1. The problem states that we have a circle with radius $r = 8$ cm and a chord of length $12$ cm. 2. We need to find the distance $d$ of the chord from the center of the circle. 3. Draw a line from the center of the circle perpendicular to the chord. This line will bisect the chord into two equal parts of length $\frac{12}{2} = 6$ cm. 4. Now, we have a right triangle formed by the radius, the distance from the center to the chord, and half the chord length. 5. Using the Pythagorean theorem on this triangle: $$r^2 = d^2 + \left(\frac{12}{2}\right)^2$$ $$8^2 = d^2 + 6^2$$ $$64 = d^2 + 36$$ 6. Solve for $d^2$: $$d^2 = 64 - 36 = 28$$ 7. Take the square root: $$d = \sqrt{28} = 2\sqrt{7}$$ Final answer: The distance of the chord from the center is $2\sqrt{7}$ cm.