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  "body": "\n## Statement\n\nLet $ABCD$ be a convex [[cyclic quadrilateral|0RCP929Y1M]] in the Euclidean plane with side lengths $a$, $b$, $c$, $d$.\nLet $s = \\frac{1}{2}(a + b + c + d)$ be its semiperimeter.\nThen the area $K$ of $ABCD$ is\n$$K = \\sqrt{(s - a)(s - b)(s - c)(s - d)}.$$\n\n## Remarks\n\nThe diagonal $AC$ splits $ABCD$ into two triangles, and the opposite angles at $B$ and $D$ add up to $\\pi$ because $ABCD$ is cyclic.\nComputing $AC^2$ by the law of cosines in both triangles and adding the two areas gives the formula after factoring.\nExample: a $3 \\times 4$ rectangle is cyclic, $s = 7$, and $\\sqrt{4 \\cdot 3 \\cdot 4 \\cdot 3} = 12$.\nThe hypothesis that $ABCD$ is cyclic is needed.\nA rhombus with side $1$ and an angle $\\theta \\ne \\pi/2$ has area $\\sin \\theta \u003c 1$, but the formula would give $1$.\nLetting $d = 0$ turns the formula into [[Heron's formula|0FRBZ2X4HJ]] for a triangle.\n\n## Explanations\n\n### L1 human\n\nFor a quadrilateral whose corners all lie on one circle, the area depends only on the four side lengths.\nBrahmagupta's formula computes it directly from them.\nIt looks just like Heron's formula for triangles, with one more factor.\n\n### L2 human\n\nFor a general quadrilateral the four sides do not fix the area, since the shape can flex.\nAmong all quadrilaterals with given sides, the cyclic one has the largest area, and Brahmagupta's formula gives that maximum.\nBretschneider's formula extends it to all convex quadrilaterals by an extra term that involves the sum of two opposite angles.\n",
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