Brahmagupta's formula

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[edit | edit source]

Brahmagupta's formula gives the area $K$ of a cyclic quadrilateral whose sides have lengths $a$ , $b$ , $c$ , $d$ as

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

where $s$ , the semiperimeter, is defined to be

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

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 $d$ 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

K = \frac{1}{4} \sqrt{(- a + b + c + d) (a - b + c + d) (a + b - c + d) (a + b + c - d)} .

Another equivalent version is

K = \frac{\sqrt{(a^{2} + b^{2} + c^{2} + d^{2})^{2} + 8 a b c d - 2 (a^{4} + b^{4} + c^{4} + d^{4})}}{4} \cdot

Proof[edit | edit source]

Trigonometric proof[edit | edit source]

Here the notations in the figure to the right are used. The area $K$ of the cyclic quadrilateral equals the sum of the areas of $△ ADB$ and $△ BDC$ :

= \frac{1}{2} p q \sin A + \frac{1}{2} r s \sin C .

But since $ABCD$ is a cyclic quadrilateral, $∠ DAB = 180° - ∠ DCB$ . Hence $sin A = sin C$ . Therefore,

K = \frac{1}{2} p q \sin A + \frac{1}{2} r s \sin A

K^{2} = \frac{1}{4} (p q + r s)^{2} \sin^{2} A

4 K^{2} = (p q + r s)^{2} (1 - \cos^{2} A)

Solving for common side $DB$ , in $△ ADB$ and $△ BDC$ , the law of cosines gives

p^{2} + q^{2} - 2 p q \cos A = r^{2} + s^{2} - 2 r s \cos C .

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

2 (p q + r s) \cos A = p^{2} + q^{2} - r^{2} - s^{2} .

Substituting this in the equation for the area,

4 K^{2} = (p q + r s)^{2} - \frac{1}{4} (p^{2} + q^{2} - r^{2} - s^{2})^{2}

16 K^{2} = 4 (p q + r s)^{2} - (p^{2} + q^{2} - r^{2} - s^{2})^{2} .

The right-hand side is of the form $a 2 - b 2 = (a - b)(a + b)$ and hence can be written as

[2 (p q + r s) - p^{2} - q^{2} + r^{2} + s^{2}] [2 (p q + r s) + p^{2} + q^{2} - r^{2} - s^{2}]

which, upon rearranging the terms in the square brackets, yields

= [(r + s)^{2} - (p - q)^{2}] [(p + q)^{2} - (r - s)^{2}]

= (q + r + s - p) (p + r + s - q) (p + q + s - r) (p + q + r - s) .

Introducing the semiperimeter $S = .mw-parser-output .sr-only{border:0;clip:rect(0,0,0,0);height:1px;margin:-1px;overflow:hidden;padding:0;position:absolute;width:1px;white-space:nowrap} p + q + r + s / 2$ ,

16 K^{2} = 16 (S - p) (S - q) (S - r) (S - s) .

Taking the square root, we get

K = \sqrt{(S - p) (S - q) (S - r) (S - s)} .

Non-trigonometric proof[edit | edit source]

An alternative, non-trigonometric proof utilizes two applications of Heron's triangle area formula on similar triangles.^[1]

Extension to non-cyclic quadrilaterals[edit | edit source]

In the case of non-cyclic quadrilaterals, Brahmagupta's formula can be extended by considering the measures of two opposite angles of the quadrilateral:

K = \sqrt{(s - a) (s - b) (s - c) (s - d) - a b c d \cos^{2} θ}

where $θ$ 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 $180° - θ$ . Since $cos(180° - θ) = -cos θ$ , we have $cos 2 (180° - θ) = cos 2 θ$ .) 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, $θ$ is 90°, whence the term

a b c d \cos^{2} θ = a b c d \cos^{2} (9 0^{\circ}) = a b c d \cdot 0 = 0,

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]

K = \sqrt{(s - a) (s - b) (s - c) (s - d) - \frac{1}{4} (a c + b d + p q) (a c + b d - p q)}

where $p$ and $q$ are the lengths of the diagonals of the quadrilateral. In a cyclic quadrilateral, $pq = ac + bd$ according to Ptolemy's theorem, and the formula of Coolidge reduces to Brahmagupta's formula.

Related theorems[edit | edit source]

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.
Increasingly complicated closed-form formulas exist for the area of general polygons on circles, as described by Maley et al.^[3]

References[edit | edit source]

↑ 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.
↑ Maley, F. Miller; Robbins, David P.; Roskies, Julie (2005). "On the areas of cyclic and semicyclic polygons". Advances in Applied Mathematics. 34 (4): 669–689. arXiv:math/0407300. doi:10.1016/j.aam.2004.09.008. S2CID 119565975.

External links[edit | edit source]

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[1] Hess, Albrecht, "A highway from Heron to Brahmagupta", Forum Geometricorum 12 (2012), 191–192.

[2] J. L. Coolidge, "A Historically Interesting Formula for the Area of a Quadrilateral", American Mathematical Monthly, 46 (1939) pp. 345-347.

[3] Maley, F. Miller; Robbins, David P.; Roskies, Julie (2005). "On the areas of cyclic and semicyclic polygons". Advances in Applied Mathematics. 34 (4): 669–689. arXiv:math/0407300. doi:10.1016/j.aam.2004.09.008. S2CID 119565975.

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