A) \[\pi {{k}^{2}}\]
B) \[3\pi {{k}^{2}}\]
C) \[5\pi {{k}^{2}}\]
D) \[2\pi {{k}^{2}}\]
Correct Answer: A
Solution :
(a): Put perpendicular from centre to the tangent, it will bisect the chord. Then, by Pythagoras theorem \[={{R}^{2}}-{{r}^{2}}={{k}^{2}}\] \[\therefore \] Area of shaded region \[=\pi {{R}^{2}}-\pi {{r}^{2}}\] \[=\pi \left( {{R}^{2}}-{{r}^{2}} \right)=\pi {{k}^{2}}\] Remarks: This is a very good question to test the intelligence of students. One has to neither calculate radius of larger circle ?R? or radius of smaller circle, ?r?. \[\left( {{R}^{2}}-{{r}^{2}} \right)\] is sufficient as we need to find n \[\pi \left( {{R}^{2}}-{{r}^{2}} \right)\] only.
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