Question

A polygon has $44$ diagonals. Then number of its sides is

• A. $10$
• B. $12$
• C. $13$
• D. $11$

Answer 1

Answer: (D)

$11$

Let the polygon has n sides.

No, of line segments obtained by joining vertices${=}_N{C}_2$

Our of these n lilne aegments are sides are sides of the polygon.

So.

No. of diagonals of the polygon ${=}^n{C}_2-n$

Given, ${}^n{C}_2-n=44$

$\begin{array}{c}\frac{n\left(n-1\right)}{2}-n=44\\ \frac{n\left(n-3\right)}{2}=44\\ n^2-3n-88=0\\ \left(n-11\right)\left(n+8\right)=0\\ n=11orn=-8\end{array}$

Since,

$n$ cannot be negative $n=11$

The polygon has $11$ sides.