The area of any polygon

There is no special formula for a general quadrilateral, and none is needed. Join the diagonal BD, and ABCD becomes two triangles. Find each area and add.

So all you need to measure is the diagonal and the two perpendicular distances from the other corners to it.

An example. In quadrilateral ABCD, the diagonal AC is 22 cm, and BM and DN (perpendiculars to AC) are 3 cm each.

∆ABC has base AC = 22 and height BM = 3, so its area is ½ × 22 × 3 = 33 cm². ∆ADC has the same base and height DN = 3, so its area is also 33 cm². The quadrilateral is 66 cm².

A pentagon? Draw two diagonals from one corner and you have three triangles. In fact any polygon can be cut into triangles, so one formula (½ × base × height) gives you the area of every polygon there is.

For a regular hexagon, six identical triangles meet at the centre, so you only need the side and the distance from the centre to a side.

Questions like this are ancient. The Śulba-Sūtras are Indian geometry texts about building altars, which had to have an exact shape and an exact area. Change the shape but keep the area: those texts give methods for many such problems. Euclid's Elements asks and solves the same kind.

Cutting a shape into pieces and rearranging them into another shape of the same area has a name: dissection. It is how the next few formulas are proved.