Daring Diagonals

Relevant wiki: Rule of Product Problem Solving

Each vertex has the same number of diagonals coming from it. Looking at a single vertex, diagonals are drawn to each vertex -- except you can't draw a diagonal to either of the two adjacent vertices, and you can't draw a diagonal to itself. So if the polygon has $n$ vertices ( $n$ sides), then the number of diagonals coming from a single vertex is $n-3.$ Now multiply that by the number of vertices and you get $n(n-3),$ except this method counts each diagonal twice. So the number of diagonals of an $n$ -gon is:

$\frac{n(n-3)}{2}.$

For a decagon, the number of diagonals is $\frac{10(7)}{2}=\boxed{35}.$

Relevant wiki: Combinations without Repetition - Intermediate

Number of ways of choosing any two vertices to make a diagonal $\dbinom{n}{2}$ and there are $n$ cases where we chose two adjacent vertices, therefore, the number of diagonals is $\dbinom{n}{2}-n$ .

Relevant wiki: Combinations without Repetition - Intermediate

We need to connect two vertices for a line. For a $n$ -sided convex polygon, there are $\displaystyle {n \choose 2}$ lines. Out of these lines there are $n$ sides of the polygon which are not the diagonals. Therefore, the number of diagonals is given by:

$n_d = {n \choose 2} - n = \frac {n(n-1)}2 - n = \frac {n(n-3)}2$

For $n= 10$ , we have $n_d = \dfrac {10(10-3)}2 = \boxed{35}$

Notation: $\displaystyle {M \choose N} = \frac {M!}{N!(M-N)!}$ denotes the binomial coefficient .

If we start off with 10 vertices, we know the first vertex chosen won't create a diagonal either with itself or the 2 adjacent vertices. So the first vertex makes 7 diagonals. The second vertex makes 7 vertices as well, because it didn't link with the first. Then it's just a process of subtracting the number of diagonals each vertex would make from the diagonals that already exist. If you add the vertices together you get; 7 +7 + 6 + 5 + 4 + 3 +2 +1+0+0 diagonals. The last two vertices are already all connected up by the time you get round the decagon, so they make no new diagonals. This is equal to 35 total diagonals.

Number of Sides Multiply by Number of Diagonals

          4             x         0.5         =               2

          5             x         1.0         =               5

          6             x         1.5         =               9

          7             x         2.0         =              14

          8             x         2.5         =              20

          9             x         3.0         =              27

          10            x         3.5         =              35

          .             x           .         =               .

          .             x           .         =               .

          .             x           .         =               .

          20            x         8.5         =             170

10C2-10= 35

=combin(10,2)-10 = 35

The formula for the number of diagonals is:

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

where n is the number of sides of the shape. When you substitute in n, you get:

$\frac{10(10-3)}{2}$ = $\frac{10(7)}{2}$ = $\frac{70}{2}$ = 35

You might want to memorize the formula because it works for every polygon

I've found that the sum of the number of sides and number of diagonals is equal to a number in the triangular number sequence( 1,3,6,10,15,21,28,36,45,.. based on the $\frac{n(n+1)}{2}$ rule).Line: 1 sides +0 diagonals=1:Triangle:3 sides + 0 diagonals= 3,Square:4 +2 =6;convex pentagon:5+5=10;convex hexagon:6+9=15......convex decagon:10+x=45 ==> x= 35

Just use this video formula, it works for a polygon with a millions sides : 0.5*S(S-3) I had to drive this equation, so I'm sure it works (: (Too bored to type it all. Wish we could attach a photo here)

You can do this question without using a formula if you consider the pattern: convex quadrilateral- 2 diagonals
convex pentagon- 5 diagonals
convex hexagon- 9 diagonals
convex heptagon- 14 diagonals convex octagon- 20 diagonals convex nonagon- 27 diagonals convex decagon- 35 diagonals

Consider the vertices without lines connecting them, and connect each vertex to every other. From the first vertex, you can draw a line to every vertex except itself: the number of lines is now $n-1$ . From the second vertex, you can draw a line to every vertex except the first (as that line already exists): the number of new lines is $n-2$ . For vertex 3, the number of new lines is $n-3$ and so on. The last vertex will have no new lines. The sum of lines is therefore

$\sum_{i=1}^{n-1}i$

Subtract from this the lines that make up the polygon, and you have the general case:

$(\sum_{i=1}^{n-1}i)-n$

The formula for No.of Diagonals in a convex polygon is n $LaTeX$

i dont know how right this is but.. i used A as how many sides there are and B as how many diagonals from one corner can be drawn.

