How this calculator works
A quadrilateral is any closed four-sided shape. Its sides and one included angle are enough to determine the rest of the figure: the angle fixes the diagonal between the two sides that form it, and that diagonal splits the quadrilateral into two triangles. The calculator adds the two triangle areas, one from the included angle and one from Heron's formula, to get the total. The angle opposite the included one follows from the law of cosines. In the cyclic mode the four vertices lie on one circle, and Brahmagupta's formula gives the area directly from the four sides. Both modes require the sides to be able to close a quadrilateral, so an impossible combination is rejected rather than producing a meaningless area.
Formula
Sides-angle mode: e = √(a² + b² - 2ab cos θ); area = ½ab sin θ + √(s₂(s₂ - c)(s₂ - d)(s₂ - e)) with s₂ = (c + d + e)/2; opposite angle φ = arccos((c² + d² - e²)/(2cd)). Cyclic mode: area = √((s - a)(s - b)(s - c)(s - d)) with s = (a + b + c + d)/2.
Worked example
Sides 5 m, 4 m, 3 m and 6 m with an 80° angle between the first two give a diagonal of about 5.8356 m. The two triangle areas total about 18.4220 m², and the angle between the 3 m and 6 m sides is about 72.30°. The same four sides as a cyclic quadrilateral enclose √(4 × 5 × 6 × 3) = √360 ≈ 18.9737 m².
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Frequently asked questions
Which angle should I enter?
Enter the angle between sides a and b, the two sides at one end of the diagonal. The label and the units are shown beside the field.
Why is the opposite angle not 180° minus the entered angle?
That relationship holds for a cyclic quadrilateral. A general quadrilateral with the sides and one angle given does not lie on a circle, so the opposite angle must be calculated from the resulting diagonal.
When can I use Brahmagupta's formula?
Only when all four vertices lie on one circle. If you know the quadrilateral is cyclic, the four sides are enough. Otherwise use the sides and angle mode.
What makes four sides impossible?
The longest side must be shorter than the sum of the other three. If it is not, the chain of sides cannot close around any quadrilateral.