$\begin{array}{1 1} 8 \\ 11 \\ 12 \\ 13 \end{array}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Let the polygon be n sided.

The no. of lines that can be drawn using the vertices = $^nC_2$

This includes n lines which are sides.

$\therefore$ No. of diagonals = $^nC_2-n=\large\frac{n(n-1)}{2}$$-n=44\:(given)$

$\Rightarrow\:n^2--3n-88=0$

$\Rightarrow\:(n+8)(n-11)=0$

$\Rightarrow\:n=11\:\:or\:\:-8\:(not\:possible.)$

$\therefore\:n=11$

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...