Brahmagupta's Formula
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Brahmagupta's Formula

In Euclidean geometry, Brahmagupta's formula is used to find the area of any cyclic quadrilateral (one that can be inscribed in a circle) given the lengths of the sides.

## Formula

Brahmagupta's formula gives the area K of a cyclic quadrilateral whose sides have lengths a, b, c, d as

${\displaystyle K={\sqrt {(s-a)(s-b)(s-c)(s-d)}}}$

where s, the semiperimeter, is defined to be

${\displaystyle s={\frac {a+b+c+d}{2}}.}$

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

${\displaystyle K={\frac {1}{4}}{\sqrt {(-a+b+c+d)(a-b+c+d)(a+b-c+d)(a+b+c-d)}}.}$

Another equivalent version is

${\displaystyle K={\frac {\sqrt {(a^{2}+b^{2}+c^{2}+d^{2})^{2}+8abcd-2(a^{4}+b^{4}+c^{4}+d^{4})}}{4}}\cdot }$

## Proof

Diagram for reference

### Trigonometric proof

Here the notations in the figure to the right are used. The area K of the cyclic quadrilateral equals the sum of the areas of ?ADB and ?BDC:

${\displaystyle ={\frac {1}{2}}pq\sin A+{\frac {1}{2}}rs\sin C.}$

But since ABCD is a cyclic quadrilateral, ?DAB = 180° - ?DCB. Hence sin A = sin C. Therefore,

${\displaystyle K={\frac {1}{2}}pq\sin A+{\frac {1}{2}}rs\sin A}$
${\displaystyle K^{2}={\frac {1}{4}}(pq+rs)^{2}\sin ^{2}A}$
${\displaystyle 4K^{2}=(pq+rs)^{2}(1-\cos ^{2}A)=(pq+rs)^{2}-(pq+rs)^{2}\cos ^{2}A.}$

Solving for common side DB, in ?ADB and ?BDC, the law of cosines gives

${\displaystyle p^{2}+q^{2}-2pq\cos A=r^{2}+s^{2}-2rs\cos C.}$

Substituting cos C = -cos A (since angles A and C are supplementary) and rearranging, we have

${\displaystyle 2(pq+rs)\cos A=p^{2}+q^{2}-r^{2}-s^{2}.}$

Substituting this in the equation for the area,

${\displaystyle 4K^{2}=(pq+rs)^{2}-{\frac {1}{4}}(p^{2}+q^{2}-r^{2}-s^{2})^{2}}$
${\displaystyle 16K^{2}=4(pq+rs)^{2}-(p^{2}+q^{2}-r^{2}-s^{2})^{2}.}$

The right-hand side is of the form a2 - b2 = (a - b)(a + b) and hence can be written as

${\displaystyle [2(pq+rs)-p^{2}-q^{2}+r^{2}+s^{2}][2(pq+rs)+p^{2}+q^{2}-r^{2}-s^{2}]}$

which, upon rearranging the terms in the square brackets, yields

${\displaystyle =[(r+s)^{2}-(p-q)^{2}][(p+q)^{2}-(r-s)^{2}]}$
${\displaystyle =(q+r+s-p)(p+r+s-q)(p+q+s-r)(p+q+r-s).}$

Introducing the semiperimeter S = ,

${\displaystyle 16K^{2}=16(S-p)(S-q)(S-r)(S-s).}$

Taking the square root, we get

${\displaystyle K={\sqrt {(S-p)(S-q)(S-r)(S-s)}}.}$

### Non-trigonometric proof

An alternative, non-trigonometric proof utilizes two applications of Heron's triangle area formula on similar triangles.[1]

## Extension to non-cyclic quadrilaterals

In the case of non-cyclic quadrilaterals, Brahmagupta's formula can be extended by considering the measures of two opposite angles of the quadrilateral:

${\displaystyle K={\sqrt {(s-a)(s-b)(s-c)(s-d)-abcd\cos ^{2}\theta }}}$

where ? is half the sum of any two opposite angles. (The choice of which pair of opposite angles is irrelevant: if the other two angles are taken, half their sum is 180° - ?. Since cos(180° - ?) = -cos ?, we have cos2(180° - ?) = cos2?.) This more general formula is known as Bretschneider's formula.

It is a property of cyclic quadrilaterals (and ultimately of inscribed angles) that opposite angles of a quadrilateral sum to 180°. Consequently, in the case of an inscribed quadrilateral, ? is 90°, whence the term

${\displaystyle abcd\cos ^{2}\theta =abcd\cos ^{2}\left(90^{\circ }\right)=abcd\cdot 0=0,}$

giving the basic form of Brahmagupta's formula. It follows from the latter equation that the area of a cyclic quadrilateral is the maximum possible area for any quadrilateral with the given side lengths.

A related formula, which was proved by Coolidge, also gives the area of a general convex quadrilateral. It is[2]

${\displaystyle K={\sqrt {(s-a)(s-b)(s-c)(s-d)-\textstyle {1 \over 4}(ac+bd+pq)(ac+bd-pq)}}}$

where p and q are the lengths of the diagonals of the quadrilateral. In a cyclic quadrilateral, pq = ac + bd according to Ptolemy's theorem, and the formula of Coolidge reduces to Brahmagupta's formula.

## Related theorems

• Heron's formula for the area of a triangle is the special case obtained by taking d = 0.
• The relationship between the general and extended form of Brahmagupta's formula is similar to how the law of cosines extends the Pythagorean theorem.
• Increasingly complicated closed-form formulas exist for the area of general polygons on circles, as described by Maley et al.[3]

## References

1. ^ Hess, Albrecht, "A highway from Heron to Brahmagupta", Forum Geometricorum 12 (2012), 191-192.
2. ^ J. L. Coolidge, "A Historically Interesting Formula for the Area of a Quadrilateral", American Mathematical Monthly, 46 (1939) pp. 345-347.
3. ^ Maley, F. Miller; Robbins, David P.; Roskies, Julie (2005). "On the areas of cyclic and semicyclic polygons" (PDF). Advances in Applied Mathematics. 34 (4): 669-689. doi:10.1016/j.aam.2004.09.008.

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