question_answer

If the difference between areas of the circumcircle and the in circle of an equilateral triangle is \[44\,\,c{{m}^{2}},\]then the area of the triangle is \[{{(A+B)}^{2}}\]

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

\[=\frac{\sqrt{3}}{4}\times \text{sid}{{\text{e}}^{2}}=\frac{\sqrt{3}}{4}\times {{(2x)}^{2}}=\sqrt{3}{{x}^{2}}=14\sqrt{3}\,\,\text{sq}\,\,\text{cm}\]