Dear LUANDATO.

I understand you because this is a hard question. But I will try to give you a good explanation.

First, the number of diagonals of a polygon is $$D=\frac{\left(n-3\right)\cdot n}{2}$$ where n is the number of sides.

Now, we have that the number of diagonals of each polygon is {11, 22, 33, 44, 55, 66, 77, 88, 99}.

So, we can set the equations $$\frac{\left(n-3\right)\cdot n}{2}=11,\ 22,\ 33,\ .\ .\ .\ ,\ 99.$$

The only solutions we are interested in are those where n is a positive integer.

The unique numbers of diagonals that satisfy this are 44 and 77. Let's prove it:

$$ \frac{\left(n-3\right)\cdot n}{2}=44\ <=>\ n^2-3n=88\ <=>\ n^2-3n-88=0,$$

The solutions of the quadratic equation above are n=11 and n=-8. We have to select n=11.

$$ \frac{\left(n-3\right)\cdot n}{2}=77\ <=>\ n^2-3n=154\ <=>\ n^2-3n-154=0,$$

The solutions of the quadratic equation above are n=14 and n=-11. We have to select n=14.

So, we have two regular polygons, one has 11 sides and the other has 14 sides.

Finally, the square root of the sum of the sides of both polygons is $$\sqrt{11+14}=\sqrt{25}=5.$$

And this is the number of sides of a Petangon.

So, the answer is option D.

I hope this can help you.

I'm available if you'd like any follow up.

Regards.