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

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 $a 2 - b 2 = (a - b)(a + b)$ 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 $S = .mw-parser-output .sfrac{white-space:nowrap}.mw-parser-output .sfrac.tion,.mw-parser-output .sfrac .tion{display:inline-block;vertical-align:-0.5em;font-size:85%;text-align:center}.mw-parser-output .sfrac .num{display:block;line-height:1em;margin:0.0em 0.1em;border-bottom:1px solid}.mw-parser-output .sfrac .den{display:block;line-height:1em;margin:0.1em 0.1em}.mw-parser-output .sr-only{border:0;clip:rect(0,0,0,0);clip-path:polygon(0px 0px,0px 0px,0px 0px);height:1px;margin:-1px;overflow:hidden;padding:0;position:absolute;width:1px}⁠p + q + r + s/2⁠$ 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.^[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:

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 $cos 2 (180° - θ) = cos 2 θ$ .) 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^[2]

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

^ Hess, Albrecht, "A highway from Heron to Brahmagupta", Forum Geometricorum 12 (2012), 191–192.
^ J. L. Coolidge, "A Historically Interesting Formula for the Area of a Quadrilateral", American Mathematical Monthly, 46 (1939) pp. 345-347.
^ Maley, F. Miller; Robbins, David P.; Roskies, Julie (2005). "On the areas of cyclic and semicyclic polygons". Advances in Applied Mathematics. 34 (4): 669–689. arXiv:math/0407300. doi:10.1016/j.aam.2004.09.008. S2CID 119565975.

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.

[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". Advances in Applied Mathematics. 34 (4): 669–689. arXiv:math/0407300. doi:10.1016/j.aam.2004.09.008. S2CID 119565975.

[1]

[2]

[3]