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Question: If centre of a regular hexagon is at origin and one of the vertex on argand diagram is \(1 + 2i,\) t...

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

A

252\sqrt{5}

B

626\sqrt{2}

C

454\sqrt{5}

D

656\sqrt{5}

Answer

656\sqrt{5}

Explanation

Solution

Sol. Let the vertices be z0,z1,.....,z5w.r.t.z_{0},z_{1},.....,z_{5}w.r.t. centre O and ∣z0∣=5|z_{0}| = \sqrt{5}

⇒\Rightarrow A0A1=∣z1−z0∣=∣z0eiθ−z0∣A_{0}A_{1} = |z_{1} - z_{0}| = |z_{0}e^{i\theta} - z_{0}| =∣z0∣∣cos⁡θ+isin⁡θ−1∣= |z_{0}||\cos\theta + i\sin\theta - 1| =5(cos⁡θ−1)2+sin⁡2θ= \sqrt{5}\sqrt{(\cos\theta - 1)^{2} + \sin^{2}\theta}

⇒=52(1−cos⁡θ)\Rightarrow = \sqrt{5}\sqrt{2(1 - \cos\theta)} =5.2sin⁡(θ/2)= \sqrt{5}.2\sin(\theta/2)

⇒\Rightarrow A0A1=5.2sin⁡(π/6)=5A_{0}A_{1} = \sqrt{5}.2\sin(\pi/6) = \sqrt{5}

(∵θ=2π6=π3)\left( \because\theta = \frac{2\pi}{6} = \frac{\pi}{3} \right) .....(i)

SimilarlyA1A2=A2A3=A3A4=A4A5=A5A0=5A_{1}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

=A0A1+A1A2+A2A3+A3A4+A4A5+A5A0=65= 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}