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(D) The area of a regular $n$-sided polygon inscribed in a circle of radius $r=1$ is given by $A_n = n 	imes \frac{1}{2} 	imes r^2 	imes \sin\left(

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$\frac{X}{5} > \frac{Y}{6} > \frac{Z}{7}$ and $X < Y < Z$

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(D) The area of a regular $n$-sided polygon inscribed in a circle of radius $r=1$ is given by $A_n = n imes \frac{1}{2} imes r^2 imes \sin\left(\frac{2\pi}{n} ight) = \frac{n}{2} \sin\left(\frac{2\pi}{n} ight)$.For a pentagon $(n=5)$,$X = \frac{5}{2} \sin\left(\frac{2\pi}{5} ight)$.For a hexagon $(n=6)$,$Y = \frac{6}{2} \sin\left(\frac{2\pi}{6} ight)$.For a heptagon $(n=7)$,$Z = \frac{7}{2} \sin\left(\frac{2\pi}{7} ight)$.Dividing by $n$,we get $\frac{X}{5} = \frac{1}{2} \sin\left(\frac{2\pi}{5} ight)$,$\frac{Y}{6} = \frac{1}{2} \sin\left(\frac{2\pi}{6} ight)$,and $\frac{Z}{7} = \frac{1}{2} \sin\left(\frac{2\pi}{7} ight)$.Since the function $f( heta) = \sin( heta)$ is decreasing for $ heta \in (0, \pi/2)$,and $\frac{2\pi}{5} > \frac{2\pi}{6} > \frac{2\pi}{7}$,it follows that $\sin\left(\frac{2\pi}{5} ight) > \sin\left(\frac{2\pi}{6} ight) > \sin\left(\frac{2\pi}{7} ight)$.Thus,$\frac{X}{5} > \frac{Y}{6} > \frac{Z}{7}$.As $n$ increases,the area of the inscribed polygon approaches the area of the circle $(\pi r^2 = \pi)$,so $X < Y < Z$.

Let $X, Y, Z$ be respectively the areas of a regular pentagon,regular hexagon,and regular heptagon which are inscribed in a circle of radius $1$. Then,

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