1. **Problem Statement:** Given two equal circles with perimeter 360 m each, intersecting at points A, B, C, and D such that AC = DB and center O is the intersection of lines AD and BC. We need to find: (i) The value of angle $D\hat{O}C = A\hat{O}B$. (ii) The length of $OA = OB = OC = OD$. (iii) The length of segment $AB$ given that $AB$ equals $DC$. 2. **Step 1: Understanding the circles and perimeter** Each circle has perimeter (circumference) $C = 360$ m. The formula for circumference of a circle is: $$C = 2\pi r$$ where $r$ is the radius. So, $$360 = 2\pi r \implies r = \frac{360}{2\pi} = \frac{180}{\pi}$$ 3. **Step 2: Radius and center distances** Since $OA = OB = OC = OD = r$, the distance from center $O$ to points $A, B, C, D$ is the radius: $$OA = OB = OC = OD = \frac{180}{\pi}$$ 4. **Step 3: Angle $D\hat{O}C = A\hat{O}B$** Because $AC = DB$ and the circles are equal and intersect symmetrically, the angle between lines $DOC$ and $AOB$ at center $O$ is the same. The two circles intersect such that the angle between the radii to the intersection points is $60^\circ$ (since the arcs subtended by these chords are equal and the circles are equal). Therefore, $$D\hat{O}C = A\hat{O}B = 60^\circ$$ 5. **Step 4: Length of $AB$ and $DC$** Since $AB$ and $DC$ are chords of the circles and are equal, Using the chord length formula: $$\text{Chord length} = 2r \sin\left(\frac{\theta}{2}\right)$$ where $\theta$ is the central angle subtending the chord. Here, $\theta = 60^\circ$ and $r = \frac{180}{\pi}$. So, $$AB = DC = 2 \times \frac{180}{\pi} \times \sin\left(\frac{60^\circ}{2}\right) = \frac{360}{\pi} \times \sin(30^\circ)$$ Since $\sin(30^\circ) = \frac{1}{2}$, $$AB = DC = \frac{360}{\pi} \times \frac{1}{2} = \frac{180}{\pi}$$ **Final answers:** (i) $D\hat{O}C = A\hat{O}B = 60^\circ$ (ii) $OA = OB = OC = OD = \frac{180}{\pi}$ meters (iii) $AB = DC = \frac{180}{\pi}$ meters