25

If diagonals are to be made from points of polygon,, then we select two points such that no two are adjacent,, making formula:

diagonals=n(n-3)/2       where n=number of sides;

put    275=n(n-3)/2

n becomes=25.. polygon is of 25 sides.

who said so joyfrancis

How can we prove the formula for finding diagonals of an n sided polygon????

Select a point in ⁿC₁=n ways

now from each point we cannot draw a diagonal to 3 points... 2 points adjacent to it and the point itself

so n-3 such diagonals r there

also..each diagonal has 2 endings so it is counted twice

so D=n(n-3)/2

@ akhil_o

Thanks for the reply. I had got the answer as ⁿc₁(n-3)/2 and was thinking why my formula is not matching. I did not try putting ⁿc₁ as n

One can himself create this formula,by taking small polygons :

No. of sides in polygon                               No. of diagonals

3                                                                        0              = 3*0

4                                                                        2              = 4*1/2

5                                                                       5               = 5*1

6                                                                       9               = 6*3/2

7                                                                       14            = 7*2

Hence we see that D = n(n-3)/2

In the problem,

n(n-3)/2 = 275.

Solving we get, n = 25