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प्रश्न
Given below is the picture of the Olympic rings made by taking five congruent circles of radius 1 cm each, intersecting in such a way that the chord formed by joining the point of intersection of two circles is also of length 1 cm. Total area of all the dotted regions assuming the thickness of the rings to be negligible is:
पर्याय
`4(pi/12-sqrt3/4)"cm"^2`
`(pi/6-sqrt3/4)"cm"^2`
`4(pi/6-sqrt3/4)"cm"^2`
`8(pi/6-sqrt3/4)"cm"^2`
MCQ
रिकाम्या जागा भरा
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उत्तर
`bb(8(pi/6-sqrt3/4)"cm"^2)`
Explanation:-
Let O be the center of the circle.
OA = OB = AB = 1 cm.
So ∆OAB is an equilateral triangle.
∴ ∠AOB = 60°
Required Area = 8 × Area of one segment with r = 1 cm, θ = 60°
= `8 xx (60^circ/360^circ xx pi xx 1^2 - sqrt(3)/4 xx 1^2)`
= `8(pi/6 - sqrt3/4)"cm"^2`
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