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/12-√3/4)  cm 2  (b) (π/6-√3/4)  cm 2 (c) 4 (π/6-√3/4)  cm 2  (d) 8 (π/6-√3/4)  cm 2

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  1. Class 10
  2. Solutions of Sample Papers for Class 10 Boards

Transcript

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.