ABCD passes through the centres of the three circles as shown in the figure. AB = 2 cm and CD = 1 cm. If the area of middle circle is the average of the areas of the other two circles, then what is the length (in cm) of BC?

Solution (By Examveda Team)

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