Question

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:

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

`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`