In a convex polygon no three diagonals are concurrent. If the total no. of points of intersection of diagonals is 70 then find the no.of diagonals in the polygon.
Should we select two diagonals at a time? Could you also tell me a general way to solve these kind of problems in P&C where we have to find points if intersection or no.of vertices etc.?
A polygon with $n$ sides has ${n}\choose{4}$ points of intersection to its diagonal, where ${n}\choose{k}$ is defined by $\frac{n!}{(n - k)! \cdot k!}$
Plugging the information provided in the problem gives us:
$\frac{n!}{(n - 4)! \cdot 4!}$ = 70
$\rightarrow n = 8$
This is a polygon with $8$ sides. The number of diagonals in a polygon with $n$ sides is given by $d = \frac{n(n-3)}{2}$, which gives us an answer of there being 20 diagonals.