question_answer

If centre of a regular hexagon is at origin and one of the vertex on arg and 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 \[|{{z}_{0}}|=\sqrt{5}.\]             \[\Rightarrow \,\,{{A}_{0}}{{A}_{1}}=|{{z}_{1}}-{{z}_{0}}|=|{{z}_{0}}{{e}^{i\theta }}-{{z}_{0}}|\]             \[=|{{z}_{0}}||\cos \theta +i\sin \theta -1|\]             \[=\sqrt{5}\,\sqrt{{{(\cos \theta -1)}^{2}}+{{\sin }^{2}}\theta }\]             \[=\sqrt{5}\,\sqrt{2(1-cos\theta )}=\sqrt{5}\,\,2\sin \theta (\theta /2)\]             \[\Rightarrow \,\,{{A}_{0}}{{A}_{1}}=\sqrt{5}.2\sin \left( \frac{\pi }{6} \right)=\sqrt{5}\]             \[\left( \because \,\,\theta =\frac{2\pi }{6}=\frac{\pi }{3} \right)\] Similarly, \[{{A}_{1}}{{A}_{2}}={{A}_{2}}{{A}_{3}}={{A}_{3}}{{A}_{4}}={{A}_{4}}{{A}_{5}}+{{A}_{5}}{{A}_{0}}=6\sqrt{5}.\] Hence the perimeter of, regular polygon is \[={{A}_{0}}{{A}_{1}}+{{A}_{1}}{{A}_{2}}+{{A}_{2}}{{A}_{3}}+{{A}_{3}}{{A}_{4}}+{{A}_{4}}{{A}_{5}}+{{A}_{5}}{{A}_{0}}=6\sqrt{5}.\]