Complex Numbers And Quadratic Equations question 642
Question: 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
Options:
A) $ 2\sqrt{5} $
B) $ 6\sqrt{2} $
C) $ 4\sqrt{5} $
D) $ 6\sqrt{5} $
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Answer:
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_0A_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_0A_1=\sqrt{5}.2\sin ( \frac{\pi }{6} )=\sqrt{5} $ $ ( \because \theta =\frac{2\pi }{6}=\frac{\pi }{3} ) $ Similarly, $ A_1A_2=A_2A_3=A_3A_4=A_4A_5+A_5A_0=6\sqrt{5}. $ Hence the perimeter of, regular polygon is $ =A_0A_1+A_1A_2+A_2A_3+A_3A_4+A_4A_5+A_5A_0=6\sqrt{5}. $
BETA
