# Brahmagupta's Formula

## Theorem

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

$\sqrt {\paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} }$

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 $\triangle ADC$

 $\displaystyle \triangle ABC$ $=$ $\displaystyle \frac 1 2 a b \sin \angle ABC$ $\displaystyle \triangle ADC$ $=$ $\displaystyle \frac 1 2 c d \sin \angle ADC$

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

 $\displaystyle \Area$ $=$ $\displaystyle \frac 1 2 a b \sin \angle ABC + \frac 1 2 c d \sin \angle ABC$ $\displaystyle \leadsto \ \$ $\displaystyle \paren {\Area}^2$ $=$ $\displaystyle \frac 1 4 \paren {a b + c d} \sin^2 \angle ABC$ $\displaystyle \leadsto \ \$ $\displaystyle 4 \paren {\Area}^2$ $=$ $\displaystyle \paren {a b + c d} \paren {1 - \cos^2 \angle ABC}$ Sum of Squares of Sine and Cosine $\displaystyle$ $=$ $\displaystyle \paren {a b + c d} - \cos^2 \angle ABC \paren {a b + c d}$

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

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

From the above:

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

Hence:

$2 \cos \angle ABC \paren {a b + c d} = a^2 + b^2 - c^2 - d^2$

Substituting this in the above equation for the area:

 $\displaystyle 4 \paren {\Area}^2$ $=$ $\displaystyle \paren {a b + c d}^2 - \frac 1 4 \paren {a^2 + b^2 - c^2 - d^2}^2$ $\displaystyle \leadsto \ \$ $\displaystyle 16 \paren {\Area}^2$ $=$ $\displaystyle 4 \paren {a b + c d}^2 - \paren {a^2 + b^2 - c^2 - d^2}^2$

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

Hence, by Difference of Two Squares, it can be written in the form $\paren {x + y} \paren {x - y}$ as:

 $\displaystyle$  $\displaystyle \paren {2 \paren {a b + c d} + a^2 + b^2 - c^2 - d^2} \paren {2 \paren {a b + c d} - a^2 - b^2 + c^2 + d^2}$ $\displaystyle$ $=$ $\displaystyle \paren {\paren {a + b}^2 - \paren {c - d}^2} \paren {\paren {c + d}^2 - \paren {a - b}^2}$ $\displaystyle$ $=$ $\displaystyle \paren {a + b + c - d} \paren {a + b + d - c} \paren {a + c + d - b} \paren {b + c + d - a}$

When we introduce the expression for the semiperimeter:

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

the above converts to:

$16 \paren {\Area}^2 = 16 \paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d}$

Taking the square root:

$\Area = \sqrt {\paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} }$

$\blacksquare$

## 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.