# Brahmagupta's Formula

## Theorem

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:

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

### Corollary

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

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

## Proof

Let $ABCD$ be a cyclic quadrilateral with sides $a, b, c, d$.

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

 $\displaystyle \triangle ABC$ $=$ $\displaystyle \frac 1 2 ab \sin \angle ABC$ $\displaystyle \triangle ADC$ $=$ $\displaystyle \frac 1 2 cd \sin \angle ADC$

From Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles, $\angle ABC + \angle ADC$ equals two right angles, that is, are supplementary.

Hence we have:

 $\displaystyle \sin \angle ABC$ $=$ $\displaystyle \sin \angle ADC$ Sine and Cosine of Supplementary Angles $\displaystyle \cos \angle ABC$ $=$ $\displaystyle -\cos \angle ADC$ Sine and Cosine of Supplementary Angles

 $\displaystyle \text {Area}$ $=$ $\displaystyle \frac 1 2 ab \sin \angle ABC + \frac 1 2 cd \sin \angle ABC$ $\displaystyle \implies \ \$ $\displaystyle \left({\text {Area} }\right)^2$ $=$ $\displaystyle \frac 1 4 \left({ab + cd}\right) \sin^2 \angle ABC$ $\displaystyle \implies \ \$ $\displaystyle 4 \left({\text {Area} }\right)^2$ $=$ $\displaystyle \left({ab + cd}\right) \left({1 - \cos^2 \angle ABC}\right)$ Sum of Squares of Sine and Cosine $\displaystyle$ $=$ $\displaystyle \left({ab + cd}\right) - \cos^2 \angle ABC \left({ab + cd}\right)$

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

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

From the above:

$\cos \angle ABC = -\cos \angle ADC$

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:

 $\displaystyle 4 \left({\text {Area} }\right)^2$ $=$ $\displaystyle \left({ab + cd}\right)^2 - \frac 1 4 \left({a^2 + b^2 - c^2 - d^2}\right)^2$ $\displaystyle \implies \ \$ $\displaystyle 16 \left({\text {Area} }\right)^2$ $=$ $\displaystyle 4 \left({ab + cd}\right)^2 - \left({a^2 + b^2 - c^2 - d^2}\right)^2$

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:

 $\displaystyle$  $\displaystyle \left({2 \left({ab + cd}\right) + a^2 + b^2 - c^2 - d^2}\right) \left({2 \left({ab + cd}\right) - a^2 - b^2 + c^2 + d^2}\right)$ $\displaystyle$ $=$ $\displaystyle \left({\left({a + b}\right)^2 - \left({c - d}\right)^2}\right) \left({\left({c + d}\right)^2 - \left({a - b}\right)^2}\right)$ $\displaystyle$ $=$ $\displaystyle \left({a + b + c - d}\right) \left({a + b + d - c}\right) \left({a + c + d - b}\right) \left({b + c + d - a}\right)$

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:

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

$\blacksquare$

## 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 $d = 0$.

## Source of Name

This entry was named for Brahmagupta.