A) \[50\frac{1}{7}\,c{{m}^{2}}\]
B) \[50\frac{2}{7}\,c{{m}^{2}}\]
C) \[75\frac{1}{7}c{{m}^{2}}\]
D) \[75\frac{2}{7}\,c{{m}^{2}}\]
Correct Answer: B
Solution :
[b] Side of an equilateral triangle = 8cm  \[\therefore \]Area of an equilateral triangle \[=\frac{\sqrt{3}}{4}\times {{(8)}^{2}}=\frac{\sqrt{3}}{4}=64=16\sqrt{3}\,c{{m}^{2}}\] Now, radius of circumcircle = side of an equilateral triangle \[=8\,\,cm\] and radius of incircle \[=\frac{16\sqrt{3}}{4}=4\sqrt{3}\,cm\] \[\therefore \]Required area \[=[{{(8)}^{2}}-{{(4\sqrt{3})}^{2}}]\,\pi \,c{{m}^{2}}\] \[=(64\,-48)\,\pi \,c{{m}^{2}}\] \[=16\,\pi \,c{{m}^{2}}=16\times \frac{22}{7}\,c{{m}^{2}}=\frac{352}{7}\,c{{m}^{2}}=50\frac{2}{7}\,c{{m}^{2}}\] |
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