# Difference between revisions of "Brahmagupta's Formula"

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− | == | + | ==Proofs== |

If we draw <math>AC</math>, we find that <math>[ABCD]=\frac{ab\sin B}{2}+\frac{cd\sin D}{2}=\frac{ab\sin B+cd\sin D}{2}</math>. Since <math>B+D=180^\circ</math>, <math>\sin B=\sin D</math>. Hence, <math>[ABCD]=\frac{\sin B(ab+cd)}{2}</math>. Multiplying by 2 and squaring, we get: | If we draw <math>AC</math>, we find that <math>[ABCD]=\frac{ab\sin B}{2}+\frac{cd\sin D}{2}=\frac{ab\sin B+cd\sin D}{2}</math>. Since <math>B+D=180^\circ</math>, <math>\sin B=\sin D</math>. Hence, <math>[ABCD]=\frac{\sin B(ab+cd)}{2}</math>. Multiplying by 2 and squaring, we get: | ||

− | <cmath>4[ABCD] | + | <cmath>4[ABCD]^2=\sin^2 B(ab+cd)^2</cmath> |

Substituting <math>\sin^2B=1-\cos^2B</math> results in | Substituting <math>\sin^2B=1-\cos^2B</math> results in | ||

<cmath>4[ABCD]^2=(1-\cos^2B)(ab+cd)^2=(ab+cd)^2-\cos^2B(ab+cd)^2</cmath> | <cmath>4[ABCD]^2=(1-\cos^2B)(ab+cd)^2=(ab+cd)^2-\cos^2B(ab+cd)^2</cmath> | ||

− | By the Law of Cosines, <math>a^2+b^2-2ab\cos B=c^2+d^2-2cd\cos D</math>. <math>\cos B=-\cos D</math>, so a little rearranging gives | + | By the [[Law of Cosines]], <math>a^2+b^2-2ab\cos B=c^2+d^2-2cd\cos D</math>. <math>\cos B=-\cos D</math>, so a little rearranging gives |

<cmath>2\cos B(ab+cd)=a^2+b^2-c^2-d^2</cmath> | <cmath>2\cos B(ab+cd)=a^2+b^2-c^2-d^2</cmath> | ||

<cmath>4[ABCD]^2=(ab+cd)^2-\frac{1}{4}(a^2+b^2-c^2-d^2)^2</cmath> | <cmath>4[ABCD]^2=(ab+cd)^2-\frac{1}{4}(a^2+b^2-c^2-d^2)^2</cmath> | ||

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<cmath>16[ABCD]^2=(a^2+2ab+b^2-c^2+2cd-d^2)(-a^2+2ab-b^2+c^2+2cd+d^2)</cmath> | <cmath>16[ABCD]^2=(a^2+2ab+b^2-c^2+2cd-d^2)(-a^2+2ab-b^2+c^2+2cd+d^2)</cmath> | ||

<cmath>16[ABCD]^2=((a+b)^2-(c-d)^2)((c+d)^2-(a-b)^2)</cmath> | <cmath>16[ABCD]^2=((a+b)^2-(c-d)^2)((c+d)^2-(a-b)^2)</cmath> | ||

− | <cmath>16[ABCD]^2=(a+b+c-d)(a+b-c+d)(c+d+a-b)(c+d-b | + | <cmath>16[ABCD]^2=(a+b+c-d)(a+b-c+d)(c+d+a-b)(c+d-a+b)</cmath> |

<cmath>16[ABCD]^2=16(s-a)(s-b)(s-c)(s-d)</cmath> | <cmath>16[ABCD]^2=16(s-a)(s-b)(s-c)(s-d)</cmath> | ||

<cmath>[ABCD]=\sqrt{(s-a)(s-b)(s-c)(s-d)}</cmath> | <cmath>[ABCD]=\sqrt{(s-a)(s-b)(s-c)(s-d)}</cmath> | ||

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Brahmagupta's formula reduces to [[Heron's formula]] by setting the side length <math>{d}=0</math>. | Brahmagupta's formula reduces to [[Heron's formula]] by setting the side length <math>{d}=0</math>. | ||

+ | |||

+ | A similar formula which Brahmagupta derived for the area of a general quadrilateral is | ||

+ | <cmath>[ABCD]^2=(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\left({\frac{B+D}{2}}\right)</cmath> | ||

+ | <cmath>[ABCD]=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\left({\frac{B+D}{2}}\right)}</cmath> | ||

+ | where <math>s=\frac{a+b+c+d}{2}</math> is the [[semiperimeter]] of the quadrilateral. What happens when the quadrilateral is cyclic? | ||

+ | |||

+ | == Problems == | ||

+ | === Intermediate === | ||

+ | *<math>ABCD</math> is a cyclic quadrilateral that has an inscribed circle. The diagonals of <math>ABCD</math> intersect at <math>P</math>. If <math>AB = 1, CD = 4,</math> and <math>BP : DP = 3 : 8,</math> then the area of the inscribed circle of <math>ABCD</math> can be expressed as <math>\frac{p\pi}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Determine <math>p + q</math>. ([[Mock AIME 3 Pre 2005 Problems/Problem 7|Source]]) | ||

[[Category:Geometry]] | [[Category:Geometry]] | ||

− | |||

[[Category:Theorems]] | [[Category:Theorems]] |

## Latest revision as of 15:57, 10 March 2022

**Brahmagupta's Formula** is a formula for determining the area of a cyclic quadrilateral given only the four side lengths.

## Definition

Given a cyclic quadrilateral with side lengths , , , , the area can be found as:

where is the semiperimeter of the quadrilateral.

## Proofs

If we draw , we find that . Since , . Hence, . Multiplying by 2 and squaring, we get: Substituting results in By the Law of Cosines, . , so a little rearranging gives

## Similar formulas

Bretschneider's formula gives a formula for the area of a non-cyclic quadrilateral given only the side lengths; applying Ptolemy's Theorem to Bretschneider's formula reduces it to Brahmagupta's formula.

Brahmagupta's formula reduces to Heron's formula by setting the side length .

A similar formula which Brahmagupta derived for the area of a general quadrilateral is where is the semiperimeter of the quadrilateral. What happens when the quadrilateral is cyclic?

## Problems

### Intermediate

- is a cyclic quadrilateral that has an inscribed circle. The diagonals of intersect at . If and then the area of the inscribed circle of can be expressed as , where and are relatively prime positive integers. Determine . (Source)