question_answer

The area of an equilateral triangle which is inscribed in a circle of radius la cm, is

A)  \[\sqrt{3{{a}^{2}}}\]sq cm     

B)  \[3\sqrt{3}{{a}^{2}}\]sq cm

C)  \[\frac{{{a}^{2}}}{3}\]sq cm  

D)  \[2\sqrt{a}\]sq cm

Correct Answer: B

Solution :

\[\sin 30{}^\circ \,=\,\frac{1}{2}\,=\,\frac{OD}{2a}\] \[OD=\frac{2a}{2}\,=\,a\] \[\frac{1}{3}(AD)=OD\] \[\Rightarrow \]   \[AD=3OD=3a\] \[\,\sin 60{}^\circ =\frac{AD}{AB}\] \[\frac{\sqrt{3}}{2}=\frac{3a}{AB}\] \[AB=2\sqrt{3a}\] Side of triangle \[=\,2\sqrt{3a}\] Area of triangle \[=\frac{\sqrt{3}}{4}\times {{(side)}^{2}}\] \[=\frac{\sqrt{3}}{4}\times {{(2\sqrt{3}a)}^{2}}\] \[=\frac{\sqrt{3}}{4}\times 4\times 3{{a}^{2}}\] \[\,=3\sqrt{3}{{a}^{2}}sq\,\,cm\]