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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.
| | It is really painful to have hemorrhoids. This can be a big hindrance for the daily routine. You cannot do all the things that you utilize to do due to the pain that you are going from.<br><br>Later I did find out he had hemorrhoidectomy operation. At the time I wondered when surgery was the number one [http://hemorrhoidtreatmentfix.com/external-hemorrhoids external hemorrhoids] for him. I didn't learn anything about hemorrhoids back then.<br><br>There are many treatments that may be used for hemorrhoid. The first and the most well known is the cream plus ointment. These are to be rubbed onto the affected piece of the anus. It helps to soothe the already inflamed blood vessels along with a momentary relief is accomplished. There is a relaxation of the tissues of the rectal column so far the hemorrhoid is not thus much bulged. If there is a bulge yet, the pain relief can not do thus much to aid.<br><br>Until recent years, most people unfortunate enough to be suffering the pain, swelling, itching plus bleeding of their hemorrhoid symptoms were left without hope of ever being healed. They had no choice however to use the same older hemorrhoids cream, hemorrhoid ointment, gels plus even suppositories. Many individuals turned to painful operation that wasn't even a permanent aid.<br><br>Thankfully, there are persons inside this world with curious minds whom do not accept the status quo. Folks whom would very reach the root of the problem rather of placing up with a temporary answer, only to have it re-appear later.<br><br>The surgery procedures commonly include scalpel, this really is normally done by cutting away the swelling plus closing up the wounds. With this procedure you're required to stay in the hospital after the process so which you are provided several painkillers. Normally you will feel the severe pain whenever the anesthesia wears off.<br><br>Walter plus I drifted apart inside the following years and I developed some minor hemorrhoid difficulties me. Every time my hemorrhoids flared-up it brought back memories of my friend Walter and his agony. I prayed that my minor flare-ups wouldn't ever cause operation. I don't understand if I may have created it from recognizing what I had enjoyed before. |
| == Formula ==
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| 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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It is really painful to have hemorrhoids. This can be a big hindrance for the daily routine. You cannot do all the things that you utilize to do due to the pain that you are going from.
Later I did find out he had hemorrhoidectomy operation. At the time I wondered when surgery was the number one external hemorrhoids for him. I didn't learn anything about hemorrhoids back then.
There are many treatments that may be used for hemorrhoid. The first and the most well known is the cream plus ointment. These are to be rubbed onto the affected piece of the anus. It helps to soothe the already inflamed blood vessels along with a momentary relief is accomplished. There is a relaxation of the tissues of the rectal column so far the hemorrhoid is not thus much bulged. If there is a bulge yet, the pain relief can not do thus much to aid.
Until recent years, most people unfortunate enough to be suffering the pain, swelling, itching plus bleeding of their hemorrhoid symptoms were left without hope of ever being healed. They had no choice however to use the same older hemorrhoids cream, hemorrhoid ointment, gels plus even suppositories. Many individuals turned to painful operation that wasn't even a permanent aid.
Thankfully, there are persons inside this world with curious minds whom do not accept the status quo. Folks whom would very reach the root of the problem rather of placing up with a temporary answer, only to have it re-appear later.
The surgery procedures commonly include scalpel, this really is normally done by cutting away the swelling plus closing up the wounds. With this procedure you're required to stay in the hospital after the process so which you are provided several painkillers. Normally you will feel the severe pain whenever the anesthesia wears off.
Walter plus I drifted apart inside the following years and I developed some minor hemorrhoid difficulties me. Every time my hemorrhoids flared-up it brought back memories of my friend Walter and his agony. I prayed that my minor flare-ups wouldn't ever cause operation. I don't understand if I may have created it from recognizing what I had enjoyed before.