Given is a rectangle with perimeter $x$ cm and side lengths in a $1:2$ ratio. Suppose that the area of the rectangle is also $x$ $\text{cm}^2$. Determine all possible values of $x$.
1. Let \( a \) and \( b \) be the side lengths of the rectangle, such that \( a = 2b \).

2. The perimeter of the rectangle is given by:
   \[
   P = 2(a + b)
   \]
   Since the perimeter is \( x \) cm, we have:
   \[
   x = 2(a + b)
   \]

3. The area of the rectangle is given by:
   \[
   A = ab
   \]
   Since the area is also \( x \) cm\(^2\), we have:
   \[
   x = ab
   \]

4. Substituting \( a = 2b \) into the perimeter equation:
   \[
   x = 2(2b + b) = 2 \cdot 3b = 6b
   \]
   Therefore:
   \[
   x = 6b
   \]

5. Substituting \( a = 2b \) into the area equation:
   \[
   x = ab = 2b \cdot b = 2b^2
   \]
   Therefore:
   \[
   x = 2b^2
   \]

6. We now have two equations:
   \[
   x = 6b
   \]
   \[
   x = 2b^2
   \]

7. Setting the two expressions for \( x \) equal to each other:
   \[
   6b = 2b^2
   \]

8. Dividing both sides by 2:
   \[
   3b = b^2
   \]

9. Rearranging the equation:
   \[
   b^2 - 3b = 0
   \]

10. Factoring out \( b \):
    \[
    b(b - 3) = 0
    \]

11. Solving for \( b \):
    \[
    b = 0 \quad \text{or} \quad b = 3
    \]

12. Since \( b = 0 \) does not make sense in the context of a rectangle, we have:
    \[
    b = 3
    \]

13. Substituting \( b = 3 \) back into the expression for \( a \):
    \[
    a = 2b = 2 \cdot 3 = 6
    \]

14. Therefore, the side lengths of the rectangle are \( a = 6 \) cm and \( b = 3 \) cm.

15. The value of \( x \) is:
    \[
    x = ab = 6 \cdot 3 = 18
    \]

The final answer is \( \boxed{18} \).