1973 AMC 12 Problem 1

The chord meets the radius halfway from the center and at a right angle

Find half the chord with the Pythagorean theorem, then double it

Let $M$ be the midpoint of the chord and also the midpoint of the radius. The distance from the center $O$ to $M$ is $6,$ while the circle’s radius is $12.$ If $A$ is one endpoint of the chord, then $$AM=\sqrt{12^2-6^2}=6\sqrt3.$$ The full chord has length $2AM=12\sqrt3.$

Therefore, the correct answer is D.