A polygon with 9 sides has how many diagonals?
The correct answer and explanation is:
The number of diagonals in a polygon with 9 sides is 27.
To calculate the number of diagonals in any polygon, the formula used is: D=n(n−3)2D = \frac{n(n – 3)}{2}
Where:
- DD is the number of diagonals
- nn is the number of sides (or vertices) of the polygon
For a polygon with 9 sides (n = 9), substituting into the formula: D=9(9−3)2D = \frac{9(9 – 3)}{2} D=9×62D = \frac{9 \times 6}{2} D=542D = \frac{54}{2} D=27D = 27
Therefore, a polygon with 9 sides has 27 diagonals.
Explanation:
A diagonal is a line segment that connects two non-adjacent vertices in a polygon. For any polygon, the total number of line segments that can be drawn between any two vertices is given by the combination formula (n2)\binom{n}{2}, which calculates the number of ways to choose two vertices from nn vertices. This gives: (n2)=n(n−1)2\binom{n}{2} = \frac{n(n – 1)}{2}
For a 9-sided polygon, the total number of line segments between any two vertices is: (92)=9(8)2=36\binom{9}{2} = \frac{9(8)}{2} = 36
However, the 9 sides of the polygon are not diagonals, so we subtract 9 from this total. Thus, the number of diagonals is: 36−9=2736 – 9 = 27
This method applies to any polygon. For a polygon with more sides, simply use the formula to calculate the number of diagonals based on the number of sides.