Given that, three circles are in such a way that each of them touches the other two.
Now, we join centre of all three circles to each other by a line segment. Since, radius of each circle is `3.5cm`.
So, AB = `2xx` Radius of circle
= `2xx3.5 = 7cm`
`rArr` AC = BC = AB = 7cm
which shows that, `DeltaABC` is an equilateral triangle with side 7 cm
we know that , each angle between two adjacent sides of an equilateral triangle is `60^(@)`
`:.` Area of sector with angle `angleA= 60^(@)`
= `(angleA)/(360^(@)) xxpir^(2)= (60^(@))/(360^(@))xx pixx(3.5)_(2)`
So, area of each sector = `3xx` Area of sector with angle A.
= `3xx(60^(@))/(360^(@))xxpixx(3.5)^(2)`
= `(1)/(2)xx(22)/(7)xx3.5xx3.5`
= `11xx(5)/(10)xx(35)/(10)=(11)/(2)xx(7)/(2)`
= `(77)/(4) = 19.25cm^(2)`
and Area of `DeltaABC = sqrt(3)/(4)xx(7)^(2)` [`because` of an equilateral triangle = `sqrt(3)/(4) ("side")^(2)`]
= `49sqrt(3)/(4)cm^(2)`
`:.` Area of shaded region enclosed between these circles = Area of `DeltaABC` - Area of each sector
= `49sqrt(3)/(4)-1925 = 1225 xx sqrt(3)-1925`
= `21.2176-1925 = 1.9676 cm^(2)`
Hence, the required area enclosed between these circles is `1.967 cm^(2)` (approx).
