Brahmagupta's Formula
Contents |
Theorem
Basic Form
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:
- $\displaystyle s = \frac{a + b + c + d} 2$
Equivalently, the area of the cyclic quadrilateral is equal to
- $\displaystyle \frac{\sqrt{\left({a^2 + b^2 + c^2 + d^2}\right)^2 + 8abcd - 2 \left({a^4 + b^4 + c^4 + d^4}\right)}} {4}$
which can be seen by making the substitutions:
| \(\displaystyle \) | \(\displaystyle s - a\) | \(=\) | \(\displaystyle \frac{-a + b + c + d} 2\) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle s - b\) | \(=\) | \(\displaystyle \frac{a - b + c + d} 2\) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle s - c\) | \(=\) | \(\displaystyle \frac{a + b - c + d} 2\) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle s - d\) | \(=\) | \(\displaystyle \frac{a + b + c - d} 2\) | \(\displaystyle \) |
Generalized Version
For a general quadrilateral, the area is given by:
- $\displaystyle \sqrt{\left({s - a}\right) \left({s - b}\right) \left({s - c}\right) \left({s - d}\right) - abcd \ \cos^2 \left({\frac {A + B} 2}\right)}$
where $A$ and $B$ are two opposite angles.
It follows from this fact that the area of a cyclic quadrilateral is the maximum possible area for any quadrilateral with the given side lengths.
Proof
Proof of Basic Form
Let $ABCD$ be a cyclic quadrilateral with sides $a, b, c, d$.
Area of $ABCD$ = Area of $\triangle ABC$ + Area of $\triangle ADC$
From the corollary to Area of a Triangle in Terms of Side and Altitude, we have:
| \(\displaystyle \) | \(\displaystyle \triangle ABC\) | \(=\) | \(\displaystyle \frac 1 2 ab \sin \angle ABC\) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \triangle ADC\) | \(=\) | \(\displaystyle \frac 1 2 cd \sin \angle ADC\) | \(\displaystyle \) |
From Opposite Angles of Cyclic Quadrilateral, $\angle ABC + \angle ADC$ equals two right angles, that is, are supplementary.
Hence we have:
| \(\displaystyle \) | \(\displaystyle \sin \angle ABC\) | \(=\) | \(\displaystyle \sin \angle ADC\) | \(\displaystyle \) | from Sine and Cosine of Supplementary Angles | ||
| \(\displaystyle \) | \(\displaystyle \cos \angle ABC\) | \(=\) | \(\displaystyle -\cos \angle ADC\) | \(\displaystyle \) | from Sine and Cosine of Supplementary Angles |
This leads to:
| \(\displaystyle \) | \(\displaystyle \text {Area}\) | \(=\) | \(\displaystyle \frac 1 2 ab \sin \angle ABC + \frac 1 2 cd \angle ABC\) | \(\displaystyle \) | |||
| \(\displaystyle \implies\) | \(\displaystyle \left({\text {Area} }\right)^2\) | \(=\) | \(\displaystyle \frac 1 4 \left({ab + cd}\right) \sin^2 \angle ABC\) | \(\displaystyle \) | |||
| \(\displaystyle \implies\) | \(\displaystyle 4 \left({\text {Area} }\right)^2\) | \(=\) | \(\displaystyle \left({ab + cd}\right) \left({1 - \cos^2 \angle ABC}\right)\) | \(\displaystyle \) | Sum of Squares of Sine and Cosine | ||
| \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \left({ab + cd}\right) - \cos^2 \angle ABC \left({ab + cd}\right)\) | \(\displaystyle \) |
Applying the Law of Cosines for $\triangle ABC$ and $\triangle ADC$ and equating the expressions for side $AC$, we have
- $a^2 + b^2 - 2ab \cos \angle ABC = c^2 + d^2 - 2cd \cos \angle ADC$
From the above, we have $\cos \angle ABC = -\cos \angle ADC$.
Hence:
- $2 \cos \angle ABC \left({ab + cd}\right) = a^2 + b^2 - c^2 - d^2$
Substituting this in the above eqn for the area:
| \(\displaystyle \) | \(\displaystyle 4 \left({\text {Area} }\right)^2\) | \(=\) | \(\displaystyle \left({ab + cd}\right)^2 - \frac 1 4 \left({a^2 + b^2 - c^2 - d^2}\right)^2\) | \(\displaystyle \) | |||
| \(\displaystyle \implies\) | \(\displaystyle 16 \left({\text {Area} }\right)^2\) | \(=\) | \(\displaystyle 4 \left({ab + cd}\right)^2 - \left({a^2 + b^2 - c^2 - d^2}\right)^2\) | \(\displaystyle \) |
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:
| \(\displaystyle \) | \(\displaystyle \) | \(\) | \(\displaystyle \left({2 \left({ab + cd}\right) + a^2 + b^2 - c^2 - d^2}\right) \left({2 \left({ab + cd}\right) - a^2 - b^2 + c^2 + d^2}\right)\) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \left({\left({a + b}\right)^2 - \left({c - d}\right)^2}\right) \left({\left({c + d}\right)^2 - \left({a - b}\right)^2}\right)\) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \left({a + b + c - d}\right) \left({a + b + d - c}\right) \left({a + c + d - b}\right) \left({b + c + d - a}\right)\) | \(\displaystyle \) |
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, we get:
- $\text {Area} = \sqrt{\left({s - a}\right) \left({s - b}\right) \left({s - c}\right) \left({s - d}\right)}$
$\blacksquare$
Proof of Generalized Version
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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 $d = 0$.
- The relationship between the general and extended form of Brahmagupta's formula is similar to how the Law of Cosines extends Pythagoras's Theorem.
- Bretschneider's Formula, which is another expression of the generalized version
Source of Name
This entry was named for Brahmagupta.
Sources
- George F. Simmons: Calculus Gems (1992), Chapter $\text {B}.1$: Appendix
