If three circles of radius a each, are drawn such that each touches the 4 other two, prove that the area included between them is equal to [Take √3 = 1.73 and π = 3.14.]

Consider the figure shown below


 



 


Here, first we join the center of all adjacent circles then the distance between the center of circles touching each other is equal to the side of an equilateral triangle formed by joining the center of adjacent circles. Therefore, we can say that the side of the equilateral triangle is equal to the twice of the radius of circle. Now by simply calculating the area of the 3 sectors and then subtracting it from the area of the equilateral triangle we can easily calculate the area of the enclosed region.


 


Given radius of each circle = “a” units


 


Central angle of each sector = θ = 60° ( ∆ABC is equilateral)


 


Side of equilateral ∆ABC = 2×a units


 



 



 



 


Area of all 3sectors = 3×Area of one sector


 



 



 



 



 



 


Area of enclosed region = Area of equilateral ∆ABC – Area of all 3 sectors


 


Area of enclosed region


 



 



 



 


(taking a2 common)


 



 



 



 



 

40