A) There is a regular polygon with \[\frac{r}{R}=\frac{1}{2}\]
B) There is a regular polygon with \[\frac{r}{R}=\frac{1}{\sqrt{2}}\]
C) There is a regular polygon with \[\frac{r}{R}=\frac{2}{3}\]
D) There is a regular polygon with \[\frac{r}{R}=\frac{\sqrt{3}}{2}\]
Correct Answer: C
Solution :
\[\tan \left( \frac{\pi }{n} \right)=\frac{\frac{x}{2}}{r}=\frac{x}{2r}\] \[r=\frac{x}{2}\cot \left( \frac{\pi }{n} \right)\] and \[\sin \frac{\pi }{n}=\frac{x}{2R}\] \[R=\frac{x}{2}\cos ec\frac{\pi }{n}\] \[\frac{r}{R}=\frac{\cot \left( \frac{\pi }{n} \right)}{\cos ec\left( \frac{\pi }{n} \right)}=\cos \left( \frac{\pi }{n} \right)\] \[n=3,\frac{r}{R}=\frac{1}{2}=.5\] \[n=4,\frac{r}{R}=\frac{1}{\sqrt{2}}=.707\] \[n=5,\frac{r}{R}=\frac{2}{3}=.6\] \[n=6,\frac{r}{R}=\frac{\sqrt{3}}{2}\] is not possible because .6 comes between \[n=3\]and \[n=4\]but no integer between\[n=3\]and\[n=4\]
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