Question

In a polygon no three diagonals are concurrent .if the total no of point of intersection of diagonal s is 70..no of diagonals of polygon

Ans 20

Pls explain in detail without giving just formula

Answer 1

I admire your intrest to understand the question and answer in detail ,i made it as detailed as possible:

A selection of four vertices of the polygon gives an interior intersection.
i.e only with 4 points can you get 1 intersection.
For example in case of square 4 points are present hence number of intersection is just one

Hence let there be n sides for the polgon
Which means choosing any 4 at a time from n gives 70 intersection (given)
that is ,
nC4=70⇒
n(n−1)(n−2)(n−3)=24×70
⇒8×7×6×5⇒8×7×6×5
⇒n=8

For apolygon with 8 sides number of diagonals is ,
nC2 - no. of sides (chosing any 2 sides at a time and subtracting n to avoid repeatation)
8C2 - 8
=28 -8
=20

I hope this helps :)