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​

Answer 1

Answer:

Correct answer is: (d) 8(π/6 - √3/4) cm2

Step-by-step explanation:

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

So ∆OAB is an equilateral triangle and

∴ ∠AOB =60°

Required Area= 8x Area of one segment with r=1cm, ѳ= 60°

Answer 2

Given:

The radius of each ring = 1cm

The length of the chord formed by joining the point of intersection of two circles = 1 cm

To Find:

The total area of all the dotted regions

Solution:

An equilateral triangle is formed by joining the two radii and a chord.

Now, let's consider each circle.

We know that :

The area of each circle is πr²

The area of an equilateral triangle = √3 / 4 X (side)²

The area of a sector of a circle = πr² X Ф / 360°

Half of the dotted region  = Area of the circular sector - Area of the equilateral triangle

So,

Area of the circular sector = $\pi X 1^2 X \frac{60}{360}$

= π / 6 cm²

Area of the equilateral triangle = $\frac{\sqrt{3} }{4} X 1^2$

= √3 / 4 cm²

Half of the dotted region = π / 6 cm² - √3 / 4 cm²

So 1 complete dotted region = 2 X (π / 6 cm² - √3 / 4 cm²)

Since there are 5 circles that form 4 dotted regions,

So the total area = 4 X 2 X (π / 6 cm² - √3 / 4 cm²)

= 8 X (π / 6 - √3 / 4) cm²

Hence, the total area is $8 X (\frac{\pi}{6} - \frac{\sqrt{3} }{4} )$ $cm^2$.