Question

How many diagonals are there if the regular convex polygon has n sides ?

Answer 1

@love_jessika15

Answer 2

@amistre64

Answer 3

(n - 3) + [(n - 3) + (n - 4) + (n - 5) + . . . + 1]

= (n - 3) + [(n - 3)(n - 2)/2]

= (n^2 - 3n)/2

Answer 4

Use a pentagon and a hexagon to get a feel for this.

All good now, @goformit100 ?

Answer 5

ok

Answer 6

since a given number of sides is equal to a given number of corners; and 2 opposite corners equals 1 diagonal .....

how are we defining a diagonal?

Answer 7

ok then ?

Answer 8

then, im wondering how we define a diagonal ....

Answer 9

If you take a hexagon you will get:

3 from the first vertex,
3 from the "next" or adjacent vertex
2 from the next (because one is already drawn)
1 from the next after that (because 2 are already drawn)
0 from the next 2 because they are already drawn

That equals 9 and fits the formula that was derived above.

Answer 10

OK

Answer 11

if all diagonals; then i see it as:

n sides are n vertexes

n + (n-1) + (n-2) + ... + 3 + 2 + 1

minus the original n sides leaves all the diagonals

Answer 12

........ok

Answer 13

pfft, mine should start at n-1

Answer 14

If you take a pentagon, you will have:

2 from the first vertex,
2 from the "next" or adjacent vertex
1 from the next (because one is already drawn)
0 from the next 2 because they are already drawn

That equals 5 and fits the formula that was derived above.

Answer 15

sum of 1 to n-1\[\frac{n(1+(n-1))}{2}-n\]
sum of 1 to n-1\[\frac{n^2}{2}-n\]
sum of 1 to n-1\[\frac{n^2-2n}{2}\]

Answer 16

So, the pattern is n-3 plus n-3 and then decreasing until you hit "1".

The formula I derived incorporates adding up from:

1 to n-3

Then you add another n-3.

Then you simplify, and then you're done!

Answer 17

(n^2 - 3n)/2

not

(n^2 - 2n)/2

Answer 18

yeah, mines off :)

Answer 19

np

Answer 20

i shoulda had n-1 and not n up there

Answer 21

\[\frac{(n-1)(n)}{2}-n\to \frac{n^2-3n}{2}\]\]

Answer 22

2 from "A", 2 from "B", 1 from "C", none from "D" and "E"

That is none that aren't drawn already.