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, \[\frac{6{{r}^{2}}}{7}=\frac{150}{847}\Rightarrow {{r}^{2}}=\frac{150}{847}\times \frac{7}{6}=\frac{25}{121}\] \[\therefore \] \[r=\sqrt{\frac{25}{121}}=\frac{5}{11}\,cm.\]You need to login to perform this action.
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