Is there something wrong with this geometry problem? Problem

In convex pentagon $ABCDE$, $AB=BC=2,CD=\sqrt{2}$,and $EA= \sqrt{3}$. If $\angle{A}=90^{\circ}$, and $\angle{B} = 120^{\circ}$, what is the area of $ABCDE$?

I had asked this question previously and almost everyone said there was an error with the question. However I added the extra word convex in there to see if that changes the question because I personally don't see how this makes the question any more valid since we can just increase $CDE$'s area and keep $ABCE$ constant.
 A: To specify a regular $n$-gon uniquely, you would need at the least $2n-3$ pieces of information (typically $n$ sides and $n-3$ angles). For instance, for a triangle, you would need $3$ pieces of information; a quadrilateral, you would need $5$ pieces of information and so on. Hence, for a pentagon you need at the least $7$ pieces of information to specify it uniquely. In your case, you have only provided $6$ pieces of information. Hence, this doesn't uniquely identify the pentagon.
Note: Though it is possible that the class of pentagons determined by $6$ pieces of information have the same area. For instance, if for a triangle, its base and height are specified then the area is uniquely determined. However, in this case, you will be able to clearly see that there is an ambiguity in choosing the point $D$, which in turn has an effect on the area.
Note that each choice of $D$ on arc of length $\sqrt2$ from $C$ gives different pentagons (including a range of convex pentagons) each with different areas. Though note that if you want to maximise the area, then choosing $D$ to be $D''$ gives you the maximum area. (It is easy to show that $\angle{AEC}$ is $90^{\circ}$. Hence, the area is maximised if the height of the triangle $EDC$ is maximised.)


The area of the pentagon is the area of the quadrilateral $ABCE$ + the area of the triangle $CDE$.
The area of the quadrilateral $ABCE$ is easy to find, since the height is $AE=\sqrt3$ and the sides are $AB=2$ and $CE=3$. Hence, the area is $\dfrac12 \cdot \sqrt3 \cdot (2+3) = \dfrac{5\sqrt3}2$.
The area of the triangle $CDE$ is
$$\Delta = \dfrac12 CD \cdot DE \cdot \sin(D) = \dfrac{DE\sin(D)}{\sqrt2} \implies DE = \dfrac{\sqrt2 \Delta}{\sin(D)}$$
From cosine rule, we have that
$$\cos(D) = \dfrac{CD^2+DE^2-EC^2}{2\cdot CD\cdot DE}$$
This gives us that
$$2+DE^2-3^2 = 2\sqrt2 DE \cos(D) \implies \dfrac{2\Delta^2}{\sin^2(D)} - 7 = \dfrac{4\Delta \cos(D)}{\sin(D)}$$
This gives us
$$2\Delta^2 - 7\sin^2(D) = 2\Delta\sin(2D) \implies 2\Delta^2 - \dfrac72\left(1-\cos(2D)\right) = 2\Delta\sin(2D)$$
$$2\Delta^2 + \Delta -\dfrac72\left(\dfrac{2+\sqrt3}2\right) = 0$$
Obtain $\Delta$ from this.
