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:


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

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


 * BrahmaguptasFormula.png

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

From Area of Triangle in Terms of Two Sides and Angle:

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:

This leads to:

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:

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:

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:


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

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.


 * Bretschneider's Formula, which extends this result to the general quadrilateral.


 * Area of Quadrilateral with Given Sides is Greatest when Quadrilateral is Cyclic