1. **Problem Statement:** We are given a circle with chords AB and CD. AB is 16 units long. CD is divided into two segments labeled $x$ and $2x$. The segment between points A and C is 4 units. We need to find the value of $x$ in simplest radical form. 2. **Relevant Theorem:** When two chords intersect inside a circle, the products of the lengths of the segments of each chord are equal. This is called the Intersecting Chords Theorem. 3. **Formula:** If chords AB and CD intersect at point E, then: $$AE \times EB = CE \times ED$$ 4. **Assigning lengths:** - Chord AB length = 16, so if E is the intersection point, then $AE$ and $EB$ are segments of AB. - The problem states the segment between A and C is 4, which suggests $AE = 4$. - Since AB = 16, $EB = 16 - AE = 16 - 4 = 12$. - Chord CD is divided into segments $x$ and $2x$, so $CE = x$ and $ED = 2x$. 5. **Apply the theorem:** $$AE \times EB = CE \times ED$$ $$4 \times 12 = x \times 2x$$ $$48 = 2x^2$$ 6. **Solve for $x$:** $$2x^2 = 48$$ $$x^2 = 24$$ $$x = \sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}$$ **Final answer:** $$x = 2\sqrt{6}$$