrilateral non-cyclic needs of just four values (the values of the four edges) or if is necesseray 6 values (the values of the four edges MORE os values of the two diagonals)?

This formula needs of 6 values (a,b,c,d,p,q):
1a04d408f073f30a1edd47b9f4501566.png


OBS: s = (a+b+c+d)/2

And this formula needs of just 4 (a,b,c,d):
3227d258f1f0437e6cad6f6000c9c479.png


But, I tested this second formula in the geogebra and it don't works...

Sources:
http://en.wikipedia.org/wiki/Trapezoid#Area
http://en.wikipedia.org/wiki/Quadrilateral#Non-trigonometric_formulas
https://en.wikipedia.org/wiki/Bretschneider's_formula#Related_formulas
 
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The four edges are obviously not sufficient. Start with a square (side length=1) - area =1. Now take a pair of diagonally opposite corners and move them toward each other, while keeping the side lengths constant. When they meet, the resultant quadrilateral area = 0.
 
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And 4 sides and 1 diagonal is not sufficient either. Consider a shape with sides 1, 1, 0.8, 0.8 in that order. Now the 0.8 v can be inside, or outside (convex or concave) with the area different.
 
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Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...