JEE Main & Advanced AIEEE Solved Paper-2003

  • question_answer
    The sum of the radii of inscribed and circumscribed circles for an n sided regular polygon of side a, is     AIEEE  Solved  Paper-2003

    A)                                         \[a\cot \,\left( \frac{\pi }{v} \right)\]

    B)       \[\frac{a}{2}\cot \left( \frac{\pi }{2n} \right)\]                    

    C) \[a\cot \,\left( \frac{\pi }{2n} \right)\]

    D) \[\frac{a}{4}\cot \left( \frac{\pi }{2n} \right)\]

    Correct Answer: B

    Solution :

    AB = a                (let)                                                            \[ON\bot AB\] and            AN = BN               In \[\Delta AON\],                     \[\tan \frac{\pi }{n}=\frac{AN}{ON}\]                     \[ON=AN\cot \frac{\pi }{n}\]                                 \[=\frac{a}{2}\cot \frac{\pi }{n}\]                              .... (i) and            \[\sin \frac{\pi }{n}=\frac{AN}{OA}\] \[OA=AN\cos ec\frac{\pi }{n}=\frac{a}{2}\cos ec\frac{\pi }{n}\]      ... (ii) Sum of the radii = ON + OA                                 \[=\frac{a}{2}\cot \frac{\pi }{n}+\frac{a}{2}\cos ec\frac{\pi }{n}\]              \[=\frac{a}{2}\left[ \frac{\cos \frac{\pi }{n}}{\sin \frac{\pi }{n}}+\frac{1}{\sin \frac{\pi }{n}} \right]\]              [from Eqs. (i) and (ii)] \[=\frac{a}{2}\left[ \frac{1+\cos \frac{\pi }{n}}{\sin \frac{\pi }{n}} \right]=\frac{a}{2}\left[ \frac{1+2{{\cos }^{2}}\frac{\pi }{2\,n}-1}{2\,\sin \frac{\pi }{2\,n}\cos \frac{\pi }{2\,n}} \right]\] \[=\frac{a}{2}\cos \frac{\pi }{2\,n}\]


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