The purpose of this activity is to engage students in recognising non-linear patterns, in this case a quadratic relationship. Students might use tables, graphs and equations to represent the relationship between sides of a polygon and the number of diagonals for the polygon.
The background knowledge presumed for this task is outlined in the diagram below:
This activity should be used in a ‘free exploration’ way with an expectation that students will justify the solutions that they find.
The procedural approach (show more)
- The student uses patterns in the numbers within a table to find a general rule.
In this approach students are likely to privilege arithmetic reasoning by creating a table of values for polygons of different numbers of sides.
They may then calculate differences between terms and note a pattern in the growth of those differences. Noting the differences will allow students to predict the number of diagonals for other polygons with more sides.
Experience with simple quadratic patterns may help students see that the second order difference of one is half that which occurs with square numbers. They may try investigating halves of squares to find the relation though this approach is complex.
You might suggest doubling the number of diagonals to look for a pattern since each diagonal connects two corners, e.g. AC is the same as CA. That allows for opposite directions and makes pattern spotting much easier. Looking for factors that produce double the number of diagonals is a fruitful approach.
Students may spot that the factors of double the diagonal totals are always three less than the side number multiplied by itself. If that product is halved it gives the general rule for diagonals. Let d represent the number of diagonals and s represent the number of sides.
The conceptual approach (show more)
- The student uses the spatial situation to generalise the number of diagonals emanating from each vertex and extend that idea to generalise the total number of vertices.
A figural (spatial) approach is potentially very fruitful for this task. Starting with the simple case of the pentagon students might notice that from each vertex two diagonals can be drawn.
That makes a total of 5 x 2 = 10 diagonals but that includes double ups. Since each diagonal is drawn twice dividing by two gives the actual number. In general for any n-gon, n-3 diagonals can be drawn from each vertex since diagonals cannot be drawn to the vertex itself or to its two adjacent vertices. Therefore, for any n-gon, the number of diagonal from each vertex is n(n-3). Since the diagonals will be double counted the product must be divided by two. That yields the general formula: