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

Given:- 

(1) Three circles of radius(r) 7 cm are placed in

such a way that each circle touches the other two

Formula used:-

(1)  Area of a Sector of Circle =   (θ /360) × πr² 

[ where, r = Radius ]

(2) Area of equilateral triangle =   (√3/4 )a²  

[where, a = Length of the side ] 

Solution:-   

We can conclude that the centers of circles form

an equilateral triangle with =  r + r = 2r   

The area enclosed between circles =
 
Area of a triangle - 3 (Area of the sector)  

The angle subtended by sector in an equilateral triangle = 60 

Area of a Sector of Circle =   (θ /360) × πr2  

⇒ ( 60/360) × (22/7) × 7²  ⇒ 77/3 
  
3 (Area of the sector)   =77 ----(1)

Area of equilateral triangle =   (√3/4 )a2   
 
a = Length of the side = 2r = 2(7) =14 cm  

⇒  (√3/4 ) × 14² = 49√3  ---(2)

Subtraction equation(2) - equation(1)

∴ The area enclosed between circles = 49√3 - 77 cm².