Brahmagupta's formula

In Euclidean geometry, Brahmagupta's formula, named after the 7th century Indian mathematician, is used to find the area of any 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 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 $d$ 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


 * $$K=\frac{\sqrt{(a^2+b^2+c^2+d^2)^2+8abcd-2(a^4+b^4+c^4+d^4)}}{4}\cdot$$

Trigonometric proof
Here the notations in the figure to the right are used. The area $K$ of the 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 $a2 − b2 = (a − b)(a + b)$ 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


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

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


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

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.

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.