question_answer

Four equal discs are placed such that each one touches two others. If the area of empty space enclosed by them is \[\frac{150}{847}\] square centi metre, then the radius of each disc is equal to

A)   \[\frac{7}{6}\,cm\]                           

B)  \[\frac{5}{6}\,cm\]

C)  \[\frac{1}{2}\,cm\]                

D)       \[\frac{5}{11}\,cm\]

Correct Answer: D

Solution :

  Radius of each disc \[=r\,cm\,\] (let)             \[\therefore \]  \[AB=BC=CD=DA=2r\,\,cm.\]             \[\therefore \]  Area of ABCD \[=4{{r}^{2}}\,sq.\,cm\]        \[\therefore \]   Area of shaded portion \[=4{{r}^{2}}-4\times \frac{{{90}^{o}}}{{{360}^{o}}}\times \pi {{r}^{2}}=4{{r}^{2}}-\frac{22}{7}{{r}^{2}}\] \[={{r}^{2}}\left( 4-\frac{22}{7} \right)={{r}^{2}}\left( \frac{28-22}{7} \right)=\frac{6{{r}^{2}}}{7}\,sq.cm\] According to the question, \[\frac{6{{r}^{2}}}{7}=\frac{150}{847}\Rightarrow {{r}^{2}}=\frac{150}{847}\times \frac{7}{6}=\frac{25}{121}\] \[\therefore \]    \[r=\sqrt{\frac{25}{121}}=\frac{5}{11}\,cm.\]