A) \[\left\{ \frac{5\pi }{3}(5\sqrt{3}+1)-5\sqrt{3} \right\}\]
B) \[\left\{ \frac{4\pi }{2}(4\sqrt{2}+1)-4\sqrt{2} \right\}\]
C) \[\left\{ \frac{4\pi }{3}(4\sqrt{3}+1)-4\sqrt{2} \right\}\]
D) \[\left\{ \frac{4\pi }{3}(4\sqrt{3}+1)-4\sqrt{3} \right\}\]
Correct Answer: D
Solution :
Clearly, \[\Delta ABC\] is an equilateral triangle of side 4 cm.
In \[\Delta BDO,\]we have \[\cos \,\angle OBD=\frac{BD}{OB}\] \[\Rightarrow \] \[\cos {{30}^{o}}=\frac{2}{OB}\] \[\Rightarrow \] \[\frac{\sqrt{3}}{2}=\frac{2}{OB}\Rightarrow \frac{\sqrt{3}}{2}=\frac{2}{OB}\Rightarrow OB=\frac{4}{\sqrt{3}}\] \[\therefore \] \[OP=OB+BP\Rightarrow R\left( \frac{4}{\sqrt{3}}+2 \right)cm\] Area of the shaded region = Area of the larger circle of radius \[R-3\times \]Area of a smaller circle of radius \[2\text{ }cm+3\](Area of a sector of angle \[{{60}^{o}}\] in a circle of radius 2 cm) - {Area of \[\Delta \text{ }ABC-3\](Area of sector of angle \[{{60}^{o}}\] in a circle of radius 2 cm)} \[\Rightarrow \] Area of the shaded region = Area of the larger circle of radius \[R-3\text{ }x\]Area of smaller circle of radius \[2\text{ }cm+6\times \]Area of a sector of angle \[{{60}^{o}}\]in a circle of radius \[2\text{ }cm-\]Area of \[\Delta ABC\] \[=\left\{ \begin{align} & \pi {{\left( \frac{4}{\sqrt{3}}+2 \right)}^{2}}-3\times \pi \times {{2}^{2}}+6\times \\ & \left( \frac{60}{360}\times \pi \times {{2}^{2}} \right)-\frac{\sqrt{3}}{4}\times {{4}^{2}} \\ \end{align} \right\}c{{m}^{2}}\] \[=\left\{ \pi \left( \frac{16}{3}+4+\frac{16}{\sqrt{3}} \right)-12\pi +4\pi -4\sqrt{3} \right\}c{{m}^{2}}\] \[=\left\{ \pi \left( \frac{4}{3}+\frac{16}{\sqrt{3}} \right)-4\sqrt{3} \right\}c{{m}^{2}}\] \[=\left\{ \frac{4\pi }{3}(4\sqrt{3}+1)-4\sqrt{3} \right\}c{{m}^{2}}\]
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