Question

In a polygon, the interior angles are in A.P with common difference $5^o$ . If the smallest angle of that polygon is ${120}^o$, then the number of sides is

• A. 6
• B. 9
• C. 16
• D. 15

Answer 1

The correct option is B 9

Let number of sides$=$n

$\Rightarrow$ Sum of angles$=(n-1)\pi$

Also sum of A.P of angle with $a=120$, $d=5$

$\Rightarrow (n-2)\pi =\frac{n}{2}(2\times 120+(n-1\left)5\right)$

$\Rightarrow 2n\pi -4\pi =240n+5n^2-4n$

$\Rightarrow 5n^2+(235-2\pi )n+4\pi =0$

$\Rightarrow 5n^2+(235-360)n+720=0$

$\Rightarrow 5n^2-125n+720=0$

$\Rightarrow n^2-25n+144=0$

$\Rightarrow n^2-16n-9n+144=0$

$\Rightarrow n(n-16)-9(n-16)=0$

$\Rightarrow (n-9)(n-16)=0$

$n=9$ or $n=16$

(For $n=16$, the angle would go $>180$ which is not possible)

$\Rightarrow \left(B\right)$.