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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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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/12−√3/4) cm2 (b) (𝜋/6−√3/4) cm2 (c) 4 (𝜋/6−√3/4) cm2 (d) 8 (𝜋/6−√3/4) cm2 Let’s consider only one circle In Δ OAB, all sides are equal ∴ Δ OAB is an equilateral triangle So, all angles are 60° Area of red shaded region = Area of sector with angle 60° and radius 1 cm − Area of equilateral triangle with side 1 cm = θ/(360° ) × 𝜋r2 − √3/4 𝑎^2 = (𝟔𝟎° )/(𝟑𝟔𝟎° ) × 𝜋 (1)2 − √𝟑/𝟒 (1)2 = ( 𝜋/(𝟔 ) − √𝟑/𝟒 ) cm2 Now, Required Area = 8 × Area of red portion = 8 × ( 𝜋/(𝟔 ) − √𝟑/𝟒 ) 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 12 years. He provides courses for Maths and Science at Teachoo.