So: (AB)/2

This is a fun different way to think of it. If you include the sides of the shapes as well, a square has 6, Pentagon has 10 and hexagon has 15. These are a nice sequence of triangular numbers. Aha! Proof-wise straight forward in that the set from each vertex shares increasing amounts of lines as you go, thus getting triangular.. Adjusting to this specific problem you just get the corresponding triangular number and subtract the number of sides. Bingo!

Plotting number of sides vs. number of diagonals and finding the best fitting curve yields a parabolic function of y = 0.5x^2 - 1.5x. The number of diagonals for an n-sided polygon can then be determined by plugging in n for x.

You can proceed iteratively, from the initial knowledge that the square has 2 diagonals. How many new diagonals would appear if we added one side, i.e. turned the square into a pentagon? Just take one side, say the one that joins vertices A and B, and put a new vertex C in the middle. We have to add new diagonals to join C to all the other vertices, except A and B of course, plus the new diagonal that joins A and B (they are no longer neighbors). So, in general, for a polygon of n sides, if we move to n+1 we have to add n-2+1=n-1 diagonals. This yields the following sequence:

4 sides => 2 diagonals

5 sides => 2+(4-1)=5 diagonals

6 sides => 5+(5-1)=9 diagonals

7 sides => 9+(6-1)=14 diagonals

8 sides => 14+(7-1)=20 diagonals

9 sides => 20+(8-1)=27 diagonals

10 sides => 27+(9-1)=35 diagonals

… and so on

nC2-n where n is the number of sides

It's a simple combination problem, the diagonal can be drawn between 2 points and there are 10 points in total, but the diagonal can't be drawn between adjacent points, so we have to account for that, hence the solution is nC2 - n, so answer for decagon is 45-10

We claim that the number of diagonals in an n-gon is given by $\frac{n-3}{2}$ .

We see that each vertex can be joined to any vertex except itself and its two adjacent vertices which means we exclude $1+2=3$ vertices. Therefore each of $n$ vertices can be joined to $(n-3)$ vertices. Therefore we have $\frac{n(n-3)}{2}$ diagonals, not forgetting that we need to divide by 2 to avoid double counting.

Alternatively, if we just consider the different conbinations of vertices, we have $\frac{n(n-1)}{2}$ combinations. But we need to discount the set of consecutive vertices, of which we have n such elements. Then the number of diagonals is $\frac{n(n-1)}{2}-n=\frac{n}{2}(n-1-2)=\frac{n(n-3)}{2}$ .

Then for a 10-gon, we have $\frac{10(10-3)}{2}=5(7)=\boxed{35}$ diagonals.

If you observe each of the given polygons you will see that every vertex has the same diagonals with every other vertex. So you calculate how many diagonals one vertex has and then multiply that by the number of the vertexes the given polygon has.

I worked this out in reverse, I looked at a single vertex and calculated how many diagonals comes from a single vertex, in the convex decagon (10 sides), there are 7 diagonals that are possible form a single vertex (as Andy Hayes mentions n-3 to arrive at this number). I manually went around each vertex and calculated how many diagonals were present but discounting the ones already counted. I came to see that:

• Vertex 1: 7 diagonals
• Vertex 2: 7 diagonals
• Vertex 3: 6 diagonals (1 was already accounted for)
• Vertex 4: 5 diagonals (2 were already accounted for)
• Vertex 5: 4 diagonals (3 were already accounted for)
• Vertex 6: 3 diagonals (4 were already accounted for)
• Vertex 7: 2 diagonals (5 were already accounted for)
• Vertex 8: 1 diagonals (6 were already accounted for)
• Vertex 9: 0 diagonals (all diagonals were already accounted for)
• Vertex 10: 0 diagonals (all diagonals were already accounted for)

This resulted in a formula that seems to hold true. Work out how many diagonals (d) are possible from a single vertex and apply the following:

d! + d

A point can make a diagonal with any point except it neighbours and itself. But diagonals can go in opposite directions. So (n(n-3))/2 is the formula. So 10×7/2=35.