A ladder is formed by removing some consecutive unit squares of a $10\times 10$ chessboard such that for each $k-$th row ($k\in \{1,2,\dots, 10\}$),  the leftmost $k-1$ unit squares are removed. How many rectangles formed by composition of unit squares does the ladder have?

$ 
\textbf{(A)}\ 625
\qquad\textbf{(B)}\ 715
\qquad\textbf{(C)}\ 1024
\qquad\textbf{(D)}\ 1512
\qquad\textbf{(E)}\ \text{None of the preceding}
$
1. **Understanding the Problem:**
   We need to count the number of rectangles formed by the composition of unit squares in a modified $10 \times 10$ chessboard. The modification is such that for each $k$-th row, the leftmost $k-1$ unit squares are removed. This forms a staircase-like structure.

2. **Visualizing the Structure:**
   The structure can be visualized as a staircase where the $k$-th row has $11-k$ unit squares. For example, the first row has 10 squares, the second row has 9 squares, and so on, until the 10th row which has 1 square.

3. **Counting Rectangles:**
   To count the number of rectangles, we need to consider all possible rectangles that can be formed by selecting two horizontal and two vertical lines from the grid.

4. **General Formula for Rectangles in a Grid:**
   In a $m \times n$ grid, the number of rectangles is given by:
   \[
   \binom{m+1}{2} \times \binom{n+1}{2}
   \]
   This is because we need to choose 2 out of $m+1$ horizontal lines and 2 out of $n+1$ vertical lines to form a rectangle.

5. **Applying the Formula to Each Row:**
   For each row $k$ (where $k$ ranges from 1 to 10), the number of unit squares is $11-k$. We need to count the rectangles that can be formed in each row and sum them up.

6. **Summing Up the Rectangles:**
   For each row $k$, the number of rectangles that can be formed is:
   \[
   \sum_{i=1}^{10} \binom{11-i+1}{2} \times \binom{11-i+1}{2}
   \]
   Simplifying, we get:
   \[
   \sum_{i=1}^{10} \binom{12-i}{2} \times \binom{12-i}{2}
   \]

7. **Calculating the Sum:**
   We need to calculate the sum of the squares of the binomial coefficients:
   \[
   \sum_{i=1}^{10} \left( \binom{12-i}{2} \right)^2
   \]
   This can be computed as:
   \[
   \sum_{i=1}^{10} \left( \frac{(12-i)(11-i)}{2} \right)^2
   \]

8. **Using the Triangular Number Formula:**
   The sum of the first $n$ triangular numbers is given by:
   \[
   \sum_{k=1}^{n} \frac{k(k+1)}{2} = \frac{n(n+1)(n+2)}{6}
   \]
   For $n=10$, this becomes:
   \[
   \frac{10 \cdot 11 \cdot 12}{6} = 220
   \]

9. **Final Calculation:**
   The number of rectangles in the staircase structure is given by:
   \[
   \binom{10+3}{4} = \binom{13}{4} = \frac{13 \cdot 12 \cdot 11 \cdot 10}{4 \cdot 3 \cdot 2 \cdot 1} = 715
   \]

The final answer is $\boxed{715}$