A) \[28\,\,c{{m}^{2}}\]
B) \[7\sqrt{3}\,\,c{{m}^{2}}\]
C) \[14\sqrt{3}\,\,c{{m}^{2}}\]
D) \[21\,\,c{{m}^{2}}\]
Correct Answer: C
Solution :
Let the each side of the equilateral triangle be 2x cm. |
\[BD=x\] |
Radius of incircle \[=OD=\frac{1}{3}AD\] |
\[=\frac{1}{3}\sqrt{{{(2x)}^{2}}-{{x}^{2}}}\] |
\[=\frac{\sqrt{3}x}{3}=\frac{x}{\sqrt{3}}\,\,cm\] |
Radius of circum circle \[BO=\sqrt{B{{D}^{2}}+O{{D}^{2}}}\] |
\[=\sqrt{{{x}^{2}}+\frac{{{x}^{2}}}{3}}=\frac{2x}{\sqrt{3}}\,\,cm\] |
According to the question, |
\[\pi {{\left( \frac{2x}{\sqrt{3}} \right)}^{2}}-\pi {{\left( \frac{x}{\sqrt{3}} \right)}^{2}}=44\] |
\[\Rightarrow \] \[\frac{4\pi {{x}^{2}}}{3}-\frac{\pi {{x}^{2}}}{3}=44\] |
\[\Rightarrow \] \[\pi {{x}^{2}}=44\] |
\[\therefore \] \[{{x}^{2}}=\frac{44\times 7}{22}=14\] |
Area of the equilateral triangle |
\[=\frac{\sqrt{3}}{4}\times \text{sid}{{\text{e}}^{2}}=\frac{\sqrt{3}}{4}\times {{(2x)}^{2}}=\sqrt{3}{{x}^{2}}=14\sqrt{3}\,\,\text{sq}\,\,\text{cm}\] |
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