### Finding the Area of a Field

It is a tricky matter to find the area of a field that has irregular or meandering boundaries. The standard method is to divide the field into triangular parts. If the boundaries are linear, this is simple. If they twist and turn, then a large number of triangles may be required.

Heron of Alexandria. Triangle of sides a, b and c and altitude h.

When we have the triangles, we calculate the area of each and sum them up. For a right-angled triangle with sides ${a}$, ${b}$ and hypotenuse ${c}$, the area is ${A = ab/2}$. For a general triangle, we learned in school that the area is equal to half the base times the height.

All very well, but constructing the height or altitude is not trivial and, if there are many triangles, it becomes arduous. A better way is to measure the length of the three sides ${a}$, ${b}$, ${c}$ and used Heron’s Theorem.

Heron was one of the great Greek mathematicians of Alexandria, following in the tradition of Euclid, Archimedes, Eratosthenes and Apollonius. He lived in the first century, from about AD 10 to AD 70.

We calculate the semi-perimiter ${s = \textstyle{\frac{1}{2}}(a+b+c)}$ and then deduce the area from

$\displaystyle A = \sqrt{s(s-a)(s-b)(s-c)} \,.$

Heron'<80><99>s result was eminently practical and was of direct use in surveying. For example, if we are measuring land area, it is a simple matter to triangulate a polygonal plot. We can easily measure the length of each side. Heron’s formula then gives the area of each triangle and the total area is the sum of the areas of the triangular elements.

Quatrilateral Fields

General quadrilateral. Sides ${a}$, ${b}$, ${c}$, ${d}$. Angles ${\alpha}$, ${\beta}$, ${\gamma}$, ${\delta}$.

The area of a rectangular field is easily found from the product of the length and breadth. For more general quadrilateral fields, we would like to obtain the area from the lengths of the four sides, ${a}$, ${b}$, ${c}$, ${d}$. In the seventh century, the great Indian mathematician Brahmagupta (c. AD 598–668) gave a formula analogous to Heron’s formula for a quadrilateral. If we define the semi-perimiter ${s = \textstyle{\frac{1}{2}}(a+b+c+d)}$ the area is given by

$\displaystyle A = \sqrt{(s-a)(s-b)(s-c)(s-d)} \,.$

But it seems that Brahmagupta did not specify that the quadrilateral must be a convex, cyclic quadrilateral. It is clear that the lengths of the four sides alone are insufficient to give the area: just consider a rhombus, which can be collapsed to have vanishing area. However, in the special case of a convex, cyclic quadrilateral, Brahmagupta’s formula is correct.

Bretschneider’s Formula

The area of a general quadrilateral is given by Bretschneider’s formula.

$\displaystyle A = \sqrt {(s-a)(s-b)(s-c)(s-d)-abcd\cdot \cos^{2}\left({\frac{\alpha +\gamma }{2}}\right)}$

Here, ${\alpha}$ and ${\gamma}$ are the internal angles at two opposite vertices. We could also use the sum of the remaining angles, ${\beta + \delta}$. This formula works for both convex and concave quadrilaterals (although not crossed ones), whether they are cyclic or otherwise.

Formulas have been obtained for cyclic pentagons, hexagons, heptagons and octagons (see Maley, et al., 2005). However, they become increasingly complicated and it is not clear whether they have any practical utility. The strategy of triangulating a polygonal field and using Heron’s formula may be the simplest and most satisfactory solution.

Sources

${\bullet}$ Maley, F. Miller; Robbins, David P.; Roskies, Julie, 2005: On the areas of cyclic and semicyclic polygons. Adv. Appl. Math., 34(4), 669–689. PDF.

${\bullet}$ Wikipedia article Bretschneider’s formula. Wiki.