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.
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.

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