# Pick’s Theorem To Find The Area Of A Polygon

I have been teaching mathematics in an Australian High School since 1982, and I am a contributing author to mathematics text books.

Sitting languidly at the front of the classroom, head down and pen in hand playing a game of ‘join the dots’, I would occasionally raise my eyes and eyeball my students to discourage improper behaviour whilst they sweated over the Area test that I was supervising.

Drawing lines through dots reminded me of a task I set my class as an introduction to Area. Students use grid paper to construct non-intersecting polygons and calculate their area by subdividing the shape into unit squares and triangles.

My students then move on to study areas of shapes by applying the appropriate formula.

## Pick’s Rule

Pick’s rule can be used to calculate the exact area of a polygon drawn on dotted grid paper. The rule (formula) requires two quantities.

1. The number of dots, P, that lie exactly on the perimeter of the shape.

2. The number of dots, I, that lie completely inside the shape.

The formula is

Pick’s Rule can be used relatively easily without the need to understand why it works. After all, most of the effort is in counting dots. Since the proof requires elementary knowledge of graph theory, I figured my students will appreciate the derivation of the formula.

The class was still labouring to complete the test, so I fell into a brown study of the proof.

## Euler’s formula

In graph theory, one of the most important relationships is Euler’s formula, which describes the relationship between the edges, vertices and faces of a flat (planar) shape.

An edge is a line, a vertex is where two or more lines meet, and a face is a region enclosed by at least three edges.

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Note

• Edges do not intersect (overlap) and need not be straight.
• The region outside the entire shape is also considered to be a face.

Euler’s formula states:

The following examples illustrate this relationship.

Euler’s formula will be used as part of the proof for Pick’s Theorem. The proof will be displayed in several steps, using a particular polygon as an example.

## The proof

Step 1

Construct the polygon on dotted grid paper and divide it into triangles so that each triangle’s vertices are on a dot. This is always possible, according to another property in graph theory.

Step 2

We find an equation involving the faces and edges.

Each face, f, inside the polygon is a triangle with three edges, so there are 3(f – 1) edges in total.

(We use f – 1and not f because the face outside the polygon has no triangle and therefore not counted).

Now let eb be the number of edges on the boundary (perimeter) of the polygon.

Then 3(f – 1) + eb= 2e, where e is the total number of edges because each triangle not adjacent to the polygon shares a side with another triangle and has been counted twice.

[This works for our polygon. v = 23, f = 30, e = 51, and 3(30 – 1) + 15 = 2 x 51]

However, 3(f – 1) + eb gives twice as many edges as there actually are on the shape. This means the number of edges counted is twice as many as the actual number of edges.

Hence we have 3(f – 1) + eb= 2e.

Step 3

We now bring Euler’s Formula into play, so there are two equations.

v + f e = 2 (Euler’s formula) and 3(f – 1) + eb= 2e.

The equation 3(f – 1) + eb= 2e can be rearranged to obtain the equivalent equation f = 2(ef) – eb+ 3.

Similarly, Euler’s Formula can be rearranged to obtain ef = v – 2.

Now substitute ef = v – 2 in f = 2(ef) – eb+ 3 to get f = 2(v – 2) – eb+ 3.

This simplifies to f = 2veb– 1 or f – 1 = 2veb– 2.

Step 4

If we let A be the area of the polygon, then f – 1 = 2A, since each face (apart from the face outside the polygon) is a triangle of area ½ and there are twice as many triangles as there are faces.

Now, v is the total number of vertices of the polygon, so we can refer to it as I + eb(Keeping the same notation as we used in Pick’s Theorem).

So f – 1 = 2veb– 2 becomes 2A = 2(I + eb ) – eb – 2.

This simplifies to 2A = 2I + eb– 2.

Finally, ebis equivalent to the number of dots, P, on the perimeter of the polygon, so we have

2A = 2I + P – 2 or 2A = P + 2I – 2.

Dividing throughout by 2 gives the required result.

Just as I proudly contemplated this last step of the proof, I looked up dreamily and saw hands up in the air.

“Sir,” from a number of students, waving test papers. “We’ve finished the test.”

“Class, I think we will look at proving Pick’s Theorem next time,” I stated quickly, trying to conceal that I had not kept track of the time.

“But Sir,” Branco began, sounding slightly confused. “You did that with us a week ago; ‘It’s good for the edification of our mathematical souls’ you said.”

“Ah, of course. Just teasing you,” I said, not too convincingly.

As students submitted their test on the way out, I considered whether there is any proof that playing ‘join the dots’ accelerates the onset of senility in an ageing mathematics teacher.

Use Pick’s Theorem to calculate the area of this polygon.

Hint: The answer is a 3-digit number whose digit sum is 14, the first two digits are odd and the third is even.

## Quiz Time

For each question, choose the best answer. The answer key is below.

1. How is Pick's Theorem used?
• To find perimeter
• To find area
• To count dots
2. Which one of the following is Euler's formula?
• v - f + e = 2
• v + f + e = 2
• v + f - e = 2
3. A polygon has 10 dots on its perimeter and 8 dots inside. The area is
• 12
• 11
• 18
4. A shape has 10 faces and and 8 vertices. How many edges are there?
• 12
• 16
• 20
5. With what types of shapes can Pick's rule be used?
• all 2-dimensional shapes
• any flat shape
• 2-dimensional polygons
6. How many faces does every polygon have?
• 1
• 2
• 3