A unit equilateral triangle is given. Divide each side into three equal  parts. Remove the equilateral triangles whose bases are middle one-third segments. Now we have a new polygon. Remove the equilateral triangles whose bases are middle one-third segments of the sides of the polygon. After repeating these steps for infinite times, what is the area of the new shape?

$ 
\textbf{(A)}\ \dfrac {1}{2\sqrt 3}
\qquad\textbf{(B)}\ \dfrac {\sqrt 3}{8}
\qquad\textbf{(C)}\ \dfrac {\sqrt 3}{10}
\qquad\textbf{(D)}\ \dfrac {1}{4\sqrt 3}
\qquad\textbf{(E)}\ \text{None of the above}
$
1. **Initial Setup**: We start with a unit equilateral triangle. The area \( S \) of an equilateral triangle with side length \( a \) is given by:
   \[
   S = \frac{a^2 \sqrt{3}}{4}
   \]
   For a unit equilateral triangle (\( a = 1 \)):
   \[
   S = \frac{1^2 \sqrt{3}}{4} = \frac{\sqrt{3}}{4}
   \]

2. **First Step**: Divide each side into three equal parts and remove the middle one-third segment's equilateral triangle. This results in removing 3 smaller equilateral triangles, each with side length \( \frac{1}{3} \). The area of each smaller triangle is:
   \[
   \text{Area of each smaller triangle} = \frac{\left(\frac{1}{3}\right)^2 \sqrt{3}}{4} = \frac{\frac{1}{9} \sqrt{3}}{4} = \frac{\sqrt{3}}{36}
   \]
   Since there are 3 such triangles removed:
   \[
   \text{Total area removed in the first step} = 3 \times \frac{\sqrt{3}}{36} = \frac{\sqrt{3}}{12}
   \]

3. **Subsequent Steps**: In each subsequent step, each side of the remaining polygon is divided into three equal parts, and the middle one-third segment's equilateral triangle is removed. The number of triangles removed and their areas follow a geometric progression.

4. **General Formula for Area Removed**: In the \( k \)-th step, the number of triangles removed is \( 3 \cdot 4^{k-1} \), and the area of each triangle removed is:
   \[
   \left(\frac{1}{3^k}\right)^2 \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{4 \cdot 3^{2k}}
   \]
   Therefore, the total area removed in the \( k \)-th step is:
   \[
   3 \cdot 4^{k-1} \cdot \frac{\sqrt{3}}{4 \cdot 3^{2k}} = 3 \cdot 4^{k-1} \cdot \frac{\sqrt{3}}{4 \cdot 9^k} = 3 \cdot 4^{k-1} \cdot \frac{\sqrt{3}}{4 \cdot 3^{2k}} = \frac{3 \cdot 4^{k-1} \cdot \sqrt{3}}{4 \cdot 9^k}
   \]

5. **Summing the Series**: The total area removed after infinite steps is the sum of the areas removed in each step:
   \[
   \sum_{k=1}^{\infty} 3 \cdot 4^{k-1} \cdot \frac{\sqrt{3}}{4 \cdot 9^k} = \frac{3 \sqrt{3}}{4} \sum_{k=1}^{\infty} \left(\frac{4}{9}\right)^k
   \]
   The series \( \sum_{k=1}^{\infty} \left(\frac{4}{9}\right)^k \) is a geometric series with the first term \( a = \frac{4}{9} \) and common ratio \( r = \frac{4}{9} \):
   \[
   \sum_{k=1}^{\infty} \left(\frac{4}{9}\right)^k = \frac{\frac{4}{9}}{1 - \frac{4}{9}} = \frac{\frac{4}{9}}{\frac{5}{9}} = \frac{4}{5}
   \]
   Therefore:
   \[
   \frac{3 \sqrt{3}}{4} \cdot \frac{4}{5} = \frac{3 \sqrt{3}}{4} \cdot \frac{4}{5} = \frac{3 \sqrt{3}}{5}
   \]

6. **Final Area Calculation**: The area of the new shape after infinite steps is:
   \[
   A = S - \text{Total area removed} = \frac{\sqrt{3}}{4} - \frac{3 \sqrt{3}}{20} = \frac{5 \sqrt{3}}{20} - \frac{3 \sqrt{3}}{20} = \frac{2 \sqrt{3}}{20} = \frac{\sqrt{3}}{10}
   \]

The final answer is \(\boxed{\frac{\sqrt{3}}{10}}\)