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Given a polygon constructed on a grid of equal-distanced grid points (i.e., points with integer coordinates):

Pick's theorem provides a simple formula for calculating the area, A, of this polygon in terms of the number, I, of interior points of the polygon and the number, B, of boundary points on the polygon.

Formula: '''A = I + ½B − 1 .

Thus, in the above example, I = 9, and B = 14. Hence, '''A = 9 + ½(14) − 1 = 9 + 7 − 1 = 15 (square units).

This is so simple that it has been correctly used by first-grade children, drawing figures on square tiles on floor or wall, or stretching strings from pegs in pegboard. They learn how to add, along with subtraction as "take away". They learn to "halve" by a one-to-one correspondence between counters.

The formula can be generalized to three dimensions and higher by Ehrhart polynomials. The formula also generalizes to surfaces of polyhedra.

### Proof

Consider a polygon P constructed from n triangles (with n+2 points), and a triangle T, with one edge in common with P. Assume Pick's theorem is true of a polygon constructed from n triangles. Since P and T share an edge, all the boundary points along the edge in common are merged to interior points, except for the two endpoints of the edge, which are merged to boundary points. So, calling the number of boundary points in common C, IPT = (IP + IT) + (C − 2) and BPT = (BP + BT) − 2(C − 2) − 2.

From the above follows (IP + IT) = IPT − (C − 2) and (BP + BT) = BPT + 2(C − 2) + 2.

Since we are assuming the theorem for P and for T separately,

APT = AP + AT
= IP + ½BP − 1 + IT + ½BT − 1
= (IP + IT) + ½(BP + BT) − 2
= IPT − (C − 2) + ½(BPT + 2(C − 2) + 2) − 2
= IPT + ½BPT − (C − 2) + ½(2(C − 2) + 2) − 2
= IPT + ½BPT − 1.

Therefore, if the theorem is true for polygons constructed from n triangles, the theorem is also true for polygons constructed from n+1 triangles.

So the theorem is true, if the theorem is true for a single triangle.

The verification for this case can be done in these short steps: