Question

A polygon has $27$ diagonals. How many sides does it have?

Answer 1

Let $n$ be the number of vertices of the polygon.
then there are $n$ sides.

Number of all connections between all the vertices $=N=\frac{n(n-1)}{2}$

Thus, number of diagonals $=N-n$
$=\frac{n(n-1)}{2}-n$
$=n\times \frac{(n-1-2)}{2}$
$=\frac{n\times (n-3)}{2}$
But actual given number of diagonals$=27$

Therefore, $\frac{n\times (n-3)}{2}=27$

$n^2-3n=54$
$n^2-3n-54=0$
$n^2-9n+6n-54=0$
$n(n-9)+6(n-9)=0$
$(n-9)(n+6)=0$
$(n-9)=0,(n+6)=0$
$\Rightarrow n=9,n=-6$
$\Rightarrow n=9$ [Since, $n$ can't be negative]

Therefore, there are $9$ sides and $9$ vertices.