Brahmagupta's Formula
Brahmagupta's Formula is a formula for determining the area of a cyclic quadrilateral given only the four side lengths, given as follows: ![\[[ABCD] = \sqrt{(s-a)(s-b)(s-c)(s-d)}\]](http://latex.artofproblemsolving.com/0/e/0/0e099a7c73655f6d04e511a6f51ff63698705e01.png)
where 
, 
, 
, 
are the four side lengths and 
.
Proofs
If we draw 
, we find that ![$[ABCD]=\frac{ab\sin B}{2}+\frac{cd\sin D}{2}=\frac{ab\sin B+cd\sin D}{2}$](http://latex.artofproblemsolving.com/9/a/7/9a78a4a4d7823c156744a4ab766dcec6dee7ad68.png)
. Since 
, 
. Hence, ![$[ABCD]=\frac{\sin B(ab+cd)}{2}$](http://latex.artofproblemsolving.com/1/6/8/1684bb2cb4c40be745baf43185216216c383eff1.png)
. Multiplying by 2 and squaring, we get:
![\[4[ABCD]^2=\sin^2 B(ab+cd)^2\]](http://latex.artofproblemsolving.com/4/9/2/492dd3aa826df33663d6a06fc9d8aabfa05a755a.png)
Substituting 
results in
![\[4[ABCD]^2=(1-\cos^2B)(ab+cd)^2=(ab+cd)^2-\cos^2B(ab+cd)^2\]](http://latex.artofproblemsolving.com/1/f/2/1f28be489b766c55eae6df424d410139c5eb3df5.png)
By the Law of Cosines, 
. 
, so a little rearranging gives
![\[2\cos B(ab+cd)=a^2+b^2-c^2-d^2\]](http://latex.artofproblemsolving.com/7/4/9/749c4e5212d14409e90c0ab2ce8f25a5a809f281.png)
![\[4[ABCD]^2=(ab+cd)^2-\frac{1}{4}(a^2+b^2-c^2-d^2)^2\]](http://latex.artofproblemsolving.com/b/c/e/bcec630c2ff8130c1a61d3c01188d931bdd856ab.png)
![\[16[ABCD]^2=4(ab+cd)^2-(a^2+b^2-c^2-d^2)^2\]](http://latex.artofproblemsolving.com/9/1/f/91f36e4160b82996311964620d4914b61bb5c9e4.png)
![\[16[ABCD]^2=(2(ab+cd)+(a^2+b^2-c^2-d^2))(2(ab+cd)-(a^2+b^2-c^2-d^2))\]](http://latex.artofproblemsolving.com/c/2/2/c22c67c3a85044b92ad940c0aefd4f63ae091e80.png)
![\[16[ABCD]^2=(a^2+2ab+b^2-c^2+2cd-d^2)(-a^2+2ab-b^2+c^2+2cd+d^2)\]](http://latex.artofproblemsolving.com/8/8/2/8826dbf9ed70539e9715150f323d75aaa5921e26.png)
![\[16[ABCD]^2=((a+b)^2-(c-d)^2)((c+d)^2-(a-b)^2)\]](http://latex.artofproblemsolving.com/5/8/7/58740b6212d38b06f97975f3789b42d7d5f18488.png)
![\[16[ABCD]^2=(a+b+c-d)(a+b-c+d)(c+d+a-b)(c+d-a+b)\]](http://latex.artofproblemsolving.com/f/0/c/f0cebe0db5acd07a2a774e4485ab80dbb825576d.png)
![\[16[ABCD]^2=16(s-a)(s-b)(s-c)(s-d)\]](http://latex.artofproblemsolving.com/f/9/0/f90de0123e7e0b19af70a51fdb63dbf701933a68.png)
![\[[ABCD]=\sqrt{(s-a)(s-b)(s-c)(s-d)}\]](http://latex.artofproblemsolving.com/e/b/7/eb7b6e6945de987f981c05f35be9b81105e76f09.png)
Similar formulas
Bretschneider's formula gives a formula for the area of a non-cyclic quadrilateral given only the side lengths; applying Ptolemy's Theorem to Bretschneider's.
Brahmagupta's formula reduces to Heron's formula by setting the side length 
.
A similar formula which Brahmagupta derived for the area of a general quadrilateral is
![\[[ABCD]^2=(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\left({\frac{B+D}{2}}\right)\]](http://latex.artofproblemsolving.com/1/e/3/1e3b37e140500efe11d17f0fdbdbec578480a2f6.png)
![\[[ABCD]=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\left({\frac{B+D}{2}}\right)}\]](http://latex.artofproblemsolving.com/2/f/3/2f36523352161ef6d906a49bf80ea44b5b181dac.png)
where 
is the semiperimeter of the quadrilateral. What happens when the quadrilateral is cyclic?
Problems
Intermediate
is a cyclic quadrilateral that has an inscribed circle. The diagonals of intersect at 
. If 
and 
then the area of the inscribed circle of can be expressed as 
, where 
and 
are relatively prime positive integers. Determine 
. (Source)
is a cyclic quadrilateral that has an inscribed circle. The diagonals of 
. If 
and 
then the area of the inscribed circle of 
, where 
and 
are relatively prime positive integers. Determine 
. (- Quadrilateral with side lengths
is inscribed in a circle. The area interior to the circle but exterior to the quadrilateral can be written in the form 
where 
and 
are positive integers such that 
and have no common prime factor. What is 
(Source)
is inscribed in a circle. The area interior to the circle but exterior to the quadrilateral can be written in the form 
where 
and 
(
