Understanding the Relationship between Angles & Diagonals in a Polygon

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SUMMARY

The discussion focuses on the mathematical relationship between angles and diagonals in polygons, specifically examining the case of pentagons. It establishes that a convex polygon with \(n\) sides has \(D_n\) diagonals calculated by the formula \(D_n=\frac{n(n-3)}{2}\) for \(n \geq 3\). The proof by induction demonstrates how adding a vertex affects the number of diagonals, leading to the formula \(D_{n+1}=\frac{(n+1)((n+1)-3)}{2}\). This mathematical exploration is relevant for presentations on polygon properties.

PREREQUISITES
  • Understanding of basic polygon properties
  • Familiarity with mathematical induction
  • Knowledge of geometric concepts related to angles and diagonals
  • Ability to interpret mathematical formulas and proofs
NEXT STEPS
  • Research the properties of convex polygons
  • Study the application of mathematical induction in geometry
  • Explore the relationship between angles and sides in various polygon types
  • Learn about advanced polygonal formulas and their proofs
USEFUL FOR

Students, educators, and anyone interested in geometry, particularly those preparing presentations on polygon properties and relationships between angles and diagonals.

highmath
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There a relationships between angles to diagonals in a polygon?
 
Mathematics news on Phys.org
Welcome to MHB!

Do you have any particular polygons in mind?
 
Can we start with pentagon?
What can I say on the pentagon itself, the angles of it, and diagonal?
I do a presentation and the topic of it is as above: "relationships between angles and diagonal". I want to show the topic and investigate it.
I don't get a mark on the presentation. It is only for adult course in the center... (community center)
So what do you say?
Thanks for any help...
 
A convex polygon with \(n\) sides has \(n\) vertices, and a diagonal can be drawn from each vertex to all but 2 of the other vertices. Iterating over all vertices, and observing the diagonals will be drawn twice, we may hypothesize that the number of diagonals \(D_n\) is given by:

$$D_n=\frac{n(n-3)}{2}$$ where \(3\le n\)

Observing the base case \(D_3=0\) is true, for a triangle has no diagonals, we may use as our inductive step, the addition of another vertex. From this new vertex, diagonals may be drawn to all but \(n-2\) of the other vertices and a new diagonal may now be drawn between the two existing vertices on either side of the new vertex, for a total of \(n-1\) new diagonals. Hence:

$$D_{n+1}=\frac{n(n-3)}{2}+n-1=\frac{n(n-3)+2(n-1)}{2}=\frac{n^2-n-2}{2}=\frac{(n+1)(n-2)}{2}=\frac{(n+1)((n+1)-3)}{2}$$

We have derived \(D_{n+1}\) from \(D_n\), thereby completing the proof by induction.
 
highmath said:
There a relationships between angles to diagonals in a polygon?
...and at how many other sites did you post this?
 

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