Question

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 ______.

Options

• `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`

Answer 1

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 `underline(8(pi/6-sqrt3/4)"cm"^2)`.

Explanation:-

Let O be the center of the circle. OA = OB = AB =1cm.

So ∆OAB is an equilateral triangle and

∴ ∠AOB =60°

Required Area = 8 × Area of one segment with

r = 1cm, θ = 60°

`=8xx(60/360xxpixx1^2-sqrt3/4xx1^2)`

`=8(pi/6-sqrt3/4)"cm"^2`