The difference between the areas of the circumcircle and the incircle of a regular polygon of $n$ sides with each side of length $2a$, is

### Concept & Incircle and Circumcircle Geometry For any regular polygon, the circumradius ($R$), the inradius ($r$), and half of the side length form a right-angled triangle. By using the Pythagorean theorem on this fundamental triangle, we can find the difference in their areas without needing the number of sides $n$. $$ R^2 = r^2 + (\text{Half-side})^2 $$ ### Step-by-Step Solution * Let the regular polygon have $n$ sides, each of length $2a$. * Let the center of the polygon be $O$. * The distance from $O$ to a vertex is the circumradius, $R$. * The perpendicular distance from $O$ to the midpoint of a side is the inradius, $r$. * This forms a right-angled triangle with the hypotenuse $R$, height $r$, and base equal to half the side length, which is $\frac{2a}{2} = a$. * By the Pythagorean theorem: $R^2 = r^2 + a^2$. * Rearranging gives: $R^2 - r^2 = a^2$. * The area of the circumcircle is $\pi R^2$. * The area of the incircle is $\pi r^2$. * The difference in their areas = $\pi R^2 - \pi r^2 = \pi (R^2 - r^2)$. * Substitute $R^2 - r^2 = a^2$ into the equation: Area Difference = $\pi a^2$. ### Exam Strategy & Shortcut Notice that the options are mostly independent of $n$ (Option A). In geometry, if a relationship holds for an $n$-sided polygon, it must hold for the simplest one. Assume a square ($n=4$) with side $2a$. Inradius $r = a$, circumradius $R = a\sqrt{2}$. Area difference = $\pi(a\sqrt{2})^2 - \pi(a)^2 = 2\pi a^2 - \pi a^2 = \pi a^2$. ### Common Pitfall Students often assume the area difference depends on the number of sides $n$ and mistakenly guess a complex option like $(2n + 1)\pi a^2$. ### Final Answer Therefore, the correct answer is **$\pi a^2$**.