question_answer

If centre of a regular hexagon is at origin and one of the vertex on argand diagram is \[1 +2i\], then its perimeter is

A) \[2\sqrt{5}\]      

B)        \[6\sqrt{2}\]

C) \[4\sqrt{5}\]      

D)        \[6\sqrt{5}\]

Correct Answer: D

Solution :

Let the vertices be \[{{z}_{0}},\,\,{{z}_{1}},...,\,{{z}_{5}}\] w.r.t. centre O at origin and \[\left| {{z}_{0}} \right|=\sqrt{5}\]  \[\Rightarrow \,\,\,\,{{\operatorname{A}}_{0}}{{A}_{1}}=\,\,\left| {{z}_{1}}-{{z}_{0}} \right|\,\,=\,\,\left| {{z}_{0}}{{e}^{i\theta }}-\,\,{{z}_{0}} \right|\] \[=\,\,\left| {{z}_{0}} \right|\left| \cos \theta +i\,\,\sin \theta -1 \right|\] \[=\,\,\sqrt{5}\sqrt{{{(cos\,\theta -1)}^{2}}+{{\sin }^{2}}\theta }\] \[=\,\,\,\sqrt{5}\sqrt{2(1-cos\theta )}\] \[=\,\,\,\sqrt{5}\,\,2\,\sin (\theta /2)\] \[\Rightarrow \,\,{{A}_{0}}{{A}_{1}}=\sqrt{5}.2\,\sin \,\left( \frac{\pi }{6} \right)=\sqrt{5}\] \[\left( \because \,\,\,\theta =\frac{2\pi }{6}=\frac{\pi }{3} \right)\] Similarly, \[{{A}_{1}}{{\operatorname{A}}_{2}}={{A}_{2}}{{A}_{3}}={{A}_{3}}{{A}_{4}}={{A}_{4}}{{A}_{5}}={{A}_{5}}{{A}_{0}}=\,\,\sqrt{5}\] Hence the perimeter of, regular polygon is \[={{A}_{0}}{{A}_{1}}+{{A}_{1}}{{A}_{2}}+{{A}_{2}}{{A}_{3}}+{{A}_{3}}{{A}_{4}}+{{A}_{4}}{{A}_{5}}+{{A}_{5}}{{A}_{0}}=6\sqrt{5}\]