In Euclidean geometry, Brahmagupta's formula, named after the 7th century Indian mathematician, is used to find the area of any cyclic quadrilateral (one that can be inscribed in a circle) given the lengths of the sides. Its generalized version, Bretschneider's formula, can be used with non-cyclic quadrilateral.Heron's formula can be thought as a special case of the Brahmagupta's formula for triangles.
Brahmagupta's formula gives the area of a cyclic quadrilateral whose sides have lengths,,, as
K=\sqrt{(s-a)(s-b)(s-c)(s-d)}
where, the semiperimeter, is defined to be
| s= | 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 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
| K= | 1 |
| 4 |
\sqrt{(-a+b+c+d)(a-b+c+d)(a+b-c+d)(a+b+c-d)}.
Another equivalent version is
| K= | \sqrt{(a2+b2+c2+d2)2+8abcd-2(a4+b4+c4+d4) |
Here the notations in the figure to the right are used. The area of the cyclic quadrilateral equals the sum of the areas of and :
K=
| 1 | |
| 2 |
pq\sinA+
| 1 | |
| 2 |
rs\sinC.
But since is a cyclic quadrilateral, . Hence . Therefore,
K=
| 1 | |
| 2 |
pq\sinA+
| 1 | |
| 2 |
rs\sinA
K2=
| 1 | |
| 4 |
(pq+rs)2\sin2A
4K2=(pq+rs)2(1-\cos2A)=(pq+rs)2-((pq+rs)\cosA)2
(using the trigonometric identity).
Solving for common side, in and, the law of cosines gives
p2+q2-2pq\cosA=r2+s2-2rs\cosC.
Substituting (since angles and are supplementary) and rearranging, we have
(pq+rs)\cosA=
| 1 | |
| 2 |
(p2+q2-r2-s2).
Substituting this in the equation for the area,
4K2=(pq+rs)2-
| 1 | |
| 4 |
(p2+q2-r2-s2)2
16K2=4(pq+rs)2-(p2+q2-r2-s2)2.
The right-hand side is of the form and hence can be written as
[2(pq+rs))-p2-q2+r2+s2][2(pq+rs)+p2+q2-r2-s2]
which, upon rearranging the terms in the square brackets, yields
16K2=[(r+s)2-(p-q)2][(p+q)2-(r-s)2]
that can be factored again into
16K2=(q+r+s-p)(p+r+s-q)(p+q+s-r)(p+q+r-s).
Introducing the semiperimeter yields
16K2=16(S-p)(S-q)(S-r)(S-s).
Taking the square root, we get
K=\sqrt{(S-p)(S-q)(S-r)(S-s)}.
An alternative, non-trigonometric proof utilizes two applications of Heron's triangle area formula on similar triangles.[1]
In the case of non-cyclic quadrilaterals, Brahmagupta's formula can be extended by considering the measures of two opposite angles of the quadrilateral:
K=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos2\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 . Since, we have .) 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
abcd\cos2\theta=abcd\cos2\left(90\circ\right)=abcd ⋅ 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]
K=\sqrt{(s-a)(s-b)(s-c)(s-d)-style{1\over4}(ac+bd+pq)(ac+bd-pq)}
where and are the lengths of the diagonals of the quadrilateral. In a cyclic quadrilateral, according to Ptolemy's theorem, and the formula of Coolidge reduces to Brahmagupta's formula.