A geometry problem regarding polygons, angles, and diagonals.

Question

The sum of the interior angles in polygon $M$ is $1980$. If the number of diagonals in polygon $N$ is 70 more than the number of diagonals in polygon $N$, then what is the sum of the number of sides on polygons on polygons $M$ and $N$?

Answer 1

The formula for the sum of the interior angles of a polygon is $180(x-2)$ where $x$ is the number of sides.

$180(x-2)=1980$
$x-2=11$
$x=13$

The formula for the number of diagonals in a polygon is $\frac {x(x-3)}2$ where x is the number of sides.

$\frac{13(13-3)}2=65$
$135=\frac{x(x-3)}2$
$x^2-3x-270=0$
$(x-18)(x+15)=0$

The negative value is extraneous, so the number of sides on polygon $N$ is 18.

Therefore, the answer is $13+18=31$.