if a regular 9-sided polygon was circumscribed around a circle, what is the difference between the areas of the polygons
Hello tomatoesYou don't say what the radius of the circle is (so I'll call it $\displaystyle r$), and I assume that you want the difference between the area of the polygon and the area of the circle.
So, imagine the polygon made up of nine isosceles triangles, with a common vertex at the centre of the circle. Then cut one of these triangles in half by a line from the centre to the point where the side of the polygon touches the circle. This line is $\displaystyle r$ units long, and the angle at the centre of the circle in this triangle is $\displaystyle \frac{360}{18}= 20^o$. From this, you should be able to see that the lines joining the centre of the circle to the vertices of the polygon have length $\displaystyle \frac{r}{\cos20^o}$.
Using the formula for the area of a triangle $\displaystyle \tfrac12bc\sin A$, this gives the area of one of the nine original triangles as $\displaystyle \frac12\Big(\frac{r}{\cos20^o}\Big)^2\sin 40^o$.
Multiply this by $\displaystyle 9$, and subtract $\displaystyle \pi r^2$, and you're done. OK?
Grandad