Is this geometry question about a pentagon correct? Problem

In 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 just need some reaffirmation that there is no solution to this problem. There are infinitely many pentagons with the properties in the question (try drawing it).
 A: It takes 7 parameters to define the shape and size of a pentagon. 
The problem gives 6.  So you should have a continuous family of pentagons meeting the conditions.  Hypothetically they could all have the same area but the discussion in comments points in the opposite direction.
The shape, and hence the area, of EABC are determined by the data.
A: You can figure out the area of the quadralateral ABCE.  D must be a point of a circle centered at C and of length $\sqrt 2$.  We can set up an equation for that.  So the area is area of quad ADCE $\pm$ area of triangle CDE.  That gives us a range of values.  Maybe 4/5 of the multiple choices are out of range.
Or maybe it was a typo.
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If CE is the base of CDE then height has a max/min of $\sqrt 2, -\sqrt 2$.
I'm too lazy to calculate CD or area ABCE.
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Actually the area ABCE is easy.  Angle CBX for some point on the perpendicular to AB at B is 30 so if we set X to form a 30-60-90 triangle, we see BX would be $\sqrt 3$ which is the same as AE so ABXE is a rectangle with area $\sqrt 3 2$.  ECX are colinear and area of BCX is $\sqrt 3/2$ so $ABCE$ has area $\sqrt 3* 5/2$.  Triangle CDE has area $3*h/2$ where h $\in [-\sqrt 2, \sqrt 2]$.
So ABCDE has area $(\sqrt 3 *5 + 3*h)/2$ where $-\sqrt 2 \le h \le \sqrt 2$.
A: For constructing a unique pentagon $ABCDE$, it needs either $\angle C$ or the side $ED$ to be known as an additional information. So there could be many pentagons constructed with the given information  
