question_answer

 

In the given figure, ABC is an equilateral triangle inscribed in a circle of radius 4 cm with centre O, then thearea of the shaded region is

A) \[\frac{5}{3}\left( 5\pi -3\sqrt{3} \right)c{{m}^{2}}\]

B)        \[\frac{4}{3}\left( 4\pi -3\sqrt{3} \right)c{{m}^{2}}\]

C) \[\frac{2}{3}\left( 2\pi -\sqrt{3} \right)c{{m}^{2}}\]

D) \[\frac{7}{3}\left( 7\pi -3\sqrt{3} \right)c{{m}^{2}}\]

Correct Answer: B

Solution :

We have, R = 4 cm \[\therefore \,\,AB=BC=CA=R\sqrt{3}=4\sqrt{3}\,cm\] \[\left[ \because \,R\frac{2}{3}h\,and\,h=\frac{\sqrt{3}}{2}a;\,\therefore \,R=\frac{a}{\sqrt{3}} \right]\] \[\therefore \]Required area \[=\frac{1}{3}\](Area of the circle Area of \[\Delta ABC\]) \[\therefore \] Required area \[=\frac{1}{3}\left\{ \pi {{R}^{2}}-\frac{\sqrt{3}}{4}\times {{\left( Side \right)}^{2}} \right\}\] \[=\frac{1}{3}\left\{ 16\pi -\frac{\sqrt{3}}{4}\times {{\left( 4\sqrt{3} \right)}^{2}} \right\}\] \[=\frac{1}{3}\left( 16\pi -12\sqrt{3} \right)\] \[=\frac{4}{3}\left( 4\pi -3\sqrt{3} \right)c{{m}^{2}}\]