# Brahmagupta's formula

In Euclidean geometry, Brahmagupta's formula, named after the 7th century Indian mathematician, is used to find the area of any convex cyclic quadrilateral (one that can be inscribed in a circle) given the lengths of the sides. Its generalized version, Bretschneider's formula, can be used with non-cyclic quadrilateral. Heron's formula can be thought as a special case of the Brahmagupta's formula for triangles.

## Formulation

Brahmagupta's formula gives the area K of a convex cyclic quadrilateral whose sides have lengths a, b, c, d as
    $K={\sqrt {(s-a)(s-b)(s-c)(s-d)}}$
where s, the semiperimeter, is defined to be
    $s={\frac {a+b+c+d}{2}}.$
This formula generalizes Heron's formula for the area of a triangle. A triangle may be regarded as a quadrilateral with one side of length zero. From this perspective, as d (or any one side) approaches zero, a cyclic quadrilateral converges into a cyclic triangle (all triangles are cyclic), and Brahmagupta's formula simplifies to Heron's formula.
If the semiperimeter is not used, Brahmagupta's formula is
    $K={\frac {1}{4}}{\sqrt {(-a+b+c+d)(a-b+c+d)(a+b-c+d)(a+b+c-d)}}.$
Another equivalent version is
    $K={\frac {\sqrt {(a^{2}+b^{2}+c^{2}+d^{2})^{2}+8abcd-2(a^{4}+b^{4}+c^{4}+d^{4})}}{4}}\cdot$

## Proof
### Trigonometric proof

Here the notations in the figure to the right are used. The area K of the convex cyclic quadrilateral equals the sum of the areas of △ADB and △BDC:
    $K={\frac {1}{2}}pq\sin A+{\frac {1}{2}}rs\sin C.$
But since □ABCD is a cyclic quadrilateral, ∠DAB = 180° − ∠DCB. Hence sin A = sin C. Therefore,
    $K={\frac {1}{2}}pq\sin A+{\frac {1}{2}}rs\sin A$
    $K^{2}={\frac {1}{4}}(pq+rs)^{2}\sin ^{2}A$
    $4K^{2}=(pq+rs)^{2}(1-\cos ^{2}A)=(pq+rs)^{2}-((pq+rs)\cos A)^{2}$
(using the trigonometric identity).
Solving for common side DB, in △ADB and △BDC, the law of cosines gives
    $p^{2}+q^{2}-2pq\cos A=r^{2}+s^{2}-2rs\cos C.$
Substituting cos C = −cos A (since angles A and C are supplementary) and rearranging, we have
    $(pq+rs)\cos A={\frac {1}{2}}(p^{2}+q^{2}-r^{2}-s^{2}).$
Substituting this in the equation for the area,
    $4K^{2}=(pq+rs)^{2}-{\frac {1}{4}}(p^{2}+q^{2}-r^{2}-s^{2})^{2}$
    $16K^{2}=4(pq+rs)^{2}-(p^{2}+q^{2}-r^{2}-s^{2})^{2}.$
The right-hand side is of the form a$^2$ − b$^2$ = (a − b)(a + b) and hence can be written as
    $[2(pq+rs))-p^{2}-q^{2}+r^{2}+s^{2}][2(pq+rs)+p^{2}+q^{2}-r^{2}-s^{2}]$
which, upon rearranging the terms in the square brackets, yields
    $16K^{2}=[(r+s)^{2}-(p-q)^{2}][(p+q)^{2}-(r-s)^{2}]$
that can be factored again into
    $16K^{2}=(q+r+s-p)(p+r+s-q)(p+q+s-r)(p+q+r-s).$
Introducing the semiperimeter S = ⁠p + q + r + s/2⁠ yields
    $16K^{2}=16(S-p)(S-q)(S-r)(S-s).$
Taking the square root, we get
    $K={\sqrt {(S-p)(S-q)(S-r)(S-s)}}.$

### Non-trigonometric proof

An alternative, non-trigonometric proof utilizes two applications of Heron's triangle area formula on similar triangles.

## Extension to non-cyclic quadrilaterals

In the case of non-cyclic quadrilaterals, Brahmagupta's formula can be extended by considering the measures of two opposite angles of the quadrilateral:
    $K={\sqrt {(s-a)(s-b)(s-c)(s-d)-abcd\cos ^{2}\theta }}$
where θ is half the sum of any two opposite angles. (The choice of which pair of opposite angles is irrelevant: if the other two angles are taken, half their sum is 180° − θ. Since cos(180° − θ) = −cos θ, we have cos$^2$(180° − θ) = cos$^2$ θ.) This more general formula is known as Bretschneider's formula.
It is a property of cyclic quadrilaterals (and ultimately of inscribed angles) that opposite angles of a quadrilateral sum to 180°. Consequently, in the case of an inscribed quadrilateral, θ is 90°, whence the term
    $abcd\cos ^{2}\theta =abcd\cos ^{2}\left(90^{\circ }\right)=abcd\cdot 0=0,$
giving the basic form of Brahmagupta's formula. It follows from the latter equation that the area of a cyclic quadrilateral is the maximum possible area for any quadrilateral with the given side lengths.
A related formula, which was proved by Coolidge, also gives the area of a general convex quadrilateral. It is
    $K={\sqrt {(s-a)(s-b)(s-c)(s-d)-\textstyle {1 \over 4}(ac+bd+pq)(ac+bd-pq)}}$
where p and q are the lengths of the diagonals of the quadrilateral. In a cyclic quadrilateral, pq = ac + bd according to Ptolemy's theorem, and the formula of Coolidge reduces to Brahmagupta's formula.

## Related theorems

* Heron's formula for the area of a triangle is the special case obtained by taking d = 0.
* The relationship between the general and extended form of Brahmagupta's formula is similar to how the law of cosines extends the Pythagorean theorem.
* Increasingly complicated closed-form formulas exist for the area of general polygons on circles, as described by Maley et al.