Answer & Solution
Answer: Option A
Solution:
Given, AB = 2CD = 1
Let BC = x
∵ Area of middle circle = Average of areas of other two circle
$$\eqalign{ & \frac{\pi }{4}{\left( {2 + x} \right)^2} = \frac{{\frac{\pi }{4}{{\left( {2 + x + 1} \right)}^2} + \frac{\pi }{4}{{\left( 2 \right)}^2}}}{2} \cr & 2\left[ {\frac{\pi }{4}{{\left( {2 + x} \right)}^2}} \right] = \frac{\pi }{4}{\left( {3 + x} \right)^2} + \frac{\pi }{4} \times 4 \cr & 2{\left( {2 + x} \right)^2} = {\left( {3 + x} \right)^2} + 4 \cr & 2\left( {4 + {x^2} + 4x} \right) = 9 + {x^2} + 6x + 4 \cr & 8 + 2{x^2} + 8x = 9 + {x^2} + 6x + 4 \cr & {x^2} + 2x - 5 = 0 \cr & x = \frac{{ - 2 \pm \sqrt {4 + 20} }}{2} \cr & x = \frac{{ - 2 \pm 2\sqrt 6 }}{2} \cr & x = - 1 \pm \sqrt 6 \cr & \therefore x = \sqrt 6 - 1 = {\text{BC}} \cr} $$
