A circle and a square are drawn on the plane so that they overlap. Together, the two shapes cover an area of $329$ square units. The area common to both shapes is $101$ square units. The area of the circle is $234$ square units. What is the perimeter of the square in units?

$\mathrm{(A) \ } 14 \qquad \mathrm{(B) \ } 48 \qquad \mathrm {(C) \ } 56 \qquad \mathrm{(D) \ } 64 \qquad \mathrm{(E) \ } 196$
1. Let the area of the square be denoted by \( A_{\text{square}} \) and the area of the circle be denoted by \( A_{\text{circle}} \). We are given:
   \[
   A_{\text{circle}} = 234 \text{ square units}
   \]
   and the total area covered by both shapes is:
   \[
   A_{\text{total}} = 329 \text{ square units}
   \]
   The area common to both shapes is:
   \[
   A_{\text{common}} = 101 \text{ square units}
   \]

2. Using the principle of inclusion and exclusion for the areas, we have:
   \[
   A_{\text{total}} = A_{\text{square}} + A_{\text{circle}} - A_{\text{common}}
   \]
   Substituting the given values:
   \[
   329 = A_{\text{square}} + 234 - 101
   \]

3. Solving for \( A_{\text{square}} \):
   \[
   329 = A_{\text{square}} + 133
   \]
   \[
   A_{\text{square}} = 329 - 133
   \]
   \[
   A_{\text{square}} = 196 \text{ square units}
   \]

4. The area of the square is given by \( s^2 \), where \( s \) is the side length of the square. Therefore:
   \[
   s^2 = 196
   \]
   Solving for \( s \):
   \[
   s = \sqrt{196}
   \]
   \[
   s = 14
   \]

5. The perimeter of the square is given by \( 4s \):
   \[
   \text{Perimeter} = 4 \times 14 = 56 \text{ units}
   \]

The final answer is \(\boxed{56}\)