Three circles of radius 7 cm are placed in such a way that each circle touches the other two. What will be the are of the portion enclosed by these circles ?

$ 49\sqrt{3} - 77 cm^2$

We can conclude that the centres of circles form an equilateral triangle with = r + r = 2r

The area enclosed between circles =

Area of a triangle - 3 (Area of sector)

The angle subtended by sector in an equilateral triangle = 60

Area of a sector of circle = \(\theta \)/360 x \(\Pi \)\( {r }^{2 } \)

⇒ \(\frac{60}{360}\) x \(\frac{22}{7}\) x \( {7 }^{2 } \) = \(\frac{77}{3}\)

3 (Area of sector)= 77    ..(1)

Area of equilateral triangle = (√3/4)\( {a }^{2 } \)

a = Length of the side = 2r = 2(7) = 14 cm

⇒ (√3/4) x \( {14}^{2 } \) = 49\(\sqrt {3 }\)       ..(2)

Subtracting equation 2 - equation 1

Therefore, the area enclosed between circles is $ 49\sqrt{3} - 77 cm^2$