7

after i did a google for the formula, found out that n(n-3)/2=14-- where n is no: of sides......


-Thx to sanju.

For a n-sided convex polygon, the number of diagonals can be draw :
+ from the first vertex is n-3 (cannot draw to 2 adjacent vertexes - the last vertex & the second & itsefl)
+ from the second vertex is n-3 too ( cannot draw to 2 adjacent vertexes - the first and the third & itself)
+ from the third is n-4 ( cannot draw to itself, the second, the first and forth vertex as second and forth vertex are adjacent to this third vertex, cannot draw to the first because there's a diagonal drawn from first vertex to the third that has been already counted as seen above in the case "draw from the first vertex" )
+ from the forth is n-5
.......
+ and so on ultil n-x equal to Zero
Total of Diagonals: (n-3) + (n-3) + (n-4) + (n-5) + .... + 1 = -1 + (n-2)+ (n-3)+ (n-4) + (n-5) + ... +1
Note that it is the sum of a series subtract to 1. And that series is equal to
{(n-2)+1}* (n-2)/2 - 1
= {(n-1)(n-2)/2} - 1
= (n^2 -3*n +2 )/2 - 1 = (n^2-3*n+2 -2)/2 =(n^2-3*n)/2 = n(n-3)/2 this is the general rule
Apply this to a 5 sided convex polygon for example: total diagonals = 5*(5-3)/2 = 5

For this question's case: n(n-3) = 14
then n must be 7.
Hope this could help