1. Identify the circle and the chord line

The circle is $x^2+y^2=4$ so its center is $O(0,0)$ and radius is $r=2.$

The chord $AB$ lies on the line $x+y=1.$

2. Find the midpoint of chord $AB$

The midpoint of a chord is the foot of the perpendicular from the center to the chord.

So we find the foot of the perpendicular from $(0,0)$ to the line $x+y-1=0.$

Using projection formula for line $ax+by+c=0$ with $a=1,b=1,c=-1$:

Foot $M$ from $(0,0)$ is $\left(\frac{-ac}{a^2+b^2},\frac{-bc}{a^2+b^2}\right)=\left(\frac{1}{2},\frac{1}{2}\right).$

Thus midpoint of chord $AB$ is $M\left(\frac12,\frac12\right).$

3. Length of chord $AB$

Distance from center to the chord line is $d=\frac{|0+0-1|}{\sqrt{1^2+1^2}}=\frac{1}{\sqrt2}.$

For a circle of radius $r$, chord length is $AB=2\sqrt{r^2-d^2}.$

Hence $AB=2\sqrt{4-\frac12}=2\sqrt{\frac72}=\sqrt{14}.$

4. Equation of the line perpendicular to $AB$ through midpoint

Since $AB$ is on $x+y=1$, its slope is $-1$. A perpendicular line has slope $1$.

Passing through $M\left(\frac12,\frac12\right)$: $y-\frac12=1\left(x-\frac12\right)$ so $y=x.$

This line meets the circle at $C$ and $D$.

5. Length of diagonal $CD$

Substitute $y=x$ into the circle: $x^2+x^2=4 \implies 2x^2=4 \implies x^2=2 \implies x=\pm \sqrt2.$

So the points are $C(\sqrt2,\sqrt2), \quad D(-\sqrt2,-\sqrt2).$

Thus $CD$ is a diameter, so $CD=2r=4.$

6. Angle between diagonals

The diagonals of quadrilateral $ABCD$ are $AB$ and $CD$. By construction, $CD \perp AB$.

Hence the area of the quadrilateral is $\text{Area} = \frac12 \times AB \times CD$ because diagonals are perpendicular.

So $\text{Area} = \frac12 \times \sqrt{14} \times 4 = 2\sqrt{14}.$

7. Final answer

Therefore, $\boxed{2\sqrt{14}}$ which corresponds to Option C.