Brahmagupta's Formula

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Brahmagupta's Formula

Basic form

In its basic and easiest-to-remember form, Brahmagupta's formula gives the area of a cyclic quadrilateral whose sides have lengths a, b, c, d as

$\sqrt{(s-a)(s-b)(s-c)(s-d)}$

where s, the semiperimeter, is determined by

$s=\frac{a+b+c+d}{2}.$

The area of a cyclic quadrilateral is the maximum possible area for any quadrilateral with the given side lengths.

Proof of Brahmagupta's formula

Diagram for reference

Area of the cyclic quadrilateral = Area of $\triangle ADB$ + Area of $\triangle BDC$

$= \frac{1}{2}pq\sin A + \frac{1}{2}rs\sin C.$

But since $A B C D$ is a cyclic quadrilateral, $\angle DAB = 180^\circ - \angle DCB.$ Hence $sin A = sin C .$ Therefore

$\mbox{Area} = \frac{1}{2}pq\sin A + \frac{1}{2}rs\sin A$

$(\mbox{Area})^2 = \frac{1}{4}\sin^2 A (pq + rs)^2$

$4(\mbox{Area})^2 = (1 - \cos^2 A)(pq + rs)^2 \,$

$4(\mbox{Area})^2 = (pq + rs)^2 - cos^2 A (pq + rs)^2. \,$

Applying law of cosines for $\triangle ADB$ and $\triangle BDC$ and equating the expressions for side $D B,$ we have

$p^2 + q^2 - 2pq\cos A = r^2 + s^2 - 2rs\cos C. \,$

Substituting $cos C = - cos A$ (since angles $A$ and $C$ are supplementary) and rearranging, we have

$2\cos A (pq + rs) = p^2 + q^2 - r^2 - s^2. \,$

Substituting this in the equation for area,

$4(\mbox{Area})^2 = (pq + rs)^2 - \frac{1}{4}(p^2 + q^2 - r^2 - s^2)^2$

$16(\mbox{Area})^2 = 4(pq + rs)^2 - (p^2 + q^2 - r^2 - s^2)^2, \,$

which is of the form $a 2 - b 2$ and hence can be written in the form $(a + b)(a - b)$ as

$(2(pq + rs) + p^2 + q^2 -r^2 - s^2)(2(pq + rs) - p^2 - q^2 + r^2 +s^2) \,$

$= ( (p+q)^2 - (r-s)^2 )( (r+s)^2 - (p-q)^2 ) \,$

$= (p+q+r-s)(p+q+s-r)(p+r+s-q)(q+r+s-p). \,$

Introducing $T = \frac{p+q+r+s}{2},$

$16(\mbox{Area})^2 = 16(T-p)(T-q)(T-r)(T-s). \,$

Taking square root, we get

$\mbox{Area} = \sqrt{(T-p)(T-q)(T-r)(T-s)}.$

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:

$\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\theta}$

where θ is half the sum of two opposite angles. (The pair is irrelevant: if the other two angles are taken, half their sum is the supplement of θ. Since cos(180° − θ) = −cosθ, we have cos²(180° − θ) = cos²θ.)

This more general formula is sometimes known as Bretschneider's formula, but according to MathWorld is apparently due to Coolidge in this form, Bretschneider's expression having been

$\sqrt{(s-a)(s-b)(s-c)(s-d)-\textstyle{1\over4}(ac+bd+pq)(ac+bd-pq)}\,$

where p and q are the lengths of the diagonals of the quadrilateral.

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 = 90°, whence the term

$abcd\cos^2\theta=abcd\cos^2 \left(90^\circ\right)=abcd\cdot0=0, \,$

giving the basic form of 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.

External links

MathWorld: Brahmagupta's formula

This article is licensed under the GNU Free Documentation License. It uses material from Wikipedia Encyclopedia article "Brahmagupta's Formula"

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Contents

Basic form

Proof of Brahmagupta's formula

Extension to non-cyclic quadrilaterals

Related theorems

External links