Find area of blue region in given figure

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Take a look at figure given below. Give it a try and when you are ready then watch the solution. 

To solve this question first we will give notation 'r' for small circle and 'R' for medium sized circle as shown in figure.

Now we will draw a diagonal and it's length will be 2r+4R (R+R+r+r+R+R) as it first act as diagonal for medium sized circle (2R) then acts as diameter of small sized circle (2r) at last it it acts as diameter of other medium sized circle (2R).
∴ 2r + 4R = 4 cm ------(1)

Now join center of 1st circle and 2nd circle as well as 1st circle and 3rd circle. It's length will be 2R and 2R. Now join center of 2nd circle and 3rd circle and it's length will be 2R√2.

Now draw a diameter to large circle passing through center of 2nd and 3rd circle. It's length will be 2R√2 + 2R and it is also diameter of large circle and hence it's length is 4.
∴ 2R√2 + 2R = 4 cm ------(2)

From (1) and (2) we can say that:
2R√2 + 2R = 2r + 4R
R(√2 - 1) = r ------(3)

From 1:
2R√2 + 2R = 4
R(√2 + 1) = 2
R = 2/(√2 + 1) cm
R = 2(√2 - 1) Put this value in (3)
2(√2 - 1)(√2 - 1) = r
r = 2(3 - 2√2) cm

Now,
Area of Medium sized circle
= πR^2
= π[2(√2 - 1)]^2
= π4(3 - 2√2)
= (12 - 8√2)π sq. cm.

And,
Area of Medium sized circle
= πr^2
= π[2(3 - 2√2)]^2
= π4(17 - 12√2)
= (68 - 48√2)π sq. cm

Now we will calculate area of blue region.
Area of blue region = Area of large circle
                                  - 4 × Area of medium sized circle
                                  - Area of small circle
= π2^2 - 4 × π[2(√2 - 1)]^2 - π[2(3 - 2√2)]^2
= 4π - 4 × (12 - 8√2)π - (68 - 48√2)π
= 4π - (48 - 32√2)π - (68 - 48√2)π
= (4 - 48 + 32√2 - 68 + 48√2)π
= (80√2 - 112)π sq. cm
= 16(5√2 - 7)π sq. cm
≈ 3.5723 sq. cm

∴ Area of blue region = 16(5√2 - 7)π sq. cm

Or
∴ Area of blue region ≈ 3.5723 sq. cm