Inside a circle with radius $6$ lie four smaller circles with centres $A,B,C$ and $D$. The circles touch each other as shown. The point where the circles with centres $A$ and $C$ touch each other is the centre of the big circle. Calculate the area of quadrilateral $ABCD$.
[img]https://1.bp.blogspot.com/-FFsiOOdcjao/XzT_oJYuQAI/AAAAAAAAMVk/PpyUNpDBeEIESMsiElbexKOFMoCXRVaZwCLcBGAsYHQ/s0/2012%2BMohr%2Bp1.png[/img]
1. **Determine the radii of circles with centers \(A\) and \(C\):**
   Since circles \(A\) and \(C\) touch each other at the center of the large circle, their radii are half the radius of the large circle. Therefore, the radii of circles \(A\) and \(C\) are:
   \[
   \frac{6}{2} = 3
   \]

2. **Identify the type of quadrilateral \(ABCD\):**
   The four sides of quadrilateral \(ABCD\) are congruent because the circles touch each other. This implies that \(ABCD\) is a rhombus. Additionally, the diagonals \(AC\) and \(BD\) intersect at right angles.

3. **Let the radii of circles \(B\) and \(D\) be \(r\):**
   We need to find the value of \(r\). Using the Pythagorean Theorem in \(\Delta AOD\), where \(O\) is the center of the large circle, we have:
   \[
   AD^2 = AO^2 + DO^2
   \]
   Substituting the known values, we get:
   \[
   (3 + r)^2 = 3^2 + (6 - r)^2
   \]

4. **Simplify the equation:**
   \[
   (3 + r)^2 = 9 + (6 - r)^2
   \]
   Expanding both sides:
   \[
   9 + 6r + r^2 = 9 + 36 - 12r + r^2
   \]
   Simplifying further:
   \[
   9 + 6r + r^2 = 45 - 12r + r^2
   \]
   Cancel out \(r^2\) from both sides:
   \[
   9 + 6r = 45 - 12r
   \]
   Combine like terms:
   \[
   18r = 36
   \]
   Solving for \(r\):
   \[
   r = 2
   \]

5. **Calculate the length of \(AD\):**
   Since \(r = 2\), the length of \(AD\) is:
   \[
   AD = 3 + 2 = 5
   \]

6. **Determine the lengths of the diagonals of the rhombus:**
   In \(\Delta AOD\), which is a right triangle, we have:
   \[
   AO = 3, \quad DO = 6 - r = 6 - 2 = 4
   \]
   Therefore, \(\Delta AOD\) is a \(3-4-5\) right triangle. The diagonals of the rhombus \(ABCD\) are:
   \[
   AC = 6 \quad \text{and} \quad BD = 8
   \]

7. **Calculate the area of the rhombus:**
   The area of a rhombus is given by:
   \[
   \text{Area} = \frac{1}{2} \times d_1 \times d_2
   \]
   Substituting the lengths of the diagonals:
   \[
   \text{Area} = \frac{1}{2} \times 6 \times 8 = 24
   \]

The final answer is \(\boxed{24}\).