How many diagonals would a 20-sided polygon have?

Question

Is there a formula to calculate this?

Answer 1

yes, there is a formula. 

A diagonal is formed when the segment is connected to a non consecutive vertex of the polygon. 

So if you have hexagon ABCDEF. the diagonal from vertex A cnnot go to A(itself), B(clockwise consecutive), or F( counter clockwise consecutive, which leaves 3 vertices that will lead to a diagonal.....AC,AD and AE.

This is the same for vertex B....and all the other vertices. So that would be 3 diagonals from each of 6 vertices making 18.

BUT...AE and EA are the same diagonal....and so is BD and DB....so you counted all the diagonals twice. So 18/2=9 diagonals for a hexagon.

 

So your 20 sided polygon(called an icosagon).

20 vertices. you can't draw from the vertex to itself or either of the 2 consecutive vertices....leaving 17 vertices from each vertex.

So 20vertices * 17 diagonals divided by 2 for the repetitions = 170 diagonals.

 

In general, if n is the number of sides, then n is also the number of vertices.

So d, the number of diagonals, d=n(n-3)/2

 

Hope this helped.

Answer 2

MIckayla,  Let's think about this problem in a general manner.  A general polygon has n vertices and n sides.  I will use the general induction method to work here.  Let n be the number of vertices, and D be the number of diagonals.  If n = 1, then D = 0 since it will be a point.  If n = 2, we will have a line segment and again D = 0.  If n = 3, still D = 0.  Now if n = 4, we know D = 2.  If n = 5 then how many diagonals are added with the extra vertex?  D = 5.  What about n = 6? Each vertex can draw a diagonal all other vertices EXCEPT the two that are adjacent to it.  So for n = 6, each vertex can have a diagonal with (n-3) other vertices for a total of n(n-3) diagonals.  But we have counted each diagonal twice, once from each end.  So we need to divide by 2 for the formula we seek.  That gives D(n) = n(n-3)/2.  We already know it is correct for n = 4.  Now we need to know that for n>= 4 our formula always works.  To avoid confusion with the references to n above, let's use i for the number of variables in our proof. We assume that it works for (i-1) diagonals and show that assumption implies that it also works for i vertices.  Assume D(i-1) = (i -1)((i -1-3)/2 = (i -1)(i - 4)/2 is correct  Now we add one vertex.  It will have diagonals to all existing vertices except the two vertices adjacent to it.  So it will add how many new diagonals? i -2 new diagonals.  Adding the number of diagonals for D(i-1) to the number of new diagonals when i is the number of vertices D(i) = (i-1)(i-4)/2 + i - 2  Simplifying D(i) = [i*i - 5i + 4]/2 + i -2 = 1/2[i*i -5i  +2i + 4 - 4] = 1/2[i*i -3i] = i(i - 3)/2.  QED

 

More than you asked but this is one method to find the formula and know it is correct.  If you understand how to do this, then you can often find a formula you need without knowing.  It does take more time however than looking up a known formula.

 

The formula is  D(n) = n(n - 3)/2.  For n = 20 we get 20(20 - 3)/2 = 20(17)/2 = 170

 

best wishes,  JAG (Lewisville)