# Brahmagupta's formula

In Euclidean geometry, Brahmagupta's formula finds the area of any cyclic quadrilateral (one that can be inscribed in a circle) given the lengths of the sides.

## Formula

Brahmagupta's formula gives the area A of a cyclic quadrilateral whose sides have lengths a, b, c, d as

$A=\sqrt{(s-a)(s-b)(s-c)(s-d)}$

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 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

Diagram for reference

### Trigonometric proof

Here the notations in the figure to the right are used. Area K of the cyclic quadrilateral = Area of $\triangle ADB$ + Area of $\triangle BDC$

$= \frac{1}{2}pq\sin A + \frac{1}{2}rs\sin C.$

But since $ABCD$ is a cyclic quadrilateral, $\angle DAB = 180^\circ - \angle DCB.$ Hence $\sin A = \sin C.$ Therefore,

$K = \frac{1}{2}pq\sin A + \frac{1}{2}rs\sin A$
$K^2 = \frac{1}{4} (pq + rs)^2 \sin^2 A$
$4K^2 = (pq + rs)^2 (1 - \cos^2 A) = (pq + rs)^2 - (pq + rs)^2 \cos^2 A.\,$

Solving for common side DB, in $\triangle$ADB and $\triangle$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

$2 (pq + rs) \cos A = p^2 + q^2 - r^2 - s^2. \,$

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

$= [ (r+s)^2 - (p-q)^2 ][ (p+q)^2 - (r-s)^2 ] \,$
$= (q+r+s-p)(p+r+s-q)(p+q+s-r)(p+q+r-s). \,$

Introducing the semiperimeter $S = \frac{p+q+r+s}{2},$

$16K^2 = 16(S-p)(S-q)(S-r)(S-s). \,$

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.[1]

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 two opposite angles. (The choice of which pair of opposite angles is irrelevant: if the other two angles are taken, half their sum is the supplement of θ. Since cos(180° − θ) = −cosθ, we have cos2(180° − θ) = cos2θ.) 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 = 90°, whence the term

$abcd\cos^2\theta=abcd\cos^2 \left(90^\circ\right)=abcd\cdot0=0, \,$

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[2]

$K=\sqrt{(s-a)(s-b)(s-c)(s-d)-\textstyle{1\over4}(ac+bd+pq)(ac+bd-pq)}\,$

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.

## References

1. ^ Hess, Albrecht, "A highway from Heron to Brahmagupta", Forum Geometricorum 12 (2012), 191–192.
2. ^ J. L. Coolidge, "A Historically Interesting Formula for the Area of a Quadrilateral", American Mathematical Monthly, 46 (1939) pp. 345-347.