Question

if three circles of radius 'a' each are drawn such that each touches the other two then prove that the area included between them is equal to 4/25a^2...plz anyone.....plz

Answer 1

According to question

Radius of each circle = a

Let the center of three circles be A, B and C

Thus joining the three circles we get a triangle ABC
in which

AB = BC = AC = 2a

Angle of triangle in each circle = 60 deg
Thus     
Are of each arc = pi/3 a^2
So are of three arcs = pi a^2

Also area of triangle ABC = root 3/4  (2a)^2
                                          = root 3  a^2

Thus Area left = root 3   a^2 - pi   a^2
                        = 3.14 - 1.73 a^2
                        = 1.41 a^2 

Answer 2

Draw three circles which Centre A B C respectively

Now

Join AB BC and CA

We know that the angle subtended by the equilateral triangle is 60 degree in all angles

Now

Required area=( area of the triangle ABC with each side is equal to 12 CM) - 3( area of sector with r = 6 CM)

= root 3/4×2a^-3 pie a^ × 60/360

=1.73a^-1.53a^

=0.20 a^

=20/100 a^

= 4/25 a^