In Euclidean geometry, Brahmagupta's formula is used to find the area of any cyclic quadrilateral (one that can be inscribed in a circle) given the lengths of the sides.
Formula
Brahmagupta's formula gives the area Template:Math of a cyclic quadrilateral whose sides have lengths Template:Math, Template:Math, Template:Math, Template:Math as
- <math>K=\sqrt{(s-a)(s-b)(s-c)(s-d)}</math>
where Template:Math, the semiperimeter, is defined to be
- <math>s=\frac{a+b+c+d}{2}.</math>
This formula generalizes Heron's formula for the area of a triangle. A triangle may be regarded as a quadrilateral with one side of length zero. From this perspective, as Template:Math approaches zero, a cyclic quadrilateral converges into a cyclic triangle (all triangles are cyclic), and Brahmagupta's formula simplifies to Heron's formula.
If the semiperimeter is not used, Brahmagupta's formula is
- <math>K=\frac{1}{4}\sqrt{(-a+b+c+d)(a-b+c+d)(a+b-c+d)(a+b+c-d)}.</math>
Another equivalent version is
- <math>K=\frac{\sqrt{(a^2+b^2+c^2+d^2)^2+8abcd-2(a^4+b^4+c^4+d^4)}}{4}\cdot</math>
Proof
Trigonometric proof
Here the notations in the figure to the right are used. The area Template:Math of the cyclic quadrilateral equals the sum of the areas of Template:Math and Template:Math:
- <math>= \frac{1}{2}pq\sin A + \frac{1}{2}rs\sin C.</math>
But since Template:Math is a cyclic quadrilateral, Template:Math. Hence Template:Math. Therefore,
- <math>K = \frac{1}{2}pq\sin A + \frac{1}{2}rs\sin A</math>
- <math>K^2 = \frac{1}{4} (pq + rs)^2 \sin^2 A</math>
- <math>4K^2 = (pq + rs)^2 (1 - \cos^2 A) </math>
Solving for common side Template:Math, in Template:Math and Template:Math, the law of cosines gives
- <math>p^2 + q^2 - 2pq\cos A = r^2 + s^2 - 2rs\cos C. </math>
Substituting Template:Math (since angles Template:Math and Template:Math are supplementary) and rearranging, we have
- <math>2 (pq + rs) \cos A = p^2 + q^2 - r^2 - s^2. </math>
Substituting this in the equation for the area,
- <math>4K^2 = (pq + rs)^2 - \frac{1}{4}(p^2 + q^2 - r^2 - s^2)^2</math>
- <math>16K^2 = 4(pq + rs)^2 - (p^2 + q^2 - r^2 - s^2)^2.</math>
The right-hand side is of the form Template:Math and hence can be written as
- <math>[2(pq + rs) - p^2 - q^2 + r^2 +s^2][2(pq + rs) + p^2 + q^2 -r^2 - s^2] </math>
which, upon rearranging the terms in the square brackets, yields
- <math>= [ (r+s)^2 - (p-q)^2 ][ (p+q)^2 - (r-s)^2 ] </math>
- <math>= (q+r+s-p)(p+r+s-q)(p+q+s-r)(p+q+r-s). </math>
Introducing the semiperimeter Template:Math,
- <math>16K^2 = 16(S-p)(S-q)(S-r)(S-s). </math>
Taking the square root, we get
- <math>K = \sqrt{(S-p)(S-q)(S-r)(S-s)}.</math>
Non-trigonometric proof
An alternative, non-trigonometric proof utilizes two applications of Heron's triangle area formula on similar triangles.[1]
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:
- <math>K=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\theta}</math>
where Template:Math is half the sum of any two opposite angles. (The choice of which pair of opposite angles is irrelevant: if the other two angles are taken, half their sum is Template:Math. Since Template:Math, we have Template:Math.) This more general formula is known as Bretschneider's formula.
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, Template:Math is 90°, whence the term
- <math>abcd\cos^2\theta=abcd\cos^2 \left(90^\circ\right)=abcd\cdot0=0, </math>
giving the basic form of Brahmagupta's formula. It follows from the latter equation that the area of a cyclic quadrilateral is the maximum possible area for any quadrilateral with the given side lengths.
A related formula, which was proved by Coolidge, also gives the area of a general convex quadrilateral. It is[2]
- <math>K=\sqrt{(s-a)(s-b)(s-c)(s-d)-\textstyle{1\over4}(ac+bd+pq)(ac+bd-pq)}</math>
where Template:Math and Template:Math are the lengths of the diagonals of the quadrilateral. In a cyclic quadrilateral, Template:Math according to Ptolemy's theorem, and the formula of Coolidge reduces to Brahmagupta's formula.
Related theorems
- Heron's formula for the area of a triangle is the special case obtained by taking Template:Math.
- The relationship between the general and extended form of Brahmagupta's formula is similar to how the law of cosines extends the Pythagorean theorem.
- Increasingly complicated closed-form formulas exist for the area of general polygons on circles, as described by Maley et al.[3]
References
- ↑ Hess, Albrecht, "A highway from Heron to Brahmagupta", Forum Geometricorum 12 (2012), 191–192.
- ↑ J. L. Coolidge, "A Historically Interesting Formula for the Area of a Quadrilateral", American Mathematical Monthly, 46 (1939) pp. 345-347.
- ↑ On the areas of cyclic and semicyclic polygons. Advances in Applied Mathematics({Template:Date).