A circle of radius $R$ is divided into $8$ equal parts. The points of division are denoted successively by $A, B, C, D, E, F , G$ and $H$. Find the area of the square formed by drawing the chords $AF$ , $BE$, $CH$ and $DG$.
1. **Understanding the Problem:**
   We are given a circle of radius \( R \) divided into 8 equal parts, creating an octagon. The points of division are \( A, B, C, D, E, F, G, \) and \( H \). We need to find the area of the square formed by the chords \( AF, BE, CH, \) and \( DG \).

2. **Identifying the Chords:**
   The chords \( AF, BE, CH, \) and \( DG \) are diagonals of the octagon that skip three vertices. These chords are equal in length and form a square.

3. **Length of the Chords:**
   To find the length of these chords, we need to determine the distance between two points on the circle that are separated by an angle of \( 135^\circ \) (since \( 360^\circ / 8 = 45^\circ \) and \( 3 \times 45^\circ = 135^\circ \)).

4. **Using the Law of Cosines:**
   The length of the chord \( AF \) can be found using the Law of Cosines in the triangle formed by the center of the circle and the points \( A \) and \( F \):
   \[
   AF^2 = R^2 + R^2 - 2R \cdot R \cdot \cos(135^\circ)
   \]
   \[
   \cos(135^\circ) = -\frac{\sqrt{2}}{2}
   \]
   \[
   AF^2 = R^2 + R^2 - 2R^2 \left(-\frac{\sqrt{2}}{2}\right)
   \]
   \[
   AF^2 = 2R^2 + R^2\sqrt{2}
   \]
   \[
   AF = R\sqrt{2 + \sqrt{2}}
   \]

5. **Area of the Square:**
   The side length of the square is \( AF \), so the area of the square is:
   \[
   \text{Area} = (AF)^2 = \left(R\sqrt{2 + \sqrt{2}}\right)^2
   \]
   \[
   \text{Area} = R^2 (2 + \sqrt{2})
   \]

The final answer is \(\boxed{R^2 (2 + \sqrt{2})}\)