As in the following diagram, square $ABCD$ and square $CEFG$ are placed side by side (i.e. $C$ is between $B$ and $E$ and $G$ is between $C$ and $D$). If $CE = 14$, $AB > 14$, compute the minimal area of $\triangle AEG$. 
[asy]
size(120); defaultpen(linewidth(0.7)+fontsize(10)); 
pair D2(real x, real y) {
 pair P = (x,y);
 dot(P,linewidth(3)); return P;
}
int big = 30, small = 14;
filldraw((0,big)--(big+small,0)--(big,small)--cycle, rgb(0.9,0.5,0.5));
draw(scale(big)*unitsquare); draw(shift(big,0)*scale(small)*unitsquare);
label("$A$",D2(0,big),NW);
label("$B$",D2(0,0),SW);
label("$C$",D2(big,0),SW);
label("$D$",D2(big,big),N);
label("$E$",D2(big+small,0),SE);
label("$F$",D2(big+small,small),NE);
label("$G$",D2(big,small),NE);
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1. Let the side length of square \(ABCD\) be \(x\). Since \(CE = 14\), the side length of square \(CEFG\) is 14.
2. The area of square \(ABCD\) is \(x^2\).
3. The area of square \(CEFG\) is \(14^2 = 196\).
4. To find the area of \(\triangle AEG\), we will subtract the areas of the other triangles from the combined areas of squares \(ABCD\) and \(CEFG\).

   - The area of \(\triangle ADG\) is:
     \[
     [ADG] = \frac{AD \cdot DG}{2} = \frac{x \cdot (x - 14)}{2} = \frac{x^2 - 14x}{2}
     \]

   - The area of \(\triangle ABE\) is:
     \[
     [ABE] = \frac{AB \cdot BE}{2} = \frac{x \cdot (x + 14)}{2} = \frac{x^2 + 14x}{2}
     \]

   - The area of \(\triangle EFG\) is:
     \[
     [EFG] = \frac{14 \cdot 14}{2} = 98
     \]

5. The combined area of squares \(ABCD\) and \(CEFG\) is:
   \[
   x^2 + 196
   \]

6. The area of \(\triangle AEG\) is:
   \[
   [AEG] = x^2 + 196 - \frac{x^2 - 14x}{2} - \frac{x^2 + 14x}{2} - 98
   \]

7. Simplify the expression for \([AEG]\):
   \[
   [AEG] = x^2 + 196 - \left(\frac{x^2 - 14x + x^2 + 14x}{2}\right) - 98
   \]
   \[
   [AEG] = x^2 + 196 - \left(\frac{2x^2}{2}\right) - 98
   \]
   \[
   [AEG] = x^2 + 196 - x^2 - 98
   \]
   \[
   [AEG] = 98
   \]

Therefore, the minimal area of \(\triangle AEG\) is 98, regardless of the value of \(x\) greater than 14.

The final answer is \(\boxed{98}\).