To solve this problem and complete the table, we need to calculate the number of diagonals in each polygon based on its number of sides.

The formula for calculating the number of diagonals $D$ in a polygon with $n$ sides is:

$D = \frac{n(n - 3)}{2}$

Where:

• $n$ is the number of sides of the polygon,
• The term $(n - 3)$ comes from the fact that three points cannot form a diagonal (i.e., the vertex itself and its two adjacent vertices),
• The division by 2 avoids counting the same diagonal twice.

Applying the formula to the given polygons:

• For a polygon with 3 sides (triangle): $D = \frac{3(3 - 3)}{2} = 0$ (Already filled in the table)

• For a polygon with 4 sides (quadrilateral): $D = \frac{4(4 - 3)}{2} = \frac{4 \times 1}{2} = 2$ (Already filled in the table)

• For a polygon with 5 sides (pentagon): $D = \frac{5(5 - 3)}{2} = \frac{5 \times 2}{2} = 5$ (Already filled in the table)

• For a polygon with 6 sides (hexagon): $D = \frac{6(6 - 3)}{2} = \frac{6 \times 3}{2} = 9$ (This fills in the next blank)

• For a polygon with 7 sides (heptagon): $D = \frac{7(7 - 3)}{2} = \frac{7 \times 4}{2} = 14$ (This fills in the next blank)

• For a polygon with 8 sides (octagon): $D = \frac{8(8 - 3)}{2} = \frac{8 \times 5}{2} = 20$ (This fills in the final blank)

Completed table:

Would you like a deeper explanation on the formula or additional information?

Here are 5 questions to expand your understanding:

1. How does the diagonal formula relate to combinations in mathematics?
2. Why doesn't a triangle have any diagonals?
3. How does the number of diagonals increase as the number of sides increases?
4. Can you derive the formula for the number of diagonals based on the reasoning behind diagonals?
5. How can you visually verify the number of diagonals in polygons with larger sides?

Tip: Always ensure the number of sides is greater than or equal to 3, as shapes with fewer sides don't form polygons!