Brahmagupta's formula

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 K of a convex cyclic quadrilateral whose sides have lengths a, b, c, d as

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

where s, the semiperimeter, is defined to be

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

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

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

Another equivalent version is

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

Proof

Diagram for reference

Trigonometric proof

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

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

But since □ABCD is a cyclic quadrilateral, ∠DAB = 180° − ∠DCB. Hence sin A = sin C. Therefore,

$K={\frac {1}{2}}pq\sin A+{\frac {1}{2}}rs\sin A$

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

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

(using the trigonometric identity).

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

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

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

$(pq+rs)\cos A={\frac {1}{2}}(p^{2}+q^{2}-r^{2}-s^{2}).$

Substituting this in the equation for the area,

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

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

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

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

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

$16K^{2}=[(r+s)^{2}-(p-q)^{2}][(p+q)^{2}-(r-s)^{2}]$

that can be factored again into

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

Introducing the semiperimeter S = ⁠p + q + r + s/2⁠ yields

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

Taking the square root, we get

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

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:

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

where θ 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° − θ. Since cos(180° − θ) = −cos θ, we have cos²(180° − θ) = cos² θ.) 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, θ is 90°, whence the term

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

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

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

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.

External links

• A geometric proof from Sam Vandervelde.
• Brahmagupta's formula at ProofWiki
• Weisstein, Eric W. "Brahmagupta's Formula". MathWorld.

This article incorporates material from proof of Brahmagupta's formula on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.