There is a square board of side '2a' units circumscribing a red  circle. Jayadev  asked to keep a dot on the above said board. The probability that he keeps the dot on the shaded region is. 

(a) π/4   (b) (4-π)/4   (c) (π - 4)/4    (d) 4/π

There is a square board of side ‘2a’ units circumscribing a red circle - CBSE Class 10 Sample Paper for 2024 Boards - Maths Standard

part 2 - Question 16 - CBSE Class 10 Sample Paper for 2024 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards - Class 10
part 3 - Question 16 - CBSE Class 10 Sample Paper for 2024 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards - Class 10

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Transcript

Probability of placing a dot on shaded region = (š“š‘Ÿš‘’š‘Ž š‘œš‘“ š‘ ā„Žš‘Žš‘‘š‘’š‘‘ š‘Ÿš‘’š‘”š‘–š‘œš‘› (š‘š‘™š‘¢š‘’))/(š‘‡š‘œš‘”š‘Žš‘™ š‘Žš‘Ÿš‘’š‘Ž) = (š‘Øš’“š’†š’‚ š’š’‡ š’”š’’š’–š’‚š’“š’† āˆ’ š‘Øš’“š’†š’‚ š’š’‡ š’„š’Šš’“š’„š’š’†)/(š‘Øš’“š’†š’‚ š’š’‡ š’„š’Šš’“š’„š’š’†) Area of square Side = 2a Area = 怖(2š‘Ž)怗^2 =怖 šŸ’š’‚ć€—^šŸ units Area of circle Radius of circle = 2š‘Ž/2 = a Area of circle = Ļ€š‘Ž^2 Now, Probability of placing a dot on shaded region = (š‘Øš’“š’†š’‚ š’š’‡ š’”š’’š’–š’‚š’“š’† āˆ’ š‘Øš’“š’†š’‚ š’š’‡ š’„š’Šš’“š’„š’š’†)/(š‘Øš’“š’†š’‚ š’š’‡ š’„š’Šš’“š’„š’š’†) = (4š‘Ž^2āˆ’šœ‹š‘Ž^2)/(4š‘Ž^2 ) = ((4 āˆ’ šœ‹ ) š‘Ž^2)/怖 4š‘Žć€—^2 = (4 āˆ’ šœ‹)/4So, the correct answer is (b)

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