Question

If a polygon of $n$ sides has $\frac{1}{2}n(n-3)$ diagonals. How many sides will a polygon having $65$ diagonal. Is there a polygon with $50$ diagonals?

Answer 1

Number of diagonals $=\frac{1}{2}n(n-3)$

If diagonals$=65$

$n^2-3n-130=0$

$n^2-13n+10n-130=0$

$n(n-13)+10(n-13)=0$

$(n-13)(n+10)=0$

$n=13$

50 diagonals are not possible because the roots are not integers.