for any polygon with 'n' number of sides, the number of diagonals will be
n(n-3) / 2
if n is 50, no. of diagonals will be 50(50-3)/2 = 47*25 = 1175
how do you get it?
take any polygon, say septagon i.e. with 7 sides
if you take 1 vertice, remainig vertices will be 7-1 that is 6.
there will be 2 consecutive vertices for any vertice, so here, this one has 2 consecutive vertices(its always like that) so the remaining vertices is 6-2 that is 4.
remember : a diagonal connects non consecutive vertices
so you know that from one vertice, there will be 4 diagonals.
multiply it by the total no. of vertices(here 7)
4*7 = 28
now divide this by 2
(since 2 vertices share the same diagonal)
therefore, 28/2 = 14 which is the no. of diagonals in a septagon.
so for an n-gon, the number of diagonals will be
n(n-3) / 2
hope it helps!
p.s. i figured it out myself during math class. then i tried it on many polygons.
It has 1,345,890 diagonals
It has 9 diagonals
It has 2 Diagonals!!!
There are 20 diagonals
It has 27 diagonals
It has 2 diagonals
In a heptagon there is 14 diagonals
A undecagon has 44 diagonals. A dodecagon has 54 diagonals. An octagon has 20 diagonals. A heptagon has 14 diagonals.
how many diagonals from a vertex a heptagon have
Improved Answer:-It has 14 diagonals
A rectangle has 2 diagonals
it has 44 diagonals :D
A pentagon has 5 diagonals.
A quadrilateral has 2 diagonals.
It ha 2 diagonals
A Nonagon has 36 diagonals
There are five diagonals in a pentagon.
0 they have no diagonals