| In the given figure, ABC is an equilateral triangle inscribed in a circle of radius 4 cm with centre O, then the area of the shaded region is: |
 |
A) \[\frac{5}{3}(5\pi -3\sqrt{3})c{{m}^{2}}\]
B) \[\frac{4}{3}(4\pi -3\sqrt{3})c{{m}^{2}}\]
C) \[\frac{2}{3}(2\pi -\sqrt{3})c{{m}^{2}}\]
D) \[\frac{7}{3}(7\pi -3\sqrt{3})c{{m}^{2}}\]
Correct Answer: B
Solution :
| [b] We have, \[R=4cm\] |
| Hence, \[AB=BC=CA=R\sqrt{3}=4\sqrt{3}\] |
| \[\left[ Since,\,\,R=\frac{2}{3}h\,\,and\,\,h=\frac{\sqrt{3}}{2}a;\,Hence,\,R=\frac{a}{\sqrt{3}} \right]\] |
| \[\angle AOC=2\angle ABC\] |
| \[=2\times 60{}^\circ =120{}^\circ \] |
| Hence, Required area |
| \[=\frac{1}{3}(Area\,of\,the\,circle\,-Area\,of\,\Delta ABC)\]Required area \[=\frac{1}{3}\left\{ \pi {{R}^{2}}-\frac{\sqrt{3}}{4}\times {{(Side)}^{2}} \right\}\] |
| \[=\frac{1}{3}\left\{ 16\pi -\frac{\sqrt{3}}{4}\times {{(4\sqrt{3})}^{2}} \right\}\] |
| \[=\frac{1}{3}(16\pi -12\sqrt{3})=\frac{4}{3}(4\pi -3\sqrt{3})c{{m}^{2}}\] |
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