Brahmagupta's formula

Statement

Let $ABCD$ be a convex cyclic quadrilateral in the Euclidean plane with side lengths $a$ , $b$ , $c$ , $d$ . Let $s = \frac{1}{2}(a + b + c + d)$ be its semiperimeter. Then the area $K$ of $ABCD$ is $K = \sqrt{(s - a)(s - b)(s - c)(s - d)}.$

Remarks

The diagonal $AC$ splits $ABCD$ into two triangles, and the opposite angles at $B$ and $D$ add up to $\pi$ because $ABCD$ is cyclic. Computing $AC^2$ by the law of cosines in both triangles and adding the two areas gives the formula after factoring. Example: a $3 \times 4$ rectangle is cyclic, $s = 7$ , and $\sqrt{4 \cdot 3 \cdot 4 \cdot 3} = 12$ . The hypothesis that $ABCD$ is cyclic is needed. A rhombus with side $1$ and an angle $\theta \ne \pi/2$ has area $\sin \theta < 1$ , but the formula would give $1$ . Letting $d = 0$ turns the formula into Heron's formula for a triangle.

Going deeper

For a student

For a general quadrilateral the four sides do not fix the area, since the shape can flex. Among all quadrilaterals with given sides, the cyclic one has the largest area, and Brahmagupta's formula gives that maximum. Bretschneider's formula extends it to all convex quadrilaterals by an extra term that involves the sum of two opposite angles.

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Cyclic quadrilateral concept
Brahmagupta's formula statement

Tag	`C0NQDQAZ2Y`
URI	`m2://C0NQDQAZ2Y`
Kind	statement, theorem
Writing	W0 stub
MSC	51M04, 51M25
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About	Cyclic quadrilateral
Uses	Cyclic quadrilateral
Same as	Wikidata Q1149348 Wikipedia Brahmagupta's formula