Sohom constructs a square $BERK$ of side length $10$. Darlnim adds points $T$, $O$, $W$, and $N$, which are the midpoints of $\overline{BE}$, $\overline{ER}$, $\overline{RK}$, and $\overline{KB}$, respectively. Lastly, Sylvia constructs square $CALI$ whose edges contain the vertices of $BERK$, such that $\overline{CA}$ is parallel to $\overline{BO}$. Compute  the area of $CALI$.
[img]https://cdn.artofproblemsolving.com/attachments/0/9/0fda0c273bb73b85f3b1bc73661126630152b3.png[/img]
1. **Identify the problem and given data:**
   - We have a square \(BERK\) with side length \(10\).
   - Points \(T\), \(O\), \(W\), and \(N\) are the midpoints of \(\overline{BE}\), \(\overline{ER}\), \(\overline{RK}\), and \(\overline{KB}\), respectively.
   - A new square \(CALI\) is constructed such that its edges contain the vertices of \(BERK\) and \(\overline{CA}\) is parallel to \(\overline{BO}\).

2. **Determine the side length of square \(CALI\):**
   - Since \(T\), \(O\), \(W\), and \(N\) are midpoints, the distance from the center of \(BERK\) to any of these points is half the side length of \(BERK\), which is \(5\).
   - The diagonal of square \(BERK\) is \(10\sqrt{2}\) (using the Pythagorean theorem: \(\sqrt{10^2 + 10^2} = 10\sqrt{2}\)).
   - The diagonal of square \(CALI\) must be twice the distance from the center of \(BERK\) to any of the midpoints, which is \(10\sqrt{2}\).

3. **Calculate the side length of square \(CALI\):**
   - The diagonal of a square is related to its side length \(s\) by the formula \(s\sqrt{2}\).
   - Therefore, for square \(CALI\), we have:
     \[
     s\sqrt{2} = 10\sqrt{2}
     \]
     \[
     s = 10
     \]

4. **Calculate the area of square \(CALI\):**
   - The area \(S\) of square \(CALI\) is given by:
     \[
     S = s^2 = 10^2 = 100
     \]

5. **Re-evaluate the problem statement and solution:**
   - The initial solution suggests that the area of \(CALI\) is \(180\), which contradicts our calculation.
   - Let's re-examine the relationship between the areas of \(BERK\) and \(CALI\):
     - The triangles outside \(BERK\) but inside \(CALI\) can be folded into \(BERK\), leaving a central square.
     - The central square has an area of \(\frac{1}{9}\) of \(CALI\).

6. **Set up the equation for the area:**
   - Let \(S\) be the area of \(CALI\).
   - The area of \(BERK\) is \(100\).
   - The equation given is:
     \[
     (S - 100) + \frac{1}{9}S = 100
     \]
     \[
     S - 100 + \frac{1}{9}S = 100
     \]
     \[
     \frac{10}{9}S - 100 = 100
     \]
     \[
     \frac{10}{9}S = 200
     \]
     \[
     S = 180
     \]

The final answer is \(\boxed{180}\)