A) \[\frac{3}{4}\]
B) \[3\sqrt{3}\]
C) 3
D) \[\frac{3\sqrt{3}}{2}\]
Correct Answer: C
Solution :
Let the vertices be \[{{z}_{0}},{{z}_{1}},........{{z}_{5}}\]w.r.t centre O as origin \[|{{z}_{0}}|=1,\] \[{{A}_{0}}{{A}_{1}}=|{{z}_{1}}-{{z}_{0}}|=|{{z}_{0}}{{e}^{i\theta }}-{{z}_{0}}|\] \[\therefore \,\,{{A}_{0}}{{A}_{1}}=|{{z}_{0}}||cos\theta +isin\theta -1|\] \[=1.\sqrt{{{(\cos \theta -1)}^{2}}+{{\sin }^{2}}\theta }=\sqrt{2(1-\cos \theta )}\] \[\therefore \,\,\,{{A}_{0}}{{A}_{1}}=\sqrt{2.2{{\sin }^{2}}\frac{\theta }{2}}=2\sin \frac{\theta }{2}\] Where \[\theta =\frac{2\pi }{6}=\frac{\pi }{3}.\] Replacing \[\theta \] by \[2\theta \] and \[4\theta \] we get, \[{{A}_{0}}{{A}_{2}}=2\sin \frac{2\theta }{2}=2\sin \theta \] & \[{{A}_{0}}{{A}_{4}}=2\sin \frac{4\theta }{2}=2\sin 2\theta \] \[\therefore \,\,({{A}_{0}}{{A}_{1}})({{A}_{0}}{{A}_{2}})({{A}_{0}}{{A}_{4}})\] \[=8\sin \frac{\pi }{6}\sin \frac{\pi }{3}\sin \frac{2\pi }{3}\] \[=8\left( \frac{1}{2} \right)\left( \frac{\sqrt{3}}{2} \right)\left( \frac{\sqrt{3}}{2} \right)=3\]
You need to login to perform this action.
You will be redirected in
3 sec
