When did brahmgupta originate cyclic quardulateral formula?
Answers
Brahmagupta's formula gives the area K of a cyclic quadrilateral whose sides have lengths a, b, c, d as
{\displaystyle K={\sqrt {(s-a)(s-b)(s-c)(s-d)}}}K={\sqrt {(s-a)(s-b)(s-c)(s-d)}}
where s, the semiperimeter, is defined to be
{\displaystyle s={\frac {a+b+c+d}{2}}.}s={\frac {a+b+c+d}{2}}.
This formula generalizes Heron's formula for the area of a triangle. A triangle may be regarded as a quadrilateral with one side of length zero. From this perspective, as d approaches zero, a cyclic quadrilateral converges into a cyclic triangle (all triangles are cyclic), and Brahmagupta's formula simplifies to Heron's formula.
If the semiperimeter is not Used, Brahmagupta's formula is
{\displaystyle K={\frac {1}{4}}{\sqrt {(-a+b+c+d)(a-b+c+d)(a+b-c+d)(a+b+c-d)}}.}K={\frac {1}{4}}{\sqrt {(-a+b+c+d)(a-b+c+d)(a+b-c+d)(a+b+c-d)}}.
Another equivalent version is
{\displaystyle K={\frac {\sqrt {(a^{2}+b^{2}+c^{2}+d^{2})^{2}+8abcd-2(a^{4}+b^{4}+c^{4}+d^{4})}}{4}}\cdot }K={\frac {{\sqrt {(a^{2}+b^{2}+c^{2}+d^{2})^{2}+8abcd-2(a^{4}+b^{4}+c^{4}+d^{4})}}}{4}}\cdot
Therefore here he originated Brahmagupta originated this cyclic quadrilateral formula in 628 AD.
Here is your answer