It is convenient to work if your figure is a polygon. You can always break it down into a finite number of triangles, and you have to remember only one formula for calculating the area of a triangle. So, the area of a triangle is half of the writing length to the length of the altitude drawn to this side. By adding up the areas of separate triangles, which are converted your will more complex form, you'll learn the desired result.
Harder to solve the puzzle and determined the area of an arbitrary shape. Such shapes can be not only straight, but curved boundaries. There are ways to approximate calculations. Simple.
First, you can use the palette. It is a tool of transparent material coated on its surface with a grid of squares or triangles with a known area. Putting the palette on top of the figures for which are looking for a square, you count the number of your units that overlap the image. Combine incompletely closed units with each other, complementing them in mind to complete. Further, by multiplying the area of one shape in the palette to the number that is calculated, you will know the approximate area your custom shapes. It is clear that the more frequent grid marked on your palette, the more accurate your result.
Secondly, you can within the boundaries of arbitrary shapes, which define the area, outline the maximum number of triangles. To determine the area of each and fold them square. This is a very approximate result. If you wish, you can also separately determine the area of the segment bounded by arcs. To do this, imagine that the segment is part of the circle. Construct this circle, and then from the centre spend the radii to the edges of the arc. The segments form an angle α. The whole area of the sector determined by the formula π*R^2*α/360. For each smaller part of your figure do you determine the area and get the overall result, adding the obtained values.
The third method is more complicated, but more accurate, and for someone easier. The area of any shapes can be determined using integral calculus. Definite integral of a function shows the area of the graph of the function to the abscissa. The area enclosed between two graphs, we can define subtraction of the definite integral, with a smaller value of integral in the same borders, but with great value. To use this method to conveniently transfer your arbitrary figure in a coordinate system and further define their functions and act in ways of higher mathematics, which here and now to delve will not.