# Brahmagupta's Fomula

**Brahmagupta's formula** is a special case of as applied to cyclic quadrilaterals.

Bretschneider's formula states that the area of a quadrilateral is given by \[\Delta^{2} = (s-a)(s-b)(s-c)(s-d) - abcd\cos^{2}\left(\frac{B+D}{2}\right)\] where \(\Delta\) is the area of the quadrilateral, \(B\) and \(D\) are the angles, \(a\), \(b\), \(c\) and \(d\) are the sides of the quadrilateral, and \(s\) is the semiperimeter of the quadrilateral, given by \(s=\frac{a+b+c+d}{2}\).

For a cyclic quadrilateral, we know that the sum of the opposite angles is \(180^{\circ}\). Hence we have \(B+D=180^{\circ}\). So the simplified version know as Brahmagupta's formula is given by:

## Brahmagupta's formula

\(\Delta^{2} = (s-a)(s-b)(s-c)(s-d)\).

Given a cyclic quadrilateral with side lengths \(a,b,c,d\), the area \(\Delta\) can be found as:

\[ \Delta =\sqrt{(s-a)(s-b)(s-c)(s-d)}\]

Alternatively without using the semiperimeter, we can use

\[ \Delta = \frac{1}{4} \sqrt{(-a+b+c+d)(a-b+c+d)(a+b-c+d)(a+b+c-d)} \]

## Proof

Assume

A,B,C and Dto be the vertices of the cyclic quadrilateral andp,q,r,sbe the lengths of sidesAD,AB,BC and CD. \[Area\quad of\quad ABCD=Area\quad of\quad \Delta ADB+Area\quad of\quad \Delta BDC\\ =\frac { pq \sin A }{ 2 } +\frac { rs \sin C }{ 2 } \]Since, the quadrilateral is cyclic, so

SinA=SinC\[A=\frac { pqsinA }{ 2 } +\frac { rssinA }{ 2 } \\ { A }^{ 2 }=\frac { 1 }{ 4 } { \left( pq+rs \right) }^{ 2 }\sin ^{ 2 }{ A } \\ 4A^{ 2 }={ \left( pq+rs \right) }^{ 2 }\left( 1-\cos ^{ 2 }{ A } \right) \\ 4A^{ 2 }={ \left( pq+rs \right) }^{ 2 }-{ \left( pq+rs \right) }^{ 2 }\left( \cos ^{ 2 }{ A } \right) \]Solving for common side

DB, in triangleADB & BDC, using laws of cosines , we get \[{ p }^{ 2 }+{ q }^{ 2 }-2pqcosA={ r }^{ 2 }+{ s }^{ 2 }-2rscosC\]As angles

A and Care supplementary ,cosA=-cosCand \[2(pq+rs)cosA={ p }^{ 2 }+{ q }^{ 2 }-{ r }^{ 2 }-{ s }^{ 2 }\]Substituting this in the equation of area, we get \[16{ A }^{ 2 }={ 4(pq+rs) }^{ 2 }-{ ({ p }^{ 2 }+{ q }^{ 2 }-{ r }^{ 2 }-{ s }^{ 2 }) }^{ 2 }\]

The R.H.S. of the equation is of the form \({ a }^{ 2 }-{ b }^{ 2 }\) , so is can be written as \[(2(pq+rs)-({ p }^{ 2 }+{ q }^{ 2 }-{ r }^{ 2 }-{ s }^{ 2 }))(2(pq+rs)+({ p }^{ 2 }+{ q }^{ 2 }-{ r }^{ 2 }-{ s }^{ 2 }))\\ =\left[ { \left( r+s \right) }^{ 2 }-{ \left( p-q \right) }^{ 2 } \right] \left[ { \left( p+q \right) }^{ 2 }-{ \left( r-s \right) }^{ 2 } \right] \\ =(p+q+r-s)(p+q-r+s)(p-q+r+s)(-p+q+r+s)\]

So, the area of cyclic quadrilateral with sides

a,b,c and dcan be written as \[A=\frac { 1 }{ 4 } \sqrt { (a+b+c-d)(a+b-c+d)(a-b+c+d)(-a+b+c+d) } \]

## See Also

**Cite as:**Brahmagupta's Fomula.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/brahmaguptas-fomula/