A) \[\frac{1}{{{\mu }^{2}}-1}\]
B) \[\frac{\pi }{2\left( {{\mu }^{2}}-1 \right)}\]
C) \[\frac{\pi }{4\left( {{\mu }^{2}}-1 \right)}\]
D) \[\frac{{{\pi }^{2}}{{\mu }^{2}}}{4\left( {{\mu }^{2}}-1 \right)}\]
Correct Answer: C
Solution :
\[\sin \,C=\frac{1}{\mu }\] |
Also, \[\frac{r}{(a/2)}=\tan \,C\] |
\[=\frac{1}{\sqrt{{{\mu }^{2}}-1}}\] |
\[\therefore \] \[r=\frac{a}{2\sqrt{{{\mu }^{2}}-1}}\] |
Area, \[\operatorname{A}=\pi {{r}^{2}}=\frac{\pi {{a}^{2}}}{4({{\mu }^{2}}-1)}\] |
Fraction \[=\frac{\pi {{r}^{2}}}{{{A}^{2}}}=\frac{\pi }{4({{\mu }^{2}}-1)}\] |
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