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| In [[Euclidean geometry]], '''[[Brahmagupta]]'s formula''' finds the [[area]] of any [[cyclic quadrilateral]] (one that can be inscribed in a circle) given the lengths of the sides.
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| == Formula == | |
| Brahmagupta's formula gives the area ''A'' of a [[cyclic quadrilateral]] whose sides have lengths ''a'', ''b'', ''c'', ''d'' as
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| : <math>A=\sqrt{(s-a)(s-b)(s-c)(s-d)}</math>
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| where ''s'', the [[semiperimeter]], is defined to be
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| : <math>s=\frac{a+b+c+d}{2}.</math>
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| 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.
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| If the semiperimeter is not used, Brahmagupta's formula is
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| : <math>K=\frac{1}{4}\sqrt{(-a+b+c+d)(a-b+c+d)(a+b-c+d)(a+b+c-d)}.</math>
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| Another equivalent version is
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| : <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>
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| Brahmagupta's formula may be seen as a formula in the half-lengths of the sides, but it also gives the area as a formula in the altitudes from the center to the sides, although if the quadrilateral does not contain the center, the altitude to the longest side must be taken as negative.{{citation needed|date=September 2012}}
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| == Proof ==
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| [[Image:Brahmaguptas formula.svg|frame|right|Diagram for reference]]
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| ===Trigonometric proof===
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| Here the notations in the figure to the right are used. Area ''K'' of the cyclic quadrilateral = Area of <math>\triangle ADB</math> + Area of <math>\triangle BDC</math>
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| :<math>= \frac{1}{2}pq\sin A + \frac{1}{2}rs\sin C.</math>
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| But since <math>ABCD</math> is a cyclic quadrilateral, <math>\angle DAB = 180^\circ - \angle DCB.</math> Hence <math>\sin A = \sin C.</math> Therefore,
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| :<math>K = \frac{1}{2}pq\sin A + \frac{1}{2}rs\sin A</math>
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| :<math>K^2 = \frac{1}{4} (pq + rs)^2 \sin^2 A</math>
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| :<math>4K^2 = (pq + rs)^2 (1 - \cos^2 A) = (pq + rs)^2 - (pq + rs)^2 \cos^2 A.\,</math>
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| Solving for common side ''DB'', in <math>\triangle</math>''ADB'' and <math>\triangle </math>''BDC'', the [[law of cosines]] gives
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| :<math>p^2 + q^2 - 2pq\cos A = r^2 + s^2 - 2rs\cos C. \,</math>
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| Substituting <math>\cos C = -\cos A</math> (since angles <math>A</math> and <math>C</math> are [[Supplementary angles|supplementary]]) and rearranging, we have
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| :<math>2 (pq + rs) \cos A = p^2 + q^2 - r^2 - s^2. \,</math> | |
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| Substituting this in the equation for the area,
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| :<math>4K^2 = (pq + rs)^2 - \frac{1}{4}(p^2 + q^2 - r^2 - s^2)^2</math>
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| :<math>16K^2 = 4(pq + rs)^2 - (p^2 + q^2 - r^2 - s^2)^2.</math>
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| The right-hand side is of the form <math>a^2-b^2 = (a-b)(a+b)</math> and hence can be written as
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| :<math>[2(pq + rs) - p^2 - q^2 + r^2 +s^2][2(pq + rs) + p^2 + q^2 -r^2 - s^2] \,</math>
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| which, upon rearranging the terms in the square brackets, yields
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| :<math>= [ (r+s)^2 - (p-q)^2 ][ (p+q)^2 - (r-s)^2 ] \,</math>
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| :<math>= (q+r+s-p)(p+r+s-q)(p+q+s-r)(p+q+r-s). \,</math>
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| Introducing the semiperimeter <math>S = \frac{p+q+r+s}{2},</math>
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| :<math>16K^2 = 16(S-p)(S-q)(S-r)(S-s). \,</math>
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| Taking the square root, we get
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| :<math>K = \sqrt{(S-p)(S-q)(S-r)(S-s)}.</math>
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| ===Non-trigonometric proof===
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| An alternative, non-trigonometric proof utilizes two applications of Heron's triangle area formula on similar triangles.<ref>Hess, Albrecht, "A highway from Heron to Brahmagupta", ''Forum Geometricorum'' 12 (2012), 191–192.</ref>
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| == Extension to non-cyclic quadrilaterals ==
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| In the case of non-cyclic quadrilaterals, Brahmagupta's formula can be extended by considering the measures of two opposite angles of the quadrilateral:
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| : <math>K=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\theta}</math>
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| where θ is ''half'' the sum of two opposite angles. (The choice of which pair of opposite angles is irrelevant: if the other two angles are taken, half ''their'' sum is the supplement of θ. Since cos(180° − θ) = −cosθ, we have cos<sup>2</sup>(180° − θ) = cos<sup>2</sup>θ.) This more general formula is known as [[Bretschneider's formula]].
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| It is a property of [[cyclic quadrilateral]]s (and ultimately of [[inscribed angle]]s) that opposite angles of a quadrilateral sum to 180°. Consequently, in the case of an inscribed quadrilateral, θ = 90°, whence the term
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| :<math>abcd\cos^2\theta=abcd\cos^2 \left(90^\circ\right)=abcd\cdot0=0, \,</math>
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| 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.
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| A related formula, which was proved by [[Julian Coolidge|Coolidge]], also gives the area of a general convex quadrilateral. It is<ref>J. L. Coolidge, "A Historically Interesting Formula for the Area of a Quadrilateral", ''American Mathematical Monthly'', '''46''' (1939) pp. 345-347.</ref>
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| : <math>K=\sqrt{(s-a)(s-b)(s-c)(s-d)-\textstyle{1\over4}(ac+bd+pq)(ac+bd-pq)}\,</math>
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| where ''p'' and ''q'' are the lengths of the diagonals of the quadrilateral. In a [[cyclic quadrilateral]], <math>pq=ac+bd</math> according to [[Ptolemy's theorem]], and the formula of Coolidge reduces to Brahmagupta's formula.
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| == Related theorems ==
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| * [[Heron's formula]] for the area of a [[triangle]] is the special case obtained by taking ''d'' = 0.
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| * The relationship between the general and extended form of Brahmagupta's formula is similar to how the [[law of cosines]] extends the [[Pythagorean theorem]].
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| == References ==
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| {{reflist}}
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| ==External links==
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| *{{mathworld|urlname=BrahmaguptasFormula|title=Brahmagupta's Formula}}
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| {{PlanetMath attribution|id=3594|title=proof of Brahmagupta's formula}}
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| {{DEFAULTSORT:Brahmagupta's Formula}}
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| [[Category:Brahmagupta]]
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| [[Category:Quadrilaterals]]
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| [[Category:Area]]
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| [[Category:Euclidean plane geometry]]
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| [[Category:Theorems in geometry]]
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