The area of the ring between two concentric circles is 1212π square inches. The length of a chord of the larger circle tangent to the smaller circle, in inches, is:

1969 AMC 12 Problem 6

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The area of the ring between two concentric circles is

12\dfrac12\pi

square inches. The length of a chord of the larger circle tangent to the smaller circle, in inches, is:

\dfrac5{\sqrt2}

5

5\sqrt2

10

10\sqrt2

If the radii are

R

and

r,

the area gives

R^2-r^2=\frac{25}{2}

Half the tangent chord, the smaller radius, and the larger radius form a right triangle

Let the radii be

R

and

r.

The ring area gives

\pi(R^2-r^2)=\frac{25}{2}\pi.

The radius to the tangent point bisects the chord. If its length is

L,

then

\left(\frac L2\right)^2+r^2=R^2.

Thus

L=2\sqrt{R^2-r^2},

L=2\sqrt{\frac{25}{2}}=5\sqrt2.

Therefore, the correct answer is C.