Three concentric circles have radii (in cm) a, b and c, where a < b < c. If a = 8 and b = 9 and the middle circle bisects the area between the other two circles, then the value of c is
It is given that $$a=8$$ cm and $$b=9$$ cm. Also, $$A_2=\frac{A_1+A_3}{2}$$ ------------(i)
To find : $$c=?$$
Solution : Area of a circle = $$\pi r^2$$
Using equation (i),
=> $$\pi (b)^2=\frac{\pi (a)^2+\pi(c)^2}{2}$$
=> $$a^2+c^2=2b^2$$
=> $$64+c^2=162$$
=> $$c^2=162-64=98$$
=> $$c=\sqrt{98}=7\sqrt2$$ cm
=> Ans - (A)