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:


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

Equivalently, the area of the cyclic quadrilateral is equal to


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

which can be seen by making the substitutions:

$$ $$ $$ $$

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


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

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

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.