Area of $\Delta ABC$ via $2$ statements 
What is the area of the $\Delta ABC$? 
Ι. $∠BAC = 45°$. 
ΙΙ. $AC = 16 cm$ and $BC = 12 cm$. 

(A)  question can be answered using  only one of the statements alone.  
(B)  question can be answered using either statement alone.  
(C)  question can be answered using Ι and ΙΙ together but not using Ι or ΙΙ alone.  
(D)  question cannot be answered even using Ι and ΙΙ together.  

My Approach:

Let $AB=x$. Then, I used the Cosine formula: $Cos A = \frac {b^2+c^2-a^2}{2bc}$
$Cos (45°)= \frac {16^2+x^2-12^2}{2(16x)}$
$\Rightarrow \frac 1 {\sqrt 2}= \frac {256+x^2-144}{2(16x)}$
$\Rightarrow x^2-16\sqrt 2x+112=0$
$\Rightarrow x=8\sqrt 2 \pm 4=AB$
Since I'm getting 2 possible triangles, so I can't distinctly find the area of the triangle.
Thus, (D).

Solution given: 

From I and II, we have $∠BAC = 45°, AC = 16 cm, BC = 12 cm$. 

 Let $D$ be the foot of the perpendicular from $C$ to $AB$. 
$AB = AD + DB$ 
$=CD+ \sqrt {{BC}^2-{CD}^2}=8\sqrt 2+\sqrt {{12}^2-{(8\sqrt 2)}^2}$ 
$=8\sqrt 2 +4$ 
Area of $\Delta ABC = \frac 12 (AB)(CD)$
$CD = AC Sin 45° = \frac {16}{\sqrt 2}=8\sqrt 2 ≃ 11.3 cm. \text { Sufficient.} $
Thus, option (C) is correct.

Could it be that the answer (thus, solution) given is wrong? Or am I making mistake somewhere? Please help.

To see that your work is correct, make the following construction:

*

*Construct $AC=16 cm$.

*Draw a line $\ell$ through $A$, inclined at an angle $45^{\circ}$ to $AC$.

*With center C, draw a circle $\Sigma$ of radius $12 cm$. You’ll notice that $\Sigma$ intersects $\ell$ at two points. These are the two possible positions of the point B. 
In the image, the triangle $\triangle ABC$ corresponds to the solution provided to you. They have assumed that the foot of perpendicular D lies between A and B, or in other words, they have taken $\angle B$ to be acute.
Obviously, the two triangles $\triangle ABC$ and $\triangle AB”C$ have different areas, so the conclusion of option D follows.
Also, to understand the algebraic solution of $8\sqrt 2\color{red}\pm 4$, imagine a perpendicular dropped from C to BB”. D bisects BB” and note that $AC=8\sqrt2,$ while $\color{blue}{B”D=BD=4}$. Thus $AB”=8\sqrt2-4$ while $AB=8\sqrt2+4$.
