If the base of a rectangle is increased by $ 10\%$ and the area is unchanged, then the altitude is decreased by:

$ \textbf{(A)}\ 9\% \qquad\textbf{(B)}\ 10\% \qquad\textbf{(C)}\ 11\% \qquad\textbf{(D)}\ 11\frac {1}{9}\% \qquad\textbf{(E)}\ 9\frac {1}{11}\%$
1. Let the original base of the rectangle be \( b \) and the original height be \( h \). The area of the rectangle is given by:
   \[
   A = b \cdot h
   \]

2. If the base is increased by \( 10\% \), the new base becomes:
   \[
   b_{\text{new}} = b + 0.1b = 1.1b
   \]

3. Let the new height be \( h_{\text{new}} \). Since the area remains unchanged, we have:
   \[
   A = b \cdot h = 1.1b \cdot h_{\text{new}}
   \]

4. Solving for \( h_{\text{new}} \):
   \[
   b \cdot h = 1.1b \cdot h_{\text{new}}
   \]
   \[
   h_{\text{new}} = \frac{b \cdot h}{1.1b} = \frac{h}{1.1}
   \]

5. Simplifying \( \frac{h}{1.1} \):
   \[
   h_{\text{new}} = \frac{h}{1.1} = \frac{h}{\frac{11}{10}} = h \cdot \frac{10}{11}
   \]

6. The new height \( h_{\text{new}} \) is \( \frac{10}{11} \) of the original height \( h \). The decrease in height is:
   \[
   h - h_{\text{new}} = h - h \cdot \frac{10}{11} = h \left(1 - \frac{10}{11}\right) = h \cdot \frac{1}{11}
   \]

7. The percentage decrease in height is:
   \[
   \frac{\frac{1}{11}h}{h} \times 100\% = \frac{1}{11} \times 100\% = 9 \frac{1}{11}\%
   \]

Conclusion:
The altitude is decreased by \( 9 \frac{1}{11}\% \).

The final answer is \( \boxed{9 \frac{1}{11}\%} \).