Diameter of the Circle

Let the distance of the larger chord from the center of the circle be $x$ . Then the distance of the smaller chord from the circle's center is $6-x$ . Let the radius of the circle be $r$ . Then we get $r^2=x^2+\dfrac{121}{4}=(6-x)^2+\dfrac{25}{4}\implies 36-12x=24\implies x=1$ . Hence $r^2=1+\dfrac{121}{4}=\dfrac{125}{4}\implies r=\dfrac{5\sqrt 5}{2}\implies d=2r=5\sqrt 5$ . Therefore the required diameter is $\boxed {5\sqrt 5}$ .