Let A_1 A_2 A_3 ldots A_9 be a nine-sided reg | vedclass

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(D) Let $O$ be the center of the circumcircle of the regular nonagon $A_1 A_2 \ldots A_9$. The side length is $s = 2$. The central angle subtended by

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$2$

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(D) Let $O$ be the center of the circumcircle of the regular nonagon $A_1 A_2 \ldots A_9$. The side length is $s = 2$. The central angle subtended by each side is $	heta = \frac{2\pi}{9}$. Let $r$ be the circumradius. In $	riangle O A_1 A_2$,by the law of cosines: $s^2 = r^2 + r^2 - 2r^2 \cos(\frac{2\pi}{9}) = 2r^2(1 - \cos(\frac{2\pi}{9})) = 4r^2 \sin^2(\frac{\pi}{9})$. Since $s = 2$,we have $4 = 4r^2 \sin^2(\frac{\pi}{9})$,so $r = \frac{1}{\sin(\frac{\pi}{9})}$. The length of a chord $A_i A_j$ subtending a central angle $\phi$ is $2r \sin(\frac{\phi}{2})$. For $A_1 A_5$,the central angle is $4 	imes \frac{2\pi}{9} = \frac{8\pi}{9}$,so $A_1 A_5 = 2r \sin(\frac{4\pi}{9})$. For $A_2 A_4$,the central angle is $2 	imes \frac{2\pi}{9} = \frac{4\pi}{9}$,so $A_2 A_4 = 2r \sin(\frac{2\pi}{9})$. The difference is $A_1 A_5 - A_2 A_4 = 2r(\sin(\frac{4\pi}{9}) - \sin(\frac{2\pi}{9}))$. Using the identity $\sin C - \sin D = 2 \sin(\frac{C-D}{2}) \cos(\frac{C+D}{2})$: $A_1 A_5 - A_2 A_4 = 2r \cdot 2 \sin(\frac{\pi}{9}) \cos(\frac{3\pi}{9}) = 4r \sin(\frac{\pi}{9}) \cos(\frac{\pi}{3})$. Substituting $r = \frac{1}{\sin(\frac{\pi}{9})}$ and $\cos(\frac{\pi}{3}) = \frac{1}{2}$: $A_1 A_5 - A_2 A_4 = 4 \cdot \frac{1}{\sin(\frac{\pi}{9})} \cdot \sin(\frac{\pi}{9}) \cdot \frac{1}{2} = 2$.

Let $A_1 A_2 A_3 \ldots A_9$ be a nine-sided regular polygon with side length $2$ units. The difference between the lengths of the diagonals $A_1 A_5$ and $A_2 A_4$ equals

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