Brahmagupta's Formula

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Theorem

The area of a cyclic quadrilateral with sides of lengths $a, b, c, d$ is:

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

where $s$ is the semiperimeter:

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


Corollary

The area of a cyclic quadrilateral with sides of lengths $a, b, c, d$ is:

$\dfrac{\sqrt{\left({a^2 + b^2 + c^2 + d^2}\right)^2 + 8 a b c d - 2 \left({a^4 + b^4 + c^4 + d^4}\right)}} 4$


Proof

Let $ABCD$ be a cyclic quadrilateral with sides $a, b, c, d$.

BrahmaguptasFormula.png

Area of $ABCD$ = Area of $\triangle ABC$ + Area of $\triangle ADC$

From Area of Triangle in Terms of Two Sides and Angle:

\(\displaystyle \triangle ABC\) \(=\) \(\displaystyle \frac 1 2 ab \sin \angle ABC\)
\(\displaystyle \triangle ADC\) \(=\) \(\displaystyle \frac 1 2 cd \sin \angle ADC\)

From Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles, $\angle ABC + \angle ADC$ equals two right angles, that is, are supplementary.


Hence we have:

\(\displaystyle \sin \angle ABC\) \(=\) \(\displaystyle \sin \angle ADC\) Sine and Cosine of Supplementary Angles
\(\displaystyle \cos \angle ABC\) \(=\) \(\displaystyle -\cos \angle ADC\) Sine and Cosine of Supplementary Angles


This leads to:

\(\displaystyle \text {Area}\) \(=\) \(\displaystyle \frac 1 2 ab \sin \angle ABC + \frac 1 2 cd \sin \angle ABC\)
\(\displaystyle \implies \ \ \) \(\displaystyle \left({\text {Area} }\right)^2\) \(=\) \(\displaystyle \frac 1 4 \left({ab + cd}\right) \sin^2 \angle ABC\)
\(\displaystyle \implies \ \ \) \(\displaystyle 4 \left({\text {Area} }\right)^2\) \(=\) \(\displaystyle \left({ab + cd}\right) \left({1 - \cos^2 \angle ABC}\right)\) Sum of Squares of Sine and Cosine
\(\displaystyle \) \(=\) \(\displaystyle \left({ab + cd}\right) - \cos^2 \angle ABC \left({ab + cd}\right)\)


Applying the Law of Cosines for $\triangle ABC$ and $\triangle ADC$ and equating the expressions for side $AC$:

$a^2 + b^2 - 2a b \cos \angle ABC = c^2 + d^2 - 2cd \cos \angle ADC$


From the above:

$\cos \angle ABC = -\cos \angle ADC$

Hence:

$2 \cos \angle ABC \left({ab + cd}\right) = a^2 + b^2 - c^2 - d^2$


Substituting this in the above equation for the area:

\(\displaystyle 4 \left({\text {Area} }\right)^2\) \(=\) \(\displaystyle \left({ab + cd}\right)^2 - \frac 1 4 \left({a^2 + b^2 - c^2 - d^2}\right)^2\)
\(\displaystyle \implies \ \ \) \(\displaystyle 16 \left({\text {Area} }\right)^2\) \(=\) \(\displaystyle 4 \left({ab + cd}\right)^2 - \left({a^2 + b^2 - c^2 - d^2}\right)^2\)


This is of the form $x^2 - y^2$.

Hence, by Difference of Two Squares, it can be written in the form $(x+y)(x-y)$ as:


\(\displaystyle \) \(\) \(\displaystyle \left({2 \left({ab + cd}\right) + a^2 + b^2 - c^2 - d^2}\right) \left({2 \left({ab + cd}\right) - a^2 - b^2 + c^2 + d^2}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({\left({a + b}\right)^2 - \left({c - d}\right)^2}\right) \left({\left({c + d}\right)^2 - \left({a - b}\right)^2}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({a + b + c - d}\right) \left({a + b + d - c}\right) \left({a + c + d - b}\right) \left({b + c + d - a}\right)\)


When we introduce the expression for the semiperimeter:

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

the above converts to:

$16 \left({\text {Area}}\right)^2 = 16 \left({s - a}\right) \left({s - b}\right) \left({s - c}\right) \left({s - d}\right)$


Taking the square root:

$\text {Area} = \sqrt{\left({s - a}\right) \left({s - b}\right) \left({s - c}\right) \left({s - d}\right)}$

$\blacksquare$


Also see

  • This formula is a generalization of Heron's Formula for the area of a triangle, which can be obtained from this by setting $d = 0$.


Source of Name

This entry was named for Brahmagupta.


Sources