# Difference between revisions of "Brahmagupta's Formula"

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

## 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)