Cyclic Quadrilateral Area Calculator (Brahmagupta's formula)
Written by Thierno Sadou Diallo, formula verified per our methodology • Last checked on 9/10/2026
Brahmagupta's formula gives the area of a cyclic quadrilateral (whose four vertices lie on a single circle) from its four sides a, b, c, d: Area = √((s−a)(s−b)(s−c)(s−d)), with s the semi-perimeter. For a rectangle with sides 3 and 4, the calculated area is exactly 12, as expected.
Explanation
A quadrilateral is called cyclic when its four vertices can all be placed on a single circle — this is not the case for all quadrilaterals (a non-square rhombus, for example, is not cyclic), but it is the case for all rectangles, all squares, and many other four-sided shapes. Brahmagupta's formula, discovered by the Indian mathematician of the same name in the 7th century, calculates the area of this quadrilateral from the length of its four sides alone — with no need to know its angles or the coordinates of its vertices, which makes it particularly useful in field surveying, where side lengths are often directly measurable while the precise corner coordinates are not always. This formula directly generalizes Heron's formula for the area of a triangle: a triangle is in fact always cyclic (three points always define a single circle containing them all), and if you consider a triangle as a quadrilateral whose fourth side has zero length, Brahmagupta's formula reduces exactly to Heron's — a 3-4-5 triangle treated as a quadrilateral (3,4,5,0) gives exactly the same area of 6 by both formulas. For a quadrilateral that is NOT cyclic, this formula does not apply: the quadrilateral area from coordinates calculator, which works for any simple quadrilateral whatever its shape, is then the appropriate tool, provided the exact positions of the four vertices are known rather than only their sides.
Example: rectangle 3 by 4
Inputs
Sides: a=3, b=4, c=3, d=4.
Calculation
s = (3+4+3+4) ÷ 2 = 7. Area = √((7−3)(7−4)(7−3)(7−4)) = √(4×3×4×3) = √144 = 12.
Result
The area of this rectangle is exactly 12 — the expected result, since a 3×4 rectangle does have area 3×4=12, and a rectangle is always a cyclic quadrilateral.
Frequently asked questions
How do you know whether my quadrilateral is cyclic before using this formula?
Some shapes are always cyclic by construction: any rectangle, any square, and more generally any quadrilateral whose opposite angles are supplementary (they sum to 180°). Beyond these simple cases, there is no way to check from the side lengths alone: four given side lengths can generally form several different quadrilaterals of different shapes, only one of which is cyclic — this formula then calculates the area of THAT specific cyclic quadrilateral, not that of another non-cyclic quadrilateral sharing the same side lengths.
Why does this formula give exactly Heron's formula when a side is zero?
Because a quadrilateral with a zero-length side is actually a disguised triangle: the two vertices connected by that zero side coincide, which brings the figure back to three distinct vertices, hence a triangle. Since a triangle is always cyclic by nature, this special case falls exactly back on Heron's formula, a nice illustration that Brahmagupta's formula is a genuine generalization of Heron's, not a mere structural coincidence.
What happens if the four sides cannot form any valid cyclic quadrilateral?
The formula requires each of the four terms (s−a), (s−b), (s−c), and (s−d) to stay strictly positive — a condition directly related to the generalized quadrilateral inequality, which requires that no side can exceed the sum of the other three. If this condition is not met, the expression under the square root becomes negative, which means no valid cyclic quadrilateral can exist with those four given side lengths, whatever their order.