A convex quadrilateral $ABCD$ is constructed out of metal rods with negligible thickness. The side lengths are $AB=BC=CD=5$ and $DA=3$. The figure is then deformed, with the angles between consecutive rods allowed to change but the rods themselves staying the same length. The resulting figure is a convex polygon for which $\angle{ABC}$ is as large as possible. What is the area of this figure?

$\text{(A) }6\qquad\text{(B) }8\qquad\text{(C) }9\qquad\text{(D) }10\qquad\text{(E) }12$
1. To maximize $\angle ABC$, we need to consider the configuration where $\angle ABC$ is as large as possible. This occurs when $AC$ is a straight line, making $ABCD$ a degenerate quadrilateral where $A$, $B$, and $C$ are collinear.

2. Given the side lengths $AB = BC = CD = 5$ and $DA = 3$, we can visualize the quadrilateral $ABCD$ as follows:
   - Place $A$ at the origin $(0, 0)$.
   - Place $B$ at $(5, 0)$.
   - Since $B$ and $C$ are collinear and $BC = 5$, place $C$ at $(10, 0)$.
   - Now, $D$ must be placed such that $DA = 3$ and $DC = 5$.

3. To find the coordinates of $D$, we solve the system of equations given by the distances:
   \[
   \sqrt{(x_D - 0)^2 + (y_D - 0)^2} = 3 \quad \text{(distance from $A$ to $D$)}
   \]
   \[
   \sqrt{(x_D - 10)^2 + (y_D - 0)^2} = 5 \quad \text{(distance from $C$ to $D$)}
   \]

4. Squaring both equations, we get:
   \[
   x_D^2 + y_D^2 = 9
   \]
   \[
   (x_D - 10)^2 + y_D^2 = 25
   \]

5. Expanding and simplifying the second equation:
   \[
   x_D^2 - 20x_D + 100 + y_D^2 = 25
   \]
   \[
   x_D^2 + y_D^2 - 20x_D + 100 = 25
   \]
   \[
   9 - 20x_D + 100 = 25 \quad \text{(substituting $x_D^2 + y_D^2 = 9$)}
   \]
   \[
   109 - 20x_D = 25
   \]
   \[
   20x_D = 84
   \]
   \[
   x_D = \frac{84}{20} = 4.2
   \]

6. Substituting $x_D = 4.2$ back into $x_D^2 + y_D^2 = 9$:
   \[
   (4.2)^2 + y_D^2 = 9
   \]
   \[
   17.64 + y_D^2 = 9
   \]
   \[
   y_D^2 = 9 - 17.64 = -8.64
   \]
   This is not possible, indicating an error in our approach. Instead, we should consider the geometric properties of the quadrilateral.

7. Since $A$, $B$, and $C$ are collinear, $D$ must be placed such that $DA = 3$ and $DC = 5$. The only possible configuration is when $D$ is directly above or below the line $ABC$.

8. The maximum area of the quadrilateral $ABCD$ occurs when $D$ forms a right triangle with $A$ and $C$. The height from $D$ to $AC$ (which is the base) is 3 units.

9. The area of the triangle $ADC$ is:
   \[
   \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 3 = 12
   \]

The final answer is $\boxed{12}$.