Brahmagupta's Formula

Basic Form
The area of a cyclic quadrilateral with sides of lengths $a, b, c, d$ is:


 * $\sqrt{\left({s - a}\right) \left({s - b}\right) \left({s - c}\right) \left({s - d}\right)}$

where $s$ is the semiperimeter:


 * $\displaystyle s = \frac{a + b + c + d} 2$

Equivalently, the area of the cyclic quadrilateral is equal to


 * $\displaystyle \frac{\sqrt{\left({a^2 + b^2 + c^2 + d^2}\right)^2 + 8abcd - 2 \left({a^4 + b^4 + c^4 + d^4}\right)}} {4}$

which can be seen by making the substitutions:

Generalized Version
For a general quadrilateral, the area is given by:


 * $\displaystyle \sqrt{\left({s - a}\right) \left({s - b}\right) \left({s - c}\right) \left({s - d}\right) - abcd \cos^2 \theta}$

where $\theta$ is half the sum of two opposite angles.

It follows from this fact that the area of a cyclic quadrilateral is the maximum possible area for any quadrilateral with the given side lengths.

Proof of Basic Form
Let $ABCD$ be a cyclic quadrilateral with sides $a, b, c, d$.


 * CyclicQuadrilateral.png

Area of $ABCD$ = Area of $\triangle ABC$ + Area of $\triangle ADC$

From the corollary to Area of a Triangle in Terms of Side and Altitude, we have:

From Opposite Angles in Cyclic Quadrilateral, $\angle ABC + \angle ADC$ equals two right angles, that is, are supplementary.

Hence we have:

This leads to:

Applying the Law of Cosines for $\triangle ABC$ and $\triangle ADC$ and equating the expressions for side $AC$, we have


 * $a^2 + b^2 - 2ab \cos \angle ABC = c^2 + d^2 - 2cd \cos \angle ADC$

From the above, we have $\cos \angle ABC = -\cos \angle ADC$.

Hence:


 * $2 \cos \angle ABC \left({ab + cd}\right) = a^2 + b^2 - c^2 - d^2$

Substituting this in the above eqn for the area:

This is of the form $x^2 - y^2$.

Hence, by Difference of Two Squares, it can be written in the form $(x+y)(x-y)$ as:

When we introduce the expression for the semiperimeter:
 * $\displaystyle s = \frac {a + b + c + d} 2$

the above converts to:
 * $16 \left({\text {Area}}\right)^2 = 16 \left({s - a}\right) \left({s - b}\right) \left({s - c}\right) \left({s - d}\right)$

Taking the square root, we get:


 * $\text {Area} = \sqrt{\left({s - a}\right) \left({s - b}\right) \left({s - c}\right) \left({s - d}\right)}$

Also see

 * This formula is a generalization of Heron's Formula for the area of a triangle, which can be obtained from this by setting $d = 0$.


 * The relationship between the general and extended form of Brahmagupta's formula is similar to how the Law of Cosines extends Pythagoras's Theorem.