Brahmagupta's Formula
Calculate the area of a cyclic quadrilateral using Brahmagupta's formula: A = sqrt((s-a)(s-b)(s-c)(s-d)).
Generalizes Heron's formula.
The Formula
where s = (a + b + c + d) / 2
Brahmagupta's formula calculates the area of a cyclic quadrilateral — a four-sided polygon whose vertices all lie on a single circle. It was discovered by the Indian mathematician and astronomer Brahmagupta in the 7th century (around 628 CE) and published in his work Brahmasphutasiddhanta. The formula is a beautiful generalization of Heron's formula for triangles.
A cyclic quadrilateral has a special property: the sum of its opposite angles equals 180 degrees. Not every quadrilateral is cyclic, but many common shapes are. All rectangles and squares are cyclic (their circumscribed circle has a diameter equal to the diagonal). Isosceles trapezoids are also cyclic. For these shapes, Brahmagupta's formula gives the exact area using only the four side lengths.
The variable s is the semi-perimeter, calculated as half the sum of all four sides. This is directly analogous to the semi-perimeter in Heron's formula. In fact, if you set one side d equal to zero, the quadrilateral degenerates into a triangle, and Brahmagupta's formula reduces exactly to Heron's formula: A = √(s(s−a)(s−b)(s−c)). This elegant connection shows that Brahmagupta's formula is the natural four-sided extension of the triangle area formula.
For a general (non-cyclic) quadrilateral, the formula needs a correction term. The generalized version, known as Bretschneider's formula, includes the two opposite angles: A = √((s−a)(s−b)(s−c)(s−d) − abcd × cos²((α+γ)/2)), where α and γ are opposite angles. For a cyclic quadrilateral, α + γ = 180°, so the cosine term vanishes and the formula simplifies to Brahmagupta's original.
Brahmagupta's work was remarkably advanced for its time. He also developed rules for arithmetic with zero and negative numbers, and his contributions to mathematics and astronomy influenced scholars across Asia and the Arab world for centuries. His area formula remains one of the most elegant results in classical geometry.
Variables
| Symbol | Meaning |
|---|---|
| A | Area of the cyclic quadrilateral (square units) |
| a, b, c, d | Lengths of the four sides (linear units) |
| s | Semi-perimeter: s = (a + b + c + d) / 2 (linear units) |
Example 1
A cyclic quadrilateral has sides of 5, 7, 8, and 10 units. Find its area.
s = (5 + 7 + 8 + 10) / 2 = 30 / 2 = 15
A = √((15−5)(15−7)(15−8)(15−10)) = √(10 × 8 × 7 × 5)
A = √(2800) = √(400 × 7)
A = 20√7 ≈ 52.92 square units
Example 2
A rectangle has sides 6 cm and 8 cm. Verify Brahmagupta's formula gives the correct area. (A rectangle is a cyclic quadrilateral.)
Sides: a = 6, b = 8, c = 6, d = 8. s = (6 + 8 + 6 + 8) / 2 = 14
A = √((14−6)(14−8)(14−6)(14−8)) = √(8 × 6 × 8 × 6)
A = √(2304) = 48
A = 48 cm² (matches the expected 6 × 8 = 48)
When to Use It
Brahmagupta's formula is used when you know the four side lengths of a cyclic quadrilateral and need to find its area.
- Calculating the area of any quadrilateral inscribed in a circle
- Verifying areas of rectangles, squares, and isosceles trapezoids using side lengths alone
- Geometry problems and competitions involving cyclic quadrilaterals
- Surveying and land measurement when four boundary lengths are known and the shape is cyclic
- Studying the mathematical connection between triangle and quadrilateral area formulas
- Architecture and design involving shapes inscribed in circular arcs