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

• A. $2\sqrt{5}$
• B. $6\sqrt{2}$
• C. $4\sqrt{5}$
• D. $6\sqrt{5}$

Answer 1

Answer: (D)

Let the vertices be $z_0,z_1,\dots ,z_5$ w.r.t. centre $O$ at origin and $\left|z_0\right|=\sqrt{5}$.

$\Rightarrow A_0{A}_1=\left|z_1-z_0\right|=\left|z_0{e}^{i\theta }-z_0\right|=\left|z_0\right||\cos \theta +i\sin \theta -1|$
$=\sqrt{5}\sqrt{(\cos \theta -1{)}^2+{\sin }^2\theta }=\sqrt{5}\sqrt{2(1-\cos \theta )}=\sqrt{5}2\sin (\theta /2)$
$\Rightarrow A_0{A}_1=\sqrt{5}.2\sin \left(\frac{\pi }{6}\right)=\sqrt{5}\left(\because \theta =\frac{2\pi }{6}=\frac{\pi }{3}\right)$
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
$=A_0{A}_1+A_1{A}_2+A_2{A}_3+A_3{A}_4+A_4{A}_5+A_5{A}_0=6\sqrt{5}$