Brahmagupta's Formula

Brahmagupta's Formula is a formula for determining the area of a cyclic quadrilateral given only the four side lengths.


Given a cyclic quadrilateral with side lengths ${a}$, ${b}$, ${c}$, ${d}$, the area ${K}$ can be found as:

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

where $s=\frac{a+b+c+d}{2}$ is the semiperimeter of the quadrilateral.


If we draw $AC$, we find that $[ABCD]=\frac{ab\sin B}{2}+\frac{cd\sin D}{2}=\frac{ab\sin B+cd\sin D}{2}$. Since $B+D=180^\circ$, $\sin B=\sin D$. Hence, $[ABCD]=\frac{\sin B(ab+cd)}{2}$. Multiplying by 2 and squaring, we get: \[4[ABCD]^2=\sin^2 B(ab+cd)^2\] Substituting $\sin^2B=1-\cos^2B$ results in \[4[ABCD]^2=(1-\cos^2B)(ab+cd)^2=(ab+cd)^2-\cos^2B(ab+cd)^2\] By the Law of Cosines, $a^2+b^2-2ab\cos B=c^2+d^2-2cd\cos D$. $\cos B=-\cos D$, so a little rearranging gives \[2\cos B(ab+cd)=a^2+b^2-c^2-d^2\] \[4[ABCD]^2=(ab+cd)^2-\frac{1}{4}(a^2+b^2-c^2-d^2)^2\] \[16[ABCD]^2=4(ab+cd)^2-(a^2+b^2-c^2-d^2)^2\] \[16[ABCD]^2=(2(ab+cd)+(a^2+b^2-c^2-d^2))(2(ab+cd)-(a^2+b^2-c^2-d^2))\] \[16[ABCD]^2=(a^2+2ab+b^2-c^2+2cd-d^2)(-a^2+2ab-b^2+c^2+2cd+d^2)\] \[16[ABCD]^2=((a+b)^2-(c-d)^2)((c+d)^2-(a-b)^2)\] \[16[ABCD]^2=(a+b+c-d)(a+b-c+d)(c+d+a-b)(c+d-a+b)\] \[16[ABCD]^2=16(s-a)(s-b)(s-c)(s-d)\] \[[ABCD]=\sqrt{(s-a)(s-b)(s-c)(s-d)}\]

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 formula reduces it to Brahmagupta's formula.

Brahmagupta's formula reduces to Heron's formula by setting the side length ${d}=0$.

A similar formula which Brahmagupta derived for the area of a general quadrilateral is \[[ABCD]^2=(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\left({\frac{B+D}{2}}\right)\] \[[ABCD]=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\left({\frac{B+D}{2}}\right)}\] where $s=\frac{a+b+c+d}{2}$ is the semiperimeter of the quadrilateral. What happens when the quadrilateral is cyclic?



  • $ABCD$ is a cyclic quadrilateral that has an inscribed circle. The diagonals of $ABCD$ intersect at $P$. If $AB = 1, CD = 4,$ and $BP : DP = 3 : 8,$ then the area of the inscribed circle of $ABCD$ can be expressed as $\frac{p\pi}{q}$, where $p$ and $q$ are relatively prime positive integers. Determine $p + q$. (Source)
Invalid username
Login to AoPS