In Euclidean geometry, Brahmagupta's formula, named after the 7th century Indian mathematician, is used to find the area of any convex 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.

Formulation

Brahmagupta's formula gives the area of a convex cyclic quadrilateral whose sides have lengths , , , as

: <math>K=\sqrt{(s-a)(s-b)(s-c)(s-d)}</math>

where , the semiperimeter, is defined to be

: <math>s=\frac{a+b+c+d}{2}.</math>

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 (or any one side) 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

: <math>K=\frac{1}{4}\sqrt{(-a+b+c+d)(a-b+c+d)(a+b-c+d)(a+b+c-d)}.</math>

Another equivalent version is

: <math>K=\frac{\sqrt{(a^2+b^2+c^2+d^2)^2+8abcd-2(a^4+b^4+c^4+d^4){4}\cdot</math>

Proof

400x400px|Diagram for reference|thumb

Trigonometric proof

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

:<math>K = \frac{1}{2}pq\sin A + \frac{1}{2}rs\sin C.</math>

But since is a cyclic quadrilateral, . Hence . Therefore,

:<math>K = \frac{1}{2}pq\sin A + \frac{1}{2}rs\sin A</math>

:<math>K^2 = \frac{1}{4} (pq + rs)^2 \sin^2 A</math>

:<math>4K^2 = (pq + rs)^2 (1 - \cos^2 A) = (pq + rs)^2 - ((pq + rs)\cos A)^2</math>

(using the trigonometric identity).

Solving for common side , in and , the law of cosines gives

:<math>p^2 + q^2 - 2pq\cos A = r^2 + s^2 - 2rs\cos C.</math>

Substituting (since angles and are supplementary) and rearranging, we have

:<math>(pq + rs) \cos A = \frac{1}{2}(p^2 + q^2 - r^2 - s^2).</math>

Substituting this in the equation for the area,

:<math>4K^2 = (pq + rs)^2 - \frac{1}{4}(p^2 + q^2 - r^2 - s^2)^2</math>

:<math>16K^2 = 4(pq + rs)^2 - (p^2 + q^2 - r^2 - s^2)^2.</math>

The right-hand side is of the form and hence can be written as

:<math>[2(pq + rs)) - p^2 - q^2 + r^2 +s^2][2(pq + rs) + p^2 + q^2 -r^2 - s^2] </math>

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

:<math>16K^2= [ (r+s)^2 - (p-q)^2 ][ (p+q)^2 - (r-s)^2 ] </math>

that can be factored again into

:<math>16K^2=(q+r+s-p)(p+r+s-q)(p+q+s-r)(p+q+r-s). </math>

Introducing the semiperimeter yields

:<math>16K^2 = 16(S-p)(S-q)(S-r)(S-s). </math>

Taking the square root, we get

:<math>K = \sqrt{(S-p)(S-q)(S-r)(S-s)}.</math>

Non-trigonometric proof

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

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:

: <math>K=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\theta}</math>

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

:<math>abcd\cos^2\theta=abcd\cos^2 \left(90^\circ\right)=abcd\cdot0=0, </math>

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

: <math>K=\sqrt{(s-a)(s-b)(s-c)(s-d)-\textstyle{1\over4}(ac+bd+pq)(ac+bd-pq)}</math>

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.

  • Heron's formula for the area of a triangle is the special case obtained by taking .
  • 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.

References

  • A geometric proof from Sam Vandervelde.