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

Last updated at April 16, 2024 by Teachoo

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
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2
}
(b) (π/6-√3/4) cm
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2
}
(c) 4 (π/6-√3/4) cm
^{
2
}
(d) 8 (π/6-√3/4) cm
^{
2
}

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/12ββ3/4) cm2 (b) (π/6ββ3/4) cm2 (c) 4 (π/6ββ3/4) cm2 (d) 8 (π/6ββ3/4) cm2
Letβs consider only one circle
In Ξ OAB, all sides are equal
β΄ 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
β Area of equilateral triangle with side 1 cm
= ΞΈ/(360Β° ) Γ πr2 β β3/4 π^2
= (ππΒ° )/(πππΒ° ) Γ π (1)2 β βπ/π (1)2
= ( π/(π ) β βπ/π ) cm2
Now,
Required Area = 8 Γ Area of red portion
= 8 Γ ( π/(π ) β βπ/π ) cm2
So, the correct answer is (d)

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, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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