Complex Numbers And Quadratic Equations question 333
Question: If centre of a regular hexagon is at origin and one of the vertex on argand diagram is 1 + 2i, then its perimeter is [RPET 1999]
Options:
A) $ 2\sqrt{5} $
B) $ 6\sqrt{2} $
C) $ 4\sqrt{5} $
D) $ 6\sqrt{5} $
Show Answer
Answer:
Correct Answer: D
Solution:
Let the vertices be $ z_0,z_1,…..,z_5w.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_{o}| $ = $ |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 /2) $
Þ $ A_0A_1 $ = $ \sqrt{5}.2\sin ( \frac{\pi }{6} ) $ = $ \sqrt{5} $ $ ( \because \theta =\frac{2\pi }{6}=\frac{\pi }{3} ) $ ….(i) Similarly, $ A_1A_2=A_2A_3=A_3A_4=A_4A_5=A_5A_0=\sqrt{5} $ . Hence the perimeter of, regular polygon is $ =A_{o}A_1+A_1A_2+A_2A_3+A_3A_4+A_4A_5+A_5A_0 $ $ =6\sqrt{5} $ .
BETA
