Let the line $$x + y = 1$$ meet the circle $$x^{2}+y^{2}=4$$ at the points A and B . If the line perpendicular to AB and passing through the mid point of the chord AB intersects the circle at $$C$$ and $$D$$, then the area of the quadrilateral ADBC is equal to :

We have the circle $$x^2 + y^2 = 4$$ (centre O at origin, radius $$r = 2$$) and the chord $$x + y = 1$$.

The perpendicular from the centre (0, 0) to the line $$x + y = 1$$ meets the chord at its midpoint M.

Distance from origin to the line $$x + y - 1 = 0$$:

The midpoint M lies on $$x + y = 1$$ and on the line $$y = x$$ (perpendicular from origin to $$x + y = 1$$ has slope 1, since the chord has slope -1). So $$2x = 1$$, giving $$M = \left(\frac{1}{2}, \frac{1}{2}\right)$$.

Half-length of AB: $$\sqrt{r^2 - d^2} = \sqrt{4 - \frac{1}{2}} = \sqrt{\frac{7}{2}}$$.

So $$AB = 2\sqrt{\frac{7}{2}} = \sqrt{14}$$.

CD is the chord along the line $$y = x$$ (the perpendicular bisector of AB passing through M). This line passes through the centre O(0,0), so CD is a diameter.

Therefore $$CD = 2r = 4$$.

The quadrilateral ADBC has diagonals AB and CD. These two chords are perpendicular to each other (AB has slope -1, CD has slope 1), and they intersect at M.

For a quadrilateral whose diagonals are perpendicular and intersect at a point, the area is:

where $$d_1$$ and $$d_2$$ are the full diagonal lengths.

The answer is $$\boxed{2\sqrt{14}}$$, which corresponds to Option 3.