question_answer

AB is a line segment of length 4 cm. P is the mid-point of AB. Circles are drawn with A, P and B as centres and radii AP = PB (see figure). The area of the shaded portion (in\[c{{m}^{2}}\]) is _____.    

A)  \[6\sqrt{3}\]               

B)  \[2\pi -6\sqrt{3}\]        

C)        \[2\pi -3\sqrt{3}\]        

D)        \[3\sqrt{3}\]

Correct Answer: C

Solution :

Join DP and CP. Now, 3 sectors are formed each subtending an angle of \[{{60}^{o}}\] at P. Area of each sector \[=\frac{\theta }{{{360}^{o}}}\times \pi {{r}^{2}}\] \[=\frac{{{60}^{o}}}{{{360}^{o}}}\times \pi \times {{(2)}^{2}}=\frac{4\pi }{6}=\frac{2\pi }{3}c{{m}^{2}}\] And area of  \[\Delta \,DCP=\frac{\sqrt{3}}{4}\,{{(PC)}^{2}}\]                         \[=\frac{\sqrt{3}}{4}\times 4=\sqrt{3}\,c{{m}^{2}}\] Therefore, area of shaded portion             \[=\left( \frac{2\pi }{3}-\sqrt{3} \right)\times 3=2\pi -3\sqrt{3}\]