Given below is the picture of the Olympic rings made by taking five congruent circles of radius 1cm each, intersecting in such a way that the chord formed by joining the point of intersection of two circles is also of length 1cm. Total area of all the dotted regions assuming the thickness of the rings to be negligible is

(a) 4 (Ο€/12-√3/4)Β  cm 2 Β (b) (Ο€/6-√3/4)Β  cm 2 (c) 4 (Ο€/6-√3/4)Β  cm 2 Β (d) 8 (Ο€/6-√3/4)Β  cm 2

Ques 37 (MCQ) - Given below is picture of the Olympic rings made by - CBSE Class 10 Sample Paper for 2022 Boards - Maths Standard [MCQ]

part 2 - Question 37 - CBSE Class 10 Sample Paper for 2022 Boards - Maths Standard [MCQ] - Solutions of Sample Papers for Class 10 Boards - Class 10
part 3 - Question 37 - CBSE Class 10 Sample Paper for 2022 Boards - Maths Standard [MCQ] - Solutions of Sample Papers for Class 10 Boards - Class 10

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Question 37 Given below is the picture of the Olympic rings made by taking five congruent circles of radius 1cm each, intersecting in such a way that the chord formed by joining the point of intersection of two circles is also of length 1cm. Total area of all the dotted regions assuming the thickness of the rings to be negligible is (a) 4 (πœ‹/12βˆ’βˆš3/4) cm2 (b) (πœ‹/6βˆ’βˆš3/4) cm2 (c) 4 (πœ‹/6βˆ’βˆš3/4) cm2 (d) 8 (πœ‹/6βˆ’βˆš3/4) cm2 Let’s consider only one circle In Ξ” OAB, all sides are equal ∴ Ξ” OAB is an equilateral triangle So, all angles are 60Β° Area of red shaded region = Area of sector with angle 60Β° and radius 1 cm βˆ’ Area of equilateral triangle with side 1 cm = ΞΈ/(360Β° ) Γ— πœ‹r2 βˆ’ √3/4 π‘Ž^2 = (πŸ”πŸŽΒ° )/(πŸ‘πŸ”πŸŽΒ° ) Γ— πœ‹ (1)2 βˆ’ βˆšπŸ‘/πŸ’ (1)2 = ( πœ‹/(πŸ” ) βˆ’ βˆšπŸ‘/πŸ’ ) cm2 Now, Required Area = 8 Γ— Area of red portion = 8 Γ— ( πœ‹/(πŸ” ) βˆ’ βˆšπŸ‘/πŸ’ ) cm2 So, the correct answer is (d)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo